Element	Symbol	Protons	Electrons	Neutrons	Atomic number	Mass number
Hydrogen	H	1	1	0	1	1
Helium	He	2	2	2	2	4
Lithium	Li	3	3	4	3	7
Beryllium	Be	4	4	5	4	9
Boron	B	5	5	6	5	11
Carbon	C	6	6	6	6	12
Nitrogen	N	7	7	7	7	14
Oxygen	O	8	8	8	8	16
Fluorine	F	9	9	10	9	19
Neon	Ne	10	10	10	10	20
Sodium	Na	11	11	12	11	23
Magnesium	Mg	12	12	12	12	24
Aluminium	Al	13	13	14	13	27
Silicon	Si	14	14	14	14	28
Phosphorus	P	15	15	16	15	31
Sulphur	S	16	16	16	16	32
Chlorine	Cl	17	17	18	17	35
Argon	Ar	18	18	22	18	40
Potassium	K	19	19	20	19	39
Calcium	Ca	20	20	20	20	40

Most atoms of elements exist as isotopes.

Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number.

By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e.

mass number

atomic number ^m _n X symbol of element

Below is the conventional method of writing the 1^st twenty elements showing the mass numbers and atomic numbers;

¹₁H ⁴₂He ⁷₃Li ⁹₄Be ¹¹₅B ¹²₆C

¹⁴₇N ¹⁶₈O ¹⁹₉F ²⁰₁₀Ne ²³₁₁Na ²⁴₁₂Mg

²⁷₁₃Al ²⁸₁₄Si ³¹₁₅P ³²₁₆S ³⁵₁₇Cl ⁴⁰₁₈Ar

³⁹₁₉K ⁴⁰₂₀C

The table below shows some common natural isotopes of some elements

Element

Isotopes

Protons

Electrons

Neutrons

Atomic

number

Mass

number

Hydrogen

¹₁H

²₁H(deuterium)

³₁H(Tritium)

Chlorine

³⁵₁₇Cl

³⁷₁₇Cl

Potassium

³⁹₁₉K

⁴⁰₁₉K

⁴¹₁₉K

Oxygen

¹⁶₈O

¹⁸₈O

Uranium

²³⁵₉₂U

²³⁸₉₂U

143

146

235

238

Neon

²²₁₀Ne

²⁰₁₀Ne

²¹₁₀Ne

The mass of an average atom is very small (10^-22 g).Masses of atoms are therefore expressed in relation to a chosen element.

The atom recommended is ¹²C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) .

All other atoms are compared to the mass of ¹²C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to ¹/₁₂ an atom of ¹²C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e;

RAM = mass of atom of an element

¹/₁₂ of one atom of ¹²C isotope

Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability.

Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples.

Chlorine occurs as 75% ³⁵₁₇Cl and 25% ³⁷₁₇Cl isotopes. Calculate the relative atomic mass of Chlorine.

Working

100 atoms of chlorine contains 75 atoms of ³⁵₁₇Cl isotopes

100 atoms of chlorine contains 75 atoms of ³⁷₁₇Cl isotopes

Therefore;

RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5

Note that:

Relative atomic mass has no units

More atoms of chlorine exist as ³⁵₁₇Cl(75%) than as ³⁷₁₇Cl(25%) therefore RAM is nearer to the more abundant isotope.

Calculate the relative atomic mass of potassium given that it exist as;

93.1% ³⁹₁₉K , 0.01% ⁴⁰₁₉K , 6.89% ⁴¹₁₉K ,

Working

100 atoms of potassium contains 93.1 atoms of ³⁹₁₉K isotopes

100 atoms of potassium contains 0.01 atoms of ⁴⁰₁₉K isotopes

100 atoms of potassium contains 6.89 atoms of ⁴¹₁₉K isotopes

Therefore;

RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39)

Note that:

Relative atomic mass has no units

More atoms of potassium exist as ³⁹₁₉K (93.1%) therefore RAM is nearer to the more abundant ³⁹₁₉K isotope.

Calculate the relative atomic mass of Neon given that it exist as;

90.92% ²⁰₁₀Ne , 0.26% ²¹₁₀Ne , 8.82% ²²₁₀Ne,

Working

100 atoms of Neon contains 90.92 atoms of ²⁰₁₀Ne isotopes

100 atoms of Neon contains 0.26 atoms of ²¹₁₀Ne isotopes

100 atoms of Neon contains 8.82 atoms of ²²₁₀Ne isotopes Therefore;

RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22)

Note that:

Relative atomic mass has no units

More atoms of Neon exist as ²⁰₁₀Ne (90.92%) therefore RAM is nearer to the more abundant ²⁰₁₀Ne isotope.

Calculate the relative atomic mass of Argon given that it exist as;

90.92% ²⁰₁₀Ne , 0.26% ²¹₁₀Ne , 8.82% ²²₁₀Ne,

The relative atomic mass is a measure of the masses of atoms. The higher the relative atomic mass, the heavier the atom.

Electrons are found in energy levels/orbital.

An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy.

By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus.

Each energy level is occupied by a fixed number of electrons:

The 1^st energy level is occupied by a maximum of two electrons

The 2^nd energy level is occupied by a maximum of eight electrons

The 3^rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available)

The 4^th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available)

This arrangement of electrons in an atom is called electron configuration / structure.

By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to

Practice examples drawing electronic configurations

a)¹₁H has – in nucleus1proton and 0 neutrons

– 1 electron in the 1^st energy levels thus:

Nucleus

Energy levels

Electrons(represented by cross(x)

Electronic structure of Hydrogen is thus: 1:

b)⁴₂He has – in nucleus 2 proton and 2 neutrons – 2 electron in the 1^st energy levels thus:

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Helium is thus: 2:

c)⁷₃Li has – in nucleus 3 proton and 4 neutrons

– 2 electron in the 1^st energy levels

–1 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Lithium is thus: 2:1

d)⁹₄Be has – in nucleus 4 proton and 5 neutrons

– 2 electron in the 1^st energy levels

–2 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Beryllium is thus: 2:2

e)¹¹₅B has – in nucleus 5 proton and 6 neutrons

– 2 electron in the 1^st energy levels

–3 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Boron is thus: 2:3

f)¹²₆C has – in nucleus 6 proton and 6 neutrons

– 2 electron in the 1^st energy levels

–4 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Carbon is thus: 2:4

g)¹⁴₇N has – in nucleus 7 proton and 7 neutrons

– 2 electron in the 1^st energy levels

–5 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Nitrogen is thus: 2:5

h)¹⁶₈O has – in nucleus 8 proton and 8 neutrons

– 2 electron in the 1^st energy levels

–6 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Oxygen is thus: 2:6

i)¹⁹₉F has – in nucleus 9 proton and 10 neutrons

– 2 electron in the 1^st energy levels

–7 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Fluorine is thus: 2:7

i)²⁰₁₀Ne has – in nucleus 10 proton and 10 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels thus

Nucleus

Energy levels

Electrons (represented by cross(x)

Electronic structure of Neon is thus: 2:8

j)²³₁₁Na has – in nucleus 11 proton and 12 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–1 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Sodium is thus: 2:8:1

k)²⁴₁₂Mg has – in nucleus 12 proton and 12 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–2 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Magnesium is thus: 2:8:2

l)²⁷₁₃Al has – in nucleus 13 proton and 14 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–3 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Aluminium is thus: 2:8:3

m)²⁸₁₄Si has – in nucleus 14 proton and 14 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–4 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Silicon is thus: 2:8:4

n)³¹₁₅P has – in nucleus 14 proton and 15 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–5 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Phosphorus is thus: 2:8:5

o)³²₁₆S has – in nucleus 16 proton and 16 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–6 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Sulphur is thus: 2:8:6

p)³⁵₁₇Cl has – in nucleus 18 proton and 17 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–7 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Chlorine is thus: 2:8:7

p)⁴⁰₁₈Ar has – in nucleus 22 proton and 18 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–8 electron in the 3^rd energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Argon is thus: 2:8:8

q)³⁹₁₉K has – in nucleus 20 proton and 19 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–8 electron in the 3^rd energy levels

–1 electron in the 4^th energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Potassium is thus: 2:8:8:1

r)⁴⁰₂₀Ca has – in nucleus 20 proton and 20 neutrons

– 2 electron in the 1^st energy levels

–8 electron in the 2^nd energy levels

–8 electron in the 3^rd energy levels

–2 electron in the 4^th energy levels thus

Nucleus

Energy levels

Electrons (represented by dot(.)

Electronic structure of Calcium is thus: 2:8:8:2

B.PERIODIC TABLE

There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table.

A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers.

This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev.

The horizontal arrangement forms period. Atoms in the same period have the same the same number of energy levels in their electronic structure. i.e.

The number of energy levels in the electronic configuration of an element determine the period to which the element is in the periodic table.

e.g.

Which period of the periodic table are the following isotopes/elements/atoms?

¹²₆C

Electron structure 2:4 => 2 energy levels used thus Period 2

²³₁₁Na

Electron structure 2:8:1 => 3 energy levels used thus Period 3

³⁹₁₉K

Electron structure 2:8:8:1 => 4 energy levels used thus Period 4

¹₁H

Electron structure 1: => 1 energy level used thus Period 1

The vertical arrangement of elements forms a group. Atoms in the same have the same the same group have the same number of outer energy level electrons as per their electronic structure. i.e.

The number of electrons in the outer energy level an element determine the group to which the element is ,in the periodic table.

¹²₆C

Electron structure 2:4 => 4 electrons in outer energy level thus Group IV

²³₁₁C

Electron structure 2:8:1 => 1 electron in outer energy level thus Group I

³⁹₁₉K

Electron structure 2:8:8:1=>1 electron in outer energy level thus Group I

¹₁H

Electron structure 1: => 1 electron in outer energy level thus Group I

By convention;

(i)Periods are named using English numerals 1,2,3,4,…

(ii)Groups are named using Roman numerals I,II,III,IV,…

There are eighteen groups in a standard periodic table.

There are seven periods in a standard periodic table.

THE STANDARD PERIODIC TABLE OF ELEMENTS

When an atom has maximum number of electrons in its outer energy level, it is said to be stable.

When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable.

All stable atoms are in group 8/18 of the periodic table. All other elements are unstable.

All unstable atoms/isotopes try to be stable through chemical reactions. A chemical reaction involves gaining or losing outer electrons (electron transfer) .When electron transfer take place, an ion is formed.

An ion is formed when an unstable atom gain or donate electrons in its outer energy level inorder to be stable. Whether an atom gain or donate electrons depend on the relative energy required to donate or gain extra electrons i.e.

Examples

¹⁹₉F has electronic structure/configuration 2:7.

It can donate the seven outer electrons to have stable electronic structure/configuration 2:.

It can gain one extra electron to have stable electronic structure/configuration 2:8. Gaining requires less energy, and thus Fluorine reacts by gaining one extra electrons.

²³₁₃Al has electronic structure/configuration 2:8:3

It can donate the three outer electrons to have stable electronic structure/configuration 2:8.

It can gain five extra electrons to have stable electronic structure/configuration 2:8:8. Donating requires less energy, and thus Aluminium reacts by donating its three outer electrons.

Elements with less than four electrons in the outer energy level donates /lose the outer electrons to be stable and form a positively charged ion called cation.

A cation therefore has more protons(positive charge) than electrons(negative charge)

Generally metals usually form cation

Elements with more than four electrons in the outer energy level gain /acquire extra electrons in the outer energy level to be stable and form a negatively charged ion called anion.

An anion therefore has less protons(positive charge) than electrons(negative charge)

Generally non metals usually form anion. Except Hydrogen

The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost.

Examples of ion formation

1.¹₁H

H -> H⁺ + e

(atom) (monovalent cation) (electrons donated/lost)

Electronic configuration 1: (No electrons remains)

²⁷₁₃Al

Al -> Al³⁺ + 3e

(atom) (trivalent cation) (3 electrons donated/lost)

Electron 2:8:3 2:8

structure (unstable) (stable)

²³₁₁Na

Na -> Na⁺ + e

(atom) (cation) ( 1 electrons donated/lost)

Electron 2:8:1 2:8

structure (unstable) (stable)

²⁴₁₂Mg

Mg -> Mg²⁺ + 2e

(atom) (cation) ( 2 electrons donated/lost)

Electron 2:8:1 2:8

structure (unstable) (stable)

¹⁶₈O

O + 2e -> O^2-

(atom) ( 2 electrons gained/acquired) (anion)

Electron 2:6 2:8

structure (unstable) (stable)

¹⁴₇N

N + 3e -> N^3-

(atom) ( 3 electrons gained/acquired) (anion)

Electron 2:5 2:8

structure (unstable) (stable)

³¹₁₅P

P + 3e -> P^3-

(atom) ( 3 electrons gained/acquired) (anion)

Electron 2:5 2:8

structure (unstable) (stable)

¹⁹₉F

F + e -> F^–

(atom) ( 1 electrons gained/acquired) (anion)

Electron 2:7 2:8

structure (unstable) (stable)

³⁵₁₇Cl

Cl + e -> Cl^–

(atom) ( 1 electrons gained/acquired) (anion)

Electron 2:8:7 2:8:8

structure (unstable) (stable)

³⁹₁₉K

K -> K⁺ + e

(atom) (cation) ( 1 electrons donated/lost)

Electron 2:8:8:1 2:8:8

structure (unstable) (stable)

When an element donate/loses its outer electrons ,the process is called oxidation. When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state.

Table showing the oxidation states of some isotopes

Element	Symbol of element / isotopes	Charge of ion	Oxidation state
Hydrogen	¹₁H ²₁H(deuterium) ³₁H(Tritium)	H⁺ H⁺ H⁺	+1 +1 +1
Chlorine	³⁵₁₇Cl ³⁷₁₇Cl	Cl^– Cl^–	-1 -1
Potassium	³⁹₁₉K ⁴⁰₁₉K ⁴¹₁₉K	K⁺ K⁺ K⁺	+1 +1 +1
Oxygen	¹⁶₈O ¹⁸₈O	O^2- O^2-	-2 -2
Magnesium	²⁴₁₂Mg	Mg²⁺	+2
sodium	²³₁₁Na	Na⁺	+1
Copper	Cu	Cu⁺ Cu²⁺	+1 +2
Iron		Fe²⁺ Fe³⁺	+2 +3
Lead		Pb²⁺ Pb⁴⁺	+2 +4
Manganese		Mn²⁺ Mn⁷⁺	+2 +7
Chromium		Cr³⁺ Cr⁶⁺	+3 +6
Sulphur		S⁴⁺ S⁶⁺	+4 +6
Carbon		C²⁺ C⁴⁺	+2 +4

Note :

Some elements can exist in more than one oxidation state.They are said to have variable oxidation state.

Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound.

Examples:

Copper (I) means Cu⁺as in Copper(I)oxide
Copper (II) means Cu²⁺as in Copper(II)oxide
Iron (II) means Fe²⁺as in Iron(II)sulphide

(iv) Iron (III) means Fe³⁺as in Iron(III)chloride

Sulphur(VI)mean S⁶⁺as in Iron(III)sulphate(VI)
Sulphur(VI)mean S⁶⁺as in sulphur(VI)oxide
Sulphur(IV)mean S⁴⁺as in sulphur(IV)oxide
Sulphur(IV)mean S⁴⁺as in sodium sulphate(IV)

(ix) Carbon(IV)mean C⁴⁺as in carbon(IV)oxide

(x) Carbon(IV)mean C⁴⁺as in Lead(II)carbonate(IV)

(xi) Carbon(II)mean C²⁺as in carbon(II)oxide

(xii) Manganese(IV)mean Mn⁴⁺as in Manganese(IV)oxide

A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae. Elements combine together to form a compound depending on their combining power.

The combining power of atoms in an element is called Valency.Valency of an element is equal to the number of:

(i)hydrogen atoms that an atom of element can combine with or displace.

(ii)electrons gained /acquired in outer energy level by non metals to be stable/attain duplet/octet.

(iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet.

(iv)charges carried by ions/cations/ions

Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency.

Table showing the valency of common radicals.

Radical name	Chemical formulae	Combining power / Valency
Ammonium	NH₄⁺	1
Hydroxide	OH^–	1
Nitrate(V)	NO₃^–	1
Hydrogen carbonate	HCO₃^–	1
Hydrogen sulphate(VI)	HSO₄^–	1
Hydrogen sulphate(IV)	HSO₃^–	1
Manganate(VII)	MnO₄^–	1
Chromate(VI)	CrO₄^2-	2
Dichromate(VI)	Cr₂O₇^2-	2
Sulphate(VI)	SO₄^2-	2
Sulphate(IV)	SO₃^2-	2
Carbonate(IV)	CO₃^2-	2
Phosphate(V)	PO₄^2-	3

Table showing the valency of some common metal and non metals

Element/metal	Valency	Element/non metal	Valency
Hydrogen	1	Florine	1
Lithium	1	Chlorine	1
Beryllium	2	Bromine	1
Boron	3	Iodine	1
Sodium	1	Carbon	4
Magnesium	2	Nitrogen	3
Aluminium	3	Oxygen	2
Potassium	1	Phosphorus	3
Calcium	2
Zinc	2
Barium	2
Mercury	2
Iron	2 and 3
Copper	1 and 2
Manganese	2 and 4
Lead	2 and 4

From the valency of elements , the chemical formular of a compound can be derived using the following procedure:

(i)Identify the elements and radicals making the compound

(ii)Write the symbol/formular of the elements making the compound starting with the metallic element

(iii)Assign the valency of each element /radical as superscript.

(iv)Interchange/exchange the valencies of each element as subscript.

(v)Divide by the smallest/lowest valency to derive the smallest whole number ratios

Ignore a valency of 1.

This is the chemical formula.

Practice examples

Write the chemical formula of

(a)Aluminium oxide

Elements making compound	Aluminium	Oxygen
Symbol of elements/radicals in compound	Al	O
Assign valencies as superscript	Al³	O²
Exchange/Interchange the valencies as subscript	Al₂	O₃
Divide by smallest valency to get whole number	–	–

Chemical formula of Aluminium oxide is thus: Al₂ O₃

This means:2atoms of Aluminium combine with 3 atoms of Oxygen

(b)Sodium oxide

Elements making compound	Sodium	Oxygen
Symbol of elements/radicals in compound	Na	O
Assign valencies as superscript	Na¹	O²
Exchange/Interchange the valencies as subscript	Na₂	O₁
Divide by smallest valency to get whole number	–	–

Chemical formula of Sodium oxide is thus: Na₂ O

This means:2atoms of Sodium combine with 1 atom of Oxygen

(c)Calcium oxide

Elements making compound	Calcium	Oxygen
Symbol of elements/radicals in compound	Ca	O
Assign valencies as superscript	Ca²	O²
Exchange/Interchange the valencies as subscript	Ca₂	O₂
Divide by two to get smallest whole number ratio	Ca₁	O₁

Chemical formula of Calcium oxide is thus: CaO

This means:1 atom of calcium combine with 1 atom of Oxygen.

(d)Lead(IV)oxide

Elements making compound	Lead	Oxygen
Symbol of elements/radicals in compound	Pb	O
Assign valencies as superscript	Pb⁴	O²
Exchange/Interchange the valencies as subscript	Pb₂	O₄
Divide by two to get smallest whole number ratio	Pb₁	O₂

Chemical formula of Lead(IV) oxide is thus: PbO₂

This means:1 atom of lead combine with 2 atoms of Oxygen.

(e)Lead(II)oxide

Elements making compound	Lead	Oxygen
Symbol of elements/radicals in compound	Pb	O
Assign valencies as superscript	Pb²	O²
Exchange/Interchange the valencies as subscript	Pb₂	O₂
Divide by two to get smallest whole number ratio	Pb₁	O₁

Chemical formula of Lead(II) oxide is thus: PbO

This means:1 atom of lead combine with 1 atom of Oxygen.

(e)Iron(III)oxide

Elements making compound	Iron	Oxygen
Symbol of elements/radicals in compound	Fe	O
Assign valencies as superscript	Fe³	O²
Exchange/Interchange the valencies as subscript	Fe₂	O₃
Divide by two to get smallest whole number ratio	–	–

Chemical formula of Iron(III) oxide is thus: Fe₂O₃

This means:2 atom of lead combine with 3 atom of Oxygen.

(f)Iron(II)sulphate(VI)

Elements making compound	Iron	sulphate(VI)
Symbol of elements/radicals in compound	Fe	SO₄
Assign valencies as superscript	Fe²	SO₄²
Exchange/Interchange the valencies as subscript	Fe₂	SO_{4 2}
Divide by two to get smallest whole number ratio	Fe₁	SO₄ ₁

Chemical formula of Iron(II) sulphate(VI) is thus: FeSO₄

This means:1 atom of Iron combine with 1 sulphate(VI) radical.

(g)Copper(II)sulphate(VI)

Elements making compound	Copper	sulphate(VI)
Symbol of elements/radicals in compound	Cu	SO₄
Assign valencies as superscript	Cu²	SO₄²
Exchange/Interchange the valencies as subscript	Cu₂	SO_{4 2}
Divide by two to get smallest whole number ratio	Cu₁	SO₄ ₁

Chemical formula of Cu(II)sulphate(VI) is thus: CuSO₄

This means:1 atom of Copper combine with 1 sulphate(VI) radical.

(h)Aluminium sulphate(VI)

Elements making compound	Aluminium	sulphate(VI)
Symbol of elements/radicals in compound	Al	SO₄
Assign valencies as superscript	Al³	SO₄²
Exchange/Interchange the valencies as subscript	Al₂	SO_{4 3}
Divide by two to get smallest whole number ratio	–	–

Chemical formula of Aluminium sulphate(VI) is thus: Al₂(SO₄)₃

This means:2 atom of Aluminium combine with 3 sulphate(VI) radical.

(i)Aluminium nitrate(V)

Elements making compound	Aluminium	nitrate(V)
Symbol of elements/radicals in compound	Al	NO₃
Assign valencies as superscript	Al³	NO₃¹
Exchange/Interchange the valencies as subscript	Al₁	NO_{3 3}
Divide by two to get smallest whole number ratio	–	–

Chemical formula of Aluminium sulphate(VI) is thus: Al (NO₃)₃

This means:1 atom of Aluminium combine with 3 nitrate(V) radical.

(j)Potassium manganate(VII)

Elements making compound	Potassium		manganate(VII)
Symbol of elements/radicals in compound	K		MnO₄
Assign valencies as superscript	K ¹		MnO₄¹
Exchange/Interchange the valencies as subscript	K₁		MnO_{4 1}
Divide by two to get smallest whole number ratio		–	–

Chemical formula of Potassium manganate(VII) is thus: KMnO₄

This means:1 atom of Potassium combine with 4 manganate(VII) radical.

(k)Sodium dichromate(VI)

Elements making compound	Sodium		dichromate(VI)
Symbol of elements/radicals in compound	Na		Cr₂O₇
Assign valencies as superscript	Na ¹		Cr₂O₇²
Exchange/Interchange the valencies as subscript	Na₂		Cr₂O_{7 1}
Divide by two to get smallest whole number ratio		–	–

Chemical formula of Sodium dichromate(VI) is thus: Na₂ Cr₂O₇

This means:2 atom of Sodium combine with 1 dichromate(VI) radical.

(l)Calcium hydrogen carbonate

Elements making compound	Calcium		Hydrogen carbonate
Symbol of elements/radicals in compound	Ca		CO₃
Assign valencies as superscript	Ca ²		HCO₃¹
Exchange/Interchange the valencies as subscript	Ca₁		HCO_{3 2}
Divide by two to get smallest whole number ratio		–	–

Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO₃)₂

This means:1 atom of Calcium combine with 2 hydrogen carbonate radical.

(l)Magnesium hydrogen sulphate(VI)

Elements making compound	Magnesium		Hydrogen sulphate(VI)
Symbol of elements/radicals in compound	Mg		HSO₄
Assign valencies as superscript	Mg ²		HSO₄¹
Exchange/Interchange the valencies as subscript	Mg₁		HSO_{4 2}
Divide by two to get smallest whole number ratio		–	–

Chemical formula of Magnesium hydrogen sulphate(VI) is thus: Mg(HSO₄)₂

This means:1 atom of Magnesium combine with 2 hydrogen sulphate(VI) radical.

Compounds are formed from chemical reactions. A chemical reaction is formed when atoms of the reactants break free to bond again and form products. A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:

Write the word equation
Write the correct chemical formula for each of the reactants and products
Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.
Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. This is called balancing.

Do not change the chemical formula of the products/reactants.

Assign in brackets, the physical state/state symbols of the reactants and products after each chemical formula as:

(i) (s) for solids

(ii) (l) for liquids

(iii) (g) for gas

(iv) (aq) for aqueous/dissolved in water to make a solution.

Practice examples

Write a balanced chemical equation for the following

Hydrogen gas is prepared from reacting Zinc granules with dilute hydrochloric acid.

Procedure

Write the word equation

Zinc + Hydrochloric acid -> Zinc chloride + hydrogen gas

Write the correct chemical formula for each of the reactants and products

Zn + HCl -> ZnCl₂ + H₂

Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.

Number of atoms of Zn on the reactant side is equal to product side

One atom of H in HCl on the reactant side is not equal to two atoms in H₂ on product side.

One atom of Cl in HCl on the reactant side is not equal to two atoms in ZnCl₂ on product side.

Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal.

Multiply HCl by “2” to get “2” Hydrogen and “2” Chlorine on product and reactant side.

Zn + 2 HCl -> ZnCl₂ + H₂

Assign in brackets, the physical state/state symbols .

Zn(s) + 2 HCl(aq) -> ZnCl₂ (aq) + H₂(g)

Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water

Procedure

Write the word equation

Hydrogen peroxide -> Water + oxygen gas

Write the correct chemical formula for each of the reactants and products

H₂O₂ -> H₂O + O₂

Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.

Number of atoms of H on the reactant side is equal to product side

Two atom of O in H₂O₂ on the reactant side is not equal to three atoms (one in H₂O and two in O₂) on product side.

Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal.

Multiply H₂O₂ by “2” to get “4” Hydrogen and “4” Oxygen on reactants

Multiply H₂O by “2” to get “4” Hydrogen and “2” Oxygen on product side

When the “2” Oxygen in O₂and the“2” in H₂O are added on product side they are equal to the“4” Oxygen on reactants side.

2H₂O₂ -> 2H₂O + O₂

Assign in brackets, the physical state/state symbols .

2H₂O₂(aq) -> 2H₂O(l) + O₂(g)

Chlorine gas is prepared from Potassium manganate(VII) reacting with hydrochloric acid to form potassium chloride solution, manganese(II) chloride solution,water and chlorine gas.

Procedure

Write the word equation

Potassium manganate(VII) + Hydrochloric acid ->

potassium chloride + manganese(II) chloride + chlorine +water

Write the correct chemical formula for each of the reactants and products

KMnO₄ + HCl -> KCl + MnCl₂ +H₂O + Cl₂

Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.

Number of atoms of K and Mn on the reactant side is equal to product side

Two atom of H in H₂O on the product side is not equal to one atom on reactant side.

Four atom of O in KMnO₄ is not equal to one in H₂O

One atom of Cl in HCl on reactant side is not equal to three (one in H₂O and two in Cl₂)

Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal.

Multiply HCl by “16” to get “16” Hydrogen and “16” Chlorine on reactants

Multiply KMnO₄ by “2” to get “2” Potassium and “2” manganese, “2 x4 =8” Oxygen on reactant side.

Balance the product side to get:

2 KMnO₄ +16 HCl -> 2 KCl + 2 MnCl₂ +8 H₂O + 5 Cl₂

Assign in brackets, the physical state/state symbols .

2KMnO₄(s) +16 HCl(aq)-> 2 KCl (aq) + 2MnCl₂(aq)+8 H₂O(l)+5 Cl₂(g)

(d)Carbon(IV)oxide gas is prepared from Calcium carbonate reacting with hydrochloric acid to form calcium chloride solution, water and carbon(IV)oxide gas.

Procedure

Write the word equation

Calcium carbonate + Hydrochloric acid ->

calcium chloride solution+ water +carbon(IV)oxide

Write the correct chemical formula for each of the reactants and products

CaCO₃ + HCl -> CaCl₂ +H₂O + CO₂

Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.

Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal.

Assign in brackets, the physical state/state symbols .

CaCO₃(s) + 2 HCl(aq) -> CaCl₂(aq) + H₂O(l) + CO₂(g)

(d)Sodium hydroxide solution neutralizes hydrochloric acid to form salt and water.

NaOH(aq) + HCl(aq) -> NaCl (aq) + H₂O(l)

(e)Sodium reacts with water to form sodium hydroxide and hydrogen gas.

2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)

(f)Calcium reacts withwater to form calcium hydroxide and hydrogen gas

Ca(s) + 2H₂O(l) -> Ca(OH)₂(aq) + H₂(g)

(g)Copper(II)Oxide solid reacts with dilute hydrochloric acid to form copper(II)chloride and water.

CuO(s) + 2HCl(aq) -> CuCl₂(aq) + H₂O(l)

(h)Hydrogen sulphide reacts with Oxygen to form sulphur(IV)Oxide and water.

2H₂S(g) + 3O₂(g) -> 2SO₂(g) + 2H₂O(l)

(i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas.

Mg(s) + 2H₂O(g) -> MgO(s) + H₂(g)

(j)Ethane(C₂H₆) gas burns in air to form Carbon(IV)Oxide and water.

2C₂H₆(g) + 7O₂(g) -> 4CO₂(g) + 6H₂O(l)

(k)Ethene(C₂H₄) gas burns in air to form Carbon(IV)Oxide and water.

