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Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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Na+ and SO42− ions are spectator ions; hence these shall not appear in the final equation.

HCO3− + 2H + → H 2O + CO2

To make equal charges on both sides, HCO3− should have a coefficient 2.

2HCO3− + 2H + → H 2O + CO2
In order to balance the hydrogen and carbon on both sides, the molecules of H2O and CO2 should have a
coefficient 2 respectively.

2HCO3− + 2H + = 2H 2O + 2CO2 or HCO3− + H + = H 2O + CO2
This is the balanced ionic equation.
Conversion of ionic equation in molecular form can be explained by following example,
Example : Write the following ionic equation in the molecular form if the reactants are chlorides.

2Fe3++Sn2+ → 2Fe2++Sn4+
For writing the reactants in molecular forms, the requisite number of chloride ions are added.

2Fe 3+ + 6Cl − + Sn2+ + 2Cl − or 2FeCl3 + SnCl2

Similarly 8 Cl − ions are added on R.H.S. to neutralise the charges.
2Fe 2+ + 4Cl − + Sn4+ + 4Cl − or 2FeCl2 + SnCl4

Thus, the balanced molecular equation is, 2FeCl3+SnCl2 = 2FeCl2 + SnCl4
Oxidation-reduction and Redox reactions.

(1) Oxidation : Oxidation is a process which involves; addition of oxygen, removal of hydrogen, addition of
non-metal, removal of metal, Increase in +ve valency, loss of electrons and increase in oxidation number.

(i) Addition of oxygen

(a) 2Mg + O2 → 2MgO (Oxidation of magnesium)
(b) S + O2 → SO2 (Oxidation of sulphur)
(c) 2CO + O2 → 2CO2 (Oxidation of carbon monoxide)
(d) Na2SO3+H2O2 → Na2SO4+H2O (Oxidation of sodium sulphite)

(ii) Removal of hydrogen

(a) H2S + Cl2 → 2HCl + S (Oxidation of hydrogen sulphide)
(b) 4HI + O2 → 2H2O + 2I2 (Oxidation of hydrogen iodide)

(c) C2H5OH Cu /300oC → CH3CHO + H2 (Oxidation of ethanol)

(d) 4HCl + MnO2 → MnCl2 + 2H2O + Cl2 (Oxidation of hydrogen chloride)

(iii) Addition of an electronegative element or addition of Non-metal

(a) Fe + S → FeS (Oxidation of iron)

(b) SnCl2 + Cl2 → SnCl4 (Oxidation of stannous chloride)

(c) 2Fe + 3F2 → 2FeF3 (Oxidation of iron)

(iv) Removal of an electropositive element or removal of metal

(a) 2KI + H2O2 → 2KOH + I2 (Oxidation of potassium iodide)

(b) 2K2MnO4 + Cl2 → 2KCl + 2KMnO4 (Oxidation of potassium manganate)

(c) 2KI + Cl2 → 2KCl + I2 (Oxidation of potassium iodide)

(v) Increase in +ve valency and Decrease in – ve valency

Increase in +ve valency

–4, –3, –2, –1, 0, +1, +2, +3, +4

Decrease in –ve valency

(a) Fe 2+ → Fe 3+ + e − (+ve valency increases)

(b) Sn2+ → Sn4+ + 2e − (+ve valency increases)

(c) [Fe (CN)6 ]4− → [Fe(CN)6 ]3− + e − (–ve valency decreases)

(d) MnO42− → MnO4− + e − (–ve valency decreases)

(vi) Loss of electrons (also known as de-electronation)

M−4 M−3 M−2 M−1 0 M+1 M+2 M+3 M+4


–e– –e– –e– –e– –e– –e– –e– –e–
Loss of electrons

(a) H 0 → H + + e − (Formation of proton)

(b) H 0 → H + + e− (De-electronation of hydrogen)
2 2 (De-electronation of Fe 2+ )

(c) Fe 2+ → Fe 3+ + e −

(d) Mg → Mg 2+ + 2e − (De-electronation of Magnesium)

(e) MnO42− → MnO4− + e − (De-electronation of MnO42− )

(f) 2Cl − → Cl2 + 2e − (De-electronation of chloride ion)
(g) 2Fe 0 → 2Fe 3+ + 6e − (De-electronation of iron)

(vii) Increase in oxidation number

(a) Mg 0 → Mg 2+ (From 0 to +2)

[ ] [ ](b) Fe +2(CN)6 4− → Fe +3 (CN)6 3− (From +2 to +3)

(c) 2Cl − → Cl 0 (From –1 to 0)

(2) Reduction : Reduction is just reverse of oxidation. Reduction is a process which involves; removal of
oxygen, addition of hydrogen, removal of non-metal, addition of metal, decrease in +ve valency, gain of electrons
and decrease in oxidation number.

(i) Removal of oxygen

(a) CuO + C → Cu + CO (Reduction of cupric oxide)

(b) H2O + C → CO+ H2 (Reduction of water)

Steam Coke Water gas

(c) Fe3O4 + 4H 2 → 3Fe + 4H 2O (Reduction of Fe3O4 )

(d) C6 H5OH + Zn → C6 H6 + ZnO (Reduction of phenol)

(ii) Addition of hydrogen

(a) Cl2 + H 2 → 2HCl (Reduction of chlorine)

(b) S + H 2 → H 2S (Reduction of sulphur)

(c) C2 H 4 + H 2 → C2 H6 (Hydrogenation of ethene)

(iii) Removal of an electronegative element or removal of Non-metal

(a) 2HgCl2 + SnCl2 → Hg 2Cl2 + SnCl4 (Reduction of mercuric chloride)

(b) 2FeCl3 + H 2 → 2FeCl 2 + 2HCl (Reduction of ferric chloride)

(c) 2FeCl3 + H 2S → 2FeCl 2 + 2HCl + S (Reduction of ferric chloride)

(iv) Addition of an electropositive element or addition of metal

(a) HgCl2 + Hg → Hg 2Cl2 (Reduction of mercuric chloride)

(b) CuCl2 + Cu → Cu2Cl2 (Reduction of cupric chloride)

(v) Decrease in +ve valency and Increase in –ve valency

Decrease in +ve valency

–4, –3, –2, –1, 0, +1, +2, +3, +4

Increase in –ve valency

(a) Fe 3+ → Fe 2+ (+ve valency decreases)
(+ve velency decreases)
(b) Sn4+ → Sn2+ (–ve valency increases)
(–ve valency increases)
(c) [Fe (CN)6 ]3− → [Fe(CN)6 ]4−
(d) MnO4− → MnO42−
(vi) Gain of electrons (also known as electronation)

M−4 M−3 M−2 M−1 0 M+1 M+2 M+3 M+4


+e– +e– +e– +e– +e– +e– +e– +e–
Gain of electrons

(a) Zn2+ (aq) + 2e − → Zn(S) (Electronation of Zn2+ )

(b) Pb 2+ + 2e − → Pb 0 (Electronation of Pb 2+ )
(c) Mn7+ + 5e − → Mn2+ (Electronation of Mn7+ )
(d) Fe 3+ + e − → Fe 2+ (Electronation of Fe 3+ )
(e) Sn4+ + 2e − → Sn2+ (Electronation of Sn4+ )

(f) Cl + e − → Cl − (Formation of chloride ion)

(g) [Fe(CN)6 ]3− + e − → [Fe(CN)6 ]4− (Eectronation of [Fe(CN)6 ]3− )
(vii) Decrease in oxidation number

(a) Mg 2+ → Mg 0 (From +2 to 0)

(b) [Fe(CN)6 ]3− → [Fe(CN)6 ]4− (From +3 to +2)
(From 0 to –1)
(c) Cl 0 → 2Cl −

(3) Redox-reactions
(i) An overall reaction in which oxidation and reduction takes place simultaneously is called redox or
oxidation-reduction reaction. These reactions involve transfer of electrons from one atom to another. Thus
every redox reaction is made up of two half reactions; One half reaction represents the oxidation and the other
half reaction represents the reduction.