C₂H₄(g) + 3O₂(g) -> 2CO₂(g) + 2H₂O(l)

(l)Ethyne(C₂H₂) gas burns in air to form Carbon(IV)Oxide and water.

2C₂H₂(g) + 5O₂(g) -> 4CO₂(g) + 2H₂O(l)

C.PERIODICITY OF CHEMICAL FAMILES/DOWN THE GROUP.

The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table.i.e

The number of occupied energy levels determine the Period and the valence electrons determine the Group.

Elements in the same group have similar physical and chemical properties. The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family.

Group I elements: Alkali metals

Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include:

Element	Symbol	Atomic number	Electron structure	Oxidation state	Valency
Lithium	Li	3	2:1	Li⁺	1
Sodium	Na	11	2:8:1	Na⁺	1
Potassium	K	19	2:8:8:1	K⁺	1
Rubidium	Rb	37	2:8:18:8:1	Rb⁺	1
Caesium	Cs	55	2:8:18:18:8:1	Cs⁺	1
Francium	Fr	87	2:8:18:32:18:8:1	Fr⁺	1

All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M⁺

The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g.

The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels).

Atomic and ionic radius

The distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron/s is called atomic radius. Atomic radius is measured in nanometers(n).The higher /bigger the atomic radius the bigger /larger the atomic size.

The distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron/s is called ionic radius. Ionic radius is also measured in nanometers(n).The higher /bigger the ionic radius the bigger /larger the size of the ion.

Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g.

The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels).

Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases.

The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level.

Table showing the atomic and ionic radius of some alkali metals

Element	Symbol	Atomic number	Atomic radius(nM)	Ionic radius(nM)
Lithium	Li	3	0.133	0.060
Sodium	Na	11	0.157	0.095
Potassium	K	19	0.203	0.133

The atomic radius of sodium is 0.157nM .The ionic radius of Na⁺ is 0.095nM. This is because sodium reacts by donating/losing the outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius.

Electropositivity

The ease of donating/losing electrons is called electropositivity. All alkali metals are electropositive. Electropositivity increase as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and can be lost/ donated easily/with ease. Francium is the most electropositive element in the periodic table because it has the highest/biggest atomic radius.

Ionization energy

The minimum amount of energy required to remove an electron from an atom of element in its gaseous state is called 1^st ionization energy. The SI unit of ionization energy is kilojoules per mole/kJmole^-1 .Ionization energy depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer electrons/energy level and thus the lower the ionization energy. For alkali metals the 1^st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease.

e.g. The 1^st ionization energy of sodium is 496 kJmole^-1 while that of potassium is 419 kJmole^-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium.

Physical properties

Soft/Easy to cut: Alkali metals are soft and easy to cut with a knife. The softness and ease of cutting increase down the group from Lithium to Francium. This is because an increase in atomic radius, decreases the strength of metallic bond and the packing of the metallic structure

Appearance: Alkali metals have a shiny grey metallic luster when freshly cut. The surface rapidly/quickly tarnishes on exposure to air. This is because the metal surface rapidly/quickly reacts with elements of air/oxygen.

Melting and boiling points: Alkali metals have a relatively low melting/boiling point than common metals like Iron. This is because alkali metals use only one delocalized electron to form a weak metallic bond/structure.

Electrical/thermal conductivity: Alkali metals are good thermal and electrical conductors. Metals conduct using the outer mobile delocalized electrons. The delocalized electrons move randomly within the metallic structure.

Summary of some physical properties of the 1^st three alkali metals

Alkali metal	Appearance	Ease of cutting	Melting point (^oC)	Boiling point (^oC)	Conductivity	1^st ionization energy
Lithium	Silvery white	Not easy	180	1330	Good	520
Sodium	Shiny grey	Easy	98	890	Good	496
Potassium	Shiny grey	Very easy	64	774	Good	419

Chemical properties

(i)Reaction with air/oxygen

On exposure to air, alkali metals reacts with the elements in the air.

Example

On exposure to air, Sodium first reacts with Oxygen to form sodium oxide.

4Na(s) + O₂(g) -> 2Na₂O(s)

The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution.

Na₂O(s) + H₂O(l) -> 2NaOH(aq)

Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate.

2NaOH(aq) + CO₂(g) -> Na₂CO₃(g) + H₂O(l)

(ii)Burning in air/oxygen

Lithium burns in air with a crimson/deep red flame to form Lithium oxide

4Li (s) + O₂(g) -> 2Li₂O(s)

Sodium burns in air with a yellow flame to form sodium oxide

4Na (s) + O₂(g) -> 2Na₂O(s)

Sodium burns in oxygen with a yellow flame to form sodium peroxide

2Na (s) + O₂(g) -> Na₂O₂ (s)

Potassium burns in air with a lilac/purple flame to form potassium oxide

4K (s) + O₂(g) -> 2K₂O (s)

(iii) Reaction with water:

Experiment

Measure 500 cm3 of water into a beaker.

Put three drops of phenolphthalein indicator.

Put about 0.5g of Lithium metal into the beaker.

Determine the pH of final product

Repeat the experiment using about 0.1 g of Sodium and Potassium.

Caution: Keep a distance

Observations

Alkali metal	Observations	Comparative speed/rate of the reaction
Lithium	-Metal floats in water -rapid effervescence/fizzing/bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14	Moderately vigorous
Sodium	-Metal floats in water -very rapid effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14	Very vigorous
Potassium	-Metal floats in water -explosive effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14	Explosive/burst into flames

Explanation

Alkali metals are less dense than water. They therefore float in water.They react with water to form a strongly alkaline solution of their hydroxides and producing hydrogen gas. The rate of this reaction increase down the group. i.e. Potassium is more reactive than sodium .Sodium is more reactive than Lithium.

The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions.

Chemical equations

2Li(s) + 2H₂O(l) -> 2LiOH(aq) + H₂(g)

2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)

2K(s) + 2H₂O(l) -> 2KOH(aq) + H₂(g)

2Rb(s) + 2H₂O(l) -> 2RbOH(aq) + H₂(g)

2Cs(s) + 2H₂O(l) -> 2CsOH(aq) + H₂(g)

2Fr(s) + 2H₂O(l) -> 2FrOH(aq) + H₂(g)

Reactivity increase down the group

(iv) Reaction with chlorine:

Experiment

Cut about 0.5g of sodium into a deflagrating spoon with a lid cover. Introduce it on a Bunsen flame until it catches fire. Quickly and carefully lower it into a gas jar containing dry chlorine to cover the gas jar.

Repeat with about 0.5g of Lithium.

Caution: This experiment should be done in fume chamber because chlorine is poisonous /toxic.

Observation

Sodium metal continues to burn with a yellow flame forming white solid/fumes.

Lithium metal continues to burn with a crimson flame forming white solid / fumes.

Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus.

Chemical equations

2Li(s) + Cl₂(g) -> 2LiCl(s)

2Na(s) + Cl₂(g) -> 2NaCl(s)

2K(s) + Cl₂(g) -> 2KCl(s)

2Rb(s) + Cl₂(g) -> 2RbCl(s)

2Cs(s) + Cl₂(g) -> 2CsCl(s)

2Fr(s) + Cl₂(g) -> 2FrCl(s) Reactivity increase down the group

The table below shows some compounds of the 1^st three alkali metals

	Lithium	sodium	Potassium
Hydroxide	LiOH	NaOH	KOH
Oxide	Li₂O	Na₂O	K₂O
Sulphide	Li₂S	Na₂S	K₂S
Chloride	LiCl	NaCl	KCl
Carbonate	Li₂CO₃	Na₂CO₃	K₂CO₃
Nitrate(V)	LiNO₃	NaNO₃	KNO₃
Nitrate(III)	–	NaNO₂	KNO₂
Sulphate(VI)	Li₂SO₄	Na₂SO₄	K₂SO₄
Sulphate(IV)	–	Na₂SO₃	K₂SO₃
Hydrogen carbonate	–	NaHCO₃	KHCO₃
Hydrogen sulphate(VI)	–	NaHSO₄	KHSO₄
Hydrogen sulphate(IV)	–	NaHSO₃	KHSO₃
Phosphate	–	Na₃PO₄	K₃PO₄
Manganate(VI)	–	NaMnO₄	KMnO₄
Dichromate(VI)	–	Na₂Cr₂O₇	K₂Cr₂O₇
Chromate(VI)	–	Na₂CrO₄	K₂CrO₄

Some uses of alkali metals include:

(i)Sodium is used in making sodium cyanide for extracting gold from gold ore.

(ii)Sodium chloride is used in seasoning food.

(iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors.

(iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents.

(v)Sodium is used as a reducing agent for the extraction of titanium from Titanium(IV)chloride.

(vi)Lithium is used in making special high strength glasses

(vii)Lithium compounds are used to make dry cells in mobile phones and computer laptops.

Group II elements: Alkaline earth metals

Group II elements are called Alkaline earth metals . The alkaline earth metals include:

Element	Symbol	Atomic number	Electron structure	Oxidation state	Valency
Beryllium	Be	4	2:2	Be²⁺	2
Magnesium	Mg	12	2:8:2	Mg²⁺	2
Calcium	Ca	20	2:8:8:2	Ca²⁺	2
Strontium	Sr	38	2:8:18:8:2	Sr²⁺	2
Barium	Ba	56	2:8:18:18:8:2	Ba²⁺	2
Radium	Ra	88	2:8:18:32:18:8:2	Ra²⁺	2

All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent. They donate /lose the two outer electrons to have oxidation state M²⁺

The number of energy levels increases down the group from Beryllium to Radium. The more the number of energy levels the bigger/larger the atomic size. e.g.

The atomic size/radius of Calcium is bigger/larger than that of Magnesium because Calcium has more/4 energy levels than Magnesium (3 energy levels).

Atomic radius and ionic radius of alkaline earth metals increase down the group as the number of energy levels increases.

The atomic radius of alkaline earth metals is bigger than the ionic radius. This is because they react by losing/donating the two outer electrons and hence lose the outer energy level.

Table showing the atomic and ionic radius of the 1^st three alkaline earth metals

Element	Symbol	Atomic number	Atomic radius(nM)	Ionic radius(nM)
Beryllium	Be	4	0.089	0.031
Magnesium	Mg	12	0.136	0.065
Calcium	Ca	20	0.174	0.099

The atomic radius of Magnesium is 0.136nM .The ionic radius of Mg²⁺ is 0.065nM. This is because Magnesium reacts by donating/losing the two outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius.

Electropositivity

All alkaline earth metals are also electropositive like alkali metals. The electropositivity increase with increase in atomic radius/size. Calcium is more electropositive than Magnesium. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The two outer electrons in calcium experience less nuclear attraction and can be lost/ donated easily/with ease because of the higher/bigger atomic radius.

Ionization energy

For alkaline earth metals the 1^st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease.

e.g. The 1^st ionization energy of Magnesium is 900 kJmole^-1 while that of Calcium is 590 kJmole^-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from magnesium to calcium.

It requires therefore less energy to donate/lose outer electron in calcium than in magnesium.

The minimum amount of energy required to remove a second electron from an ion of an element in its gaseous state is called the 2^nd ionization energy.

The 2^nd ionization energy is always higher /bigger than the 1^st ionization energy.

This because once an electron is donated /lost form an atom, the overall effective nuclear attraction on the remaining electrons/energy level increase. Removing a second electron from the ion require therefore more energy than the first electron.

The atomic radius of alkali metals is higher/bigger than that of alkaline earth metals.This is because across/along the period from left to right there is an increase in nuclear charge from additional number of protons and still additional number of electrons entering the same energy level. Increase in nuclear charge increases the effective nuclear attraction on the outer energy level which pulls it closer to the nucleus. e.g.

Atomic radius of Sodium (0.157nM) is higher than that of Magnesium (0.137nM). This is because Magnesium has more effective nuclear attraction on the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus.

Physical properties

Soft/Easy to cut: Alkaline earth metals are not soft and easy to cut with a knife like alkali metals. This is because of the decrease in atomic radius of corresponding alkaline earth metal, increases the strength of metallic bond and the packing of the metallic structure. Alkaline earth metals are

(i)ductile(able to form wire/thin long rods)

(ii)malleable(able to be hammered into sheet/long thin plates)

(iii)have high tensile strength(able to be coiled without breaking/ not brittle/withstand stress)

Appearance: Alkali earth metals have a shiny grey metallic luster when their surface is freshly polished /scrubbed. The surface slowly tarnishes on exposure to air. This is because the metal surface slowly undergoes oxidation to form an oxide. This oxide layer should be removed before using the alkaline earth metals.

Melting and boiling points: Alkaline earth metals have a relatively high melting/ boiling point than alkali metals. This is because alkali metals use only one delocalized electron to form a weaker metallic bond/structure. Alkaline earth metals use two delocalized electrons to form a stronger metallic bond /structure.

The melting and boiling points decrease down the group as the atomic radius/size increase reducing the strength of metallic bond and packing of the metallic structure. e.g.

Beryllium has a melting point of 1280^oC. Magnesium has a melting point of 650^oC.Beryllium has a smaller atomic radius/size than magnesium .The strength of metallic bond and packing of the metallic structure is thus stronger in beryllium.

Electrical/thermal conductivity: Alkaline earth metals are good thermal and electrical conductors. The two delocalized valence electrons move randomly within the metallic structure.

Electrical conductivity increase down the group as the atomic radius/size increase making the delocalized outer electrons less attracted to nucleus. Alkaline earth metals are better thermal and electrical conductors than alkali metals because they have more/two outer delocalized electrons.e.g.

Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium. The more delocalized electrons the better the electrical conductor.

Calcium is a better conductor than magnesium.

Calcium has bigger/larger atomic radius than magnesium because the delocalized electrons are less attracted to the nucleus of calcium and thus more free /mobile and thus better the electrical conductor

Summary of some physical properties of the 1^st three alkaline earth metals

Alkaline earth metal	Appearance	Ease of cutting	Melting point (^oC)	Boiling point (^oC)	Conduct- ivity	1^st ionization energy	2^nd ionization energy
Beryllium	Shiny grey	Not easy	1280	3450	Good	900	1800
Magnesium	Shiny grey	Not Easy	650	1110	Good	736	1450
calcium	Shiny grey	Not easy	850	1140	Good	590	970

Chemical properties

(i)Reaction with air/oxygen

On exposure to air, the surface of alkaline earth metals is slowly oxidized to its oxide on prolonged exposure to air.

Example

On exposure to air, the surface of magnesium ribbon is oxidized to form a thin film of Magnesium oxide

. 2Mg(s) + O₂(g) -> 2MgO(s)

(ii)Burning in air/oxygen

Experiment

Hold a about 2cm length of Magnesium ribbon on a Bunsen flame. Stop heating when it catches fire/start burning.

Caution: Do not look directly at the flame

Put the products of burning into 100cm3 beaker. Add about 5cm3 of distilled water. Swirl. Test the mixture using litmus papers.
Repeat with Calcium

Observations

-Magnesium burns with a bright blindening flame

-White solid /ash produced

-Solid dissolves in water to form a colourless solution

-Blue litmus paper remain blue

-Red litmus paper turns blue

-colourless gas with pungent smell of urine

Explanation

Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride.

2Mg (s) + O₂(g) -> 2MgO(s)

3Mg (s) + N₂(g) -> Mg₃N₂ (s)

Magnesium oxide dissolves in water to form magnesium hydroxide.

MgO(s) + H₂O (l) -> Mg(OH)₂(aq)

Magnesium nitride dissolves in water to form magnesium hydroxide and produce ammonia gas.

Mg₃N₂ (s) + 6H₂O(l) -> 3Mg(OH)₂(aq) + 2NH₃ (g)

Magnesium hydroxide and ammonia are weakly alkaline with pH 8/9/10/11 and turns red litmus paper blue.

Calcium burns in air with faint orange/red flame to form a mixture of both Calcium oxide and calcium nitride.

2Ca (s) + O₂(g) -> 2CaO(s)

3Ca (s) + N₂(g) -> Ca₃N₂ (s)

Calcium oxide dissolves in water to form calcium hydroxide.

CaO(s) + H₂O(l) -> Ca(OH)₂(aq)

Calcium nitride dissolves in water to form calcium hydroxide and produce ammonia gas.

Ca₃N₂ (s) + 6H₂O(l) -> 3Ca(OH)₂(aq) + 2NH₃ (g)

Calcium hydroxide is also weakly alkaline solution with pH 8/9/10/11 and turns red litmus paper blue.

(iii)Reaction with water

Experiment

Measure 50 cm3 of distilled water into a beaker.

Scrub/polish with sand paper 1cm length of Magnesium ribbon

Place it in the water. Test the product-mixture with blue and red litmus papers.

Repeat with Calcium metal.

Observations

-Surface of magnesium covered by bubbles of colourless gas.

-Colourless solution formed.

-Effervescence/bubbles/fizzing takes place in Calcium.

-Red litmus paper turns blue.

-Blue litmus paper remains blue.

Explanations

Magnesium slowly reacts with cold water to form Magnesium hydroxide and bubbles of Hydrogen gas that stick on the surface of the ribbon.

Mg(s) + 2H₂O (l) -> Mg(OH)₂(aq) + H₂ (g)

Calcium moderately reacts with cold water to form Calcium hydroxide and produce a steady stream of Hydrogen gas.

Ca(s) + 2H₂O (l) -> Ca(OH)₂(aq) + H₂ (g)

(iv)Reaction with water vapour/steam

Experiment

Put some cotton wool soaked in water/wet sand in a long boiling tube.

Coil a well polished magnesium ribbon into the boiling tube.

Ensure the coil touches the side of the boiling tube. Heat the cotton wool/sand slightly then strongly heat the Magnesium ribbon .

Set up of apparatus

Observations

-Magnesium glows red hot then burns with a blindening flame.

-Magnesium continues to glow/burning even without more heating.

-White solid/residue.

-colourless gas collected over water.

Explanation

On heating wet sand, steam is generated which drives out the air that would otherwise react with /oxidize the ribbon.

Magnesium burns in steam/water vapour generating enough heat that ensures the reaction goes to completion even without further heating. White Magnesium oxide is formed and hydrogen gas is evolved.

To prevent suck back, the delivery tube should be removed from the water before heating is stopped at the end of the experiment.

Mg(s) + H₂O (l) -> MgO(s) + H₂ (g)

(v)Reaction with chlorine gas.

Experiment

Lower slowly a burning magnesium ribbon/shavings into a gas jar containing Chlorine gas. Repeat with a hot piece of calcium metal.

Observation

-Magnesium continues to burn in chlorine with a bright blindening flame.

-Calcium continues to burn for a short time.

-White solid formed .

-Pale green colour of chlorine fades.

Explanation

Magnesium continues to burn in chlorine gas forming white magnesium oxide solid.

Mg(s) + Cl₂ (g) -> MgCl₂ (s)

Calcium burns slightly in chlorine gas to form white calcium oxide solid. Calcium oxide formed coat unreacted Calcium stopping further reaction

Ca(s) + Cl₂ (g) -> CaCl₂ (s)

(v)Reaction with dilute acids.

Experiment

Place about 4.0cm3 of 0.1M dilute sulphuric(VI)acid into a test tube. Add about 1.0cm length of magnesium ribbon into the test tube. Cover the mouth of the test tube using a thumb. Release the gas and test the gas using a burning splint.

Repeat with about 4.0cm3 of 0.1M dilute hydrochloric/nitric(V) acid.

Repeat with 0.1g of Calcium in a beaker with all the above acid

Caution: Keep distance when using calcium

Observation

-Effervescence/fizzing/bubbles with dilute sulphuric(VI) and nitric(V) acids

-Little Effervescence/fizzing/bubbles with calcium and dilute sulphuric(VI) acid.

-Colourless gas produced that extinguishes a burning splint with an explosion/ “pop” sound.

-No gas is produced with Nitric(V)acid.

-Colourless solution is formed.

Explanation

Dilute acids react with alkaline earth metals to form a salt and produce hydrogen gas.

Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water.

Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used.

Chemical equations

Mg(s) + H₂SO₄ (aq) -> MgSO₄(aq) + H₂ (g)

Mg(s) + 2HNO₃ (aq) -> Mg(NO₃)₂(aq) + H₂ (g)

Mg(s) + 2HCl (aq) -> MgCl₂(aq) + H₂ (g)

Ca(s) + H₂SO₄ (aq) -> CaSO₄(s) + H₂ (g)

(insoluble CaSO₄(s) coat/cover Ca(s))

Ca(s) + 2HNO₃ (aq) -> Ca(NO₃)₂(aq) + H₂ (g)

Ca(s) + 2HCl (aq) -> CaCl₂(aq) + H₂ (g)

Ba(s) + H₂SO₄ (aq) -> BaSO₄(s) + H₂ (g)

(insoluble BaSO₄(s) coat/cover Ba(s))

Ba(s) + 2HNO₃ (aq) -> Ba(NO₃)₂(aq) + H₂ (g)

Ba(s) + 2HCl (aq) -> BaCl₂(aq) + H₂ (g)

The table below shows some compounds of some alkaline earth metals

	Beryllium	Magnesium	Calcium	Barium
Hydroxide	Be(OH)₂	Mg(OH)₂	Ca(OH)₂	Ba(OH)₂
Oxide	BeO	MgO	CaO	BaO
Sulphide	–	MgS	CaS	BaS
Chloride	BeCl₂	MgCl₂	CaCl₂	BaCl₂
Carbonate	BeCO₃	MgCO₃	CaCO₃	BaCO₃
Nitrate(V)	Be(NO₃)₂	Mg(NO₃)₂	Ca(NO₃)₂	Ba(NO₃)₂
Sulphate(VI)	BeSO₄	MgSO₄	CaSO₄	BaSO₄
Sulphate(IV)	–	–	CaSO₃	BaSO₃
Hydrogen carbonate	–	Mg(HCO₃)₂	Ca(HCO₃)₂	–
Hydrogen sulphate(VI)	–	Mg(HSO₄)₂	Ca(HSO₄)₂	–

Some uses of alkaline earth metals include:

(i)Magnesium hydroxide is a non-toxic/poisonous mild base used as an anti acid medicine to relieve stomach acidity.

(ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light.

(iii) Making plaster of Paris-Calcium sulphate(VI) is used in hospitals to set a fractures bone.

(iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building.

(iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms.

(v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth.

(vi)In the blast furnace-Limestone is added to the blast furnace to produce more reducing agent and remove slag in the blast furnace for extraction of Iron.

(c)Group VII elements: Halogens

Group VII elements are called Halogens. They are all non metals. They include:

Element

Symbol

Atomic number

Electronicc configuration

Charge of ion

Valency

State at Room Temperature

Fluorine

Chlorine

Bromine

Iodine

Astatine

2:7

2:8:7

2:8:18:7

2:8:18:18:7

2:8:18:32:18:7

F^–

Cl^–

Br^–

I^–

At^–

Pale yellow gas

Pale green gas

Red liquid

Grey Solid

Radioactive

All halogen atoms have seven electrons in the outer energy level. They acquire/gain one electron in the outer energy level to be stable. They therefore are therefore monovalent .They exist in oxidation state X^–

The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g.

The atomic size/radius of Chlorine is bigger/larger than that of Fluorine because Chlorine has more/3 energy levels than Fluorine (2 energy levels).

Atomic radius and ionic radius of Halogens increase down the group as the number of energy levels increases.

The atomic radius of Halogens is smaller than the ionic radius. This is because they react by gaining/acquiring extra one electron in the outer energy level. The effective nuclear attraction on the more/extra electrons decreases. The incoming extra electron is also repelled causing the outer energy level to expand to reduce the repulsion and accommodate more electrons.

Table showing the atomic and ionic radius of four Halogens

Element	Symbol	Atomic number	Atomic radius(nM)	Ionic radius(nM)
Fluorine	F	9	0.064	0.136
Chlorine	Cl	17	0.099	0.181
Bromine	Br	35	0.114	0.195
Iodine	I	53	0.133	0.216

The atomic radius of Chlorine is 0.099nM .The ionic radius of Cl^– is 0.181nM. This is because Chlorine atom/molecule reacts by gaining/acquiring extra one electrons. The more/extra electrons/energy level experience less effective nuclear attraction /pull towards the nucleus .The outer enegy level expand/increase to reduce the repulsion of the existing and incoming gained /acquired electrons.

Electronegativity

The ease of gaining/acquiring extra electrons is called electronegativity. All halogens are electronegative. Electronegativity decreases as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius.

The outer electrons experience less nuclear attraction and thus ease of gaining/acquiring extra electrons decrease.

It is measured using Pauling’s scale.

Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron.

Table showing the electronegativity of the halogens.

Halogen	F	Cl	Br	I	At
Electronegativity (Pauling scale)	4.0	3.0	2.8	2.5	2.2

The electronegativity of the halogens decrease down the group from fluorine to Astatine. This is because atomic radius increases down the group and thus decrease electron – attracting power down the group from fluorine to astatine.

Fluorine is the most electronegative element in the periodic table because it has the small atomic radius.

Electron affinity

The minimum amount of energy required to gain/acquire an extra electron by an atom of element in its gaseous state is called 1^st electron affinity. The SI unit of electron affinity is kilojoules per mole/kJmole^-1 . Electron affinity depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer energy level electrons and thus the lower the electron affinity. For halogens the 1^st electron affinity decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. Due to its small size/atomic radius Fluorine shows exceptionally low electron affinity. This is because a lot of energy is required to overcome the high repulsion of the existing and incoming electrons.

Table showing the election affinity of halogens for the process

X + e -> X^–

Halogen	F	Cl	Br	I
Electron affinity kJmole^-1	-333	-364	-342	-295

The higher the electron affinity the more stable theion.i.e

Cl^– is a more stable ion than Br^–because it has a more negative / exothermic electron affinity than Br^–

Electron affinity is different from:

(i) Ionization energy.

Ionization energy is the energy required to lose/donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state.

(ii) Electronegativity.

-Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process:

X(g) + e -> X^–(g)

Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.

It does not involve use of energy but theoretical arbitrary Pauling’ scale of measurements.

Physical properties

State at room temperature

Fluorine and Chlorine are gases, Bromine is a liquid and Iodine is a solid. Astatine is radioactive .

All halogens exist as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces.

Melting/Boiling point

The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius.

Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid.

Iodine sublimes when heated to form (caution: highly toxic/poisonous) purple vapour.

This is because Iodine molecules are held together by weak van-der-waals/intermolecular forces which require little heat energy to break.

Electrical conductivity

All Halogens are poor conductors of electricity because they have no free delocalized electrons.

Solubility in polar and non-polar solvents

All halogens are soluble in water(polar solvent).

When a boiling tube containing either chlorine gas or bromine vapour is separately inverted in a beaker containing distilled water and tetrachloromethane (non-polar solvent), the level of solution in boiling tube rises in both water and tetrachloromethane.

This is because halogen are soluble in both polar and non-polar solvents. Solubility of halogens in water/polar solvents decrease down the group. Solubility of halogens in non-polar solvent increase down the group.

The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine.

Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous.

Table showing the physical properties of Halogens

Halogen	Formula of molecule	Electrical conductivity	Solubility in water	Melting point(^oC)	Boiling point(^oC)
Fluorine	F₂	Poor	Insoluble/soluble in tetrachloromethane	-238	-188
Chlorine	Cl₂	Poor	Insoluble/soluble in tetrachloromethane	-101	-35
Bromine	Br₂	Poor	Insoluble/soluble in tetrachloromethane	7	59
Iodine	I₂	Poor	Insoluble/soluble in tetrachloromethane	114	sublimes

Chemical properties

(i)Displacement

Experiment

Place separately in test tubes about 5cm³ of sodium chloride, Sodium bromide and Sodium iodide solutions.

Add 5 drops of chlorine water to each test tube:

Repeat with 5 drops of bromine water instead of chlorine water

Observation

Using Chlorine water

-Yellow colour of chlorine water fades in all test tubes except with sodium chloride.

-Coloured Solution formed.

Using Bromine water

Yellow colour of bromine water fades in test tubes containing sodium iodide.

-Coloured Solution formed.

Explanation

The halogens displace each other from their solution. The more electronegative displace the less electronegative from their solution.

Chlorine is more electronegative than bromine and iodine.

On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine.

Bromine is more electronegative than iodide but less 6than chlorine.

On adding Bromine water, iodine is displaced from its solution but not chlorine.

Table showing the displacement of the halogens

(V) means there is displacement (x ) means there is no displacement

Halogen ion in solution Halogen	F^–	Cl^–	Br^–	I^–
F₂	X
Cl₂	X	X
Br₂	X	X	X
I₂	X	X	X	X

Chemical /ionic equations

With Fluorine

F₂(g) + 2NaCl^–(aq) -> 2NaF(aq) + Cl₂(aq)

F₂(g) + 2Cl^–(aq) -> 2F^–(aq) + Cl₂(aq)

F₂(g) + 2NaBr^–(aq) -> 2NaF(aq) + Br₂(aq)

F₂(g) + 2Br^–(aq) -> 2F^–(aq) + Br₂(aq)

F₂(g) + 2NaI^–(aq) -> 2NaF(aq) + I₂(aq)

F₂(g) + 2I^–(aq) -> 2F^–(aq) + I₂(aq)

With chlorine

Cl₂(g) + 2NaCl^–(aq) -> 2NaCl(aq) + Br₂(aq)

Cl₂(g) + 2Br^–(aq) -> 2Cl^–(aq) + Br₂(aq)

Cl₂(g) + 2NaI^–(aq) -> 2NaCl(aq) + I₂(aq)

Cl₂(g) + 2I^–(aq) -> 2Cl^–(aq) + I₂(aq)

With Bromine

Br₂(g) + 2NaI^–(aq) -> 2NaBr(aq) + I₂(aq)

Br₂(g) + 2I^–(aq) -> 2Br^–(aq) + I₂(aq)

Uses of halogens

Florine – manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber.
Reduce tooth decay when added in small amounts/quantities in tooth paste.

NB –large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis.

Hydrogen fluoride is used to engrave words /pictures in glass.

Bromine – Silver bromide is used to make light sensitive photographic paper/films.

Iodide – Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine.

The table below to show some compounds of halogens.

Element Halogen	H	Na	Mg	Al	Si	C	P
F	HF	NaF	MgH₂	AlF₃	SiF₄	CF₄	PF₃
Cl	HCl	NaCl	MgCl	AlCl₃	SiCl₃	CCl₄	PCl₃
Br	HBr	NaBr	MgBr₂	AlBr₃	SiBr₄	CBr₄	PBr₃
I	Hl	Nal	Mgl₂	All₃	SiI₄	Cl₂	PBr₃

Below is the table showing the bond energy of four halogens.

Bond Bond energy k J mole^-1

Cl-Cl 242

Br-Br 193

I-I 151

What do you understand by the term “bond energy”

Bond energy is the energy required to break/ form one mole of chemical bond

Explain the trend in bond Energy of the halogens above:

–Decrease down the group from chlorine to Iodine

-Atomic radius increase down the group decreasing the energy required to break the covalent bonds between the larger atom with reduced effective nuclear @ charge an outer energy level that take part in bonding.

(c)Group VIII elements: Noble gases

Group VIII elements are called Noble gases. They are all non metals. Noble gases occupy about 1.0% of the atmosphere as colourless gaseous mixture. Argon is the most abundant with 0.9%.

They exists as monatomic molecules with very weak van-der-waals /intermolecular forces holding the molecules.

They include:

Element	Symbol	Atomic number	Electron structure	State at room temperature
Helium	He	2	2:	Colourless gas
Neon	Ne	10	2:8	Colourless gas
Argon	Ar	18	2:8:8	Colourless gas
Krypton	Kr	36	2:8:18:8	Colourless gas
Xenon	Xe	54	2:8:18:18:8	Colourless gas
Radon	Rn	86	2:8:18:32:18:8	Radioctive

All noble gas atoms have a stable duplet(two electrons in the 1^st energy level) or octet(eight electrons in other outer energy level)in the outer energy level. They therefore do not acquire/gain extra electron in the outer energy level or donate/lose. They therefore are therefore zerovalent .

The number of energy levels increases down the group from Helium to Randon. The more the number of energy levels the bigger/larger the atomic size/radius. e.g.

The atomic size/radius of Argon is bigger/larger than that of Neon because Argon has more/3 energy levels than Neon (2 energy levels).

Atomic radius noble gases increase down the group as the number of energy levels increases.

The effective nuclear attraction on the outer electrons thus decrease down the group.

The noble gases are generally unreactive because the outer energy level has the stable octet/duplet. The stable octet/duplet in noble gas atoms lead to a comparatively very high 1^st ionization energy. This is because losing /donating an electron from the stable atom require a lot of energy to lose/donate and make it unstable.

As atomic radius increase down the group and the 1^st ionization energy decrease, very electronegative elements like Oxygen and Fluorine are able to react and bond with lower members of the noble gases.e.g

Xenon reacts with Fluorine to form a covalent compound XeF₆.This is because the outer electrons/energy level if Xenon is far from the nucleus and thus experience less effective nuclear attraction.

Noble gases have low melting and boiling points. This is because they exist as monatomic molecules joined by very weak intermolecular/van-der-waals forces that require very little energy to weaken and form liquid and break to form a gas.

The intermolecular/van-der-waals forces increase down the group as the atomic radius/size increase from Helium to Radon. The melting and boiling points thus increase also down the group.

Noble gases are insoluble in water and are poor conductors of electricity.

Element	Formula of molecule	Electrical conductivity	Solubility in water	Atomic radius(nM)	1^st ionization energy	Melting point(⁰C)	Boiling point(⁰C)
Helium	He	Poor	Insoluble	0.128	2372	-270	-269
Neon	Ne	Poor	Insoluble	0.160	2080	-249	-246
Argon	Ar	Poor	Insoluble	0.192	1520	-189	-186
Krypton	Kr	Poor	Insoluble	0.197	1350	-157	-152
Xenon	Xe	Poor	Insoluble	0.217	1170	-112	-108
Radon	Rn	Poor	Insoluble	0.221	1134	-104	-93

Uses of noble gases

Argon is used in light bulbs to provide an inert environment to prevent oxidation of the bulb filament

Argon is used in arch welding as an insulator.

Neon is used in street and advertisement light

Helium is mixed with Oxygen during deep sea diving and mountaineering.

Helium is used in weather balloon for meteorological research instead of Hydrogen because it is unreactive/inert.Hydrogen when impure can ignite with an explosion.

Helium is used in making thermometers for measuring very low temperatures.

PERIODICITY OF ACROSS THE PERIOD.

(See Chemical bonding and Structure)

UPGRADE

CHEMISTRY

FORM 2

Periodicity of CHEMICAL FAMILIES

Comprehensive tutorial notes

MUTHOMI S.G

www.kcselibrary.info

0720096206

PERIODICITY OF CHEMICAL FAMILES

(Patterns down the group)

The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table. i.e

The number of occupied energy levels determine the Period and the valence electrons determine the Group.

Group I elements: Alkali metals

Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include:

Element	Symbol	Atomic number	Electron structure	Oxidation state	Valency
Lithium	Li	3	2:1	Li⁺	1
Sodium	Na	11	2:8:1	Na⁺	1
Potassium	K	19	2:8:8:1	K⁺	1
Rubidium	Rb	37	2:8:18:8:1	Rb⁺	1
Caesium	Cs	55	2:8:18:18:8:1	Cs⁺	1
Francium	Fr	87	2:8:18:32:18:8:1	Fr⁺	1

All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M⁺

The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g.

The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels).

Atomic and ionic radius

Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g.

The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels).

Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases.

The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level.

Table showing the atomic and ionic radius of some alkali metals

Element	Symbol	Atomic number	Atomic radius(nM)	Ionic radius(nM)
Lithium	Li	3	0.133	0.060
Sodium	Na	11	0.157	0.095
Potassium	K	19	0.203	0.133

Electropositivity

Ionization energy

Physical properties

Summary of some physical properties of the 1^st three alkali metals

Alkali metal	Appearance	Ease of cutting	Melting point (^oC)	Boiling point (^oC)	Conductivity	1^st ionization energy
Lithium	Silvery white	Not easy	180	1330	Good	520
Sodium	Shiny grey	Easy	98	890	Good	496
Potassium	Shiny grey	Very easy	64	774	Good	419

Chemical properties

(i)Reaction with air/oxygen

On exposure to air, alkali metals reacts with the elements in the air.

Example

On exposure to air, Sodium first reacts with Oxygen to form sodium oxide.

4Na(s) + O₂(g) -> 2Na₂O(s)

The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution.

Na₂O(s) + H₂O(l) -> 2NaOH(aq)

Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate.

2NaOH(aq) + CO₂(g) -> Na₂CO₃(g) + H₂O(l)

(ii)Burning in air/oxygen

Lithium burns in air with a crimson/deep red flame to form Lithium oxide

4Li (s) + O₂(g) -> 2Li₂O(s)

Sodium burns in air with a yellow flame to form sodium oxide

4Na (s) + O₂(g) -> 2Na₂O(s)

Sodium burns in oxygen with a yellow flame to form sodium peroxide

2Na (s) + O₂(g) -> Na₂O₂ (s)

Potassium burns in air with a lilac/purple flame to form potassium oxide

4K (s) + O₂(g) -> 2K₂O (s)

(iii) Reaction with water:

Experiment

Measure 500 cm3 of water into a beaker.

Put three drops of phenolphthalein indicator.

Put about 0.5g of Lithium metal into the beaker.

Determine the pH of final product

Repeat the experiment using about 0.1 g of Sodium and Potassium.

Caution: Keep a distance

Observations

Alkali metal	Observations	Comparative speed/rate of the reaction
Lithium	-Metal floats in water -rapid effervescence/fizzing/bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14	Moderately vigorous
Sodium	-Metal floats in water -very rapid effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14	Very vigorous
Potassium	-Metal floats in water -explosive effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14	Explosive/burst into flames

Explanation

Chemical equations

2Li(s) + 2H₂O(l) -> 2LiOH(aq) + H₂(g)

2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)

2K(s) + 2H₂O(l) -> 2KOH(aq) + H₂(g)

2Rb(s) + 2H₂O(l) -> 2RbOH(aq) + H₂(g)

2Cs(s) + 2H₂O(l) -> 2CsOH(aq) + H₂(g)

2Fr(s) + 2H₂O(l) -> 2FrOH(aq) + H₂(g)

Reactivity increase down the group

(iv) Reaction with chlorine:

Experiment

Repeat with about 0.5g of Lithium.

Caution: This experiment should be done in fume chamber because chlorine is poisonous /toxic.

Observation

Sodium metal continues to burn with a yellow flame forming white solid/fumes.

Lithium metal continues to burn with a crimson flame forming white solid / fumes.

Chemical equations

2Li(s) + Cl₂(g) -> 2LiCl(s)

2Na(s) + Cl₂(g) -> 2NaCl(s)

2K(s) + Cl₂(g) -> 2KCl(s)

2Rb(s) + Cl₂(g) -> 2RbCl(s)

2Cs(s) + Cl₂(g) -> 2CsCl(s)

2Fr(s) + Cl₂(g) -> 2FrCl(s) Reactivity increase down the group

The table below shows some compounds of the 1^st three alkali metals

	Lithium	sodium	Potassium
Hydroxide	LiOH	NaOH	KOH
Oxide	Li₂O	Na₂O	K₂O
Sulphide	Li₂S	Na₂S	K₂S
Chloride	LiCl	NaCl	KCl
Carbonate	Li₂CO₃	Na₂CO₃	K₂CO₃
Nitrate(V)	LiNO₃	NaNO₃	KNO₃
Nitrate(III)	–	NaNO₂	KNO₂
Sulphate(VI)	Li₂SO₄	Na₂SO₄	K₂SO₄
Sulphate(IV)	–	Na₂SO₃	K₂SO₃
Hydrogen carbonate	–	NaHCO₃	KHCO₃
Hydrogen sulphate(VI)	–	NaHSO₄	KHSO₄
Hydrogen sulphate(IV)	–	NaHSO₃	KHSO₃
Phosphate	–	Na₃PO₄	K₃PO₄
Manganate(VI)	–	NaMnO₄	KMnO₄
Dichromate(VI)	–	Na₂Cr₂O₇	K₂Cr₂O₇
Chromate(VI)	–	Na₂CrO₄	K₂CrO₄

Some uses of alkali metals include:

(i)Sodium is used in making sodium cyanide for extracting gold from gold ore.

(ii)Sodium chloride is used in seasoning food.

(iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors.

(iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents.

(v)Sodium is used as a reducing agent for the extraction of titanium from Titanium(IV)chloride.

(vi)Lithium is used in making special high strength glasses

(vii)Lithium compounds are used to make dry cells in mobile phones and computer laptops.

Group II elements: Alkaline earth metals

Group II elements are called Alkaline earth metals . The alkaline earth metals include:

Element	Symbol	Atomic number	Electron structure	Oxidation state	Valency
Beryllium	Be	4	2:2	Be²⁺	2
Magnesium	Mg	12	2:8:2	Mg²⁺	2
Calcium	Ca	20	2:8:8:2	Ca²⁺	2
Strontium	Sr	38	2:8:18:8:2	Sr²⁺	2
Barium	Ba	56	2:8:18:18:8:2	Ba²⁺	2
Radium	Ra	88	2:8:18:32:18:8:2	Ra²⁺	2

All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent. They donate /lose the two outer electrons to have oxidation state M²⁺

The number of energy levels increases down the group from Beryllium to Radium. The more the number of energy levels the bigger/larger the atomic size. e.g.

The atomic size/radius of Calcium is bigger/larger than that of Magnesium because Calcium has more/4 energy levels than Magnesium (3 energy levels).

Atomic radius and ionic radius of alkaline earth metals increase down the group as the number of energy levels increases.

The atomic radius of alkaline earth metals is bigger than the ionic radius. This is because they react by losing/donating the two outer electrons and hence lose the outer energy level.

Table showing the atomic and ionic radius of the 1^st three alkaline earth metals

Element	Symbol	Atomic number	Atomic radius(nM)	Ionic radius(nM)
Beryllium	Be	4	0.089	0.031
Magnesium	Mg	12	0.136	0.065
Calcium	Ca	20	0.174	0.099

Electropositivity

Ionization energy

For alkaline earth metals the 1^st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease.

It requires therefore less energy to donate/lose outer electron in calcium than in magnesium.

The minimum amount of energy required to remove a second electron from an ion of an element in its gaseous state is called the 2^nd ionization energy.

The 2^nd ionization energy is always higher /bigger than the 1^st ionization energy.

Physical properties

(i)ductile(able to form wire/thin long rods)

(ii)malleable(able to be hammered into sheet/long thin plates)

(iii)have high tensile strength(able to be coiled without breaking/ not brittle/withstand stress)

The melting and boiling points decrease down the group as the atomic radius/size increase reducing the strength of metallic bond and packing of the metallic structure. e.g.

Electrical/thermal conductivity: Alkaline earth metals are good thermal and electrical conductors. The two delocalized valence electrons move randomly within the metallic structure.

Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium. The more delocalized electrons the better the electrical conductor.

Calcium is a better conductor than magnesium.

Summary of some physical properties of the 1^st three alkaline earth metals

Alkaline earth metal	Appearance	Ease of cutting	Melting point (^oC)	Boiling point (^oC)	Conduct- ivity	1^st ionization energy	2^nd ionization energy
Beryllium	Shiny grey	Not easy	1280	3450	Good	900	1800
Magnesium	Shiny grey	Not Easy	650	1110	Good	736	1450
calcium	Shiny grey	Not easy	850	1140	Good	590	970

Chemical properties

(i)Reaction with air/oxygen

On exposure to air, the surface of alkaline earth metals is slowly oxidized to its oxide on prolonged exposure to air.

Example

On exposure to air, the surface of magnesium ribbon is oxidized to form a thin film of Magnesium oxide

. 2Mg(s) + O₂(g) -> 2MgO(s)

(ii)Burning in air/oxygen

Experiment

Hold a about 2cm length of Magnesium ribbon on a Bunsen flame. Stop heating when it catches fire/start burning.

Caution: Do not look directly at the flame

Put the products of burning into 100cm3 beaker. Add about 5cm3 of distilled water. Swirl. Test the mixture using litmus papers.
Repeat with Calcium

Observations

-Magnesium burns with a bright blindening flame

-White solid /ash produced

-Solid dissolves in water to form a colourless solution

-Blue litmus paper remain blue

-Red litmus paper turns blue

-colourless gas with pungent smell of urine

Explanation

Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride.

2Mg (s) + O₂(g) -> 2MgO(s)

3Mg (s) + N₂(g) -> Mg₃N₂ (s)

Magnesium oxide dissolves in water to form magnesium hydroxide.

MgO(s) + H₂O (l) -> Mg(OH)₂(aq)

Magnesium nitride dissolves in water to form magnesium hydroxide and produce ammonia gas.

Mg₃N₂ (s) + 6H₂O(l) -> 3Mg(OH)₂(aq) + 2NH₃ (g)

Magnesium hydroxide and ammonia are weakly alkaline with pH 8/9/10/11 and turns red litmus paper blue.

Calcium burns in air with faint orange/red flame to form a mixture of both Calcium oxide and calcium nitride.

2Ca (s) + O₂(g) -> 2CaO(s)

3Ca (s) + N₂(g) -> Ca₃N₂ (s)

Calcium oxide dissolves in water to form calcium hydroxide.

CaO(s) + H₂O(l) -> Ca(OH)₂(aq)

Calcium nitride dissolves in water to form calcium hydroxide and produce ammonia gas.

Ca₃N₂ (s) + 6H₂O(l) -> 3Ca(OH)₂(aq) + 2NH₃ (g)

Calcium hydroxide is also weakly alkaline solution with pH 8/9/10/11 and turns red litmus paper blue.

(iii)Reaction with water

Experiment

Measure 50 cm3 of distilled water into a beaker.

Scrub/polish with sand paper 1cm length of Magnesium ribbon

Place it in the water. Test the product-mixture with blue and red litmus papers.

Repeat with Calcium metal.

Observations

-Surface of magnesium covered by bubbles of colourless gas.

-Colourless solution formed.

-Effervescence/bubbles/fizzing takes place in Calcium.

-Red litmus paper turns blue.

-Blue litmus paper remains blue.

Explanations

Magnesium slowly reacts with cold water to form Magnesium hydroxide and bubbles of Hydrogen gas that stick on the surface of the ribbon.

Mg(s) + 2H₂O (l) -> Mg(OH)₂(aq) + H₂ (g)

Calcium moderately reacts with cold water to form Calcium hydroxide and produce a steady stream of Hydrogen gas.

Ca(s) + 2H₂O (l) -> Ca(OH)₂(aq) + H₂ (g)

(iv)Reaction with water vapour/steam

Experiment

Put some cotton wool soaked in water/wet sand in a long boiling tube.

Coil a well polished magnesium ribbon into the boiling tube.

Ensure the coil touches the side of the boiling tube. Heat the cotton wool/sand slightly then strongly heat the Magnesium ribbon .

Set up of apparatus

Observations

-Magnesium glows red hot then burns with a blindening flame.

-Magnesium continues to glow/burning even without more heating.

-White solid/residue.

-colourless gas collected over water.

Explanation

On heating wet sand, steam is generated which drives out the air that would otherwise react with /oxidize the ribbon.

Magnesium burns in steam/water vapour generating enough heat that ensures the reaction goes to completion even without further heating. White Magnesium oxide is formed and hydrogen gas is evolved.

To prevent suck back, the delivery tube should be removed from the water before heating is stopped at the end of the experiment.

Mg(s) + H₂O (l) -> MgO(s) + H₂ (g)

(v)Reaction with chlorine gas.

Experiment

Lower slowly a burning magnesium ribbon/shavings into a gas jar containing Chlorine gas. Repeat with a hot piece of calcium metal.

Observation

-Magnesium continues to burn in chlorine with a bright blindening flame.

-Calcium continues to burn for a short time.

-White solid formed .

-Pale green colour of chlorine fades.

Explanation

Magnesium continues to burn in chlorine gas forming white magnesium oxide solid.

Mg(s) + Cl₂ (g) -> MgCl₂ (s)

Calcium burns slightly in chlorine gas to form white calcium oxide solid. Calcium oxide formed coat unreacted Calcium stopping further reaction

Ca(s) + Cl₂ (g) -> CaCl₂ (s)

(v)Reaction with dilute acids.

Experiment

Repeat with about 4.0cm3 of 0.1M dilute hydrochloric/nitric(V) acid.

Repeat with 0.1g of Calcium in a beaker with all the above acid

Caution: Keep distance when using calcium

Observation

-Effervescence/fizzing/bubbles with dilute sulphuric(VI) and nitric(V) acids

-Little Effervescence/fizzing/bubbles with calcium and dilute sulphuric(VI) acid.

-Colourless gas produced that extinguishes a burning splint with an explosion/ “pop” sound.

-No gas is produced with Nitric(V)acid.

-Colourless solution is formed.

Explanation

Dilute acids react with alkaline earth metals to form a salt and produce hydrogen gas.

Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water.

Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used.

Chemical equations

Mg(s) + H₂SO₄ (aq) -> MgSO₄(aq) + H₂ (g)

Mg(s) + 2HNO₃ (aq) -> Mg(NO₃)₂(aq) + H₂ (g)

Mg(s) + 2HCl (aq) -> MgCl₂(aq) + H₂ (g)

Ca(s) + H₂SO₄ (aq) -> CaSO₄(s) + H₂ (g)

(insoluble CaSO₄(s) coat/cover Ca(s))

Ca(s) + 2HNO₃ (aq) -> Ca(NO₃)₂(aq) + H₂ (g)

Ca(s) + 2HCl (aq) -> CaCl₂(aq) + H₂ (g)

Ba(s) + H₂SO₄ (aq) -> BaSO₄(s) + H₂ (g)

(insoluble BaSO₄(s) coat/cover Ba(s))

Ba(s) + 2HNO₃ (aq) -> Ba(NO₃)₂(aq) + H₂ (g)

Ba(s) + 2HCl (aq) -> BaCl₂(aq) + H₂ (g)

The table below shows some compounds of some alkaline earth metals

	Beryllium	Magnesium	Calcium	Barium
Hydroxide	Be(OH)₂	Mg(OH)₂	Ca(OH)₂	Ba(OH)₂
Oxide	BeO	MgO	CaO	BaO
Sulphide	–	MgS	CaS	BaS
Chloride	BeCl₂	MgCl₂	CaCl₂	BaCl₂
Carbonate	BeCO₃	MgCO₃	CaCO₃	BaCO₃
Nitrate(V)	Be(NO₃)₂	Mg(NO₃)₂	Ca(NO₃)₂	Ba(NO₃)₂
Sulphate(VI)	BeSO₄	MgSO₄	CaSO₄	BaSO₄
Sulphate(IV)	–	–	CaSO₃	BaSO₃
Hydrogen carbonate	–	Mg(HCO₃)₂	Ca(HCO₃)₂	–
Hydrogen sulphate(VI)	–	Mg(HSO₄)₂	Ca(HSO₄)₂	–

Some uses of alkaline earth metals include:

(i)Magnesium hydroxide is a non-toxic/poisonous mild base used as an anti acid medicine to relieve stomach acidity.

(ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light.

(iii) Making plaster of Paris-Calcium sulphate(VI) is used in hospitals to set a fractures bone.

(iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building.

(iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms.

(v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth.

(vi)In the blast furnace-Limestone is added to the blast furnace to produce more reducing agent and remove slag in the blast furnace for extraction of Iron.

(c)Group VII elements: Halogens

Group VII elements are called Halogens. They are all non metals. They include:

Element

Symbol

Atomic number

Electronicc configuration

Charge of ion

Valency

State at Room Temperature

Fluorine

Chlorine

Bromine

Iodine

Astatine

2:7

2:8:7

2:8:18:7

2:8:18:18:7

2:8:18:32:18:7

F^–

Cl^–

Br^–

I^–

At^–

Pale yellow gas

Pale green gas

Red liquid

Grey Solid

Radioactive

The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g.

The atomic size/radius of Chlorine is bigger/larger than that of Fluorine because Chlorine has more/3 energy levels than Fluorine (2 energy levels).

Atomic radius and ionic radius of Halogens increase down the group as the number of energy levels increases.

Table showing the atomic and ionic radius of four Halogens

Element	Symbol	Atomic number	Atomic radius(nM)	Ionic radius(nM)
Fluorine	F	9	0.064	0.136
Chlorine	Cl	17	0.099	0.181
Bromine	Br	35	0.114	0.195
Iodine	I	53	0.133	0.216

Electronegativity

The outer electrons experience less nuclear attraction and thus ease of gaining/acquiring extra electrons decrease.

It is measured using Pauling’s scale.

Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron.

Table showing the electronegativity of the halogens.

Halogen	F	Cl	Br	I	At
Electronegativity (Pauling scale)	4.0	3.0	2.8	2.5	2.2

Fluorine is the most electronegative element in the periodic table because it has the small atomic radius.

Electron affinity

Table showing the election affinity of halogens for the process

X + e -> X^–

Halogen	F	Cl	Br	I
Electron affinity kJmole^-1	-333	-364	-342	-295

The higher the electron affinity the more stable theion.i.e

Cl^– is a more stable ion than Br^–because it has a more negative / exothermic electron affinity than Br^–

Electron affinity is different from:

(i) Ionization energy.

(ii) Electronegativity.

-Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process:

X(g) + e -> X^–(g)

Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.

It does not involve use of energy but theoretical arbitrary Pauling’ scale of measurements.

Physical properties

State at room temperature

Fluorine and Chlorine are gases, Bromine is a liquid and Iodine is a solid. Astatine is radioactive .

All halogens exist as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces.

Melting/Boiling point

The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius.

Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid.

Iodine sublimes when heated to form (caution: highly toxic/poisonous) purple vapour.

This is because Iodine molecules are held together by weak van-der-waals/intermolecular forces which require little heat energy to break.

Electrical conductivity

All Halogens are poor conductors of electricity because they have no free delocalized electrons.

Solubility in polar and non-polar solvents

All halogens are soluble in water(polar solvent).

The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine.

Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous.

Table showing the physical properties of Halogens

Halogen	Formula of molecule	Electrical conductivity	Solubility in water	Melting point(^oC)	Boiling point(^oC)
Fluorine	F₂	Poor	Insoluble/soluble in tetrachloromethane	-238	-188
Chlorine	Cl₂	Poor	Insoluble/soluble in tetrachloromethane	-101	-35
Bromine	Br₂	Poor	Insoluble/soluble in tetrachloromethane	7	59
Iodine	I₂	Poor	Insoluble/soluble in tetrachloromethane	114	sublimes

Chemical properties

(i)Displacement

Experiment

Place separately in test tubes about 5cm³ of sodium chloride, Sodium bromide and Sodium iodide solutions.

Add 5 drops of chlorine water to each test tube:

Repeat with 5 drops of bromine water instead of chlorine water

Observation

Using Chlorine water

-Yellow colour of chlorine water fades in all test tubes except with sodium chloride.

-Coloured Solution formed.

Using Bromine water

Yellow colour of bromine water fades in test tubes containing sodium iodide.

-Coloured Solution formed.

Explanation

The halogens displace each other from their solution. The more electronegative displace the less electronegative from their solution.

Chlorine is more electronegative than bromine and iodine.

On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine.

Bromine is more electronegative than iodide but less 6than chlorine.

On adding Bromine water, iodine is displaced from its solution but not chlorine.

Table showing the displacement of the halogens

(V) means there is displacement (x ) means there is no displacement

Halogen ion in solution Halogen	F^–	Cl^–	Br^–	I^–
F₂	X
Cl₂	X	X
Br₂	X	X	X
I₂	X	X	X	X

Chemical /ionic equations

With Fluorine

F₂(g) + 2NaCl^–(aq) -> 2NaF(aq) + Cl₂(aq)

F₂(g) + 2Cl^–(aq) -> 2F^–(aq) + Cl₂(aq)

F₂(g) + 2NaBr^–(aq) -> 2NaF(aq) + Br₂(aq)

F₂(g) + 2Br^–(aq) -> 2F^–(aq) + Br₂(aq)

F₂(g) + 2NaI^–(aq) -> 2NaF(aq) + I₂(aq)

F₂(g) + 2I^–(aq) -> 2F^–(aq) + I₂(aq)

With chlorine

Cl₂(g) + 2NaCl^–(aq) -> 2NaCl(aq) + Br₂(aq)

Cl₂(g) + 2Br^–(aq) -> 2Cl^–(aq) + Br₂(aq)

Cl₂(g) + 2NaI^–(aq) -> 2NaCl(aq) + I₂(aq)

Cl₂(g) + 2I^–(aq) -> 2Cl^–(aq) + I₂(aq)

With Bromine

Br₂(g) + 2NaI^–(aq) -> 2NaBr(aq) + I₂(aq)

Br₂(g) + 2I^–(aq) -> 2Br^–(aq) + I₂(aq)

Uses of halogens

Florine – manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber.
Reduce tooth decay when added in small amounts/quantities in tooth paste.

NB –large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis.

Hydrogen fluoride is used to engrave words /pictures in glass.

Bromine – Silver bromide is used to make light sensitive photographic paper/films.

Iodide – Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine.

The table below to show some compounds of halogens.

Element Halogen	H	Na	Mg	Al	Si	C	P
F	HF	NaF	MgF₂	AlF₃	SiF₄	CF₄	PF₃
Cl	HCl	NaCl	MgCl₂	AlCl₃	SiCl ₄	CCl₄	PCl₃
Br	HBr	NaBr	MgBr₂	AlBr₃	SiBr₄	CBr₄	PBr₃
I	Hl	Nal	Mgl₂	All₃	SiI₄	C l ₄	PBr₃

Below is the table showing the bond energy of four halogens.

Bond Bond energy k J mole^-1

Cl-Cl 242

Br-Br 193

I-I 151

What do you understand by the term “bond energy”

Bond energy is the energy required to break/ form one mole of chemical bond

Explain the trend in bond Energy of the halogens above:

–Decrease down the group from chlorine to Iodine

(c)Group VIII elements: Noble gases

Group VIII elements are called Noble gases. They are all non metals. Noble gases occupy about 1.0% of the atmosphere as colourless gaseous mixture. Argon is the most abundant with 0.9%.

They exists as monatomic molecules with very weak van-der-waals /intermolecular forces holding the molecules.

They include:

Element	Symbol	Atomic number	Electron structure	State at room temperature
Helium	He	2	2:	Colourless gas
Neon	Ne	10	2:8	Colourless gas
Argon	Ar	18	2:8:8	Colourless gas
Krypton	Kr	36	2:8:18:8	Colourless gas
Xenon	Xe	54	2:8:18:18:8	Colourless gas
Radon	Rn	86	2:8:18:32:18:8	Radioctive

The number of energy levels increases down the group from Helium to Randon. The more the number of energy levels the bigger/larger the atomic size/radius. e.g.

The atomic size/radius of Argon is bigger/larger than that of Neon because Argon has more/3 energy levels than Neon (2 energy levels).