(ii) The redox reactions are of following types

(a) Direct redox reaction : The reactions in which oxidation and reduction takes place in the same vessel
are called direct redox reactions.

(b) Indirect redox reaction : The reactions in which oxidation and reduction takes place in different vessels
are called indirect redox reactions. Indirect redox reactions are the basis of electro-chemical cells.

(c) Intermolecular redox reactions : In which one substance is oxidised while the other is reduced. For

2 Al + Fe2O3 → Al2O3 + 2Fe
Here, Al is oxidised to Al2O3 while Fe2O3 is reduced to Fe.
(d) Intramolecular redox reactions : In which one element of a compound is oxidised while the other is
reduced. For example,

2 KClO3 ∆ → 2 KCl + 3 O2

Here, Cl +5 in KClO3 is reduced to Cl −1 in KCl while O 2− in KClO3 is oxidised to O 0 .

(iii) To see whether the given chemical reaction is a redox reaction or not, the molecular reaction is written in
the form of ionic reaction and now it is observed whether there is any change in the valency of atoms or ions. If
there is a change in valency, the chemical reaction will be a redox reaction otherwise not. For example,

(a) BaO2 + H 2SO4 → BaSO4 + H 2O2

(b) CuSO4 + 4 NH 3 → [Cu(NH 3 )4 ]SO4

In above examples there is no change in the valency of any ion or atom, thus these are not redox reactions.

(iv) Some examples of redox reactions are,

(a) –2e–

2HgCl 2 + SnCl 2 → Hg 2Cl2 + SnCl 4

+ 2e –

Here mercuric ion is reduced to mercurous ion and stannous ion is oxidised to stannic ion, i.e., mercuric ion
acts as an oxidising agent while stannous ion acts as a reducing agent.

(b) +2e–

SnCl2 + 2FeCl3 → 2FeCl2 + SnCl4

– 2e–


Here ferric ion is reduced to ferrous ion by gain of one electron while stannous ion is oxidised to stannic ion
by loss of two electrons. The ferric ion acts as an oxidising agent while stannous ion acts as a reducing agent.

(c) +2e –

2Na2S2O3 + I 2 → Na2S4 O6 + 2NaI


Here thiosulphate ion is oxidised to tetrathionate ion by loss of electrons while iodine is reduced to iodide ion
by gain of electrons. Thiosulphate ion acts as a reducing agent and iodine acts as an oxidising agent.

(d) –2e–


0 C2+u SO4 → ZnSO4 0

Zn+ + Cu

+2e –

(e) –2e –

00 +− (Where X = F, Cl, Br, I)
H2+ X2 → 2H− X


(f) 00 → Z+2n −I12
Zn+ I 2


(g) +2e–

0 0 Z+2n(CN)2
2 A+1g CN → 2
Zn+ Ag +



(h) +2e–

0 2 A+1–g2eN– O3 → C+2u(NO3 )2 + 2 0

Cu+ Ag


Oxidising and Reducing agents (Oxidants and Reductants).

(1) Definition : The substance (atom, ion or molecule) that gains electrons and is thereby reduced to a low
valency state is called an oxidising agent, while the substance that loses electrons and is thereby oxidised to a
higher valency state is called a reducing agent.


An oxidising agent is a substance the oxidation number of whose atom or atoms decreases while a reducing
agent is a substance the oxidation number of whose atom increases.

(2) Important oxidising agents
(i) Molecules made up of electronegative elements, e.g. O2, O3 and X2 (halogens).
(ii) Compounds containing an element which is in the highest oxidation state e.g. KMnO4 , K2Cr2O7 ,

Na2Cr2O7 , CrO3, H 2SO4 , HNO3 , NaNO3 , FeCl3 , HgCl2 , KClO4 , SO3 , CO2 , H 2O2 etc.

(iii) Oxides of elements e.g. MgO, CuO, CrO3 , CO2, P4 O10 , etc.
(iv) Fluorine is the strongest oxidising agent.

(3) Important reducing agents
(i) All metals e.g. Na, Zn, Fe, Al, etc.
(ii) A few non-metals e.g. C, H2, S etc.
(iii) Hydracids : HCl, HBr, HI, H2S etc.

(iv) A few compounds containing an-element in the lower oxidation state (ous), e.g. FeCl2, FeSO4 ,

SnCl2 , Hg 2Cl2 , Cu2O etc.

(v) Metallic hydrides e.g. NaH, LiH etc.

(vi) Organic compounds like HCOOH and (COOH)2 and their salts, aldehydes, alkanes etc.
(vii) Lithium is the strongest reducing agent in solution.

(viii) Cesium is the strongest reducing agent in absence of water. Other reducing agents are Na2S2O3 and KI.

(ix) Hypo prefix indicates that central atom of compound has the minimum oxidation state so it will act as a
reducing agent. e.g., H3 PO2 (hypophosphorous acid).

(4) Substances which act as oxidising as well as reducing agents
Example : H2O2, SO2, H2SO3, HNO2, NaNO2, Na2SO3, O3 etc.

(5) Tips for the identification of oxidising and reducing agents
(i) If an element is in its highest possible oxidation state in a compound, the compound can function as an
oxidising agent, e.g. KMnO4 , K2Cr2O7 , HNO3, H2SO4 , HClO4 etc.
(ii) If an element is in its lowest possible oxidation state in a compound, the compound can function only as a
reducing agent, e.g. H 2S, H 2C2O4 , FeSO4 , Na2S2O3 , SnCl2 etc.
(iii) If an element is in its intermediate oxidation state in a compound, the compound can function both as an
oxidising agent as well as reducing agent, e.g. H 2O2, H 2SO3 , HNO2, SO2 etc.
(iv) If a highly electronegative element is in its highest oxidation state in a compound, that compound can
function as a powerful oxidising agent, e.g. KClO4 , KClO3 , KBrO3 , KIO3 etc.
(v) If an electronegative element is in its lowest possible oxidation state in a compound or in free state, it can
function as a powerful reducing agent, e.g. I − , Br − , N 3− etc.

(6) Tests for oxidising agents
(i) Aqueous solutions of oxidising agents react with,

(a) Hydrogen sulphide to give a milky yellow precipitate of sulphur.