Atomic radius noble gases increase down the group as the number of energy levels increases.

The effective nuclear attraction on the outer electrons thus decrease down the group.

Xenon reacts with Fluorine to form a covalent compound XeF₆.This is because the outer electrons/energy level if Xenon is far from the nucleus and thus experience less effective nuclear attraction.

The intermolecular/van-der-waals forces increase down the group as the atomic radius/size increase from Helium to Radon. The melting and boiling points thus increase also down the group.

Noble gases are insoluble in water and are poor conductors of electricity.

Element	Formula of molecule	Electrical conductivity	Solubility in water	Atomic radius(nM)	1^st ionization energy	Melting point(⁰C)	Boiling point(⁰C)
Helium	He	Poor	Insoluble	0.128	2372	-270	-269
Neon	Ne	Poor	Insoluble	0.160	2080	-249	-246
Argon	Ar	Poor	Insoluble	0.192	1520	-189	-186
Krypton	Kr	Poor	Insoluble	0.197	1350	-157	-152
Xenon	Xe	Poor	Insoluble	0.217	1170	-112	-108
Radon	Rn	Poor	Insoluble	0.221	1134	-104	-93

Uses of noble gases

Argon is used in light bulbs to provide an inert environment to prevent oxidation of the bulb filament

Argon is used in arch welding as an insulator.

Neon is used in street and advertisement light

Helium is mixed with Oxygen during deep sea diving and mountaineering.

Helium is used in weather balloon for meteorological research instead of Hydrogen because it is unreactive/inert. Hydrogen when impure can ignite with an explosion.

Helium is used in making thermometers for measuring very low temperatures.

UPGRADE

CHEMISTRY

FORM 2

STRUCTURE & BONDING

Comprehensive tutorial notes

MUTHOMI S.G

www.kcselibrary.info

0720096206

ssssss

CHEMICAL BONDING AND STRUCTURE

CHEMICAL BONDING

A chemical bond is formed when atoms of the same or different elements share, gain, donate or delocalize their outer energy level electrons to combine during chemical reactions inorder to be stable.

Atoms have equal number of negatively charged electrons in the energy levels and positively charged protons in the nucleus.

Atoms are chemically stable if they have filled outer energy level. An energy level is full if it has duplet (2) or octet (8) state in outer energy level.

Noble gases have duplet /octet. All other atoms try to be like noble gases through chemical reactions and forming molecules.

Only electrons in the outer energy level take part in formation of a chemical bond. There are three main types of chemical bonds formed by atoms:

(i) covalent bond

(ii) ionic/electrovalent bond

(iii) metallic bond

(i)COVALENT BOND

A covalent bond is formed when atoms of the same or different element share some or all the outer energy level electrons to combine during chemical reactions inorder to attain duplet or octet.

A shared pair of electrons is attracted by the nucleus (protons) of the two atoms sharing.

Covalent bonds are mainly formed by non-metals to form molecules. A molecule is a group of atoms of the same or different elements held together by a covalent bond. The number of atoms making a molecule is called atomicity. Noble gases are monatomic because they are stable and thus do not bond with each other or other atoms. Most other gases are diatomic

The more the number of electrons shared, the stronger the covalent bond.

A pair of electrons that do not take part in the formation of a covalent bond is called a lone pair of electrons.

Mathematically, the number of electrons to be shared by an atom is equal to the number of electrons remaining for the atom to be stable/attain duplet/octet /have maximum electrons in outer energy level.

The following diagrams illustrate the formation of covalent bonds:

a)hydrogen molecule is made up of two hydrogen atoms in the outer energy level each requiring one electron to have a stable duplet.

To show the formation of covalent bonding in the molecule then the following data/information is required;

Symbol of atom/element taking part in bonding H H

Number of protons/electrons 1 1

Electron configuration/structure 1: 1:

Number of electron in outer energy level 1 1

Number of electrons remaining to be stable/shared 1 1

Number of electrons not shared(lone pairs) 0 0

Diagram method 1

Diagram method 2

●x

H H

Note:

After bonding the following intramolecular forces exist:

(i)the attraction of the shared electrons by both nucleus /protons of the atoms

(ii) the repulsion of the nucleus of one atom on the other.

(iii)balance of the attraction and repulsion is maintained inside/intramolecular/within the molecule as follows;

E₁

P₁P₁

E₁

(iv)Protons(P₁) from nucleus of atom 1 repel protons (P₂) from nucleus of atom 2.

(v)Electron (E₁) in the energy levels of atom 1 repel electron (E₂) in the energy levels of atom 2.

(vi) Protons(P₁) from nucleus of atom 1 attract electron (E₂) in the energy levels of atom 2.

(vii) protons (P₂) from nucleus of atom 2 attract electron (E₂) in the energy levels of atom 2.

b) Fluorine, chlorine, bromine and iodine molecules are made up also of two atoms sharing the outer energy level electrons to have a stable octet.

To show the formation of covalent bonding in the molecule then the following data/information is required;

(i) fluorine

Symbol of atom/element taking part in bonding F F

Number of protons/electrons 9 9

Electron configuration/structure 2:7 2:7

Number of electron in outer energy level 7 7

Number of electrons remaining to be stable/shared 1 1

Number of outer electrons not shared( 3-lone pairs) 6 6

Diagram method 1

Diagram method 2

(ii) chlorine

Symbol of atom/element taking part in bonding Cl Cl

Number of protons/electrons 17 17

Electron configuration/structure 2:8:7 2:8:7

Number of electron in outer energy level 7 7

Number of electrons remaining to be stable/shared 1 1

Number of outer electrons not shared( 3-lone pairs) 6 6

Diagram method 1

Diagram method 2

(iii) Bromine

Symbol of atom/element taking part in bonding Br Br

Number of protons/electrons 35 35

Electron configuration/structure 2:8:18:7 2:8:18:7

Number of electron in outer energy level 7 7

Number of electrons remaining to be stable/shared 1 1

Number of outer electrons not shared( 3-lone pairs) 6 6

Diagram method 1

Diagram method 2

(iv) Iodine

Symbol of atom/element taking part in bonding I I

Number of protons/electrons 53 53

Electron configuration/structure 2:8:18:18:7 2:8:18:18:7

Number of electron in outer energy level 7 7

Number of electrons remaining to be stable/shared 1 1

Number of outer electrons not shared( 3-lone pairs) 6 6

Diagram method 1

Diagram method 2

c) Oxygen molecule is made up of two atoms sharing each two outer energy level electrons to have a stable octet as shown below;

Symbol of atom/element taking part in bonding O O

Number of protons/electrons 8 8

Electron configuration/structure 2:6 2:6

Number of electron in outer energy level 6 6

Number of electrons remaining to be stable/shared 2 2

Number of outer electrons not shared( 2-lone pairs) 4 4

Diagram method 1

Diagram method 2

d) Nitrogen and phosphorus molecule is made up of two atoms sharing each three outer energy level electrons to have a stable octet as shown below;

(i) Nitrogen

Symbol of atom/element taking part in bonding N N

Number of protons/electrons 7 7

Electron configuration/structure 2:5 2:5

Number of electron in outer energy level 5 5

Number of electrons remaining to be stable/shared 3 3

Number of outer electrons not shared ( 3-lone pairs) 2 2

Diagram method 1

Diagram method 2

(ii) Phosphorus

Symbol of atom/element taking part in bonding P P

Number of protons/electrons 15 15

Electron configuration/structure 2:8:5 2:8:5

Number of electron in outer energy level 5 5

Number of electrons remaining to be stable/shared 3 3

Number of outer electrons not shared ( 3-lone pairs) 2 2

Diagram method 1

Diagram method 2

e) Water molecule is made up of hydrogen and oxygen. Hydrogen requires to share one electron with oxygen to be stable/attain duplet. Oxygen requires to share two electrons to be stable/attain octet. Two hydrogen atoms share with one oxygen atom for both to be stable as shown below;

Symbol of atoms/elements taking part in bonding O H

Number of protons/electrons 8 1

Electron configuration/structure 2:6 1

Number of electron in outer energy level 6 1

Number of electrons remaining to be stable/shared 2 1

Number of electrons not shared( 2-Oxygen lone pairs) 4 0

Diagram method 1

Diagram method 2

f) Ammonia molecule is made up of Hydrogen and Nitrogen. Hydrogen requires to share one electron with Nitrogen to be stable/attain duplet. Nitrogen requires to share three electrons to be stable/attain octet. Three hydrogen atoms share with one nitrogen atom for both to be stable as shown below;

Symbol of atoms/elements taking part in bonding N H

Number of protons/electrons 7 1

Electron configuration/structure 2:5 1:

Number of electron in outer energy level 5 1

Number of electrons remaining to be stable/shared 3 1

Number of electrons not shared( 1-Nitrogen lone pairs) 2 0

Diagram method 1

Diagram method 2

g)Carbon(IV) oxide molecule is made up of carbon and oxygen. Carbon requires to share four electrons with oxygen to be stable/attain octet. Oxygen requires to share two electrons to be stable/attain octet. Two oxygen atoms share with one carbon atom for both to be stable as shown below;

Symbol of atoms/elements taking part in bonding O C

Number of protons/electrons 8 6

Electron configuration/structure 2:6 2:4

Number of electron in outer energy level 6 4

Number of electrons remaining to be stable/shared 2 4

2-lone pairs from each Oxygen atom) 2 0

Diagram method 1

Diagram method 2

h) Methane molecule is made up of hydrogen and carbon. Hydrogen requires sharing one electron with carbon to be stable/attain duplet. Carbon requires sharing four electrons to be stable/attain octet. Four hydrogen atoms share with one carbon atom for both to be stable as shown below;

Symbol of atoms/elements taking part in bonding C H

Number of protons/electrons 6 1

Electron configuration/structure 2:4 1

Number of electron in outer energy level 4 1

Number of electrons remaining to be stable/shared 4 1

Number of electrons not shared ( No lone pairs) 0 0

Diagram method 1

Diagram method 2

i) Tetrachloromethane molecule is made up of chlorine and carbon. Chlorine requires sharing one electron with carbon to be stable/attain octet. Carbon requires sharing four electrons to be stable/attain octet. Four chlorine atoms share with one carbon atom for both to be stable as shown below;

Symbol of atoms/elements taking part in bonding C Cl

Number of protons/electrons 6 17

Electron configuration/structure 2:4 2:8:7

Number of electron in outer energy level 4 7

Number of electrons remaining to be stable/shared 4 1

3-lone pairs from each Chlorine atom(24 electrons) 0 6

Diagram method 1

Diagram method 2

j) Ethane molecule is made up of six hydrogen and two carbon atoms. Hydrogen requires to share one electron with carbon to be stable/attain duplet. Carbon requires to share four electrons to be stable/attain octet. Three hydrogen atoms share with one carbon atom while another three hydrogen atoms share with a different carbon atom. The two carbon atoms bond by sharing a pair of the remaining electrons as shown below;

Symbol of atoms/elements taking part in bonding C H

Number of protons/electrons 6 1

Electron configuration/structure 2:4 1

Number of electron in outer energy level 4 1

Number of electrons remaining to be stable/shared 4 1

Number of electrons not shared( No lone pairs) 0 0

Diagram method 1

Diagram method 2

k) Ethene molecule is made up of four hydrogen and two carbon atoms. Hydrogen requires to share one electron with carbon to be stable/attain duplet. Carbon requires to share four electrons to be stable/attain octet. Two hydrogen atoms share with one carbon atom while another two hydrogen atoms share with a different carbon atom. The two carbon atoms bond by sharing two pairs of the remaining electrons as shown below;

Symbol of atoms/elements taking part in bonding C H

Number of protons/electrons 6 1

Electron configuration/structure 2:4 1

Number of electron in outer energy level 4 1

Number of electrons remaining to be stable/shared 4 1

Number of electrons not shared( No lone pairs) 0 0

Diagram method 1

Diagram method 2

l) Ethyne molecule is made up of two hydrogen and two carbon atoms. Hydrogen requires to share one electron with carbon to be stable/attain duplet. Carbon requires to share four electrons to be stable/attain octet. One hydrogen atoms share with one carbon atom while another hydrogen atoms share with a different carbon atom. The two carbon atoms bond by sharing three pairs of the remaining electrons as shown below;

Symbol of atoms/elements taking part in bonding C H

Number of protons/electrons 6 1

Electron configuration/structure 2:4 1

Number of electron in outer energy level 4 1

Number of electrons remaining to be stable/shared 4 1

Number of electrons not shared( No lone pairs) 0 0

Diagram method 1

Diagram method 2

j) Ethanol molecule is made up of six hydrogen one Oxygen

atom two carbon atoms.

Five Hydrogen atoms share their one electron each with carbon to be stable/attain duplet. One Hydrogen atoms share one electron with Oxygen for both to attain duplet/octet

Each Carbon uses four electrons to share with “O”and “H”attain octet/duplet.

NB: Oxygen has two lone pairs

j)Ethanoic molecule is made up of four hydrogen two Oxygen atom two carbon atoms.

Three Hydrogen atoms share their one electron each with carbon to be stable/attain duplet. One Hydrogen atoms share one electron with Oxygen for both to attain duplet/octet

Each Carbon uses four electrons to share with “O”and “H”attain octet/duplet.

NB: Each Oxygen atom has two lone pairs

By convention (as a rule), a

(i) single covalent bond made up of two shared( a pair) electrons is represented by a dash(—)

(ii) double covalent bond made up of four shared( two pairs) electrons is represented by a double dash(==)

(iii) triple covalent bond made up of six shared( three pairs) electrons is represented by a triple dash(==)

The representation below show the molecules covered in (a) to (k) above:

Hydrogen molecule(H₂) H–H

Fluorine molecule(F₂) F–F

Chlorine molecule(Cl₂) Cl–Cl

Bromine molecule(Br₂) Br–Br

Iodine molecule(I₂) I–I

Oxygen molecule(O₂) O=O

Nitrogen molecule(N₂) N=N

Phosphorus molecule(P₂) P=P

Water molecule (H₂O) H–O–H

j Ammonia molecule(NH₃) H–N–H

k)Carbon(IV) oxide molecule(CO₂) O==C==O

l)Methane molecule(CH₄) H–C–H

m)Tetrachloromethane molecule(CCl₄ Cl–C–Cl

H H

n)Ethane molecule(C₂H₆) H–C—C–H

H H

p)Ethene molecule(C₂H₄) H-C==C-H

H H

q)Ethyne molecule(C₂H₆) H-C—C-H

Dative /coordinate bond

A dative/coordinate bond is a covalent bond formed when a lone pair of electrons is donated then shared to an electron-deficient species/ion/atom.

During dative/coordinate bonding, all the shared pair of electrons are donated by one of the combining/bonding species/ ion/atom.

Like covalent bonding, coordinate /dative bond is mainly formed by non-metals.

Illustration of coordinate /dative bond

a)Ammonium ion(NH₄⁺)

The ammonium ion is made up of ammonia (NH₃) molecule and hydrogen (H⁺) ion. (H⁺) ion has no electrons. NH₃ is made up of covalent bonding from Nitrogen and Hydrogen. One lone pair of electrons is present in Nitrogen atom after the bonding. This lone pair is donated and shared with an electron-deficient H⁺ ion

Diagram method 1

Diagram method 2

b)Phosphine ion (PH₄⁺)

The Phosphine ion is made up of phosphine(NH₃) molecule and hydrogen (H⁺) ion. (H⁺) ion has no electrons. PH₃ is made up of covalent bonding from Phosphorus and Hydrogen. One lone pair of electrons is present in Phosphorus atom. After the bonding this lone pair is donated and shared with the electron-deficient H⁺ ion

Diagram method 1

Diagram method 2

c) Hydroxonium (H₃O⁺) ion

The hydroxonium ion is made up of water (H₂O) molecule and hydrogen (H⁺) ion. (H⁺) ion has no electrons. The H₂O molecule is made up of covalent bonding from Oxygen and Hydrogen. One lone pair of electrons out of the two present in Oxygen atom after the bonding is donated and shared with the electron-deficient H⁺ ion

Diagram method 1

Diagram method 2

d) Carbon (II) oxide (CO)

Carbon (II) oxide is made up of carbon and Oxygen atoms sharing each two outer electron and not sharing each two electrons. Oxygen with an extra lone pair of electrons donates and share with the carbon atom for both to be stable.

Diagram method 1

Diagram method 2

e) Aluminium (III) chloride (AlCl₃/Al₂Cl₆)

Aluminium (III) chloride is made up of aluminium and chlorine. One aluminium atom shares its outer electrons with three separate chlorine atoms. All chlorine atoms attain stable octet but aluminium does not. Another molecule of aluminium chloride shares its chlorine lone pair of electrons with the aluminium atom for both to be stable. This type of bond exists only in vapour phase after aluminium chloride sublimes.

Diagram method 1

Diagram method 2

A dative/coordinate bond is by convention represented by an arrow (→) heading from the donor of the shared pair of electrons.

Below is the representation of molecules in the above examples;

a)Ammonium ion.

H− N→H

b)Phosphine ion H

H− P→H

c)Hydroxonium ion

H− O→H

d)Carbon(II) oxide O→C

d) Aluminium(III)chloride Cl Cl Cl

Al Al

Cl Cl Cl

(ii)IONIC/ELECTROVALENT BOND

An ionic/electrovalent bond is extreme of a covalent bond.

During ionic/electrovalent bonding there is complete transfer of valence electrons to one electronegative atom from an electropositive atom.

All metals are electropositive and easily/readily donate/lose their valence electrons.

All non-metals are electronegative and easily/readily gain/acquire extra electrons.

Ionic/electrovalent bonding therefore mainly involves transfer of electrons from metal/metallic radical to non-metallic radical.

When an electropositive atom donates /loses the valence electrons, it forms a positively charged cation to attain stable octet/duplet.

When an electronegative atom gains /acquires extra valence electrons, it forms a negatively charged anion to attain stable octet/duplet.

The electrostatic attraction force between the stable positively charged cation and the stable negatively charged anion with opposite charges constitute the ionic bond.

Like in covalent/dative/coordinate bonding, only the outer energy level electrons take part in the formation of ionic/electrovalent bond

Like in covalent/dative/coordinate bonding, the more electrons taking part / involved in the formation of ionic/electrovalent bond, the stronger the ionic /electrovalent bond.

Illustration of ionic /electrovalent bond

a)Sodium chloride(NaCl)

Sodium chloride(NaCl) is formed when a sodium atom donate its outer valence electrons to chlorine atom for both to attain stable octet:

Symbol of atoms/elements taking part in bonding Na Cl

Number of protons/electrons 11 17

Electron configuration/structure 2:8:1 2:8:7

Number of electron in outer energy level 11 7

Number of electrons donated and gained to be stable 1 1

New electron configuration/structure 2:8: 2:8:

Symbol of cation/anion after bonding Na⁺ Cl^–

Diagram

b)Magnesium chloride(MgCl₂)

Magnesium chloride (MgCl₂) is formed when a magnesium atom donate its two outer valence electrons to chlorine atoms. Two chlorine atoms are required to gain each one electron. All the ions (cations and anions) attain stable octet:

Symbol of atoms/elements taking part in bonding Mg Cl

Number of protons/electrons 11 17

Electron configuration/structure 2:8:2 2:8:7

Number of electron in outer energy level 2 7

Number of electrons donated and gained to be stable 2 1

New electron configuration/structure 2:8: 2:8:

Symbol of cation/anion after bonding Mg²⁺ Cl^–

Diagram

c)Lithium oxide(Li₂O)

Lithium oxide(Li₂O)is formed when a Lithium atom donate its outer valence electrons to Oxygen atom. Two Lithium atoms are required to donate/lose each one electron and attain stable duplet. Oxygen atom acquires the two electrons and attain stable octet:

Symbol of atoms/elements taking part in bonding Li O

Number of protons/electrons 3 8

Electron configuration/structure 2:1 2:6

Number of electron in outer energy level 1 6

Number of electrons donated and gained to be stable 1 2

New electron configuration/structure 2: 2:8:

Symbol of cation/anion after bonding Li⁺ O^2-

Diagram

d)Aluminium(III) oxide(Al₂O₃)

Aluminium(III) oxide(Al₂O₃)is formed when a Aluminium atom donate its three outer valence electrons to Oxygen atom. Two Aluminium atoms are required to donate/lose each three electron and attain stable octet. Three Oxygen atoms gain/ acquire the six electrons and attain stable octet:

Symbol of atoms/elements taking part in bonding Al O

Number of protons/electrons 13 8

Electron configuration/structure 2:8:3 2:6

Number of electron in outer energy level 3 6

Number of electrons donated and gained to be stable 3 2

New electron configuration/structure 2:8: 2:8:

Symbol of cation/anion after bonding Al³⁺ O^2-

Diagram

e)Calcium oxide(CaO)

Calcium oxide(CaO)is formed when a Calcium atom donate its two outer valence electrons to Oxygen atom. Both attain stable octet:

Symbol of atoms/elements taking part in bonding Ca O

Number of protons/electrons 20 8

Electron configuration/structure 2:8:8:2 2:6

Number of electron in outer energy level 2 6

Number of electrons donated and gained to be stable 2 2

New electron configuration/structure 2:8:8: 2:8:

Symbol of cation/anion after bonding Ca²⁺ O^2-

Diagram

Some compounds can be formed from ionic/electrovalent, covalent and dative/coordinate bonding within their atoms/molecules:

a)Formation of ammonium chloride:

Ammonium chloride is formed from the reaction of ammonia gas and hydrogen chloride gas. Both ammonia and hydrogen chloride gas are formed from covalent bonding. During the reaction of ammonia and hydrogen chloride gas to form Ammonium chloride;

-ammonia forms a dative/coordinate bond with electron deficient H⁺ ion from Hydrogen chloride to form ammonium ion(NH₄⁺)ion.

-the chloride ion Cl^–and ammonium ion(NH₄⁺)ion bond through ionic / electrovalent bond from the electrostatic attraction between the opposite/unlike charges.

Diagram

b) Dissolution/dissolving of hydrogen chloride:

Hydrogen chloride is formed when hydrogen and chlorine atoms form a covalent bond. Water is formed when hydrogen and Oxygen atoms also form a covalent bond. When hydrogen chloride gas is dissolved in water;

-water molecules forms a dative/coordinate bond with electron deficient H⁺ ion from Hydrogen chloride to form hydroxonium ion(H₃O⁺)ion.

-the chloride ion Cl^–and hydroxonium ion(H₃O⁺)ion bond through ionic / electrovalent bond from the electrostatic attraction between the opposite/unlike charges.

Diagram

c)Dissolution/dissolving of ammonia gas:

Ammmonia gas is formed when hydrogen and Nitrogen atoms form a covalent bond. Water is formed when hydrogen and Oxygen atoms also form a covalent bond. When Ammonia gas is dissolved in water;

-ammonia forms a dative/coordinate bond with electron deficient H⁺ ion from a water molecule to form ammonium ion(NH₄⁺)ion.

-the hydroxide ion OH^–and ammonium ion(NH₄⁺)ion bond through ionic / electrovalent bond from the electrostatic attraction between the opposite/unlike charges.

Diagram

(iii)METALLIC BOND

A metallic bond is formed when metallic atoms delocalize their outer electrons inorder to be stable.

Metals delocalize their outer electrons to form positively charged cation .

The electrostatic attraction force between the metallic cation and the negatively charged electrons constitute the metallic bond.

The more delocalized electrons the stonger the metallic bond.

Illustration of ionic /electrovalent bond

a) Sodium (Na) is made of one valence electron. The electron is donated to form Na⁺ The electron is delocalized /free within many sodium ions.

Symbol of atoms/elements taking part in bonding Na Na Na

Number of protons/electrons 11 11 11

Electron configuration/structure 2:8:1 2:8:1 2:8:1

Number of electron in outer energy level 1 1 1

Number of electrons delocalized/free within 1 1 1

New electron configuration/structure 2:8: 2:8: 2:8:

Symbol of cation after metallic bonding Na⁺ Na⁺ Na⁺

Diagram

(three)Metallic cations attract

(three) free/delocalized electrons

b) Aluminium (Al) is made of three valence electron. The three electrons are donated to form Al³⁺ The electrons are delocalized /free within many aluminium ions.

Symbol of atoms/elements taking part in bonding Al Al Al

Number of protons/electrons 13 13 13

Electron configuration/structure 2:8:3 2:8:3 2:8:3

Number of electron in outer energy level 3 3 3

Number of electrons delocalized/free within 3 3 3

New electron configuration/structure 2:8: 2:8: 2:8:

Symbol of cation after metallic bonding Al³⁺ Al³⁺ Al³⁺

Diagram

(three)Metallic cations attract

(nine) free/delocalized electrons

c)Calcium (Ca) is made of two valence electron.The two electrons are donated to form Ca²⁺ ion.The electrons are delocalized /free within many Calcium ions.

Symbol of atoms/elements taking part in bonding Ca Ca Ca

Number of protons/electrons 20 20 20

Electron configuration/structure 2:8:8:2 2:8:8:2 2:8:8:2

Number of electron in outer energy level 2 2 2

Number of electrons delocalized/free within 2 2 2

New electron configuration/structure 2:8:8: 2:8:8: 2:8:8:

Symbol of cation after metallic bonding Ca²⁺ Ca²⁺ Ca²⁺

Diagram

(three)Metallic cations attract

(six) free/delocalized electrons

d) Magnesium (Mg) is made of two valence electron. The two electrons are donated to form Mg²⁺The electrons are delocalized /free within many Magnesium ions.

Symbol of atoms/elements taking part in bonding Mg Mg

Number of protons/electrons 12 12

Electron configuration/structure 2:8:2 2:8:2

Number of electron in outer energy level 2 2

Number of electrons delocalized/free within 2 2

New electron configuration/structure 2:8: 2:8:

Symbol of cation after metallic bonding Mg²⁺ Mg²⁺

Diagram

(two)Metallic cations attract

(four) free/delocalized electrons

e)Lithium (Li) is made of one valence electron.The electron is donated to form Li⁺ion.The electron is delocalized /free within many Lithium ions.ie;

Symbol of atoms/elements taking part in bonding Li Li Li Li

Number of protons/electrons 3 3 3 3

Electron configuration/structure 2:1 2:1 2:1 2:1

Number of electron in outer energy level 1 1 1 1

Number of electrons delocalized/free within 1 1 1 1

New electron configuration/structure 2:1: 2:1: 2:1: 2:1:

Symbol of cation after metallic bonding Li⁺ Li⁺ Li⁺ Li⁺

Diagram

(four)Metallic cations attract

(four) free/delocalized electrons

B.CHEMICAL STRUCTURE

Chemical structure is the pattern/arrangement of atoms after they have bonded. There are two main types of chemical structures:

(i)simple molecular structure

(ii) giant structures

(i)Simple molecular structure

Simple molecular structure is the pattern formed after atoms of non-metals have covalently bonded to form simple molecules.

Molecules are made of atoms joined together by weak intermolecular forces called Van-der-waals forces.The Van-der-waals forces hold the molecules together while the covalent bonds hold the atoms in the molecule.

Illustration of simple molecular structure

a)Hydrogen molecule(H₂)

Hydrogen gas is made up of strong covalent bonds/intramolecular forces between each hydrogen atom making the molecule. Each molecule is joined to another by weak Van-der-waals forces/ intermolecular forces.

Illustration of simple molecular structure

a)Hydrogen molecule(H₂)

b)Oxygen molecule(O₂)

Oxygen gas is made up of strong covalent bonds/intramolecular forces between each Oxygen atom making the molecule. Each molecule is joined to another by weak Van-der-waals forces/ intermolecular forces.

Strong intramolecular forces/covalent bond

O=O:::: O=O:::: O=O:::: O=O

: : : : : : : : : : : : weak intermolecular

O=O:::: O=O:::: O=O:::: O=O forces/van-der-waals forces

c)Iodine molecule(I₂)

Iodine solid crystals are made up of strong covalent bonds/intramolecular forces between each iodine atom making the molecule.Each molecule is joined to another by weak Van-der-waals forces/ intermolecular forces.

Strong intramolecular forces/covalent bond

I— I:::: I — I:::: I — I:::: I — I

: : : : : : : : : : : : : : weak intermolecular

I — I:::: I — I:::: I — I:::: I — I forces/van-der-waals forces

d)Carbon(IV) oxide molecule(CO₂)

Carbon(IV) oxide gas molecule is made up of strong covalent bonds/intramolecular forces between each Carbon and oxygen atoms making the molecule. Each molecule is joined to another by weak Van-der-waals forces/ intermolecular forces.

Strong intramolecular forces/covalent bond

O=C=O:::: O=C=O:::: O=C=O

: : : : : : weak intermolecular

O=C=O:::: O=C=O:::: O=C=O forces/van-der-waals forces

The following are the main characteristic properties of simple molecular structured compounds:

a)State

Most simple molecular substances are gases, liquid or liquids or solid that sublimes or has low boiling/melting points at room temperature (25^oC) and pressure (atmospheric pressure).

Examples of simple molecular substances include:

-all gases eg Hydrogen, oxygen, nitrogen, carbon (IV) oxide,

–Petroleum fractions eg Petrol, paraffin, diesel, wax,

-Solid non-metals eg Sulphur, Iodine

-Water

b) Low melting/boiling points

Melting is the process of weakening the intermolecular/ van-der-waal forces/ of attraction between the molecules that holding the substance/compound.

Note;

(i)Melting and boiling does not involve weakening/breaking the strong intramolecular force/covalent bonds holding the atoms in the molecule.

(ii) Melting and boiling points increase with increase in atomic radius/size of the atoms making the molecule as the intermolecular forces / van-der-waal forces of attraction between the molecules increase. e.g.

Iodine has a higher melting/boiling point than chlorine because it has a higher /bigger atomic radius/size than chlorine, making the molecule to have stronger intermolecular force/ van-der-waal forces of attraction between the molecules than chlorine. Iodine is hence a solid and chlorine is a gas.

(c)Insoluble in water/soluble in organic solvents

Polar substances dissolve in polar solvents. Water is a polar solvent .Molecular substances do not thus dissolve in water because they are non-polar. They dissolve in non-polar solvents like methylbenzene, benzene, tetrachloromethane or propanone.

d)Poor conductors of heat and electricity

Substances with free mobile ions or free mobile/delocalized electrons conduct electricity. Molecular substances are poor conductors of heat/electricity because their molecules have no free mobile ions/electrons. This makes them very good insulators.

Hydrogen bonds

A hydrogen bond is an intermolecular force of attraction in which a very electronegative atom attracts hydrogen atom of another molecule.

The most electronegative elements are Fluorine, Oxygen and Nitrogen .Molecular compounds made up of these elements usually have hydrogen bonds.

Hydrogen bonds are stronger than van-der-waals forces but weaker than covalent bonds. Molecular compounds with hydrogen bonds thus have higher melting/boiling points than those with van-der-waals forces.

Illustration of Hydrogen bonding

a)Water molecule

During formation of covalent bond, the oxygen atom attract/pull the shared electrons more to itself than Hydrogen creating partial negative charges(δ^–)in Oxygen and partial positive charges(δ⁺)in Hydrogen.

Two molecules attract each other at the partial charges through Hydrogen bonding.

The hydrogen bonding in water makes it;

(i)a liquid with higher boiling and melting point than simple molecular substances with higher molecular mass. e.g. Hydrogen sulphide as in the table below;

Influence of H-bond in water (H₂O) in comparison to H₂S

Substance	Water/ H₂O	Hydrogen sulphide/ H₂S
Relative molecular mass	18	34
Melting point(^oC)	0	-85
Boiling point(^oC)	100	-60

(ii)have higher volume in solid (ice) than liquid (water) and thus ice is less dense than water. Ice therefore floats above liquid water.

b)Ethanol molecule

Like in water, the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen.

This creates a partial negative charge (^δ-) on oxygen and partial positive charge(^δ+) on hydrogen.

Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer.

A dimmer is a molecule formed when two molecules join together as below:

Hydrogen bonds covalent bonds

R₁ O^{δ-…………………….…}H^δ+ O^δ-

H^δ+ R₂

R₁and R₂ are extensions of the molecule.

For ethanol it is made up of CH₃CH₂– to make the structure:

Hydrogen bonds covalent bonds

CH₃CH₂ O^{δ-………………………….…}H^δ+ O^δ-

H^δ+ CH₂CH₃

b)Ethanoic acid molecule

Like in water and ethanol above, the oxygen atom attracts/pulls the shared electrons in the covalent bond in ethanoic acid more to itself than Hydrogen.

This creates a partial negative charge (^δ-)on oxygen and partial positive charge(^δ+) on hydrogen.

Two ethanoic acid molecules attract each other at the partial charges through Hydrogen-bonding forming a dimer.

Hydrogen bonds covalent bonds

R₁ C O^{δ-………………………….…}H^δ+ O^δ-

O^δ- H^{δ+………………..….}O^δ- C R₂

R₁and ₂ are extensions of the molecule.

For ethanoic acid the extension is made up of CH₃– to make the structure;

Hydrogen bonds covalent bonds

CH₃ C O^{δ-…………………………………….…}H^δ+ O^δ-

O^δ- H^{δ+…………………..……..………}O^δ- C CH₃

Ethanoic acid like ethanol exists as a dimer.

Ethanoic acid has a higher melting/boiling point than ethanol .This is because ethanoic acid has two/more hydrogen bond than ethanol.

d) Proteins and sugars in living things also have multiple/complex hydrogen bonds in their structures.

(ii) Giant structure

This is the pattern formed after substances /atoms /ions bond to form a long chain network.

Giant structures therefore extend in all directions to form a pattern that continues repeating itself.

There are three main giant structures.

a) giant covalent/atomic structure b)giant ionic structure

c)giant metallic structure

a) giant covalent/atomic structure

Giant covalent/atomic structure is the pattern formed after atoms have covalently bonded to form long chain pattern consisting of indefinite number of atoms covalently bonded together.

The strong covalent bonds hold all the atoms together to form a very well packed structure. Examples of substances with giant covalent/atomic structure include:

(i) carbon-diamond

(ii) carbon-graphite

(iii)silicon

(iv) silicon(IV) oxide/sand

Carbon-graphite and carbon-diamond are allotropes of carbon.

Allotropy is the existence of an element in more than one stable physical form at the same temperature and pressure.

Allotropes are atoms of the same element existing in more than one stable physical form at the same temperature and pressure.

Other elements that exhibit/show allotropy include;

-Sulphur as monoclinic sulphur and rhombic sulphur

-Phosphorus as white phosphorus and red phosphorus

The structure of carbon-diamond

Carbon has four valence electrons. The four valence electrons are used to form covalent bonds.

During the formation of diamond, one carbon atom covalently bond with four other carbon atoms.

C C

_{x x}.

_x C _x —–> C ._x C_x. C ——> C C C

_{x x}.

C C

After the bonding, the atoms rearrange to form a regular tetrahedral in which one carbon is in the centre while four are at the apex/corners.

C C

This pattern repeats itself to form a long chain number of atoms covalently bonded together indefinitely. The pattern is therefore called giant tetrahedral structure. It extends in all directions where one atom of carbon is always a centre of four others at the apex/corner of a regular tetrahedral.

C C

The giant tetrahedral structure of carbon-diamond is very well/closely packed and joined/bonded together by strong covalent bond.

This makes carbon-diamond to have the following properties:

a) High melting/boiling point.

The giant tetrahedral structure is very well packed and joined together by strong covalent bonds.

This requires a lot of energy/heat to weaken for the element to melt and break for the element to boil.

b) High density.

Carbon diamond is the hardest known natural substance.

This is because the giant tetrahedral structure is a very well packed pattern/structure and joined together by strong covalent bonds.

This makes Carbon diamond be used to make drill for drilling boreholes/oil wells.

The giant tetrahedral structure of carbon diamond is a very closely packed pattern /structure such that heat transfer by conduction is possible. This makes carbon diamond a good thermal conductor.

c) Poor conductor of electricity.

Carbon-diamond has no free/delocalized electrons within its structure and thus do not conduct electricity.

d) Insoluble in water.

Carbon-diamond is insoluble in water because it is non-polar and do not bond with water molecules.

e) Is abrasive/Rough.

The edges of the closely well packed pattern/structure of Carbon-diamond make its surface rough/abrasive and thus able to smoothen /cut metals and glass.

f) Have characteristic luster.

Carbon-diamond has a high optical dispersion and thus able to disperse light to different colours .This makes Carbon-diamond one of the most popular gemstone for making jewellery.

The structure of carbon-graphite

During the formation of graphite, one carbon atom covalently bond with three other carbon atoms leaving one free/delocalized electron.

C C

_{x x}.

_x C _x —–> C ._x C_x ——> C C _x free/delocalized electron

_{x x}.

C C

After the bonding, the atoms rearrange and join together to form a regular hexagon in which six carbon atoms are at the apex/corners.

The regular hexagon is joined to another in layers on the same surface by van-der-waals forces.

Each layer extends to form a plane in all directions.

The fourth valence electron that does not form covalent bonding is free/mobile /delocalized within the layers.

This structure/pattern is called giant hexagonal planar structure.

The giant hexagonal planar structure of carbon-graphite is closely packed and joined/bonded together by strong covalent bonds. This makes carbon-graphite to have the following properties:

a) High melting/boiling point.

The giant hexagonal planar structure of carbon-graphite is well packed and joined together by strong covalent bonds.

This requires a lot of energy/heat to weaken for the element to melt and break for the element to boil.

b) Good conductor of electricity.

Carbon-graphite has free/delocalized 4^th valence electrons within its structure and thus conducts electricity.

c) Insoluble in water.

Carbon-graphite is insoluble in water because it is non-polar and do not bond with water molecules.

d) Soft.

Layers of giant hexagonal planar structure of carbon graphite are held together by van-der-waals forces.

The van-der-waals forces easily break when pressed and reform back on releasing/reducing pressure/force thus making graphite soft.

e) Smooth and slippery.

When pressed at an angle the van-der-waals forces easily break and slide over each other making graphite soft and slippery.

It is thus used as a dry lubricant instead of oil.

f)Some uses of carbon-graphite.

As a dry lubricant– carbon graphite is smooth and slippery and thus better lubricant than oil.Oil heat up when reducing friction.
Making Lead-pencils- When pressed at an angle on paper the van-der-waals forces easily break and slide smoothly over contrasting background producing its characteristic black background.
As moderator in nuclear reactors to reduce the rate of decay/disintegration of radioactive nuclides/atoms/isotopes.
As electrode in dry/wet cells/battery- carbon graphite is inert and good conductor of electricity. Current is thus able to move from one electrode/terminal to the other in dry and wet cells/batteries. Carbon graphite is also very cheap.

b) giant ionic structure

Giant ionic structure is the pattern formed after ions have bonded through ionic/electrovalent bonding to form a long chain consisting of indefinite number of ions.

The strong ionic/electrovalent bond holds all the cations and anions together to form a very well packed structure.

Substances with giant ionic structure are mainly crystals of salts e.g. sodium chloride, Magnesium chloride, Sodium iodide, Potassium chloride, copper (II) sulphate(VI).

The structure of sodium chloride

Sodium chloride is made up of sodium (Na⁺) and chloride (Cl^–)ions.

Sodium (Na⁺) ion is formed when a sodium atom donate /loose/donate an electron. Chloride (Cl^–) ion is formed when a chlorine atom gain /acquire an extra electron from sodium atom.

Many Na⁺ and Cl^– ions then rearrange such that one Na⁺ion is surrounded by six Cl^–ions and one Cl^–ion is surrounded by six Na⁺ions.

The pattern formed is a giant cubic structure where Cl^– ion is sand witched between Na⁺ions and the same to Na⁺ions.

This pattern forms a crystal.

A crystal is a solid form of a substance in which particles are arranged in a definite pattern regularly repeated in three dimensions.

The structure of sodium chloride

The giant cubic structure/crystal of sodium chloride is as below;

The giant cubic structure/crystal of sodium chloride is very well packed and joined by strong ionic/electrovalent bonds. This makes sodium chloride and many ionic compounds to have the following properties:

a) Have high melting /boiling points.

The giant cubic lattice structure of sodium chloride is very closely packed into a crystal that requires a lot of energy/heat to weaken and melt/boil. This applies to all crystalline ionic compounds.

b) Are good conductors of electricity in molten and aqueous state but poor conductor of electricity in solid.

Ionic compounds have fused ions in solid crystalline state.

On heating and dissolving in water, the crystal is broken into free mobile ions (Na⁺ and Cl^– ions).

The free mobile ions are responsible for conducting electricity in ionic compounds in molten and aqueous states.

c)Soluble in water

Ionic compounds are polar and dissolve in polar water molecules.

On dissolving, the crystal breaks to free the fused ions which are then surrounded by water molecules.

b) giant metallic structure

This is the pattern formed after metallic atoms have bonded through metallic bond.

The pattern formed is one where the metallic cations rearrange to form a cubic structure.

The cubic structure is bound together by the free delocalized electrons that move freely within.

The more delocalized electrons, the stronger the metallic bond.

The structure of sodium and aluminium.

Sodium has one valence electrons.

Aluminium has three valence electrons.

After delocalizing the valence electrons ,the metal cations (Na⁺ and Al³⁺) rearrange to the apex /corners of a regular cube that extend in all directions.

The delocalized electrons remain free and mobile as shown below:

The giant cubic structure makes metals to have the following properties:

a) Have high melting/boiling point

The giant cubic structure is very well packed and joined/bonded together by the free delocalized electrons.

The more delocalized electrons the higher the melting/boiling point.

The larger/bigger the metallic cation ,the weaker the packing of the cations and thus the lower the melting/boiling point. e.g.

(i) Sodium and potassium have both one valence delocalized electron.

Atomic radius of potassium is larger/bigger than that of sodium and hence less well packed in its metallic structure.

Sodium has therefore a higher melting/boiling point than potassium.

(ii) Sodium has one delocalized electron.

Aluminium has three delocalized electrons.

Atomic radius of sodium is larger/bigger than that of aluminium and hence less well packed in its metallic structure.

Aluminium has therefore a higher melting/boiling point than sodium because of the smaller well packed metallic (Al³⁺)ions and bonded/joined by more/three delocalized electrons.

The table below shows the comparative melting/boiling points of some metals:

Metal	Electronic structure	Atomic radius(nM)	Melting point(^oC)	Boiling point(^oC)
Sodium	2:8:1	0.155	98	890
Potassium	2:8:8:1	0.203	64	774
Magnesium	2:8:2	0.136	651	1110
Aluminium	2:8:3	0.125	1083	2382

b) Good electrical and thermal conductor/electricity.

All metals are good conductors of heat and electricity including Mercury which is a liquid.

The mobile delocalized electrons are free within the giant metallic structure to move from one end to the other transmitting heat/electric current.

The more delocalized electrons the better the thermal/electrical conductivity.

High temperatures/heating lowers the thermal/electrical conductivity of metals because the delocalized electrons vibrate and move randomly hindering transfer of heat

From the table above:

Compare the electrical conductivity of;

(i)Magnesium and sodium

Magnesium is a better conductor than sodium.

Magnesium has more/two delocalized electrons than sodium. The more delocalized electrons the better the electrical conductor.

(ii)Potassium and sodium

Potassium is a better conductor than sodium.

Potassium has bigger/larger atomic radius than sodium. The delocalized electrons are less attracted to the nucleus of the atom and thus more free /mobile and thus better the electrical conductor.

c) Insoluble in water

All metals are insoluble in water because they are non polar and thus do not bond with water.

Metals higher in the reactivity/electrochemical series like; Potassium, sodium, Lithium and Calcium reacts with cold water producing hydrogen gas and forming an alkaline solution of their hydroxides.ie

2K(s) + 2H₂O(l) -> 2KOH(aq) + H₂(g)

2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)

2Li(s) + 2H₂O(l) -> 2LiOH(aq) + H₂(g)

Ca(s) + 2H₂O(l) -> Ca(OH)₂(aq)+ H₂(g)

Heavy metal like Magnesium, Aluminium, Iron, Zinc and Lead react with steam/water vapour to produce hydrogen gas and form the corresponding oxide.

Mg(s) + H₂O(g) -> MgO(s) + H₂(g)

Fe(s) + H₂O(g) -> FeO(s) + H₂(g)

Zn(s) + H₂O(g) -> ZnO(s) + H₂(g)

Pb(s) + H₂O(g) -> PbO(s) + H₂(g)

2Al(s) + 3H₂O(g) -> Al₂O₃(s) + 3H₂(g)

Metals lower in the reactivity/electrochemical series than hydrogen like; copper, Mercury, Gold Silver and Platinum do not react with water/vapour.

d) Shiny metallic-lustre

All metals have a shiny grey metallic luster except copper which is brown.

When exposed to sunlight, the delocalized electrons gain energy, they vibrate on the metal surface scattering light to appear shiny.

With time, most metals corrode and are covered by a layer of the metal oxide.

The delocalized electrons are unable to gain and scatter light and the metal surface tarnishes/become dull.

e) Ductile and malleable

All metals are malleable (can be made into thin sheet) and ductile (can be made into wire.

When beaten/hit/pressed lengthwise the metallic cations extend and is bound /bonded by the free/mobile electrons to form a sheet.

When beaten/hit/pressed lengthwise and bredthwise the metallic cations extend and is bound /bonded by the free/mobile electrons to form a wire/thin strip.

f) Have high tensile strength

Metals are not brittle. The free delocalized electrons bind the metal together when it is bent /coiled at any angle.

The meta thus withstand stress/coiling

g) Form alloys

An alloy is a uniform mixture of two or more metals.

Some metals have spaces between their metallic cations which can be occupied by another metal cation with smaller atomic radius.

Common alloys include:

Brass(Zinc and Copper alloy)

Bronze(Copper and Tin alloy)

German silver

Summary of Bonding and structure

	Simple molecular structure	Giant covalent /atomic structure	Giant ionic structure	Giant metallic structure
(i)Examples	I₂,S₈,HCl,O₂,CH₄	Graphite,diamond Si,SiO₂	NaCl, KCl, CaO,CuSO₄	Na,Fe,Cr,Hg,K
Constituent particles making structure	molecules	Atoms (of non-metals)	Ions (cation and anions)	Atoms (of metals)
Type of substance	Non-metal element/non-metal molecule/non-metal compound(electronegative elements)	Group IV non-metals and some of their oxides	Metal-non metal compounds(compounds of electropositive and electronegative compounds)	Metallic compounds Metallic elements (with low electonegativity and high electropositivity)
Bonding in solid state	-Strong covalent bonds hold atoms together within separate molecules (intramolecular forces) -Weak van-der-waals forces hold separate molecules together (intermolecular forces)	Atoms are linked through the whole structure by very strong covalent bonds.	Electrostatic attraction of cations and anions link the whole structure through strong ionic bond.	EEElectrostatic Electrostatic attraction of outer mobile electrons for positive nuclei binds atoms together though metallic bond
Properties (i) Volatility	-Highly volatile with low melting/boiling point -Low latent heat of fusion/vaporization	-Non volatile with very high melting/boiling points -Low latent heat of fusion / vaporization	-Non volatile with very high melting/boiling points -Low latent heat of fusion / vaporization	-Non volatile with very high melting/boiling points -Low latent heat of fusion / vaporization
(ii) State at room temperature /pressure	Usually gases,volatile liquids or solids that sublimes	solids	solids	Solids except Mercury(liquid)
(iii) Hardness	Soft and brittle(low tensile strength)	Hard and brittle(low tensile strength)	Hard and brittle(low tensile strength)	Hard, malleable, ductile and have high tensile strength
(iv) Thermal /electrical conductivity	Poor thermal and electrical conductor when solid ,liquid or aqueous solutions but some dissolve and react to form electrolytes e.g. Hydrogen chloride and ammonia gases.	Poor thermal and electrical conductor when solid ,liquid or aqueous solutions but -Carbon-graphite is a good electrical conductor while -Carbon-diamond is a good thermal conductor.	Poor thermal and electrical conductor when solid. Good thermal and electrical conductor in liquid/molten and aqueous states when the ions are not fused	Good thermal and electrical conductor in solid and liquid/molten states due to the free mobile /delocalized electrons
(v) Solubility	Insoluble in polar solvents e.g. Water Soluble in non-polar solvents e.g. tetrachloromethane, benzene, methylbenzene	Insoluble in all solvents	Soluble in polar solvents e.g. Water Insoluble in non-polar solvents e.g. tetrachloromethane, benzene, methylbenzene	Insoluble in polar/non-polar colvents. -Some react with polar solvents -Some metal dissolve in other metals to form alloys e.g. Brass is formed when Zinc dissolve in copper.

PERIODICITY OF BONDING AND STRUCTURE

The periodic table does not classify elements as metals and non-metals. The table arranges

them in terms of atomic numbers.

However, based on structure and bonding of the elements in the periodic table;

(i)-the top right hand corner of about twenty elements are non-metals

(ii)-left of each non-metal is an element which shows characteristics of both metal and non-metal.

These elements are called semi-metals/metalloids. They include Boron, silicon, Germanium, Arsenic, and Terullium

(iii)-all other elements in the periodic table are metal.

(iv)-Hydrogen is a non-metal with metallic characteristic/property of donating/losing outer electron to form cation/H⁺ ion.

(v) –bromine is the only known natural liquid non-metal element at room temperature and pressure.

(vi) –mercury is only known natural liquid metal element at room temperature and pressure.

(vii) Carbon-graphite is a semi metals/metalloids. Carbon-diamond is a pure non-metal yet both are allotropes of carbon (same element)

a) Sketch of the periodic table showing metals ,metalloid and non-metals

Metals Metalloids Non-metals

H								He
Li	Be		B	C	N	O	F	Ne
Na	Mg		Al	Si	P	S	Cl	Ar
K	Ca	Transition metals	Ga	Ge	As	Se	Br	Kr
Rb	Sr		In	Sn	Sb	Te	I	Xe
Cs	Ba		Tl	Pb	Bi	Po	At	Rn
Fr	Ra

b)Periodicity in the physical properties of elements across period 2 and 3

Study table I and II below:

Table I(period 2)

Property	Li	Be	B	C	N	O	F	Ne
Melting point(^oC)	180	1280	2030	3700 (graphite) 3550 (diamond)	-210	-219	-220	-250
Boiling point(^oC)	1330	2480	3930	Graphite sublimes 4830 (diamond)	-200	-180	-190	-245
Density at room temperature (gcm^-3)	0.50	1.85	2.55	2.25 (graphite) 3.53 (diamond)	0.81	0.14	0.11	0.021
Type of element	Metal	Metal	Metal	Metalloid	Non-metal	Non-metal	Non-metal	Non-metal
Chemical structure	Giant metallic	Giant metallic	Giant atomic/ covalent	Giant atomic/ covalent	Simple molecula or molecule/ N₂	Simple molecula or molecules /O₂	Simple molecula or molecule/F₂	Simple molecula or molecule/Ne
State at room temperature	Solid	Solid	Solid	Solid	gas	gas	gas	gas
Electron structure	2:1	2:2	2:3	2:4	2:5	2:6	2:7	2:8
Valency	1	2	3	4	3	2	1	–
Formular of ion	Li⁺	Be²⁺	B³⁺	–	N^3-	O^2-	F^–	–

Table II (period 3)

Property	Na	Mg	Al	Si	P(white)	S(Rhombic)	Cl	Ar
Melting point(^oC)	98	650	660	1410	44	114	-101	-189
Boiling point(^oC)	890	1120	2450	2680	280	445	-34	-186
Density at room temperature (gcm^-3)	0.97	1.74	2.70	2.33 (graphite) 3.53 (diamond)	1.82	2.07	0.157	0.011
Type of element	Metal	Metal	Metal	Metalloid	Non-metal	Non-metal	Non-metal	Non-metal
Chemical structure	Giant metallic	Giant metallic	Giant metallic	Giant atomic/ covalent	Simple molecula or molecule/ P₄	Simple molecula or molecules /S₈	Simple molecula or molecule/Cl₂	Simple molecula or molecule/Ar
State at room temperature	Solid	Solid	Solid	Solid	Solid	Solid	gas	gas
Electron structure	2:8:1	2:8:2	2:8:3	2:8:4	2:8:5	2:8:6	2:8:7	2:8:8
Valency	1	2	3	4	3	2	1	–
Formular of ion	Na⁺	Mg²⁺	Al³⁺	–	P^3-	S^2-	Cl^–	–

From table I and II above:

Explain the trend in atomic radius along /across a period in the periodic table

Observation

Atomic radius of elements in the same period decrease successively across/along a period from left to right.

Explanation

Across/along the period from left to right there is an increase in nuclear charge from additional number of protons and still additional number of electrons entering the same energy level.

Increase in nuclear charge increases the effective nuclear attraction on the outer energy level pulling it closer to the nucleus successively across the period .e.g.

(i)From the table 1and 2 above, atomic radius of Sodium (0.157nM) is higher than that of Magnesium(0.137nM). This is because Magnesium has more effective nuclear attraction on the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus than sodium.

(ii)The rate of decrease in the atomic radius become smaller as the atom become heavier e.g. Atomic radius of Magnesium from sodium falls by(0.157nM- 0.137nM) =0.02

Atomic radius of Chlorine from sulphur falls by(0.104nM- 0.099nM) =0.005

This is because gaining/adding one more proton to 11 already present cause greater proportional change in nuclear attraction power to magnesium than gaining/adding one more proton to 16 already present in sulphur to chlorine.

(iii)Period 3 elements have more energy levels than Period 2 elements. They have therefore bigger/larger atomic radius/size than corresponding period 2 elements in the same group.

2.Explain the trend in ionic radius along/across a period in the periodic table

Observation

Ionic radius of elements in the same period decrease successively across/along a period from left to right for the first three elements then increase drastically then slowly successively decrease.

Explanation

Across/along the period from left to right elements change form electron donors/losers (reducing agents) to electron acceptors (oxidizing agents).

(i)An atom form stable ion by either gaining/acquiring/ accepting extra electron or donating/losing outer electrons.

(ii)Metals form stable ions by donating/losing all the outer energy level electrons and thus also the outer energy level .i.e.

-Sodium ion has one less energy level than sodium atom. The ion is formed by sodium atom donating/losing (all) the outer energy level electron and thus also the outer energy level making the ion to have smaller ionic radius than atom.

(iii)Ionic radius therefore decrease across/along the period from Lithium to Boron in period 2 and from Sodium to Aluminium in period 3.This is because the number of electrons donated/lost causes increased effective nuclear attraction on remaining electrons /energy levels.

(iv)Non-metals form stable ion by gaining/acquiring/accepting extra electron in the outer energy level. The extra electron/s increases the repulsion among electrons and reduces the effective nuclear attraction on outer energy level. The outer energy level therefore expand/enlarge/increase in order to accommodate the extra repelled electrons .The more electrons gained/accepted/acquired the more repulsion and the more expansion to accommodate them and hence bigger/larger atomic radius. e.g.

–Nitrogen ion has three electrons more than Nitrogen atom. The outer energy level expand/enlarge/increase to accommodate the extra repelled electrons. Nitrogen atom thus has smaller atomic radius than the ionic radius of nitrogen ion.

(v) Ionic radius decrease from group IV onwards from left to right. This because the number of electrons gained to form ion decrease across/along the period from left to right. e.g. Nitrogen ion has bigger/larger ionic radius than Oxygen.

3.Explain the trend in melting and boiling point of elements in a period in the periodic table.

Observation

The melting and boiling point of elements rise up to the elements in Group IV(Carbon/Silicon) along/across the period then continuously falls.

Explanation

Melting/boiling points depend on the packing of the structure making the element and the strength of the bond holding the atoms/molecules together.

Across/along the period (2 and 3) the structure changes from giant metallic, giant atomic/covalent to simple molecular.

(i)For metals, the number of delocalized electrons increases across/along the period and hence stronger metallic bond/structure thus requiring a lot of heat/energy to weaken.

The strength of a metallic bond also depends on the atomic radius/size. The melting /boiling point decrease as the atomic radius/size of metals increase due to decreased packing of larger atoms. e.g.

-The melting /boiling point of Lithium is lower than that of Beryllium because Beryllium has two/more delocalized electrons and hence stronger metallic structure/bond.

– The melting /boiling point of Lithium is higher than that of Sodium because the atomic radius/size Lithium is smaller and hence better packed and hence forms stronger metallic structure/bond.

(ii)Carbon-graphite/carbon-diamond in period 2 and Silicon in period 3 form very well packed giant atomic/covalent structures held together by strong covalent bonds. These elements have therefore very high melting/boiling points.

Both Carbon-graphite/ carbon-diamond have smaller atomic radius/size than Silicon in period 3 and thus higher melting/boiling points due to better/closer packing of smaller atoms in their well packed giant atomic/covalent structures.

(ii)Non-metals from group V along/across the period form simple molecules joined by weak intermolecular /van-der-waals force. The weak intermolecular /van-der-waals force require little energy/heat to weaken leading to low melting/boiling points. The strength of the intermolecular /van-der-waals forces decrease with decrease in atomic radius/ size lowering the melting/boiling points along/across the period (and raising the melting/boiling points down the group).e.g.

-The melting /boiling point of Nitrogen is higher than that of Oxygen. This is because the atomic radius/ size of Nitrogen is higher than that of Oxygen and hence stronger intermolecular /van-der-waals forces between Nitrogen molecules.

-The melting /boiling point of Chlorine is higher than that of Fluorine. This is because the atomic radius/ size of Chlorine is higher than that of Fluorine and hence stronger intermolecular /van-der-waals forces between Chlorine molecules.

(iii)Rhombic sulphur exists as a puckered ring of S₈atoms which are well packed. Before melting the ring break and join to very long chains that entangle each other causing the unusually high melting/boiling point of Rhombic sulphur.

(iv)Both sulphur and phosphorus exists as allotropes.

Sulphur exists as Rhombic-sulphur and monoclinic-sulphur. Rhombic-sulphur is the stable form of sulphur at room temperature and pressure.

Phosphorus exists as white-phosphorus and red-phosphorus.

White-phosphorus is the stable form of Phosphorus at room temperature and pressure.

State and explain the trend in density of elements in a period in the periodic table.

Observation: Density increase upto the elements in group IV then falls across/along the period successively

Explanation:

Density is the mass per unit volume occupied by matter/particles/atoms/molecules of element.

(i)For metals ,the stronger metallic bond and the more delocalized electrons ensure a very well packed giant metallic structure that occupy less volume and thus higher density.

The more the number of delocalized electrons along/across the period, the higher the density. e.g.

(i)Aluminium has a higher density than sodium. This is because aluminium has more /three delocalized electrons than /one sodium thus forms a very well packed giant metallic structure that occupy less volume per given mass/density.

(ii)Carbon-graphite ,carbon-diamond and silicon in group IV form a well packed giant atomic/covalent structure that is continuously joined by strong covalent bonds hence occupy less volume per given mass/density.