H 2S + Oxidising agent → S (milky yellow ppt.)

(b) Potassium iodide solution and evolve iodine which gives intense blue colour with starch solution

KI + Oxidising agent → I 2 Starch solution → Intense blue colour

(c) Freshly prepared solution of ferrous ammonium sulphate in presence of dil. H 2SO4 . Ferric ions (Fe 3+ )
can be detected by adding ammonium thiocyanate solution when a deep red colouration is produced.

Oxidising agent + Fe 2+ → Fe 3+ CNS− → Fe(CNS)3 (deep red clouration)
(ii) Insoluble oxidising agents on,
(a) Strong heating evolve oxygen which relights a glowing splinter.
(b) Warming with concentrated hydrochloric acid evolve chlorine which bleaches the moist litmus paper.

(7) Tests for reducing agents
(i) Aqueous solutions of reducing agents react with,

(a) Acidified potassium permanganate solution and decolourise it.

(b) Few drops of acidified potassium dichromate solution, green colouration is produced.

(c) Few drops of ferric chloride solution. The ferrous ions thus formed give a deep blue colouration with
potassium ferricyanide (K3[Fe(CN)6 ]).

Insoluble reducing agents on,

(d) Heating with concentrated nitric acid, evolve brown fumes of nitrogen dioxide.

(e) Heating with powdered cupric salt, form a red deposit of copper which does not dissolve in warm dilute
sulphuric acid.

(8) Equivalent weight of oxidising and reducing agents

(i) Equivalent weight of a substance (oxidant or reductant) is equal to molecular weight divided by number of
electrons lost or gained by one molecule of the substance in a redox reaction.

Equivalent weight of oxidising agent = Molecular weight
No.of electrons gained by one molecule

Equivalent weight of reducing agent = Molecular weight
No.of electrons lost by one molecule

(ii) In other words, it is equal to the molecular weight of oxidant or reductant divided by the change in
oxidation number.

Equivalent weight of oxidising agent = Molecular weight
Change in O.N. per mole

Equivalent weight of reducing agent = Molecular weight
Change in O.N. per mole

Equivalent weight of few oxidising/reducing agents

Agents O. N. Product O. N. Change in Total Eq. wt.
+3 O. N. per Change in
Cr2O7 2− +6 Cr 3+ O. N. per mole Mol. wt./6
C2O4 2− atom 3×2=6 Mol. wt./2

+ 3 CO2 +4 1 1×2=2

S2O32− +2 S4 O6 2 − + 2.5 0.5 0.5 × 2 = 1 Mol. wt./1
H2O2 –1 H2O –2 1 1×2=2 Mol. wt./2
H2O2 –1 O2 0 1 1×2=2 Mol. wt./2
+7 Mn2+ +2 5 5×1=5 Mol. wt./5
(Acidic medium) +7 MnO2 +4 3 3×1=3 Mol. wt./3

MnO4− +7 MnO42− +6 1 1×1=1 Mol. wt./1
(Neutral medium)

(Alkaline medium)

Oxidation number or Oxidation state.

(1) Definition : Charge on an atom produced by donating or accepting electrons is called oxidation number
or oxidation state. It is the number of effective charges on an atom.

(2) Valency and oxidation number : Valency and oxidation number concepts are different. In some cases
(mainly in the case of electrovalent compounds), valency and oxidation number are the same but in other cases
they may have different values. Points of difference between the two have been tabulated below

Valency Oxidation number

It is the combining capacity of the element. No plus O.N. is the charge (real or imaginary) present on the atom
or minus sign is attached to it. of the element when it is in combination. It may have plus
or minus sign.

Valency of an element is usually fixed. O.N. of an element may have different values. It depends
on the nature of compound in which it is present.

Valency is always a whole number. O.N. of the element may be a whole number or fractional.
O.N. of the element may be zero.
Valency of the element is never zero except of noble

(3) Oxidation number and Nomenclature
(i) When an element forms two monoatomic cations (representing different oxidation states), the two ions are
distinguished by using the ending-ous and ic. The suffix – ous is used for the cation with lower oxidation state and
the suffix – ic is used for the cation with higher oxidation state.

For example : Cu+ (oxidation number +1) cuprous : Cu2+ (oxidation number +2) cupric

(ii) Albert Stock proposed a new system known as Stock system. In this system, the oxidation states are
indicated by Roman numeral written in parentheses immediately after the name of the element. For example,

Cu2O Copper (I) oxide SnO Tin (II) oxide
Iron (II) chloride Manganess (VII) oxide
FeCl2 Potassium dichromate (VI) Mn2O7 Sodium chromate (VI)
K2Cr2O7 Vanadium (V) oxide Na2CrO4 Copper (II) oxide
V2O5 Tin (IV) oxide CuO Iron (III) chloride
SnO2 FeCl3

Note :  Stock system is not used for non-metals.

(4) Rules for the determination of oxidation number of an atom
The following rules are followed in ascertaining the oxidation number of an atom,
(i) If there is a covalent bond between two same atoms then oxidation numbers of these two atoms will be
zero. Bonded electrons are symmetrically distributed between two atoms. Bonded atoms do not acquire any charge.
So oxidation numbers of these two atoms are zero.

A : A or A – A → A* + A*
For e.g. Oxidation number of Cl in Cl2, O in O2 and N and N2 is zero.
(ii) If covalent bond is between two different atoms then electrons are counted towards more electronegative
atom. Thus oxidation number of more electronegative atom is negative and oxidation number of less
electronegative atom is positive. Total number of charges on any element depends on number of bonds.

A – B → A+ + B : –
A – B → A+2 + : B : –2
The oxidation number of less electronegative element (A) is + 1 and + 2 respectively.
(iii) If there is a coordinate bond between two atoms then oxidation number of donor atom will be + 2 and of
acceptor atom will be – 2.
A → B → A2+ + B-2
(iv) The oxidation number of all the atoms of different elements in their respective elementary states is taken to
be zero. For example, in N2, Cl2, H2, P4, S8, O2, Br2, Na, Fe, Ag etc. the oxidation number of each atom is zero.
(v) The oxidation number of a monoatomic ion is the same as the charge on it. For example, oxidation
numbers of Na+ , Mg 2+ and Al 3+ ions are + 1, + 2 and + 3 respectively while those of Cl−,S2− and N 3− ions are

–1, –2 and –3 respectively.
(vi) The oxidation number of hydrogen is + 1 when combined with non-metals and is –1 when combined

with active metals called metal hydrides such as LiH, KH, MgH2, CaH2 etc.
(vii) The oxidation number of oxygen is – 2 in most of its compounds, except in peroxides like H2O2, BaO2

etc. where it is –1. Another interesting exception is found in the compound OF2 (oxygen difluoride) where the
oxidation number of oxygen is + 2. This is due to the fact that fluorine being the most electronegative element
known has always an oxidation number of –1.

(viii) In compounds formed by union of metals with non-metals, the metal atoms will have positive oxidation

numbers and the non-metals will have negative oxidation numbers. For example,

(a) The oxidation number of alkali metals (Li, Na, K etc.) is always +1 and those of alkaline earth metals (Be,
Mg, Ca etc) is + 2.