Carbon-graphite form a less well packed giant hexagonal planar structure joined by Van-der-waals forces. Its density (2.25gcm^-3) is therefore less than that of Carbon-diamond(3.53gcm^-3) and silicon(2.33gcm^-3).Both diamond and silicon have giant tetrahedral structure that is better packed. Carbon-diamond has smaller atomic radius/size than silicon. Its density is thus higher because of better packing and subsequently higher density. Carbon-diamond is the hardest known natural substance by having the highest density.

(iii)For non-metals, the strength of the intermolecular /van-der-waals forces decreases with decrease in atomic radius/size along/across the period. This decreases the mass occupied by given volume of atoms in a molecule from group VI onwards. e.g.

Phosphorus has a higher atomic radius/size than chlorine and Argon and thus stronger intermolecular/van-der-waals forces that ensure a given mass of phosphorus occupy less volume than chlorine and neon.

5.State and explain the trend in thermal/electrical conductivity of elements in a period in the periodic table.

Observation:

Increase along/across the period from group I, II, and III then decrease in Group IV to drastically decrease in group V to VIII (O).

Explanation

(i)Metals have free delocalized electrons that are responsible for thermal/electrical conductivity.Thermal/electrical conductivity increase with increase in number of delocalized electrons. The thermal conductivity decrease with increase in temperature/heating.

e.g.

Aluminium with three delocalized electrons from each atom in its metallic structure has the highest electrical /thermal conductivity in period 3.

(ii)Carbon-graphite has also free 4^th valency electrons that are delocalized within its layers of giant hexagonal planar structure. They are responsible for the electrical conductivity of graphite.

(iii)Silicon and carbon diamond do not conduct electricity but conducts heat. With each atom too close to each other in their very well packed giant tetrahedral structure, heat transfer /radiate between the atoms. The thermal conductivity increase with increase in temperature/heating.

(iv)All other non-metals are poor /non-conductor of heat and electricity. They are made of molecules with no free /mobile delocalized electrons in their structure.

Periodicity of the oxides of elements along/across period 3

The table below summarizes some properties of the oxides of elements in period 3 of the periodic table.

Formular of oxide/ Property	Na₂O	MgO	Al₂O₃	SiO₂	P₂O₅ P₄O₆	SO₂ SO₃	Cl₂O₇ Cl₂O



Melting point(^oC)	1193	3075	2045	1728	563	-76	-60
Boiling point(^oC)	1278	3601	2980	2231	301	-10	-9
Bond type	Ionic	Ionic	Ionic	Covalent	Covalent	Covalent	Covalent
Chemical structure	Giant ionic structure	Giant ionic structure	Giant ionic structure	Giant atomic/ covalent	Simple molecula or molecule	Simple molecula or molecules	Simple molecula or molecule
State at room temperature	Solid	Solid	Solid	Solid	Solid	gas	Gas (Cl₂O₇is a liquid)
Nature of Oxide	Basic/ alkaline	Basic/ alkaline	Amphotellic oxide	2:8:4	2:8:5	2:8:6	2:8:7
Reaction with water	React to form NaOH /alkaline solution	React to form MgOH)₂ /weakly alkaline solution	Don’t react with water.	Don’t react with water.	React to form H₂PO₄ /weakly acidic solution	-SO₂ react to form H₂SO₃ . H₂SO₃is quickly oxidized to H₂SO₄ -SO₂ react to form H₂SO₄/ strongly acidic	-Cl₂O₇ reacts to form HClO₄ /weakly acidic solution
Reaction with dilute acids	Reacts to form salt and water	Reacts to form salt and water	Reacts to form salt and water	No reaction	No reaction	No reaction	No reaction

All the oxides of elements in period 3 except those of sulphur and chlorine are solids at room temperature and pressure.
Across/along the period, bonding of the oxides changes from ionic in sodium oxide magnesium oxide and aluminium oxide (show both ionic and covalent properties) to covalent in the rest of the oxides.

Across/along the period, the structure of the oxides changes from giant ionic structure in sodium oxide, magnesium oxide and aluminium oxide to giant atomic/covalent structure in silicon (IV) oxide. The rest of the oxides form simple molecules/molecular structure.

Sodium oxide and magnesium oxide are basic /alkaline in nature. Aluminium oxide is amphotellic in nature (shows both acidic and basic characteristics). The rest of the oxides are acidic in nature.

Ionic compounds/oxides have very high melting/boiling points because of the strong electrostatic attraction joining the giant ionic crystal lattice.

The melting/boiling points increase from sodium oxide to aluminium oxide as the number of electrons involved in bonding increase, increasing the strength of the ionic bond/structure.

Silicon (IV) oxide is made of a well packed giant atomic/covalent structure joined by strong covalent bonds.

This results in a solid with very high melting/boiling point.

7.Phosphorus (V) oxide, sulphur(IV) oxide/ sulphur (VI) oxide and dichloride heptoxide exist as simple molecules/molecular structure joined by weak van-der-waals/intermolecular forces.

This results in them existing as low melting /boiling point solids/gases.

Ionic oxide conducts electricity in molten and aqueous states but not in solid.

In solid state the ions are fused/fixed but on heating to molten state and when dissolved in water, the ions are free / mobile.

Sodium oxide, magnesium oxide and aluminium oxide are therefore good conductors in molten and aqueous states.

Covalent bonded oxides do not conduct electricity in solid, molten or in aqueous states.

This is because they do not have free / mobile ion. Phosphorus (V) oxide, sulphur(IV) oxide/ sulphur (VI) oxide and dichloride heptoxide are thus non-conductors/insulators.

Silicon (IV) oxide is a poor/weak conductor of heat in solid state. This is because it has very closely packed structure for heat to radiate conduct along its structure.

Electopositivity decrease across the period while electronegativity increase across the period. The oxides thus become less ionic and more covalent along/across the period.

12.The steady change from giant ionic structure to giant atomic/ covalent structure then simple molecular structure lead to profound differences in the reaction of the oxides with water,acids and alkalis/bases:

(i) Reaction with water

a) Ionic oxides react with water to form alkaline solutions e.g.;

I.Sodium oxide reacts/dissolves in water forming an alkaline solution of sodium hydroxide.

Chemical equation: Na₂O(s) + H₂O (l) -> 2NaOH(aq)

Magnesium oxide slightly/ slowly reacts/dissolves in water forming an alkaline solution of magnesium hydroxide

Chemical equation: MgO(s) + 2H₂O (l) -> Mg(OH)₂ (aq)

III. Aluminium oxide does reacts/dissolves in water.

b) Non-metallic oxides are acidic. They react with water to form weakly acidic solutions:
Phosphorus (V) oxide readily reacts/dissolves in water forming a weak acidic solution of phosphoric (V) acid.

Chemical equation: P₄O₁₀ (s) + 6H₂O (l) -> 4H₃PO₄ (aq)

Chemical equation: P₂O₅ (s) + 3H₂O (l) -> 2H₃PO₄ (aq)

Sulphur (IV) oxide readily reacts/dissolves in water forming a weak acidic solution of sulphuric (IV) acid.

Chemical equation: SO₂ (g) + H₂O (l) -> H₂SO₃ (aq)

Sulphur (VI) oxide quickly fumes in water to form concentrated sulphuric (VI) acid which is a strong acid.

Chemical equation: SO₃ (g) + H₂O (l) -> H₂SO₄ (aq)

III. Dichlorine oxide reacts with water to form weak acidic solution of chloric(I) acid/hypochlorous acid.

Chemical equation: Cl₂O (g) + H₂O (l) -> 2HClO (aq)

Dichlorine heptoxide reacts with water to form weak acidic solution of chloric(VII) acid.

Chemical equation: Cl₂O₇ (l) + H₂O (l) -> 2HClO₄ (aq)

c) Silicon (IV) oxide does not react with water.

It reacts with hot concentrated alkalis forming silicate (IV) salts. e.g.

Silicon (IV) oxide react with hot concentrated sodium hydroxide to form sodium silicate (IV) salt.

Chemical equation: SiO₂ (s) + 2NaOH (aq) -> Na₂SiO₃ (aq) + H₂O (l)

(ii) Reaction with dilute acids

a) Ionic oxides react with dilute acids to form salt and water only. This is a neutralization e.g.

Chemical equation: Na₂O(s) + H₂SO₄ (aq) -> Na₂SO₄ (aq) + H₂O(l)

Chemical equation: MgO(s) + 2HNO₃(aq) -> Mg (NO₃)₂ (aq) + H₂O(l)

Chemical equation: Al₂O₃ (s) + 6HCl(aq) -> 2AlCl₃ (aq) + 3H₂O(l)

Aluminium oxide is amphotellic and reacts with hot concentrated strong alkalis sodium/potassium hydroxides to form complex sodium aluminate(III) and potassium aluminate(III) salt.

Chemical equation: Al₂O₃ (s) + 2NaOH(aq) + 3H₂O(l) -> 2 NaAl(OH)₄ (aq)

Chemical equation: Al₂O₃ (s) + 2KOH(aq) + 3H₂O(l) -> 2 KAl(OH)₄ (aq)

b) Acidic oxides do not react with dilute acids.

c)Periodicity of the Chlorides of elements along/across period 3

The table below summarizes some properties of the chlorides of elements in period 3 of the periodic table.

Formular of chloride/ Property	NaCl	MgCl₂	AlCl₃	SiCl₄	PCl₅ PCl₃	SCl₂ S₂Cl₂	Cl₂



Melting point(^oC)	801	714	Sublimes at 180 ^oC	-70	PCl₅ Sublimes at -94 ^oC	-78	-101
Boiling point(^oC)	1465	1418	423(as Al₂Cl₆ vapour	57	74(as P₂Cl₆ Vapour 164 (as PCl₅)	decomposes at 59^oC	-34
Bond type	Ionic	Ionic	Ionic/ Covalent/ dative	Covalent	Covalent	Covalent	Covalent
Chemical structure	Giant ionic structure	Giant ionic structure	Molecular/ dimerizes	Simple molecula or molecule	Simple molecula or molecule	Simple molecula or molecules	Simple molecula or molecule
State at room temperature	Solid	Solid	Solid	liquid	Liquid PCl₅ is solid	liquid	Gas
Nature of Chloride	Neutral	Neutral	Strongly acidic	Strongly acidic	Strongly acidic	Strongly acidic	Strongly acidic
pH of solution	7.0	7.0	3.0	3.0	3.0	3.0	3.0
Reaction with water	Dissolve	Dissolve	-Hydrolysed by water -Acidic hydrogen chloride fumes produced	-Hydrolysed by water -Acidic hydrogen chloride fumes produced	Hydrolysed by water -Acidic hydrogen chloride fumes produced	Hydrolysed by water -Acidic hydrogen chloride fumes produced	Forms HCl and HClO
Electrical conductivity in molten/aqueous state	good	good	poor	nil	nil	nil	nil

Sodium Chloride, Magnesium chloride and aluminium chloride are solids at room temperature and pressure.

Silicon(IV) chloride, phosphorus(III)chloride and disulphur dichloride are liquids. Phosphorus(V)chloride is a solid. Both chlorine and sulphur chloride are gases.

Across/along the period bonding changes from ionic in Sodium Chloride and Magnesium chloride to covalent in the rest of the chlorides.

Anhydrous aluminium chloride is also a molecular compound .Each aluminium atom is covalently bonded to three chlorine atoms.

In vapour/gaseous phase/state two molecules dimerizes to Al₂O₆ molecule through coordinate/dative bonding.

Across/along the period the structure changes from giant ionic in Sodium Chloride and Magnesium chloride to simple molecules/molecular structure in the rest of the chlorides.

Ionic chlorides have very high melting /boiling points because of the strong ionic bond/electrostatic attraction between the ions in their crystal lattice.The rest of the chlorides have low melting /boiling points because of the weak van-der-waal /intermolecular forces.

Sodium Chloride and Magnesium chloride in molten and aqueous state have free/mobile ions and thus good electrical conductors. Aluminium chloride is a poor conductor. The rest of the chlorides do not conduct because they have no free/mobile ions.

Ionic chloride form neutral solutions with pH =7. These chlorides ionize/dissociate completely into free cations and anions.i.e;

Sodium Chloride and Magnesium chloride have pH=7 because they are fully/completely ionized/dissociated into free ions.

Chemical equation NaCl (s) -> Na⁺(aq) + Cl^–(aq)

Chemical equation MgCl₂ (s) -> Mg²⁺(aq) + 2Cl^–(aq)

8 Across/along the period from aluminium chloride, hydrolysis of the chloride takes place when reacting/dissolved in water.

Hydrolysis is the reaction of a compound when dissolved in water.

a)Aluminium chloride is hydrolyzed by water to form aluminium hydroxide and fumes of hydrogen chloride gas. Hydrogen chloride gas dissolves in water to acidic hydrochloric acid. Hydrochloric acid is a strong acid with low pH and thus the mixture is strongly acidic.

Chemical equation AlCl₃ (s) + 3H₂O(l)-> Al(OH)₃(s) + 3HCl(g)

b)Silicon(IV) chloride is hydrolyzed by water to form silicon(IV)oxide and fumes of hydrogen chloride gas. Hydrogen chloride gas dissolves in water to acidic hydrochloric acid. Hydrochloric acid is a strong acid with low pH and thus the mixture is strongly acidic.

Chemical equation SiCl₄ (l) + 2H₂O(l)-> SiO₂(s) + 4HCl(g)

This reaction is highly exothermic producing /evolving a lot of heat that cause a rise in the temperature of the mixture.

c) Both phosphoric (V) chloride and phosphoric (III) chloride are hydrolyzed by water to form phosphoric (V) acid and phosphoric (III) acid Fumes of hydrogen chloride gas are produced. Hydrogen chloride gas dissolves in water to acidic hydrochloric acid. Hydrochloric acid is a strong acid with low pH and thus the mixture is strongly acidic.

Chemical equation PCl₅ (s) + 4H₂O(l)-> H₃PO₄(aq) + 5HCl(g)

Chemical equation PCl₃ (s) + 3H₂O(l)-> H₃PO₄(aq) + 3HCl(g)

This reaction is also highly exothermic producing /evolving a lot of heat that cause a rise in the temperature of the mixture.

d) Disulphur dichloride similarly hydrolyzes in water to form yellow deposits of sulphur and produce a mixture of sulphur (IV) oxide and hydrogen chloride Hydrogen chloride gas dissolves in water to acidic hydrochloric acid. Hydrochloric acid is a strong acid with low pH and thus the mixture is strongly acidic.

Chemical equation 2S₂Cl₂ (l) + 2H₂O(l)-> 3S(s) + SO₂(g) + 4HCl(g)

COMPREHENSIVE REVISION QUESTIONS

1.The grid below represents periodic table. Study it and answer the questions that follow. The letters do not represent the actual symbols of the elements.

						A
B			G	H	E	C
	J	I	L
D	N				M

(a) (I) Indicate on the grid the position of an element represented by letter N whose electronic configuration of a divalent cation is 2:8:8 . ( 1 mark )

(II) Name the bond formed between D and H react. Explain your answer.(2 marks )

Ionic/electrovalent

D is electropositive thus donates two electrons to electronegative H

(III) Write an equation for the reaction between B and water. ( 1 mark )

Chemical equation 2B (s) + 2H₂O(l) -> 2BOH(aq) + H₂ (g)

(IV) How do the atomic radii of I and L compare. Explain. ( 2 marks )

(V) In terms of structure and bonding explain why the oxide of G has lower melting point than oxide of L. ( 2 marks )

(b) Study the information given below and answer the question that follow.

Formula of compound N	NaCl	MgCl₂	Al₂Cl₆	SiCl₄	PCl₃	SCl₂
B.P(⁰C)	1470	1420	Sublimes	60	75	60
M.P(⁰C)	800	710	At 800⁰C	-70	90	-80

( I)Why is the formula of aluminium chloride given as Al₂Cl₆ and not AlCl₃ ? ( 1 mark )

(II) Give two chlorides that are liquid at room temperature. Give a reason for the answer. (2 marks )

(III) Give a reason why Al₂Cl₆ has a lower melting point than MgCl₂although both Al and Mg are metals. (1 mark )

(IV) Which of the chlorides would remain in liquid state for the highest temperature range explain why ? (2 mark )

(Kakamega)

a) Study the information given below and answer the questions that follow.

Element

Atomic radius (nm)

Ionic radius (nm)

Formula of oxide

Melting point of oxide (‘C)

0.364

0.830

0.592

0.381

0.762

0.421

0.711

0.485

0.446

0.676

A₂O

BO₂

E₂O₃

G₂O₅

-119

837

1466

242

1054

(i) Which elements are non-metals? Give a reason. (2mks)

(ii) Explain why the melting point of the oxide of R is higher than that of the oxide of S. (2mks)

(iii) Give two elements that would react vigorously with each other. Explain your answer. (2mks)

b) Study the information in the table below and answer the questions that follow (The letters do not represent the actual symbols of the elements)

		Ionization Energy_kJ/Mole
Element	Electronic configuration	1^st ionization energy	2^nd ionization energy
A	2.2	900	1800
B	2.8.2	736	1450
C	2.8.8.2	590	1150

(i) What chemical family do the elements A, B and C belong? (1mk)

(ii) What is meant by the term ionization energy? (1mk)

iii) The 2^nd ionization energy is higher that the 1^st ionization energy of each. Explain

(1mk)

(iv)When a piece of element C is placed in cold water, it sinks to the bottom and an effervescence of a colourless gas that burns explosively is produced. Use a simple diagram to illustrate how this gas can be collected during this experiment. (3mks)

The grid below represents part of the periodic table. The letters do not represent the actual symbols.

						A
B		X	G	Z	E	V
	J	I	L	T
D	N				M

a) Select the most reactive non-metal. (1mk)

b) Write the formula of the compound consisting of

I.D and Z only. (2mk)

X and Z

c) Select an element that can form an ion of change +2 (1mk)

d) Which element has the least ionization energy? Explain (2mks)
e) Suggest with reasons a likely pH value of an aqueous solution of the chlorine of:(3mks)

f) To which chemical family do the following elements belong? (2mk)

g) An element K has relative atomic mass of 40.2.It has two isotopes of masses 39 and 42. Calculate the relative abundance of each isotope. (3mks)

4.The grid below shows part of the periodic table study it and answer the questions that follow. The letters do not represent the true symbols.




				A
	B	C	D		E
F	G
					H

\(a) Which element forms ions with charge of 2-. Explain (2mks)

(b) What is the nature of the oxide formed by C. (1mk)

(d)Write down a balanced equation between B and Chlorine. (1mk)

(e) Explain how the atomic radii of F and G compare. (1mk)

(f) If the oxides of F and D are separately dissolved in water, state and explain the effects of their aqueous solutions on litmus. (3mks)

(a) The grid below show part of the periodic table.(The letter do not represent the actual symbols).Use it to answer the questions that follow.

T							Q
			S		R	K
A	J	Y		U		L
W						M	B
	C					N
P

(i)Select the most reactive non-metal. (1mk)

(ii)Select an element that forms a divalent cation. (1mk)

(iii)Element Z has atomic number 14.Show its position in the grid. (1mk)

(iv)How do the atomic radii of U and J compare? (2mks)

(v)How do electrical conductivity of A and Y compare? (2mks)

(vi)How does the boiling point of elements K, L and M vary? Explain (2mks

(b) The table below gives information on four elements by letters K, L, M and N. Study it and answer the questions that follow. The letters do not represent the actual symbols of the elements.

Element	Electron arrangement	Atomic radius	Ionic radius
K	2:8:2	0.136	0.065
L	2:8:7	0.099	0.181
M	2:8:8:1	0.203	0.133
N	2:8:8:2	0.174	0.099

(a) Which two elements have similar properties? Explain. (2mks)

(b) Which element is a non-metal? Explain. (1mk)

The grid given below represents part of the periodic table study it and answer the questions that follow. (The letters do not represent the actual symbols of the elements.)

				A
		B
C	D		E
F

(i) What name is given to the group of elements to which C and F belong? (1mk)

(ii) Which letter represents the element that is the least reactive? (1mk)

(iii) What type of bond is formed when B and E react? Explain (2mks)

(iv)Write formula of the compound formed where elements D and oxygen gas react. (1mk)

(v) On the grid indicate the a tick (√) the position of element G which is in the third period of the periodic table and forms G^3- ions. (1mk)

(b) Study the information in the table below and answer the questions that follow. (The letter do not represents the actual symbols of the substance).

Substance	Melting point ^oC	Boiling point ^oC	Solubility in water	Density at room. Temp/g/cm³
H	-117	78.5	Very soluble	0.8
J	-78	-33	Very soluble	0.77x 1^-3
K	-23	77	Insoluble	1.6
L	– 219	-183	Slightly Soluable	1.33 x 10^-3

I.(i) Which substance would dissolve in water and could be separated from the solution by fractional distillation. (1mk)

(ii) Which substances is a liquid at room temperature and when mixed with water two layers would be formed? (1mk)

Which letter represents a substance that is a gas at room temperature and which can be collected ;

(i) Over water? (1mk)

(ii) By downward displacement of air? Density of air at room temperature = 1.29 x 10^-3 g/C

(1mk)

UPGRADE

CHEMISTRY

FORM 2

Introduction to SALTS

Comprehensive tutorial notes

MUTHOMI S.G

www.kcselibrary.info

0720096206

INTRODUCTION TO SALTS

1.(a) A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid.

A salt is therefore formed when the hydrogen ions in an acid are replaced wholly/fully or partially/partly ,directly or indirectly by a metal or ammonium radical.

(b) The number of ionizable/replaceable hydrogen in an acid is called basicity of an acid.

Some acids are therefore:

(i)monobasic acids generally denoted HX e.g.

HCl, HNO₃,HCOOH,CH3COOH.

(ii)dibasic acids ; generally denoted H₂X e.g.

H₂SO₄, H₂SO₃, H₂CO₃,HOOCOOH.

(iii)tribasic acids ; generally denoted H₃X e.g.

H₃PO₄.

(i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.

(ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.

Table showing normal and acid salts derived from common acids

Acid name	Chemical formula	Basicity	Normal salt	Acid salt
Hydrochloric acid	HCl	Monobasic	Chloride(Cl^–)	None
Nitric(V)acid	HNO₃	Monobasic	Nitrate(V)(NO₃^–)	None
Nitric(III)acid	HNO₂	Monobasic	Nitrate(III)(NO₂^–)	None
Sulphuric(VI)acid	H₂SO₄	Dibasic	Sulphate(VI) (SO₄^2-)	Hydrogen sulphate(VI) (HSO₄^–)
Sulphuric(IV)acid	H₂SO₃	Dibasic	Sulphate(IV) (SO₃^2-)	Hydrogen sulphate(IV) (HSO₃^–)
Carbonic(IV)acid	H₂CO₃	Dibasic	Carbonate(IV)(CO₃^2-)	Hydrogen carbonate(IV) (HCO₃^–)
Phosphoric(V) acid	H₃PO₄	Tribasic	Phosphate(V)(PO₄^3-)	Dihydrogen phosphate(V) (H₂PO₄^2-) Hydrogen diphosphate(V) (HP₂O₄^2-)

The table below show shows some examples of salts.

Base/alkali	Cation	Acid	Anion	Salt	Chemical name of salts
NaOH	Na⁺	HCl	Cl^–	NaCl	Sodium(I)chloride
Mg(OH)₂	Mg²⁺	H₂SO₄	SO₄^2-	MgSO₄ Mg(HSO₄)₂	Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI)
Pb(OH)₂	Pb²⁺	HNO₃	NO₃^–	Pb(NO₃)₂	Lead(II)nitrate(V)
Ba(OH)₂	Ba²⁺	HNO₃	NO₃^–	Ba(NO₃)₂	Barium(II)nitrate(V)
Ca(OH)₂	Ba²⁺	H₂SO₄	SO₄^2-	MgSO₄	Calcium sulphate(VI)
NH₄OH	NH₄⁺	H₃PO₄	PO₄^3-	(NH₄ )₃PO₄ (NH₄ )₂HPO₄ NH₄ H₂PO₄	Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V)
KOH	K⁺	H₃PO₄	PO₄^3-	K₃PO₄	Potassium phosphate(V)
Al(OH)₃	Al³⁺	H₂SO₄	SO₄^2-	Al₂(SO₄)₂	Aluminium(III)sulphate(VI)
Fe(OH)₂	Fe²⁺	H₂SO₄	SO₄^2-	FeSO₄	Iron(II)sulphate(VI)
Fe(OH)₃	Fe³⁺	H₂SO₄	SO₄^2-	Fe₂(SO₄)₂	Iron(III)sulphate(VI)

(d) Some salts undergo hygroscopy, deliquescence and efflorescence.

(i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution.

Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt.

(ii)Deliquescent salts /compounds are those that absorb water from the atmosphere and form a solution.

Some salts which are deliquescent include: Sodium nitrate(V),Calcium chloride, Sodium hydroxide, Iron(II)chloride, Magnesium chloride.

(iii)Efflorescent salts/compounds are those that lose their water of crystallization to the atmosphere.

Some salts which effloresces include: sodium carbonate decahydrate, Iron(II)sulphate(VI)heptahydrate, sodium sulphate (VI)decahydrate.

(e)Some salts contain water of crystallization.They are hydrated.Others do not contain water of crystallization. They are anhydrous.

Table showing some hydrated salts.

Name of hydrated salt	Chemical formula
Copper(II)sulphate(VI)pentahydrate	CuSO₄.5H₂O
Aluminium(III)sulphate(VI)hexahydrate	Al₂ (SO₄)₃.6H₂O
Zinc(II)sulphate(VI)heptahydrate	ZnSO₄.7H₂O
Iron(II)sulphate(VI)heptahydrate	FeSO₄.7H₂O
Calcium(II)sulphate(VI)heptahydrate	CaSO₄.7H₂O
Magnesium(II)sulphate(VI)heptahydrate	MgSO₄.7H₂O
Sodium sulphate(VI)decahydrate	Na₂SO₄.10H₂O
Sodium carbonate(IV)decahydrate	Na₂CO₃.10H₂O
Potassium carbonate(IV)decahydrate	K₂CO₃.10H₂O
Potassium sulphate(VI)decahydrate	K₂SO₄.10H₂O

(f)Some salts exist as a simple salt while some as complex salts. Below are some complex salts.

Table of some complex salts

Name of complex salt	Chemical formula	Colour of the complex salt
Tetraamminecopper(II)sulphate(VI)	Cu(NH₃)₄ SO₄ H₂O	Royal/deep blue solution
Tetraamminezinc(II)nitrate(V)	Zn(NH₃)₄ (NO₃ )₂	Colourless solution
Tetraamminecopper(II) nitrate(V)	Cu(NH₃)₄ (NO₃ )₂	Royal/deep blue solution
Tetraamminezinc(II)sulphate(VI)	Zn(NH₃)₄ SO₄	Colourless solution

(g)Some salts exist as two salts in one. They are called double salts.

Table of some double salts

Name of double salts	Chemical formula
Trona(sodium sesquicarbonate)	Na₂CO₃ NaHCO₃.2H₂O
Ammonium iron(II)sulphate(VI)	FeSO₄(NH₄)₂SO₄.2H₂O
Ammonium aluminium(III)sulphate(VI)	Al₂(SO₄)₃(NH₄)₂SO₄.H₂O

(h)Some salts dissolve in water to form a solution. They are said to be soluble. Others do not dissolve in water. They form a suspension/precipitate in water.

Table of solubility of salts

Soluble salts	Insoluble salts
All nitrate(V)salts
All sulphate(VI)/SO₄^2- salts except	Barium(II) sulphate(VI)/BaSO₄ Calcium(II) sulphate(VI)/CaSO₄ Lead(II) sulphate(VI)/PbSO₄
All sulphate(IV)/SO₃^2- salts except	Barium(II) sulphate(IV)/BaSO₃ Calcium(II) sulphate(IV)/CaSO₃ Lead(II) sulphate(IV)/PbSO₃
All chlorides/Cl^–except	Silver chloride/AgCl Lead(II)chloride/PbCl₂(dissolves in hot water)
All phosphate(V)/PO₄^3-
All sodium,potassium and ammonium salts
All hydrogen carbonates/HCO₃^–
All hydrogen sulphate(VI)/ HSO₄^–
Sodium carbonate/Na₂CO₃, potassium carbonate/ K₂CO₃, ammonium carbonate (NH₄)₂CO₃	except All carbonates
All alkalis(KOH,NaOH, NH₄OH)	except All bases

13 Salts can be prepared in a school laboratory by a method that uses its solubility in water.

Soluble salts may be prepared by using any of the following methods:

(i)Direct displacement/reaction of a metal with an acid.

By reacting a metal higher in the reactivity series than hydrogen with a dilute acid,a salt is formed and hydrogen gas is evolved.

Excess of the metal must be used to ensure all the acid has reacted.

When effervescence/bubbling /fizzing has stopped ,excess metal is filtered.

The filtrate is heated to concentrate then allowed to crystallize.

Washing with distilled water then drying between filter papers produces a sample crystal of the salt. i.e.

M(s) + H₂X -> MX(aq) + H₂(g)

Examples

Mg(s) + H₂SO₄(aq) -> MgSO₄(aq) + H₂(g)

Zn(s) + H₂SO₄(aq) -> ZnSO₄(aq) + H₂(g)

Pb(s) + 2HNO₃(aq) -> Pb(NO₃)₂(aq) + H₂(g)

Ca(s) + 2HNO₃(aq) -> Ca(NO₃)₂(aq) + H₂(g)

Mg(s) + 2HNO₃(aq) -> Mg(NO₃)₂(aq) + H₂(g)

Mg(s) + 2HCl(aq) -> MgCl₂(aq) + H₂(g)

Zn(s) + 2HCl(aq) -> ZnCl₂(aq) + H₂(g)

(ii)Reaction of an insoluble base with an acid

By adding an insoluble base (oxide/hydroxide )to a dilute acid until no more dissolves, in the acid,a salt and water are formed. Excess of the base is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter papers e.g.