(b) The oxidation number of halogens (F, Cl, Br, I) is always –1 in metal halides such as KF, AlCl3, MgBr2,
CdI2. etc.

(ix) In compounds formed by the union of different elements, the more electronegative atom will have negative
oxidation number whereas the less electronegative atom will have positive oxidation number. For example,

(a) N is given an oxidation number of –3 when it is bonded to less electronegative atom as in NH3 and NI3, but
is given an oxidation number of + 3 when it is bonded to more electronegative atoms as in NCl3.

(b) Since fluorine is the most electronegative element known so its oxidation number is always –1 in its
compounds i.e. oxides, interhalogen compounds etc.

(c) In interhalogen compounds of Cl, Br, and I; the more electronegative of the two halogens gets the
oxidation number of –1. For example, in BrCl3, the oxidation number of Cl is –1 while that of Br is +3.

(x) For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero. For example, in
NH3 the sum of the oxidation numbers of nitrogen atom and 3 hydrogen atoms is equal to zero. For a complex ion,

the sum of the oxidation numbers of all the atoms is equal to charge on the ion. For example, in SO42− ion, the sum

of the oxidation numbers of sulphur atom and 4 oxygen atoms must be equal to –2.

(xi) It may be noted that oxidation number is also frequently called as oxidation state. For example, in H2O,
the oxidation state of hydrogen is +1 and the oxidation state of oxygen is – 2. This means that oxidation number
gives the oxidation state of an element in a compound.

(xii) In the case of representative elements, the highest oxidation number of an element is the same as its
group number while highest negative oxidation number is equal to (8 – Group number) with negative sign with a
few exceptions. The most common oxidation states of the representative elements are shown in the following table,

Group Outer shell Common oxidation numbers (states)
configuration except zero in free state

I A ns1 +1

II A ns2 +2

III A ns2np1 +3, +1

IV A ns2np2 +4,+3,+2,+1, –1, –2, –3, –4

VA ns2np3 +5,+3,+1, –1, –3

VI A ns2np4 +6,+4,+2,–2

VII A ns2np5 +7,+5,+3, +1, –1

(xiii) Transition metals exhibit a large number of oxidation states due to involvement of (n –1) d electron
besides ns electron.

(xiv) Oxidation number of a metal in carbonyl complex is always zero, e.g. Ni has zero oxidation state in [Ni(CO)4 ].

(xv) Those compounds which have only C, H and O the oxidation number of carbon can be calculated by
following formula,

Oxidation number of ' C' = (nO ×2 - nH )

Where, nO is the number of oxygen atom, nH is the number of hydrogen atom, nC is the number of carbon atom.


For example, (a) CH 3OH ; nH = 4,nC = 1,nO = 1

Oxidation number of ‘C’ = (1 × 2− 4) = −2


(b) HCOOH ; nH = 2, nO = 2,nc = 1

Oxidation number of carbon = (2 × 2 − 2) = +2

(5) Procedure for calculation of oxidation number : By applying the above rules, we can calculate the
oxidation numbers of elements in the molecules/ions by the following steps.

(i) Write down the formula of the given molecule/ion leaving some space between the atoms.

(ii) Write oxidation number on the top of each atom. In case of the atom whose oxidation number has to be
calculated write x.

(iii) Beneath the formula, write down the total oxidation numbers of each element. For this purpose, multiply
the oxidation numbers of each atom with the number of atoms of that kind in the molecule/ion. Write the product in
a bracket.

(iv) Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge on the ion.

(v) Solve for the value of x.

Example: 1 The oxidation state of Mn in K2MnO4 is [CPMT 1982; 83; 84; Delhi PMT 1982; NCERT 1973; AMU 2000]

(a) + 2 (b) + 7 (c) – 2 (d) + 6

Solution : (d) K2 MnO4 2+ x − 2×4 = 0
Example: 2
Solution : (a) x = 8 − 2 = +6
So, oxidation state of Mn is +6.
Example: 3
Solution : (d) The oxidation number of Cr in K2Cr2O7 is [CPMT 1981; 85, 90, 93, 99; KCET 1992; BHU 1988, 98;

AFMC 1991, 99; EAMCET 1986; MP PMT 1996, 99, 2002; MP PET/PMT 1998; Bihar CEE 1995; RPET 2000]

(a) + 6 (b) – 7 (c) + 2 (d) – 2

K2 Cr 2O7

2+ 2x − 2×7 = 0 or 2x − 14 +2 = 0 or 2x = 12 or x = 12 = +6

So, oxidation state of Cr is +6.

Oxidation number of N in (NH4 )2 SO4 is [CPMT 1996]

(a) –1/3 (b) – 1 (c) + 1 (d) – 3

(NH4 )2 SO4 ⇌ 2 N H + + SO4−2


x + 4 = +1 or x = 1 − 4 = −3

Example: 4 The oxidation number of Sb in K2H2Sb2O7 is
Solution : (b)
(a) + 2 (b) + 5 (c) – 2 (d) – 5
Example: 5
Solution : (a) As K2H2Sb2O7 is a neutral species, therefore the sum of the oxidation number of all the atoms present in
Example: 6
K2H2Sb2O7 = 0

Let oxidation number per Sb atom = x

∴ 2 + 2 + 2x + (−2)7 = 0 [ Oxidation number of K, H and O are +1, +1 and –2 respectively]

or 4 + 2x − 14 = 0 or 2x − 10 = 0 or x = + 5

The oxidation number of Al in Ca3 (AlO3 )2 is

(a) + 3 (b) – 1 (c) + 2 (d) 0

Ca3(AlO3 )2

2 × 3 + 2x – 6 × 2 = 0 or 6 + 2x − 12 = 0 or 2x = 6 or x = + 3

The oxidation state of Cl in Ca(OCl) Cl is

(a) – 1 (b) + 2 (c) + 3 (d) 0

Solution : (a) OCl is known as hypochlorite radical having contribution towards = – 1.
Let the oxidation state of underlined chlorine atom = x
Example: 7
Solution : (a) ∴ 2 + (− 1) + x = 0 or x = −1

The oxidation state of underlined Cl = –1.
While the oxidation state of Cl present in OCl radical = + 1.

Oxidation state of oxygen in hydrogen peroxide is

[DPMT 1984; 91; CPMT 1988; MNR 1994; UPSEAT 2001; RPMT 2002]

(a) –1 (b) + 1 (c) 0 (d) – 2

In all peroxides oxygen shows –1 oxidation state.

Example: 8 The brown ring complex compound is formulated as [Fe(H 2O)5 NO + ]SO4 . The oxidation state of iron is
Solution : (b)
[EAMCET 1987; IIT 1987; MP PMT 1994; AIIMS 1997; DCE 2000]

(a) + 1 (b) + 2 (c) + 3 (d) 0

[ ]Fe (H 2O)5 NO SO4 [Oxidation number of H2O = 0 ; Oxidation number of NO = 0; Oxidation number of SO42− =–2]

x + 0 + 0 − 2 = 0 or x − 2 = 0 or x = + 2

Example: 9 The oxidation state of V in Rb4 Na [H V10O28 ] is
Solution : (b)
(a) + 2 (b) + 5 (c) – 2 (d) – 5

The oxidation state of Rb = +1 (Since it is alkali metals)

Let oxidation state of V = x

∴ 4 × 1 + 1 + 1 + 10x + 28 × ( − 2) = 0

or 6 + 10x − 56 = 0 or 10x − 50 = 0 or x = 5
∴ The oxidation state of V = +5

Example: 10 The oxidation state of Fe in FeS2 is
Solution : (b)
(a) – 2 (b) + 2 (c) 0 (d) + 1


The structure of FeS2 is Fe


Therefore FeS2 contain one per disulphide linkage, which is similar to that of peroxide linkages, having
contribution of – 2 units towards the oxidation state
Let the oxidation state of Fe = x
∴ x − 2 = 0 or x = 2

The oxidation state of Fe in FeS2 is + 2.
Oxidation number of some elements in compounds, ions or chemical species

Element Oxidation Compounds, ions or chemical species
Sulphur (s) Number H2S, ZnS, NaHS, (SnS3)2–, BaS, CS2

Nitrogen 0 S, S4, S8, SCN–
(N) +1 S2, F2, S2Cl2
+4 SO2, H2SO3, (SO3)2–, SOCl2, NaHSO3, Ca[HSO3]2, [HSO3]–, SF4
+6 H2SO4, (SO4)2-, [HSO4]–, BaSO4, KHSO4, SO3, SF6, H2S2O7, (S2O7)2–
–3 NH3, (NH4)+, AlN, Mg3N2, (N)3–, Ca3N2, CN–

–2 N2H4, (N2H5)+
–1 NH2OH
–1/3 NaN3, N3H
0 N2
+1 N2O
+2 NO

Chlorine +3 HNO2, (NO2)–, NaNO2, N2O3, NF3
(Cl) +4 NO2
+5 HNO3, (NO3)–, KNO3, N2O5
0 HCl, NaCl, CaCl2, AlCl3, ICl, ICl5, SOCl2, CrO2Cl2, KCl, K2PtCl6, HAuCl4, CCl4
+1 Cl, Cl2
+3 HOCl, NaOCl, (OCl)–, Cl2O
+4 KClO2, (ClO2)–, HClO2

Hydrogen +5 (ClO3)–, KClO3, NaClO3, HClO3
(H) +7 HClO4, Cl2O7, KClO4, (ClO4)–
–1 NaH, CaH2, LiAlH4, LiH
+1 NH3, PH3, HF
PH3, (PH4)+, Ca3P2
Phosphoru –3
s P4
0 H3PO2, KH2PO2, BaH4P2O4
(P) +1 PI3, PBr3, PCl3, P2O3, H3PO3
+3 (PO4)3–, H3PO4, Ca3(PO4)2, H4P2O7, P4O10, PCl5, (P2O7)4–, Mg2P2O7, ATP
Oxygen +5 H2O, PbO2, (CO3)2–, (PO4)2–, SO2, (C2O4)2–, HOCl, (OH)–, (O)2-
(O) –2 Na2O2, BaO2, H2O2, (O2)2–, Peroxides
–1 KO2
Carbon – 1/2 O, O2, O3
(C) 0 O2F2
+1 OF2
+2 CH4
–4 C2H6
–3 CH3Cl, C2H4
–2 CaC2, C2H2
–1 Diamond, Graphite, C6H12O6, C2H4O2, HCHO, CH2Cl2
0 CO, CHCl3, HCN
+2 H2C2O4, (C2O4)2–
+3 CO2, H2CO3, (HCO3)–, CCl4, Na2CO3, Ca2CO3, CS2, CF4, (CO3)–2

+4 Cr2(SO4)3, CrCl3, Cr2O3, [Cr(H2O)4Cl3]
K2CrO4, (CrO4)2–, K2Cr2O7, (Cr2O7)2–, KCrO3Cl, CrO2Cl2, Na2Cr3O10, CrO3
Chromium + 3 MnO, MnSO4, MnCl2, Mn(OH)2

(Cr) + 6 Mn3O4
Manganese + 2 MnO2, K2MnO3
K2MnO4, (MnO4)2–
(Mn) + 8/3 KMnO4, (MnO4)–, HMnO4
Silicon +3 SiH4, Mg2Si
+4 SiO2, K2SiO3, SiCl4
+6 Fe3O4
–4 FeSO4 (Ferrous ammonium sulphate), K4Fe(CN)6, FeCl2
K3[Fe(CN)6], FeCl3
(Si) +4

Iron + 8
(Fe) 3



Iodine +7 H4IO6− , KIO4
(I) +8 OsO4
+6 XeO3, XeF6


(6) Exceptional cases of evaluation of oxidation numbers : The rules described earlier are usually
helpful in determination of the oxidation number of a specific atom in simple molecules but these rules fail in the
following cases. In these cases, the oxidation numbers are evaluated using the concepts of chemical bonding

Type I. In molecules containing peroxide linkage in addition to element-oxygen bonds. For example,

(i) Oxidation number of S in H2SO5 (Permonosulphuric acid or Caro's acid).
By usual method; H2SO5

2 × 1 + x + 5 × (−2) = 0 or x = + 8

But this cannot be true as maximum oxidation number for S cannot exceed + 6. Since S has only 6 electrons

in its valence shell. This exceptional value is due to the fact that two oxygen atoms in H2SO5 shows peroxide linkage
as shown below,

O Peroxide linkage


Therefore the evaluation of o.n. of sulphur here should be made as follows,

2 × (+1) + x + 3 × (–2) + 2 × (–1)

(for H) (for S) (for O) (for O–O)

or 2 + x – 6 – 2 = 0 or x = + 6.

(ii) Oxidation number of S in H2S2O8 (Peroxidisulphuric acid or Marshall's acid)
By usual method ; H2S2O8

1 × 2 + 2x + 8 (–2) = 0

2x = + 16 – 2 = 14 or x = + 7

Similarly Caro's acid, Marshall's acid also has a peroxide linkage so that in which S shows +6 oxidation state.

Peroxide linkage



Therefore the evaluation of oxidation state of sulphur should be made as follow,

2 × (+1) + 2 × (x) + 6 × (–2) + 2 × (–1) = 0

(for H) (for S) (for O) (for O–O)

or 2 + 2x – 12 – 2 = 0 or x = + 6.

(iii) Oxidation number of Cr in CrO5 (Blue perchromate)
By usual method CrO5 ; x – 10 = 0 or x = + 10

This cannot be true as maximum o.n. of Cr cannot be more than + 6. Since Cr has only five electrons in 3d

orbitals and one electron in 4s orbital. This exceptional value is due to the fact that four oxygen atoms in CrO5 are
in peroxide linkage. The chemical structure of CrO5 is

Peroxide linkage OO Peroxide linkage


Therefore, the evaluation of o.n. of Cr should be made as follows

x + 1 × (– 2) + 4 (–1) = 0

(for Cr) (for O) (for O–O)

or x – 2 – 4 = 0 or x = + 6.