PbO(s) + 2HNO₃(aq) -> Pb(NO₃)₂(aq) + H₂O (l)

Pb(OH)₂(s) + 2HNO₃(aq) -> Pb(NO₃)₂(aq) + 2H₂O (l)

CaO (s) + 2HNO₃(aq) -> Ca(NO₃)₂(aq) + H₂O (l)

MgO (s) + 2HNO₃(aq) -> Mg(NO₃)₂(aq) + H₂O (l)

MgO (s) + 2HCl(aq) -> MgCl₂(aq) + H₂O (l)

ZnO (s) + 2HCl(aq) -> ZnCl₂(aq) + H₂O (l)

Zn(OH)₂(s) + 2HNO₃(aq) -> Zn(NO₃)₂(aq) + 2H₂O (l)

CuO (s) + 2HCl(aq) -> CuCl₂(aq) + H₂O (l)

CuO (s) + H₂SO₄(aq) -> CuSO₄(aq) + H₂O (l)

Ag₂O(s) + 2HNO₃(aq) -> 2AgNO₃(aq) + H₂O (l)

Na₂O(s) + 2HNO₃(aq) -> 2NaNO₃(aq) + H₂O (l)

(iii)reaction of insoluble /soluble carbonate /hydrogen carbonate with an acid.

By adding an excess of a soluble /insoluble carbonate or hydrogen carbonate to adilute acid, effervescence /fizzing/bubbling out of carbon(IV)oxide gas shows the reaction is taking place. When effervescence /fizzing/bubbling out of the gas is over, excess of the insoluble carbonate is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter paper papers e.g.

PbCO₃ (s) + 2HNO₃(aq) -> Pb(NO₃)₂(aq) + H₂O (l)+ CO₂(g)

ZnCO₃ (s) + 2HNO₃(aq) -> Zn(NO₃)₂(aq) + H₂O (l)+ CO₂(g)

CaCO₃ (s) + 2HNO₃(aq) -> Ca(NO₃)₂(aq) + H₂O (l)+ CO₂(g)

MgCO₃ (s) + H₂SO₄(aq) -> MgSO₄(aq) + H₂O (l)+ CO₂(g)

Cu CO₃ (s) + H₂SO₄(aq) -> CuSO₄(aq) + H₂O (l) + CO₂(g)

Ag₂CO₃ (s) + 2HNO₃(aq) -> 2AgNO₃(aq) + H₂O (l) + CO₂(g)

Na₂CO₃ (s) + 2HNO₃(aq) -> 2NaNO₃(aq) + H₂O (l) + CO₂(g)

K₂CO₃ (s) + 2HCl(aq) -> 2KCl(aq) + H₂O (l) + CO₂(g)

NaHCO₃ (s) + HNO₃(aq) -> NaNO₃(aq) + H₂O (l) + CO₂(g)

KHCO₃ (s) + HCl(aq) -> KCl(aq) + H₂O (l) + CO₂(g)

(iv)neutralization/reaction of soluble base/alkali with dilute acid

By adding an acid to a burette into a known volume of an alkali with 2-3 drops of an indicator, the colour of the indicator changes when the acid has completely reacted with an alkali at the end point. The procedure is then repeated without the indicator .The solution mixture is then heated to concentrate , allowed to crystallize ,washed with distilled water before drying with filter papers. e.g.

NaOH (aq) + HNO₃(aq) -> NaNO₃(aq) + H₂O (l)

KOH (aq) + HNO₃(aq) -> KNO₃(aq) + H₂O (l)

KOH (aq) + HCl(aq) -> KCl(aq) + H₂O (l)

2KOH (aq) + H₂SO₄(aq) -> K₂SO₄(aq) + 2H₂O (l)

2 NH₄OH (aq) + H₂SO₄(aq) -> (NH₄)₂SO₄(aq) + 2H₂O (l)

NH₄OH (aq) + HNO₃(aq) -> NH₄NO₃(aq) + H₂O (l)

(iv)direct synthesis/combination.

When a metal burn in a gas jar containing a non metal , the two directly combine to form a salt. e.g.

2Na(s) + Cl₂(g) -> 2NaCl(s)

2K(s) + Cl₂(g) -> 2KCl(s)

Mg(s) + Cl₂(g) -> Mg Cl₂ (s)

Ca(s) + Cl₂(g) -> Ca Cl₂ (s)

Some salts once formed undergo sublimation and hydrolysis. Care should be taken to avoid water/moisture into the reaction flask during their preparation.Such salts include aluminium(III)chloride(AlCl₃) and iron (III)chloride(FeCl₃)

Heated aluminium foil reacts with chlorine to form aluminium(III)chloride that sublimes away from the source of heating then deposited as solid again

2Al(s) + 3Cl₂(g) -> 2AlCl₃ (s/g)

Once formed aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas.

AlCl₃(s)+ 3H₂ O(g) -> Al(OH)₃ (aq) + 3HCl(g)

Heated iron filings reacts with chlorine to form iron(III)chloride that sublimes away from the source of heating then deposited as solid again

2Fe(s) + 3Cl₂(g) -> 2FeCl₃ (s/g)

Once formed , aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas.

FeCl₃(s)+ 3H₂ O(g) -> Fe(OH)₃ (aq) + 3HCl(g)

(b)Insoluble salts can be prepared by reacting two suitable soluble salts to form one soluble and one insoluble. This is called double decomposition or precipitation. The mixture is filtered and the residue is washed with distilled water then dried.

CuSO₄(aq) + Na₂CO₃ (aq) -> CuCO₃ (s) + Na₂ SO₄(aq)

BaCl₂(aq) + K₂SO₄ (aq) -> BaSO₄ (s) + 2KCl (aq)

Pb(NO₃)₂(aq) + K₂SO₄ (aq) -> PbSO₄ (s) + 2KNO₃ (aq)

2AgNO₃(aq) + MgCl₂ (aq) -> 2AgCl(s) + Mg(NO₃)₂ (aq)

Pb(NO₃)₂(aq) + (NH₄)₂SO₄ (aq) -> PbSO₄ (s) + 2NH₄NO₃(aq)

BaCl₂(aq) + K₂SO₃ (aq) -> BaSO₃ (s) + 2KCl (aq)

Salts may lose their water of crystallization , decompose ,melt or sublime on heating on a Bunsen burner flame.

The following shows the behavior of some salts on heating gently /or strongly in a laboratory school burner:

(a)effect of heat on chlorides

All chlorides have very high melting and boiling points and therefore are not affected by laboratory heating except ammonium chloride. Ammonium chloride sublimes on gentle heating. It dissociate into the constituent ammonia and hydrogen chloride gases on strong heating.

NH₄Cl(s) NH₄Cl(g) NH₃(g) + HCl(g)

(sublimation) (dissociation)

(b)effect of heat on nitrate(V)

(i) Potassium nitrate(V)/KNO₃ and sodium nitrate(V)/NaNO₃ decompose on heating to form Potassium nitrate(III)/KNO₂ and sodium nitrate(III)/NaNO₂ and producing Oxygen gas in each case.

2KNO₃ (s) -> 2KNO₂(s) + O₂(g)

2NaNO₃ (s) -> 2NaNO₂(s) + O₂(g)

(ii)Heavy metal nitrates(V) salts decompose on heating to form the oxide and a mixture of brown acidic nitrogen(IV)oxide and oxygen gases. e.g.

2Ca(NO₃)₂ (s) -> 2CaO(s) + 4NO₂(g) + O₂(g)

2Mg(NO₃)₂(s) -> 2MgO(s) + 4NO₂(g) + O₂(g)

2Zn(NO₃)₂(s) -> 2ZnO(s) + 4NO₂(g) + O₂(g)

2Pb(NO₃)₂(s) -> 2PbO(s) + 4NO₂(g) + O₂(g)

2Cu(NO₃)₂(s) -> 2CuO(s) + 4NO₂(g) + O₂(g)

2Fe(NO₃)₂(s) -> 2FeO(s) + 4NO₂(g) + O₂(g)

(iii)Silver(I)nitrate(V) and mercury(II) nitrate(V) are lowest in the reactivity series. They decompose on heating to form the metal(silver and mercury)and the Nitrogen(IV)oxide and oxygen gas. i.e.

2AgNO₃(s) -> 2Ag (s) + 2NO₂(g) + O₂(g)

2Hg(NO₃)₂(s) -> 2Hg (s) + 4NO₂(g) + O₂(g)

(iv)Ammonium nitrate(V) and Ammonium nitrate(III) decompose on heating to Nitrogen(I)oxide(relights/rekindles glowing splint) and nitrogen gas respectively.Water is also formed.i.e.

NH₄NO₃(s) -> N₂O (g) + H₂O(l)

NH₄NO₂(s) -> N₂ (g) + H₂O(l)

(c) effect of heat on nitrate(V)

Only Iron(II)sulphate(VI), Iron(III)sulphate(VI) and copper(II)sulphate(VI) decompose on heating. They form the oxide, and produce highly acidic fumes of acidic sulphur(IV)oxide gas.

2FeSO₄ (s) -> Fe₂O₃(s) + SO₃(g) + SO₂(g)

Fe₂(SO₄)₃(s) -> Fe₂O₃(s) + SO₃(g)

CuSO₄ (s) -> CuO(s) + SO₃(g)

(d) effect of heat on carbonates(IV) and hydrogen carbonate(IV).

(i)Sodium carbonate(IV)and potassium carbonate(IV)do not decompose on heating.

(ii)Heavy metal nitrate(IV)salts decompose on heating to form the oxide and produce carbon(IV)oxide gas. Carbon (IV)oxide gas forms a white precipitate when bubbled in lime water. The white precipitate dissolves if the gas is in excess. e.g. CuCO₃ (s) -> CuO(s) + CO₂(g)

CaCO₃ (s) -> CaO(s) + CO₂(g)

PbCO₃ (s) -> PbO(s) + CO₂(g)

FeCO₃ (s) -> FeO(s) + CO₂(g)

ZnCO₃ (s) -> ZnO(s) + CO₂(g)

(iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e.

2NaHCO₃(s) -> Na₂CO₃(s) + CO₂(g) + H₂O(l)

2KHCO₃(s) -> K₂CO₃(s) + CO₂(g) + H₂O(l)

(iii) Calcium hydrogen carbonate (IV) and Magnesium hydrogen carbonate(IV) decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i. e.

Ca(HCO₃)₂(aq) -> CaCO₃(s) + CO₂(g) + H₂O(l)

Mg(HCO₃)₂(aq) -> MgCO₃(s) + CO₂(g) + H₂O(l)

UPGRADE

CHEMISTRY

FORM 2

Introduction to ELECTROLYSIS

Comprehensive tutorial notes

MUTHOMI S.G

www.kcselibrary.info

0720096206

INTRODUCTION TO ELECTROLYSIS (ELECTROLYTIC CELL)

1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity.

A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong.

2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions. Common strong electrolytes include:

(i)all mineral acids

(ii)all strong alkalis/sodium hydroxide/potassium hydroxide.

(iii)all soluble salts

3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions.

Common weak electrolytes include:

(i)all organic acids

(ii)all bases except sodium hydroxide/potassium hydroxide.

(iii)Water

4. A compound that is not decomposed by an electric current is called non-electrolyte.

Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions .

Common non-electrolytes include:

(i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol)

(ii)all hydrocarbons(alkanes /alkenes/alkynes)

(iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar)

5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state.

6.During electrolysis, the free ions are attracted to the electrodes.

An electrode is a rod through which current enter and leave the electrolyte during electrolysis.

An electrode that does not influence/alter the products of electrolysis is called an inert electrode.

Common inert electrodes include:

(i)Platinum

(ii)Carbon graphite

Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells).

7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte

8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte

9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e.

M(l) -> M⁺(l) + e (for cations from molten electrolytes)

M(s) -> M⁺(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water)

The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode

During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.

X⁺ (aq) + 2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water)

2X⁺ (l) + 2e -> X (l) (for cations from molten electrolytes)

The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode.

The below set up shows an electrolytic cell.

For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples:

a)To determine the products of electrolysis of molten Lead(II)chloride

(i)Decomposition of electrolyte into free ions;

PbCl₂ (l) -> Pb ²⁺(l) + 2Cl^–(l)

(Compound decomposed into free cation and anion in liquid state)

(ii)At the cathode/negative electrode(-);

Pb ²⁺(l) + 2e -> Pb (l)

(Cation / Pb ²⁺gains / accepts / acquires electrons to form free atom)

(iii)At the anode/positive electrode(+);

2Cl^–(l) -> Cl₂ (g) + 2e

(Anion / Cl^–donate/lose electrons to form free atom then a gas molecule)

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid lead metal.

II.At the anode pale green chlorine gas.

b)To determine the products of electrolysis of molten Zinc bromide

(i)Decomposition of electrolyte into free ions;

ZnBr₂ (l) -> Zn ²⁺(l) + 2Br^–(l)

(Compound decomposed into free cation and anion in liquid state)

(ii)At the cathode/negative electrode(-);

Zn ²⁺(l) + 2e -> Zn(l)

(Cation / Zn²⁺gains / accepts / acquires electrons to form free atom)

(iii)At the anode/positive electrode(+);

2Br^–(l) -> Br₂ (g) + 2e

(Anion / Br^–donate/lose electrons to form free atom then a liquid molecule which change to gas on heating)

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid Zinc metal.

II.At the anode red bromine liquid / red/brown bromine gas.

c)To determine the products of electrolysis of molten sodium chloride

(i)Decomposition of electrolyte into free ions;

NaCl (l) -> Na ⁺(l) + Cl^–(l)

(Compound decomposed into free cation and anion in liquid state)

(ii)At the cathode/negative electrode(-);

2Na⁺(l) + 2e -> Na (l)

(Cation / Na⁺gains / accepts / acquires electrons to form free atom)

(iii)At the anode/positive electrode(+);

2Cl^–(l) -> Cl₂ (g) + 2e

(Anion / Cl^–donate/lose electrons to form free atom then a gas molecule)

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid sodium metal.

II.At the anode pale green chlorine gas.

d)To determine the products of electrolysis of molten Aluminium (III)oxide

(i)Decomposition of electrolyte into free ions;

Al₂O₃ (l) -> 2Al ³⁺(l) + 3O^2-(l)

(Compound decomposed into free cation and anion in liquid state)

(ii)At the cathode/negative electrode(-);

4Al ³⁺ (l) + 12e -> 4Al (l)

(Cation / Al ³⁺gains / accepts / acquires electrons to form free atom)

(iii)At the anode/positive electrode(+);

6O^2-(l) -> 3O₂ (g) + 12e

(Anion /6O^2- donate/lose 12 electrons to form free atom then three gas molecule)

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid aluminium metal.

II.At the anode colourless gas that relights/rekindles glowing splint.

13.In industries electrolysis has the following uses/applications:

(a)Extraction of reactive metals from their ores.

Potassium, sodium ,magnesium, and aluminium are extracted from their ores using electrolytic methods.

(b)Purifying copper after exraction from copper pyrites ores.

Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there:

(i)At the cathode; Cu²⁺ (aq) + 2e -> Cu(s) (Pure copper deposits on the strip

(ii)At the anode; Cu(s) ->Cu²⁺ (aq) + 2e (impure copper erodes/dissolves)

(c)Electroplating

The label EPNS(Electro Plated Nickel Silver) on some steel/metallic utensils mean they are plated/coated with silver and/or Nickel to improve their appearance(add their aesthetic value)and prevent/slow corrosion(rusting of iron). Electroplating is the process of coating a metal with another metal using an electric current. During electroplating, the cathode is made of the metal to be coated/impure.

Example:

During the electroplating of a spoon with silver

(i)the spoon/impure is placed as the cathode(negative terminal of battery)

(ii)the pure silver is placed as the anode(positive terminal of battery)

(iii)the pure silver erodes/ionizes/dissociates to release electrons:

Ag(s) ->Ag⁺ (aq) + e (impure silver erodes/dissolves)

(iv) silver (Ag⁺)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure

Ag⁺ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon)

UPGRADE

CHEMISTRY

FORM 2

Chemistry of CARBON

Comprehensive tutorial notes

MUTHOMI S.G

www.kcselibrary.info

0720096206

A: CARBON

Carbon is an element in Group IV(Group 4)of the Periodic table .It has atomic number 6 and electronic configuration 2:4 and thus has four valence electrons(tetravalent).It does not easily ionize but forms strong covalent bonds with other elements including itself.

(a)Occurrence

Carbon mainly naturally occurs as:

(i)allotropes of carbon i.e graphite, diamond and fullerenes.

(ii)amorphous carbon in coal, peat ,charcoal and coke.

(iii)carbon(IV)oxide gas accounting 0.03% by volume of normal air in the atmosphere.

(b)Allotropes of Carbon

Carbon naturally occur in two main crystalline allotropic forms, carbon-graphite and carbon-diamond

Carbon-diamond	Carbon-graphite
Shiny crystalline solid	Black/dull crystalline solid
Has a very high melting/boiling point because it has a very closely packed giant tetrahedral structure joined by strong covalent bonds	Has a high melting/boiling point because it has a very closely packed giant hexagonal planar structure joined by strong covalent bonds
Has very high density(Hardest known natural substance)	Soft
Abrassive	Slippery
Poor electrical conductor because it has no free delocalized electrons	Good electrical conductor because it has free 4^th valency delocalized electrons
Is used in making Jewels, drilling and cutting metals	Used in making Lead-pencils,electrodes in batteries and as a lubricant
Has giant tetrahedral structure	Has giant hexagonal planar structure

c)Properties of Carbon

(i)Physical properties of carbon

Carbon occur widely and naturally as a black solid

It is insoluble in water but soluble in carbon disulphide and organic solvents.

It is a poor electrical and thermal conductor.

(ii)Chemical properties of carbon

Burning

Experiment

Introduce a small piece of charcoal on a Bunsen flame then lower it into a gas jar containing Oxygen gas. Put three drops of water. Swirl. Test the solution with blue and red litmus papers.

Observation

-Carbon chars then burns with a blue flame

-Colourless and odourless gas produced

-Solution formed turn blue litmus paper faint red. Red litmus paper remains red.

Explanation

Carbon burns in air and faster in Oxygen with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas. Carbon burns in limited supply of air with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas. Carbon (IV) oxide gas dissolve in water to form weak acidic solution of Carbonic (IV)acid.

Chemical Equation

C(s) + O₂(g) -> CO₂(g) (in excess air)

2C(s) + O₂(g) -> 2CO(g) (in limited air)

CO₂(g) + H₂O (l) -> H₂CO₃ (aq) (very weak acid)

Reducing agent

Experiment

Mix thoroughly equal amounts of powdered charcoal and copper (II)oxide into a crucible. Heat strongly.

Observation

Colour change from black to brown

Explanation

Carbon is a reducing agent. For ages it has been used to reducing metal oxide ores to metal, itself oxidized to carbon(IV)oxide gas. Carbon reduces black copper(II)oxide to brown copper metal

Chemical Equation

2CuO(s) + C(s) -> 2Cu(s) + CO₂(g)

(black) (brown)

2PbO(s) + C(s) -> 2Pb(s) + CO₂(g)

(brown when hot/ (grey)

yellow when cool)

2ZnO(s) + C(s) -> 2Zn(s) + CO₂(g)

(yellow when hot/ (grey)

white when cool)

Fe₂O₃(s) + 3C(s) -> 2Fe(s) + 3CO₂(g)

(brown when hot/cool (grey)

Fe₃O₄ (s) + 4C(s) -> 3Fe(s) + 4CO₂(g)

(brown when hot/cool (grey)

B: COMPOUNDS OF CARBON

The following are the main compounds of Carbon

(i)Carbon(IV)Oxide(CO₂)

(ii)Carbon(II)Oxide(CO)

(iii)Carbonate(IV) (CO₃^2-)and hydrogen carbonate(IV(HCO₃^–)

(iv)Sodium carbonate(Na₂CO₃)

(i) Carbon(IV)Oxide (CO₂)

(a)Occurrence

Carbon(IV)oxide is found:

-in the air /atmosphere as 0.03% by volume.

-a solid carbon(IV)oxide mineral in Esageri near Eldame Ravine and Kerita near Limuru in Kenya.

(b)School Laboratory preparation

In the school laboratory carbon(IV)oxide can be prepared in the school laboratory from the reaction of marble chips(CaCO₃)or sodium hydrogen carbonate(NaHCO₃) with dilute hydrochloric acid.

(c)Properties of carbon(IV)oxide gas(Questions)

1.Write the equation for the reaction for the school laboratory preparation of carbon (IV)oxide gas.

Any carbonate reacted with dilute hydrochloric acid should be able to generate carbon (IV)oxide gas.

Chemical equations

CaCO₃(s) + 2HCl(aq) -> CaCO₃ (aq) + H₂O(l) + CO₂(g)

ZnCO₃(s) + 2HCl(aq) -> ZnCO₃ (aq) + H₂O(l) + CO₂(g)

MgCO₃(s) + 2HCl(aq) -> MgCO₃ (aq) + H₂O(l) + CO₂(g)

CuCO₃(s) + 2HCl(aq) -> CuCO₃ (aq) + H₂O(l) + CO₂(g)

NaHCO₃(s) + HCl(aq) -> Na₂CO₃ (aq) + H₂O(l) + CO₂(g)

KHCO₃(s) + HCl(aq) -> K₂CO₃ (aq) + H₂O(l) + CO₂(g)

2.What method of gas collection is used in preparation of Carbon(IV)oxide gas. Explain.

Downward delivery /upward displacement of air/over mercury

Carbon(IV)oxide gas is about 1½ times denser than air.

3.What is the purpose of :

(a)water?

To absorb the more volatile hydrogen chloride fumes produced during the vigorous reaction.

(b)sodium hydrogen carbonate?

To absorb the more volatile hydrogen chloride fumes produced during the vigorous reaction and by reacting with the acid to produce more carbon (IV)oxide gas .

Chemical equation

NaHCO₃(s) + HCl(aq) -> Na₂CO₃ (aq) + H₂O(l) + CO₂(g)

(c)concentrated sulphuric(VI)acid?

To dry the gas/as a drying agent

4.Describe the smell of carbon(IV)oxide gas

Colourless and odourless

Effect on lime water.

Experiment

Bubbled carbon(IV)oxide gas into a test tube containing lime water for about three minutes

Observation

White precipitate is formed.

White precipitate dissolved when excess carbon(IV)oxide gas is bubbled .

Explanation

Carbon(IV)oxide gas reacts with lime water(Ca(OH)₂) to form an insoluble white precipitate of calcium carbonate. Calcium carbonate reacts with more Carbon(IV) oxide gas to form soluble Calcium hydrogen carbonate.

Chemical equation

Ca(OH)₂(aq) + CO₂(g) -> CaCO₃ (s) + H₂O(l)

CaCO₃ (aq) + H₂O(l) + CO₂(g) -> Ca(HCO₃)₂ (aq)

Effects on burning Magnesium ribbon

Experiment

Lower a piece of burning magnesium ribbon into a gas jar containing carbon (IV)oxide gas.

Observation

The ribbon continues to burn with difficulty

White ash/solid is formed.

Black speck/solid/particles formed on the side of gas jar.

Explanation

Carbon(IV)oxide gas does not support combustion/burning.Magnesium burn to produce/release enough heat energy to decompose Carbon(IV) oxide gas to carbon and oxygen.Magnesium continues to burn in Oxygen forming white Magnesium Oxide solid/ash.Black speck/particle of carbon/charcoal residue forms on the sides of reaction flask. During the reaction Carbon(IV) oxide is reduced(Oxidizing agent)to carbon while Magnesium is Oxidized to Magnesium Oxide.

Chemical equation

2Mg(s) + CO₂(g) -> C (s) + 2MgO(l)

Dry and wet litmus papers were separately put in a gas jar containing dry carbon (IV)oxide gas. State and explain the observations made.

Observation

Blue dry litmus paper remain blue

Red dry litmus paper remain Red

Blue wet/damp/moist litmus paper turn red

Red wet/damp/moist litmus paper remain red

Explanation

Dry Carbon (IV) oxide gas is a molecular compound that does not dissociate/ionize to release H⁺ and thus has no effect on litmus papers.

Wet/damp/moist litmus papers contains water that dissolves/react with dry carbon (IV) oxide gas to form the weak solution of carbonic (IV) acid(H₂CO₃).

Carbonic (IV) acid dissociate/ionizes to a few /little free H⁺ and CO₃^2-.

The few H⁺ (aq) ions are responsible for turning blue litmus paper to faint red showing the gas is very weakly acidic.

Chemical equation

H₂CO₃(aq) -> 2H⁺ (aq) + CO₃^2-(aq)

Explain why Carbon (IV)oxide cannot be prepared from the reaction of:

(i) marble chips with dilute sulphuric(VI)acid.

Explanation

Reaction forms insoluble calcium sulphate(VI)that cover/coat unreacted marble chips stopping further reaction

Chemical equation

CaCO₃(s) + H₂SO₄ (aq) -> CaSO₄ (s) + H₂O(l) + CO₂(g)

PbCO₃(s) + H₂SO₄ (aq) -> PbSO₄ (s) + H₂O(l) + CO₂(g)

BaCO₃(s) + H₂SO₄ (aq) -> BaSO₄ (s) + H₂O(l) + CO₂(g)

(ii) Lead(II)carbonate with dilute Hydrochloric acid.

Reaction forms insoluble Lead(II)Chloride that cover/coat unreacted Lead(II) carbonate stopping further reaction unless the reaction mixture is heated. Lead(II)Chloride is soluble in hot water.

Chemical equation

PbCO₃(s) + 2HCl (aq) -> PbCl₂ (s) + H₂O(l) + CO₂(g)

Describe the test for the presence of Carbon (IV)oxide.

Using burning splint

Lower a burning splint into a gas jar suspected to contain Carbon (IV)oxide gas.The burning splint is extinguished.

Using Lime water.

Bubble the gas suspected to be Carbon (IV)oxide gas.A white precipitate that dissolve in excess bubbling is formed.

Chemical equation

Ca(OH)₂(aq) + CO₂(g) -> CaCO₃ (s) + H₂O(l)

CaCO₃ (aq) + H₂O(l) + CO₂(g) -> Ca(HCO₃)₂ (aq)

10.State three main uses of Carbon (IV)oxide gas

(i)In the Solvay process for the manufacture of soda ash/sodium carbonate

(ii)In preservation of aerated drinks

(iii)As fire extinguisher because it does not support combustion and is denser than air.

(iv)In manufacture of Baking powder.

(ii) Carbon(II)Oxide (CO)

(a)Occurrence

Carbon(II)oxide is found is found from incomplete combustion of fuels like petrol charcoal, liquefied Petroleum Gas/LPG.

(b)School Laboratory preparation

In the school laboratory carbon(II)oxide can be prepared from dehydration of methanoic acid/Formic acid(HCOOH) or Ethan-1,2-dioic acid/Oxalic acid(HOOCCOOH) using concentrated sulphuric(VI) acid. Heating is necessary.

METHOD 1:Preparation of Carbon (IV)Oxide from dehydration of Oxalic/ethan-1,2-dioic acid

METHOD 2:Preparation of Carbon (IV)Oxide from dehydration of Formic/Methanoic acid

(c)Properties of Carbon (II)Oxide(Questions)

1.Write the equation for the reaction for the preparation of carbon(II)oxide using;

(i)Method 1;

Chemical equation

HOOCCOOH(s) –Conc.H₂SO₄–> CO(g) + CO₂ (g) + H₂O(l)

H₂C₂O₄(s) –Conc.H₂SO₄–> CO(g) + CO₂ (g) + H₂O(l)

(ii)Method 2;

Chemical equation

HCOOH(s) –Conc.H₂SO₄–> CO(g) + H₂O(l)

H₂CO₂(s) –Conc.H₂SO₄–> CO(g) + H₂O(l)

2.What method of gas collection is used during the preparation of carbon (II) oxide.

Over water because the gas is insoluble in water.

Downward delivery because the gas is 1 ½ times denser than air .

3.What is the purpose of :

(i) Potassium hydroxide/sodium hydroxide in Method 1

To absorb/ remove carbon (II) oxide produced during the reaction.

2KOH (aq) + CO₂(g) -> K₂CO₃ (s) + H₂O(l)

2NaOH (aq) + CO₂(g) -> Na₂CO₃ (s) + H₂O(l)

(ii) Concentrated sulphuric(VI)acid in Method 1 and 2.

Dehydrating agent –removes the element of water (Hydrogen and Oxygen in ratio 2:1) present in both methanoic and ethan-1,2-dioic acid.

Describe the smell of carbon(II)oxide.

Colourless and odourless.

State and explain the observation made when carbon(IV)oxide is bubbled in lime water for a long time.

No white precipitate is formed.

Dry and wet/moist/damp litmus papers were separately put in a gas jar containing dry carbon(IV)oxide gas. State and explain the observations made.

Observation

-blue dry litmus paper remains blue

-red dry litmus paper remains red

– wet/moist/damp blue litmus paper remains blue

– wet/moist/damp red litmus paper remains red

Explanation

Carbon(II)oxide gas is a molecular compound that does not dissociate /ionize to release H+ ions and thus has no effect on litmus papers. Carbon(II)oxide gas is therefore a neutral gas.

Carbon (II)oxide gas was ignited at the end of a generator as below.

Flame K

Dry carbon(II)oxide

(i)State the observations made in flame K.