Type II. In molecules containing covalent and coordinate bonds, following rules are used for evaluating the
oxidation numbers of atoms.

(i) For each covalent bond between dissimilar atoms the less electronegative element is assigned the oxidation
number of + 1 while the atom of the more electronegative element is assigned the oxidation number
of –1.

(ii) In case of a coordinate-covalent bond between similar or dissimilar atoms but the donor atom is less
electronegative than the acceptor atom, an oxidation number of +2 is assigned to the donor atom and an oxidation
number of –2 is assigned to the acceptor atom.

Conversely, if the donor atom is more electronegative than the acceptor atom, the contribution of the
coordinate bond is neglected.

Example :

(a) Oxidation number of C in HC ≡ N and HN → C


The evaluation of oxidation number of C cannot be made directly by usual rules since no standard rule exists

for oxidation numbers of N and C.

In such cases, evaluation of oxidation number should be made using indirect concept or by the original

concepts of chemical bonding.

(b) Oxidation number of carbon in H–N → C


The contribution of coordinate bond is neglected since the bond is directed from a more electronegative N

atom (donor) to a less electronegative carbon atom (acceptor).

Therefore the oxidation number of N in HN → C remains – 3 as it has three covalent bonds.


1 × (+ 1) + 1 × (– 3) + x = 0

(for H) (for N) (for C)

or 1 + x – 3 = 0 or x = + 2.

(c) Oxidation number of carbon in HC ≡ N

In HC ≡ N , N is more electronegative than carbon, each bond gives an oxidation number of –1 to N. There
are three covalent bonds, the oxidation number of N is HC ≡ N is taken as – 3

Now HC ≡ N ∴ +1 + x – 3 = 0 ⇒ x = + 2

Type III. In a molecule containing two or more atoms of same or different elements in different oxidation states.

(i) Oxidation number of S in Na2S2O3
By usual method Na2S2O3

∴ 2 × (+1) + 2 × x + 3 (–2) = 0 or 2 + 2x – 6 = 0 or x = 2.

But this is unacceptable as the two sulphur atoms in Na2S2O3 cannot have the same oxidation number
because on treatment with dil. H2SO4, one sulphur atom is precipitated while the other is oxidised to SO2.

Na2S2O3 + H 2SO4 → Na2SO4 + SO2 + S + H 2O

In this case, the oxidation number of sulphur is evaluated from concepts of chemical bonding. The chemical

structure of Na2S2O3 is


Na + O − S O − Na +


Due to the presence of a co-ordinate bond between two sulphur atoms, the acceptor sulphur atom has

oxidation number of – 2 whereas the other S atom gets oxidation number of + 2.

2 × (+1) + 3 × (–2) + x × 1 + 1 × (– 2) = 0

(for Na) (for O) (for S) (for coordinated S)

or + 2 – 6 + x – 2 = 0 or x = + 6

Thus two sulphur atoms in Na2S2O3 have oxidation number of – 2 and +6.
(ii) Oxidation number of chlorine in CaOCl2 (bleaching powder)
In bleaching powder, Ca(OCl)Cl, the two Cl atoms are in different oxidation states i.e., one Cl– having

oxidation number of –1 and the other as OCl– having oxidation number of +1.

(iii) Oxidation number of N in NH4NO3 (wrong)
By usual method N2H4O3 ; 2x + 4 × (+1) + 3 × (–1) = 0
2x + 4 – 3 = 0 or 2x = + 1

No doubt NH4NO3 has two nitrogen atoms but one N has negative oxidation number (attached to H) and the

other has positive oxidation number (attached to O). Hence the evaluation should be made separately for NH +

and NO3−

NH + x + 4 × (+1) = +1 or x = – 3

NO3− x + 3 (– 2) = –1 or x = + 5.

(iv) Oxidation number of Fe in Fe3O4

In Fe3O4, Fe atoms are in two different oxidation states. Fe3O4 can be considered as an equimolar mixture of

FeO (iron (II) oxide) and Fe2O3 (iron (III) oxide). Thus in one molecule of Fe3O4, two Fe atoms are in + 3 oxidation

state and one Fe atom is in + 2 oxidation state.

(v) Oxidation number of S in sodium tetrathionate (Na2S4O6). Its structure can be represented as

follows : Na + O S S S S O Na +


The two S-atoms which are linked to each other have oxidation number of zero. The oxidation number of

other S-atoms can be calculated as follows

Let oxidation number of S = x.

∴ 2 × x + 2 × 0 + 6 × ( – 2) = – 2

(for S) (for S–S) (for O)

x = + 5.

Balancing of oxidation-reduction reactions.

Though there are a number of methods for balancing oxidation – reduction reactions, two methods are very
important. These are,(1) Oxidation number method, (2) Ion – electron method

(1) Oxidation number method : The method for balancing redox reactions by oxidation number change
method was developed by Johnson. In a balanced redox reaction, total increase in oxidation number must be
equal to the total decrease in oxidation number. This equivalence provides the basis for balancing redox reactions.

This method is applicable to both molecular and ionic equations. The general procedure involves the following

(i) Write the skeleton equation (if not given, frame it) representing the chemical change.

(ii) Assign oxidation numbers to the atoms in the equation and find out which atoms are undergoing
oxidation and readuction. Write separate equations for the atoms undergoing oxidation and reduction.

(iii) Find the change in oxidation number in each equation. Make the change equal in both the equations by
multiplying with suitable integers. Add both the equations.

(iv) Complete the balancing by inspection. First balance those substances which have undergone change in
oxidation number and then other atoms except hydrogen and oxygen. Finally balance hydrogen and oxygen by
putting H2O molecules wherever needed.

The final balanced equation should be checked to ensure that there are as many atoms of each element on the
right as there are on the left.

(v) In ionic equations the net charges on both sides of the equation must be exactly the same. Use H+
ion/ions in acidic reactions and OH– ion/ions in basic reactions to balance the charge and number of hydrogen and
oxygen atoms.

The following example illustrate the above rules,

Step : I Cu + HNO3 → Cu(NO3 )2 + NO2 + H 2O (Skeleton equation)

Step: II Writing the oxidation number of all the atoms.

0 + H+1 N+5 −2 3 → C+2u( N+5 O−2 3 )2 + N+4 O−2 2 + H+1 2O−2

Cu O

Step: III Change in oxidation number has occurred in copper and nitrogen.

0 → C+2u ( NO3 )2 ......(i)


H N+5 O3 → N+4 O2 ......(ii)

Increase in oxidation number of copper = 2 units per molecule Cu

Decrease in oxidation number of nitrogen = 1 unit per molecule HNO3
Step: IV To make increase and decrease equal, equation (ii) is multiplied by 2.

Cu + 2HNO3 → Cu(NO3 )2 + 2NO2 + H 2O
Step: V Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained.

Cu + 4HNO3 → Cu(NO3 )2 + 2NO2 + 2H 2O
This is the balanced equation.

(2) Ion-electron method (half reaction method)
The method for balancing redox-reactions by ion electron method was developed by Jette and LaMev

in 1927. It involves the following steps

(i) Write down the redox reaction in ionic form.

(ii) Split the redox reaction into two half reactions, one for oxidation and other for reduction.

(iii) Balance each half reaction for the number of atoms of each element. For this purpose,

(a) Balance the atoms other than H and O for each half reaction using simple multiples.

(b) Add water molecules to the side deficient in oxygen and H+ to the side deficient in hydrogen. This is done
in acidic or neutral solutions.

(c) In alkaline solution, for each excess of oxygen, add one water molecule to the same side and 2OH– ions to
the other side. If hydrogen is still unbalanced, add one OH– ion for each excess hydrogen on the same side and one
water molecule to the other side.

(iv) Add electrons to the side deficient in electrons as to equalise the charge on both sides.

(v) Multiply one or both the half reactions by a suitable number so that number of electrons become equal in
both the equations.

(vi) Add the two balanced half reactions and cancel any term common to both sides.

The following example illustrate the above rules

Step: I I 2 + OH − → IO3− + I − + H 2O (Ionic equation)

Step: II Splitting into two half reactions, I 2 + OH − → IO3− + H 2O ; I 2 → I −
(Oxidation half reaction) (Reduction half reaction)

Step: III Adding OH − ions, I 2 + 12OH − → 2IO3− + 6H 2O
Step: IV Adding electrons to the sides deficient in electrons,

I 2 + 12OH − → 2IO3− + 6H 2O + 10e − ; I 2 + 2e − → 2I −
Step: V Balancing electrons in both the half reactions.

I 2 + 12OH − → 2IO3− + 6H 2O + 10e − ; 5[I 2 + 2e − → 2I − ]
Step: VI Adding both the half reactions.

6 I 2 + 12OH − → 2IO3− + 6H 2O + 10I − ; Dividing by 2, 3 I 2 + 6OH − → IO3− + 5I − + 3H 2O


(1) Turpentine and numerous other olefinic compounds, phosphorus and certain metals like Zn and Pb can

absorb oxygen from the air in presence of water. The water is oxidised to hydrogen peroxide. This phenomenon of
formation of H2O2 by the oxidation of H2O is known as autoxidation. The substance such as turpentine or
phosphorus or lead which can activate the oxygen is called activator. The activator is supposed to first combine
with oxygen to form an addition compound, which acts as an autoxidator and reacts with water or some other

acceptor so as to oxidise the latter. For example;

Pb + O2 → PbO2 PbO2 + H 2O → PbO + H 2O2
(activator ) (acceptor)

(2) The turpentine or other unsaturated compounds which act as activators are supposed to take up oxygen
molecule at the double bond position to form unstable peroxide called moloxide, which then gives up the oxygen
to water molecule or any other acceptor.


RHC CHR + 2H 2O → RCH = CHR + 2H 2O2

2KI + H 2O2 → 2KOH + I 2
The evolution of iodine from KI solution in presence of turpentine can be confirmed with starch solution which
turns blue.

(3) The concept of autoxidation help to explain the phenomenon of induced oxidation. Na2SO3 solution is
oxidised by air but Na3 AsO3 solution is not oxidised by air. If a mixture of both is taken, it is observed both are
oxidised. This is induced oxidation.

Na2SO3 + O2 → Na2SO5


Na2SO5 + Na3 AsO3 → Na3 AsO4 + Na2SO4
Na2SO3 + Na3 AsO3 + O2 → Na2SO4 + Na3 AsO4


One and the same substance may act simultaneously as an oxidising agent and as a reducing agent with the

result that a part of it gets oxidised to a higher state and rest of it is reduced to lower state of oxidation. Such a

reaction, in which a substance undergoes simultaneous oxidation and reduction is called disproportionation and

the substance is said to disproportionate.

Following are the some examples of disproportionation, decrease


(1) H 2O2 + H2 O−1 2 = 0 (2) 4 K C+5l O3 → 3K C+7l O4 + K−C1 l

H2O + O2 increase
–1 –2

decrease decrease

(3) 0 +1 −3 (4) 0 + 6 NaOH hot → 5 Na C−1l + Na +5 + 3H2O
4P+ 3NaOH + 3H2O → 3NaH2 PO2 + PH3 3Cl2 ClO3

increase increase

Important applications of redox-reactions.

Many applications are based on redox reactions which are occuring in environment. Some important
examples are listed below;

(1) Many metal oxides are reduced to metals by using suitable reducing agents. For example Al2O3 is reduced
to aluminium by cathodic reduction in electrolytic cell. Fe2O3 is reduced to iron in a blast furnace using coke.

(2) Photosynthesis is used to convert carbon dioxide and water by chlorophyll of green plants in the presence
of sunlight to carbohydrates.

6CO2(g) + 6H 2O(l) CShulonrloigphhtyll → C6 H12O6(aq.) + 6O2(g)
In this case, CO2 is reduced to carbohydrates and water is oxidised to oxygen. The light provides the energy
required for the reaction.
(3) Oxidation of fuels is an important source of energy which satisfies our daily need of life.

Fuels + O2 → CO2 + H 2O + Energy

In living cells, glucose (C6 H12O6 ) is oxidised to CO2 and H 2O in the presence of oxygen and energy is

released, C6 H12O6(aq.) + 6O2 (g) → 6CO2 (g) + 6H 2O(l) + Energy

(4) The electrochemical cells involving reaction between hydrogen and oxygen using hydrogen and oxygen
electrodes in fuel cells meet our demand of electrical energy in space capsule.

(5) Respiration in animals and humans is also an important application of redox reactions.


s and p block elements

Hydrogen and Its Compounds.

(1) Position of hydrogen in the periodic table
Hydrogen is the first element in the periodic table. Hydrogen is placed in no specific group due to its
property of giving electron (When H − is formed) and also losing electron (When H + is formed).
(i) Hydrogen is placed in group I (Alkali metals) as,
(a) It has one electron in its (Outer) Shell- 1s1 like other alkali metals which have (inert gas) ns1
(b) It forms monovalent H + ion like Li+ , Na + 
(c) It valency is also 1.
(d) Its oxide (H 2O) is stable as Li2O, Na2O .
(e) It is a good reducing agent (In atomic as well as molecular state) like Na, Li
(ii) Hydrogen also resembles halogens (Group VIII A) as,
(a) It is also diatomic (H 2 ) like F2, Cl2 
(b) It also forms anion H − like F − , Cl −  by gain of one electron.
(c) H − has stable inert gas (He) configuration as CH 4 , C2 H6 like halogens CCl4 , SF2Cl2 etc.
(d) H is one electron short of duplet (Stable configuration) like F, Cl, which are also one electron
deficient than octet, F − 2s 2 2p5 ; Cl − 3s 2 3p5 .

(e) (IE) of H(1312 kJ mol −1) is of the same order as that of halogens.

(iii) (IE) of H is very high in comparison with alkali metals. Also size of H + is very small compared to
that of alkali metal ion. H forms stable hydride only with strongly electropositive metals due to smaller
value of its electron affinity (72.8 kJ mol −1) .