Gas burns with a blue flame

(ii)Write the equation for the reaction taking place at flame K.

2CO(g) + O₂(g) -> 2CO₂(g)

Carbon(II)oxide is a reducing agent. Explain

Experiment

Pass carbon(II)oxide through glass tube containing copper (II)oxide. Ignite any excess poisonous carbon(II)oxide.

Observation

Colour change from black to brown. Excess carbon (II)oxide burn with a blue flame.

Explanation

Carbon is a reducing agent. It is used to reduce metal oxide ores to metal, itself oxidized to carbon(IV)oxide gas. Carbon(II)Oxide reduces black copper(II)oxide to brown copper metal

Chemical Equation

CuO(s) + CO(g) -> Cu(s) + CO₂(g)

(black) (brown)

PbO(s) + CO(g) -> Pb(s) + CO₂(g)

(brown when hot/ (grey)

yellow when cool)

ZnO(s) + CO(g) -> Zn(s) + CO₂(g)

(yellow when hot/ (grey)

white when cool)

Fe₂O₃(s) + 3CO(s) -> 2Fe(s) + 3CO₂(g)

(brown when hot/cool (grey)

Fe₃O₄ (s) + 4CO(g) -> 3Fe(s) + 4CO₂(g)

(brown when hot/cool (grey)

These reaction are used during the extraction of many metals from their ore.

Carbon (II) oxide is a pollutant. Explain.

Carbon(II)oxide is highly poisonous/toxic.It preferentially combine with haemoglobin to form stable carboxyhaemoglobin in the blood instead of oxyhaemoglobin.This reduces the free haemoglobin in the blood causing nausea , coma then death.

10.The diagram below show a burning charcoal stove/burner/jiko. Use it to answer the questions that follow.

Explain the changes that take place in the burner

Explanation

Charcoal stove has air holes through which air enters. Air oxidizes carbon to carbon(IV)oxide gas at region I. This reaction is exothermic(-∆H) producing more heat.

Chemical equation

C(s) + O₂(g) -> CO₂(g)

Carbon(IV)oxide gas formed rises up to meet more charcoal which reduces it to Carbon(II)oxide gas.

Chemical equation

2CO₂ (g) + O₂(g) -> 2CO (g)

At the top of burner in region II, Carbon (II)oxide gas is further oxidized to Carbon(IV)oxide gas if there is plenty of air but escape if the air is limited poisoning the living things around.

Chemical equation

2CO (g) + O₂(g) -> 2CO₂ (g)

(excess air)

11.Describe the test for the presence of carbon(II)oxide gas.

Experiment

Burn/Ignite the pure sample of the gas. Pass/Bubble the products into lime water/Calcium hydroxide .

Observation

Colourless gas burns with a blue flame. A white precipitate is formed that dissolve on further bubbling of the products.

Chemical equation

2CO (g) + O₂(g) -> 2CO₂ (g) (gas burns with blue flame)

Chemical equation

Ca(OH)₂ (aq) + CO₂ (g) -> CaCO₃ (s) + H₂O(l)

Chemical equation

CO₂ (g) + CaCO₃ (s) + H₂O(l) -> Ca(HCO₃)₂ (aq)

State the main uses of carbon (II)oxide gas.

(i) As a fuel /water gas

(ii)As a reducing agent in the blast furnace for extracting iron from iron ore(Magnetite/Haematite)

(iii)As a reducing agent in extraction of Zinc from Zinc ore/Zinc blende

(iv) As a reducing agent in extraction of Lead from Lead ore/Galena

(v) As a reducing agent in extraction of Copper from Copper iron sulphide/Copper pyrites.

(iii)Carbonate(IV) (CO₃^2-)and hydrogen carbonate(IV(HCO₃^–)

1.Carbonate (IV) (CO₃^2-) are normal salts derived from carbonic(IV)acid (H₂CO₃) and hydrogen carbonate (IV) (HCO₃^–) are acid salts derived from carbonic(IV)acid.

Carbonic(IV)acid(H₂CO₃) is formed when carbon(IV)oxide gas is bubbled in water. It is a dibasic acid with two ionizable hydrogens.

H₂CO₃(aq) ->2H⁺(aq) + CO₃^2-(aq)

H₂CO₃(aq) -> H⁺(aq) + HCO₃^–(aq)

2.Carbonate (IV) (CO₃^2-) are insoluble in water except Na₂CO₃ , K₂CO₃ and (NH₄)₂CO₃

3.Hydrogen carbonate (IV) (HCO₃^–) are soluble in water. Only five hydrogen carbonates exist. Na HCO₃ , KHCO₃ ,NH₄HCO₃Ca(HCO₃)₂ and Mg(HCO₃)₂

Ca(HCO₃)₂ and Mg(HCO₃)₂ exist only in aqueous solutions.

3.The following experiments show the effect of heat on Carbonate (IV) (CO₃^2-) and Hydrogen carbonate (IV) (HCO₃^–) salts:

Experiment

In a clean dry test tube place separately about 1.0 of the following:

Zinc(II)carbonate(IV), sodium hydrogen carbonate(IV), sodium carbonate(IV), Potassium carbonate(IV) ammonium carbonate(IV), potassium hydrogen carbonate(IV), Lead(II)carbonate(IV), Iron(II)carbonate(IV), and copper(II)carbonate(IV). Heat each portion gently the strongly. Test any gases produced with lime water.

Observation

(i)Colorless droplets form on the cooler parts of test tube in case of sodium carbonate(IV) and Potassium carbonate(IV).

(ii)White residue/solid left in case of sodium hydrogen carbonate(IV), sodium carbonate(IV), Potassium carbonate(IV) and potassium hydrogen carbonate(IV).

(iii)Colour changes from blue/green to black in case of copper(II)carbonate(IV).

(iv) Colour changes from green to brown/yellow in case of Iron (II)carbonate(IV).

(v) Colour changes from white when cool to yellow when hot in case of Zinc (II) carbonate(IV).

(vi) Colour changes from yellow when cool to brown when hot in case of Lead (II) carbonate(IV).

(vii)Colourless gas produced that forms a white precipitate with lime water in all cases.

Explanation

Sodium carbonate(IV) and Potassium carbonate(IV) exist as hydrated salts with 10 molecules of water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid.

Chemical equation

Na₂CO₃ .10H₂O(s) -> Na₂CO₃ (s) + 10H₂O(l)

K₂CO₃ .10H₂O(s) -> K₂CO₃ (s) + 10H₂O(l)

Carbonate (IV) (CO₃^2-) and Hydrogen carbonate (IV) (HCO₃^–) salts decompose on heating except Sodium carbonate(IV) and Potassium carbonate(IV).

(a) Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV) decompose on heating to form sodium carbonate(IV) and Potassium carbonate(IV).Water and carbon(IV)oxide gas are also produced.

Chemical equation

2NaHCO₃ (s) -> Na₂CO₃ (s) + H₂O(l) + CO₂ (g)

(white) (white)

2KHCO₃ (s) -> K₂CO₃ (s) + H₂O(l) + CO₂ (g)

(white) (white)

(b) Calcium hydrogen carbonate(IV) and Magnesium hydrogen carbonate(IV) decompose on heating to form insoluble Calcium carbonate(IV) and Magnesium carbonate(IV).Water and carbon(IV)oxide gas are also produced.

Chemical equation

Ca(HCO₃)₂ (aq) -> CaCO₃ (s) + H₂O(l) + CO₂ (g)

(Colourless solution) (white)

Mg(HCO₃)₂ (aq) -> MgCO₃ (s) + H₂O(l) + CO₂ (g)

(Colourless solution) (white)

(c) Ammonium hydrogen carbonate(IV) decompose on heating to form ammonium carbonate(IV) .Water and carbon(IV)oxide gas are also produced.

Chemical equation

2NH₄HCO₃ (s) -> (NH₄)₂CO₃ (s) + H₂O(l) + CO₂ (g)

(white) (white)

(d)All other carbonates decompose on heating to form the metal oxide and produce carbon(IV)oxide gas e.g.

Chemical equation

MgCO₃ (s) -> MgO (s) + CO₂ (g)

(white solid) (white solid)

Chemical equation

BaCO₃ (s) -> BaO (s) + CO₂ (g)

(white solid) (white solid)

Chemical equation

CaCO₃ (s) -> CaO (s) + CO₂ (g)

(white solid) (white solid)

Chemical equation

CuCO₃ (s) -> CuO (s) + CO₂ (g)

(blue/green solid) (black solid)

Chemical equation

ZnCO₃ (s) -> ZnO (s) + CO₂ (g)

(white solid) (white solid when cool/

Yellow solid when hot)

Chemical equation

PbCO₃ (s) -> PbO (s) + CO₂ (g)

(white solid) (yellow solid when cool/

brown solid when hot)

4.The following experiments show the presence of Carbonate (IV) (CO₃^2-) and Hydrogen carbonate (IV) (HCO₃^–) ions in sample of a salt:

(a)Using Lead(II) nitrate(V)

Using a portion of salt solution in a test tube .add four drops of Lead(II)nitrate(V)solution.Preserve.

Observation

inference

White precipitate/ppt

CO₃^2- ,SO₃^2- ,SO₄^2- ,Cl^–

^–

To the preserved solution ,add six drops of dilutte nitric(V)acid. Preserve.

Observation

inference

White precipitate/ppt persists

White precipitate/ppt dissolves

SO₄^2- ,Cl^–

CO₃^2- ,SO₃^2-

To the preserved sample( that forms a precipitate ),heat to boil.

Observation

inference

White precipitate/ppt persists

White precipitate/ppt dissolves

SO₄^2-

Cl^–

To the preserved sample( that do not form a precipitate ),add three drops of acidified potassium manganate(VII)/lime water

Observation

inference

Effervescence/bubbles/fizzing colourless gas produced

Acidified KMnO₄ decolorized/no white precipitate on lime water

Effervescence/bubbles/fizzing colourless gas produced

Acidified KMnO₄ not decolorized/ white precipitate on lime water

SO₃^2-

CO₃^2-

Experiments/Observations:

(b)Using Barium(II)nitrate(V)/ Barium(II)chloride

To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve.

Observation	Inference
White precipitate/ppt	SO₄^2- , SO₃^2- , CO₃^2- ions

To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve.

Observation 1

Observation	Inference
White precipitate/ppt persists	SO₄^2- , ions

Observation 2

Observation	Inference
White precipitate/ppt dissolves	SO₃^2- , CO₃^2- , ions

III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).

Observation 1

Observation

Inference

(i)acidified potassium manganate(VII)decolorized

(ii)Orange colour of acidified potassium

dichromate(VI) turns to green

SO₃^2-ions

Observation 2

Observation

Inference

(i)acidified potassium manganate(VII) not decolorized

(ii)Orange colour of acidified potassium

dichromate(VI) does not turns to green

CO₃^2-ions

Explanations

Using Lead(II)nitrate(V)

(i)Lead(II)nitrate(V) solution reacts with chlorides(Cl^–), Sulphate (VI) salts (SO₄^2- ), Sulphate (IV)salts (SO₃^2-) and carbonates(CO₃^2-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV).

Chemical/ionic equation:

Pb²⁺(aq) + Cl^– (aq) -> PbCl₂(s)

Pb²⁺(aq) + SO₄²⁺ (aq) -> PbSO₄ (s)

Pb²⁺(aq) + SO₃²⁺ (aq) -> PbSO₃ (s)

Pb²⁺(aq) + CO₃²⁺ (aq) -> PbCO₃ (s)

(ii)When the insoluble precipitates are acidified with nitric(V) acid,

– Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists.

– Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.

. Chemical/ionic equation:

PbSO₃ (s) + 2H⁺(aq) -> H₂ O (l) + Pb²⁺(aq) + SO₂ (g)

PbCO₃ (s) + 2H⁺(aq) -> H₂ O (l) + Pb²⁺(aq) + CO₂ (g)

(iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed;

– Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling)

– Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling.

(iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced;

– sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.

Chemical equation:

5SO₃^2-(aq)+ 2MnO₄^– (aq) +6H+(aq) -> 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3H₂O(l)

(purple) (colourless)

3SO₃^2-(aq)+ Cr₂O₇^2-(aq) +8H+(aq) -> 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

(Orange) (green)

– Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.

Chemical equation:

Ca(OH)₂(aq)+ CO₂ (g) -> CaCO₃(s) + H₂O(l)

These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.

Using Barium(II)nitrate(V)/ Barium(II)Chloride

(i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO₄^2- ), Sulphate (IV)salts (SO₃^2-) and carbonates(CO₃^2-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV).

Chemical/ionic equation:

Ba²⁺(aq) + SO₄²⁺ (aq) -> BaSO₄ (s)

Ba²⁺(aq) + SO₃²⁺ (aq) -> BaSO₃ (s)

Ba²⁺(aq) + CO₃²⁺ (aq) -> BaCO₃ (s)

(ii)When the insoluble precipitates are acidified with nitric(V) acid,

– Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists.

– Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.

. Chemical/ionic equation:

BaSO₃ (s) + 2H⁺(aq) -> H₂ O (l) + Ba²⁺(aq) + SO₂ (g)

BaCO₃ (s) + 2H⁺(aq) -> H₂ O (l) + Ba²⁺(aq) + CO₂ (g)

(iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced;

– sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.

Chemical equation:

5SO₃^2-(aq)+ 2MnO₄^– (aq) +6H+(aq) -> 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3H₂O(l)

(purple) (colourless)

3SO₃^2-(aq)+ Cr₂O₇^2-(aq) +8H+(aq) -> 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

(Orange) (green)

Chemical equation:

Ca(OH)₂(aq)+ CO₂ (g) -> CaCO₃(s) + H₂O(l)

These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.

(iii) Sodium carbonate(IV) (Na₂CO₃)

(a)Extraction of sodium carbonate from soda ash

Sodium carbonate naturally occurs in Lake Magadi in Kenya as Trona.trona is the double salt ; sodium sesquicarbonate. NaHCO₃ .Na₂CO₃ .H₂O.It is formed from the volcanic activity that takes place in Lake Naivasha, Nakuru ,Bogoria and Elementeita .All these lakes drain into Lake Magadi through underground rivers. Lake Magadi has no outlet.

Solubility of Trona decrease with increase in temperature.High temperature during the day causes trona to naturally crystallize .It is mechanically scooped/dredged/dug and put in a furnace.

Inside the furnace, trona decompose into soda ash/sodium carbonate.

Chemical equation

2NaHCO₃ .Na₂CO₃ .H₂O (s) -> 3Na₂CO₃ (s) + 5H₂O(l) + CO₂ (g)

(trona) (soda ash)

Soda ash is then bagged and sold as Magadi soda.It is mainly used:

(i)in making glass to lower the melting point of raw materials (sand/SiO₂ from 1650^oC and CaO from 2500^oC to around 1500^oC)

(ii)in softening hard water

(iii)in the manufacture of soapless detergents.

(iv)Swimming pool “pH increaser”

Sodium chloride is also found dissolved in the lake. Solubility of sodium chloride decrease with decreases in temperature/ sodium chloride has lower solubility at lower temperatures. When temperatures decrease at night it crystallize out .The crystals are then mechanically dug/dredged /scooped then packed for sale as animal/cattle feeds and seasoning food.

Summary flow diagram showing the extraction of Soda ash from Trona

Sodium chloride and Trona dissolved in the sea

Natural fractional crystallization

Crystals of Trona (Day time)

Crystals of sodium chloride(At night)

Dredging /scooping/ digging

Crushing

Furnace (Heating)

Carbon(IV) oxide

Soda ash

Bagging

NaCl(s)

Bagging Na₂CO₃ (s)

b)The Solvay process for industrial manufacture of sodium carbonate(IV)

(i)Raw materials.

–Brine /Concentrated Sodium chloride from salty seas/lakes.

–Ammonia gas from Haber.

–Limestone /Calcium carbonate from chalk /limestone rich rocks.

–Water from rivers/lakes.

(ii)Chemical processes

Ammonia gas is passed up to meet a downward flow of sodium chloride solution / brine to form ammoniated brine/ammoniacal brine mixture in the ammoniated brine chamber

The ammoniated brine mixture is then pumped up, atop the carbonator/ solvay tower.

In the carbonator/ solvay tower, ammoniated brine/ammoniacal brine mixture slowly trickle down to meet an upward flow of carbon(IV)oxide gas.

The carbonator is shelved /packed with quartz/broken glass to

(i) reduce the rate of flow of ammoniated brine/ammoniacal brine mixture.

(ii)increase surface area of the liquid mixture to ensure a lot of ammoniated brine/ammoniacal brine mixture react with carbon(IV)oxide gas.

Insoluble sodium hydrogen carbonate and soluble ammonium chloride are formed from the reaction.

Chemical equation

CO₂(g) + H₂O(l) + NaCl (aq) + NH₃(g) -> NaHCO₃(s) + NH₄Cl(aq)

The products are then filtered. Insoluble sodium hydrogen carbonate forms the residue while soluble ammonium chloride forms the filtrate.

Sodium hydrogen carbonate itself can be used:

(i) as baking powder and preservation of some soft drinks.

(ii) as a buffer agent and antacid in animal feeds to improve fibre digestion.

(iii) making dry chemical fire extinguishers.

In the Solvay process Sodium hydrogen carbonate is then heated to form Sodium carbonate/soda ash, water and carbon (IV) oxide gas.

Chemical equation

2NaHCO₃ (s) -> Na₂CO₃(s) + CO₂(g) + H₂O(l)

Sodium carbonate is stored ready for use in:

(i) during making glass/lowering the melting point of mixture of sand/SiO₂ from 1650^oC and CaO from 2500^oC to around 1500^oC

(ii) in softening hard water

(iii) in the manufacture of soapless detergents.

(iv) swimming pool “pH increaser”.

Water and carbon(IV)oxide gas are recycled back to the ammoniated brine/ammoniacal brine chamber.

More carbon(IV)oxide is produced in the kiln/furnace. Limestone is heated to decompose into Calcium oxide and carbon(IV)oxide.

Chemical equation

CaCO₃ (s) -> CaO(s) + CO₂(g)

Carbon(IV)oxide is recycled to the carbonator/solvay tower. Carbon (IV)oxide is added water in the slaker to form Calcium hydroxide. This process is called slaking.

Chemical equation

CaO(s) + H₂O (l) -> Ca(OH)₂ (aq)

Calcium hydroxide is mixed with ammonium chloride from the carbonator/solvay tower in the ammonia regeneration chamber to form Calcium chloride , water and more ammonia gas.

Chemical equation

Ca(OH)₂ (aq) +2NH₄Cl (aq) -> CaCl₂(s) + 2NH₃(g) + H₂O(l)

NH₃(g) and H₂O(l) are recycled.

Calcium chloride may be used:

(i)as drying agent in the school laboratory during gas preparation (except ammonia gas)

(ii)to lower the melting point of solid sodium chloride / rock salt salts during the Downs process for industrial extraction of sodium metal.

Detailed Summary flow diagram of Solvay Process

Practice

The diagram below shows part of the Solvay process used in manufacturing sodium carbonate. Use it to answer the questions that follow.

Carbon (IV)oxide

Ammonia

Process II

Sodium carbonate

Saturated sodium chloride solution

Sodium hydrogen carbonate

Ammonium chloride

Process I

(a)Explain how Sodium Chloride required for this process is obtained from the sea.

Sea water is pumped /scooped into shallow pods. Evaporation of most of the water takes place leaving a very concentrated solution.

(b)(i) Name process:

Filtration
Decomposition

(ii) Write the equation for the reaction in process:

Process I

Chemical equation

CO₂(g) + H₂O(l) + NaCl (aq) + NH₃(g) -> NaHCO₃(s) + NH₄Cl(aq)

Process II

Chemical equation

2NaHCO₃ (s) -> Na₂CO₃(s) + CO₂(g) + H₂O(l)

(c)(i) Name two substances recycled in the solvay process

Ammonia gas , Carbon(IV)Oxide and Water.

(ii)Which is the by-product of this process?

Calcium(II)Chloride /CaCl₂

(iii)State two uses that the by-product can be used for:

As a drying agent in the school laboratory preparation of gases.
In the Downs cell/process for extraction of Sodium to lower the melting point of rock salt.

(iv)Write the chemical equation for the formation of the by-products in the Solvay process.

Chemical equation

Ca(OH)₂ (aq) +2NH₄Cl (aq) -> CaCl₂(s) + 2NH₃(g) + H₂O(l)

(d)In an experiment to determine the % purity of Sodium carbonate produced in the Solvay process ,2.15g of the sample reacted with exactly 40.0cm3 of 0.5M Sulphuric(VI)acid.

(i)Calculate the number of moles of sodium carbonate that reacted.

Chemical equation

Na₂CO₃ (aq) +H₂SO₄ (aq) -> Na₂SO₄ (aq)+ CO₂(g) + H₂O(l)

Mole ratio Na₂CO₃ :H₂SO₄ => 1:1

Moles H₂SO₄ = Molarity x Volume => 0.5 x 40.0 = 0.02 Moles

1000 1000

Moles of Na₂CO₃ = 0.02 Moles

(ii)Determine the % of sodium carbonate in the sample.

Molar mass of Na₂CO₃ = 106g

Mass of Na₂CO₃ = moles x Molar mass => 0.02 x 106 = 2.12 g

% of Na₂CO₃=( 2.12 g x 100) = 98.6047%

2.15

(e) State two uses of soda ash.

(i) during making glass/lowering the melting point of mixture of sand/SiO₂ from 1650^oC and CaO from 2500^oC to around 1500^oC

(ii) in softening hard water

(iii) in the manufacture of soapless detergents.

(iv) swimming pool “pH increaser”.

(f)The diagram below shows a simple ammonia soda tower used in manufacturing sodium carbonate .Use it to answer the questions that follow:

Substance A

Sodium hydrogen carbonate

Metal plates

Excess Carbon(IV)oxide

Raw materials

(i)Name the raw materials needed in the above process

-Ammonia

-Water

-Carbon(IV)oxide

-Limestone

-Brine/ Concentrated sodium chloride

(ii)Identify substance A

Ammonium chloride /NH₄Cl

(iii) Write the equation for the reaction taking place in:

I.Tower.

Chemical equation

CO₂(g) + NaCl (aq) + H₂O(l) + NH₃(g) -> NaHCO₃(s) + NH₄Cl(aq)

Production of excess carbon (IV)oxide.

Chemical equation

CaCO₃ (s) -> CaO(s) + CO₂(g)

III. The regeneration of ammonia

Chemical equation

Ca(OH)₂ (aq) +2NH₄Cl (aq) -> CaCl₂(s) + 2NH₃(g) + H₂O(l)

(iv)Give a reason for having the circular metal plates in the tower.

-To slow the downward flow of brine.

-To increase the rate of dissolving of ammonia.

-To increase the surface area for dissolution

(v)Name the gases recycled in the process illustrated above.

Ammonia gas , Carbon(IV)Oxide and Water.

Describe how you would differentiate between carbon (IV)oxide and carbon(II)oxide using chemical method.

Method I

-Bubble both gases in lime water/Ca(OH)₂

-white precipitate is formed if the gas is carbon (IV) oxide

– No white precipitate is formed if the gas is carbon (II) oxide

Method II

-ignite both gases

– Carbon (IV) oxide does not burn/ignite

– Carbon (II) oxide burn with a blue non-sooty flame.

Method III

-Lower a burning splint into a gas containing each gas separately.

-burning splint is extinguished if the gas is carbon (IV) oxide

-burning splint is not extinguished if the gas is carbon (II) oxide.

3.Using Magnesium sulphate(VI)solution ,describe how you can differentiate between a solution of sodium carbonate from a solution of sodium hydrogen carbonate

-Add Magnesium sulphate(VI) solution to separate portions of a solution of sodium carbonate and sodium hydrogen carbonate in separate test tubes

-White precipitate is formed in test tube containing sodium carbonate

-No white precipitate is formed in test tube containing sodium hydrogen carbonate.

Chemical equation

Na₂CO₃ (aq) +MgSO₄ (aq) -> Na₂SO₄ (aq) + MgCO₃(s)

(white ppt)

Ionic equation

CO₃^2- (aq) + Mg²⁺ (aq) -> MgCO₃(s)

(white ppt)

Chemical equation

2NaHCO₃ (aq) +MgSO₄ (aq) -> Na₂SO₄ (aq) + Mg(HCO₃)₂ (aq)

(colourless solution)

The diagram below shows a common charcoal burner .Assume the burning take place in a room with sufficient supply of air.

(a)Explain what happens around:

(i)Layer A

Sufficient/excess air /oxygen enter through the air holes into the burner .It reacts with/oxidizes Carbon to carbon(IV)oxide

Chemical equation

C(s) + O₂(g) -> CO₂ (g)

(ii)Layer B

Hot carbon(IV)oxide rises up and is reduced by more carbon/charcoal to carbon (II)oxide.

Chemical equation

C(s) + CO₂(g) -> 2CO (g)

(ii)Layer C

Hot carbon(II)oxide rises up and burns with a blue flame to be oxidized by the excess air to form carbon(IV)oxide.

2CO (g) + O₂(g) -> 2CO₂(g)

(b)State and explain what would happen if the burner is put in an enclosed room.

The hot poisonous /toxic carbon(II)oxide rising up will not be oxidized to Carbon(IV)oxide.

(c)Using a chemical test , describe how you would differentiate two unlabelled black solids suspected to be charcoal and copper(II)oxide.

Method I

-Burn/Ignite the two substances separately.

-Charcoal burns with a blue flame

– Copper(II)oxide does not burn

Method II

-Add dilute sulphuric(VI)acid/Nitric(V)acid/Hydrochloric acid separately.

-Charcoal does not dissolve.

– Copper(II)oxide dissolves to form a colourless solution.

Excess Carbon(II)oxide was passed over heated copper(II)oxide as in the set up shown below for five minutes.

(a)State and explain the observations made in the combustion tube.

Observation

Colour change from black to brown

Explanation

Carbon (II)oxide reduces black copper(II)oxide to brown copper metal itself oxidized to Carbon(IV)oxide.

Chemical equation

CO(g) + CuO (s) -> Cu(s) + CO₂(g)

(black) (brown)

(b) (i)Name the gas producing flame A

Carbon(II)oxide

(ii)Why should the gas be burnt?

It is toxic/poisonous

(iii)Write the chemical equation for the production of flame A

2CO(g) + O₂(g) -> 2CO₂(g)

(c)State and explain what happens when carbon(IV)oxide is prepared using Barium carbonate and dilute sulphuric(VI)acid.

Reaction starts then stops after sometime producing small/little quantity of carbon(IV)oxide gas.

Barium carbonate react with dilute sulphuric(VI)acid to form insoluble Barium sulphate(VI) that cover/coat unreacted Barium carbonate stopping further reaction to produce more Carbon(IV)oxide.

(d) Using dot () and cross(x) to represent electrons show the bonding in a molecule of :

(i) Carbon(II)oxide

(ii) Carbon(IV)Oxide.

(e) Carbon (IV)oxide is an environmental pollutant of global concern. Explain.

-It is a green house gas thus causes global warming.

-It dissolves in water to form acidic carbonic acid which causes “acid rain”

(f)Explain using chemical equation why lime water is used to test for the presence of Carbon (IV) oxide instead of sodium hydroxide.

Using lime water/calcium hydroxide:

– a visible white precipitate of calcium carbonate is formed that dissolves on bubbling excess Carbon (IV) oxide gas

Chemical equation

Ca(OH)₂(aq) + CO₂(g) -> CaCO₃ (s) + H₂O(l)

(white precipitate)

CaCO₃ (aq) + H₂O(l) + CO₂(g) -> Ca(HCO₃)₂ (aq)

Using sodium hydroxide:

– No precipitate of sodium carbonate is formed Both sodium carbonate and sodium hydrogen carbonate are soluble salts/dissolves.

Chemical equation

2NaOH (aq) + CO₂(g) -> Na₂CO₃ (s) + H₂O(l)

(No white precipitate)

Na₂CO₃ (s) + H₂O(l) + CO₂(g) -> 2NaHCO₃ (s)

(g)Ethan-1,2-dioic acid and methanoic acid may be used to prepare small amount of carbon(II)oxide in a school laboratory.

(i) Explain the modification in the set up when using one over the other.

Before carbon(II)oxide is collected:

-when using methanoic acid, no concentrated sodium/potassium hydroxide is needed to absorb Carbon(IV)oxide.

-when using ethan-1,2-dioic acid, concentrated sodium/potassium hydroxide is needed to absorb Carbon(IV)oxide.

(ii)Write the equation for the reaction for the formation of carbon(II)oxide from:

I.Methanoic acid.

Chemical equation HCOOH(aq) -> CO(g) + H₂O(l)

Ethan-1,2-dioic acid

Chemical equation HOOCCOOH(aq) -> CO₂(g)+CO(g)+H₂O(l)

(h)Both carbon(II)oxide and carbon(IV)oxide affect the environment. Explain why carbon(II)oxide is more toxic/poisonous.

-Both gases are colourless,denser than water and odourless.

-Carbon(II)oxide is preferentially absorbed by human/mammalian haemoglobin when inhaled forming stable carboxyhaemoglobin instead of oxyhaemoglobin.This reduces the free haemoglobin in the blood leading to suffocation and quick death. —Carbon(IV)oxide is a green house gas that increases global warming.

-Carbon(II)oxide is readily oxidized to carbon(IV)oxide

6.Study the flow chart below and use it to answer the questions that follow.

(a)Name:

(i)the white precipitate A

Calcium carbonate

(ii) solution B

Calcium hydrogen carbonate

(iii) gas C

Carbon(IV)oxide

(iv) white residue B

Calcium oxide

(v) solution D