(iv) In view of the anomalous behavior of hydrogen, it is difficult to assign any definite position to it
in the periodic table. Hence it is customary to place it in group I (Along with alkali metals) as well as in
group VII (Along with halogens).

(2) Discovery and occurrence : It was discovered by Henry Cavendish in 1766. Its name
hydrogen was proposed by Lavoisier. Hydrogen is the 9th most abundant element in the earth’s crust.

(3) Preparation of Dihydrogen : Dihydrogen can be prepared by the following methods,
(i) Laboratory method : In the laboratory, dihydrogen can be prepared by the action of dil.
H 2SO4 on granulated Zinc, Zn + H 2SO4 (dil.) → ZnSO4 + H 2
(ii) Industrial method
(a) By the electrolysis of water : The hydrogen prepared by this method is highly pure. Dihydrogen is
collected at cathode. 2H 2O (l) Electrolysis → 2H 2(g) + O2(g)
(b) Hydrocarbon steam process : H 2 is prepared by the action of steam on hydrocarbon. e.g.

CH 4 + H 2O 1170K → CO + 3H 2

(c) Bosch process : H2 + CO+ H2O 773K→ CO2 + 2H2
water Fe2O3 , Cr2O3
gas steam

(d) Lane’s process : H 2 is prepared by passing alternate currents of steam and water gas over red
hot iron. The method consists of two stages,

Oxidation stage : 3Fe + 4H2O 1025−1075K → Fe 3 O4 + 4H2 + 161KJ

Iron filings (Steam) Magnetic oxide of iron

Reduction stage : 2Fe3O4 + 4CO + 4H 2 → 6Fe + 4CO2 + 4H 2O
water gas

(4) Physical properties of dihydrogen : It is a colourless, tasteless and odourless gas. It is slightly

soluble in water. It is highly combustible. The Physical constants of atomic hydrogen are,

Atomic radius (pm) – 37 ; Ionic radius of H − ion (pm) – 210; Ionisation energy (kJ mol −1) – 1312;

Electron affinity (kJ mol −1) –72.8; Electronegativity – 2.1.

(5) Chemical properties of dihydrogen : Dihydrogen is quite stable and dissociates into
hydrogen atoms only when heated above 2000 K, H 2 2000K → H + H . Its bond dissociation energy is

very high, H 2 → H + H ; ∆H = 435.9 kJ mol −1 . Due to its high bond dissociation energy, it is not very
reactive. However, it combines with many elements or compounds.

(i) Action with metals : To forms corresponding hydrides. 2Na + H 2 Heat → 2NaH ;

Ca + H 2 Heat → CaH 2 .

With transition metals (elements of d – block) such as Pd, Ni, Pt etc. dihydrogen forms interstitial
hydrides in which the small molecules of dihydrogen occupy the interstitial sites in the crystal lattices of
these hydrides. As a result of formation of interstitial hydrides, these metals adsorb large volume of
hydrogen on their surface. This property of adsorption of a gas by a metal is called occlusion. The
occluded hydrogen can be liberated from the metals by strong heating.

(ii) Reaction with Non-metals : 2H 2 + O2 970K → 2H 2O ; Fe, Mo

N 2 + 3H 2 750K,Pressure → 2NH 3

H 2 + F2 Dark → 2HF ; Sunlight

H 2 + Cl 2 673K,Pressure → 2HCl

H 2 + Br2 → 2HBr ; H2 + I2 673K → 2HI

The reactivity of halogen towards dihydrogen decreases as, F2 > Cl2 > Br2 > I 2

As a result, F2 reacts in dark, Cl2 in the presence of sunlight, Br2 reacts only upon heating while the

reaction with I 2 occurs in the presence of a catalyst.

(iii) Reaction with unsaturated hydrocarbons : H2 reacts with unsaturated hydrocarbons such

as ethylene and acetylene to give saturated hydrocarbons.

H 2C = CH 2 + H2 Ni or Ptor Pd → CH 3 − CH 3 ; HC ≡ CH+ 2H 2 Ni or Ptor Pd → CH 3 − CH 3
Ethylene 473 K Acetylene 473 K
Ethane Ethane

This reaction is used in the hydrogenation or hardening of oils. The vegetable oils such as
groundnut oil or cotton-seed oil are unsaturated in nature because they contain at least one double bond

in their molecules. Dihydrogen is passed through the oils at about 473 K in the presence of catalyst to form
solid fats. The vegetable ghee such as Dalda, Rath, etc. are usually prepared by this process.

Vegetable oil+ H2 Ni Fat

(liquid) 473K → (solid)

(6) Uses of Dihydrogen
(i) As a reducing agent, (ii) In the hydrogenation of vegetable oils, (iii) As a rocket fuel in the form of
liquid H 2 (iv) In the manufacture of synthetic petrol, (v) In the preparation of many compounds such as
NH 3 , CH 3OH, Urea etc, (vi) It is used in the oxy-hydrogen torch for welding if temperature around
2500°C is required. It is also used in atomic hydrogen torch for welding purposes in which temperature of
the order of 4000°C is required.

Different forms of hydrogen

(1) Atomic hydrogen : It is obtained by the dissociation of hydrogen molecules. The atomic

hydrogen is stable only for a fraction of a second and is extremely reactive. It is obtained by passing

dihydrogen gas at atmospheric pressure through an electric

arc struck between two tungsten rods.

The electric arc maintains a temperature around 4000 – Tungsten rod
4500°C. As the molecules of dihydrogen gas pass through the
electric arc, these absorb energy and get dissociated into H2 98
atoms as

H (g) Electric → 2H(g) : ∆H 435.90KJ mol −1 Tungstem rod
arc Atomic hydrogen torch

2 =

This arrangement is also called atomic hydrogen torch.

(2) Nascent hydrogen : The hydrogen gas prepared in the reaction mixture in contact with the

substance with which it has to react, is called nascent hydrogen. It is also called newly born hydrogen. It is

more reactive than ordinary hydrogen. For example, if ordinary hydrogen is passed through acidified

KMnO4 (pink in colour), its colour is not discharged. On the other hand, if zinc pieces are added to the

same solution, bubbles of hydrogen rise through the solution and the colour is discharged due to the

reduction on KMnO4 by nascent hydrogen.

KMnO4 + H 2 + H 2SO4 → No Re action ; Zn + H 2SO4 → ZnSO4 + 2[H] × 5
Molecular Nascent hydrogen

2KMnO4 + 3H 2SO4 + 10H → K2SO4 + 2MnSO4 + 8H 2O

(3) Ortho and para hydrogen : A molecule of dihydrogen contains two atoms. The nuclei of both
the atoms in each molecule of dihydrogen are spinning. Depending upon the direction of the spin of the
nuclei, the hydrogen is of two types,

(i) Molecules of hydrogen in which the spins of both the nuclei are in the same directions, called ortho

(ii) Molecules of hydrogen in which the spins of both the nuclei are in the opposite directions, called
para hydrogen.

Ordinary dihydrogen is an equilibrium mixture of ortho and para hydrogen. Ortho hydrogen ⇌ Para
hydrogen. The amount of ortho and para hydrogen varies with temperature as,

(a) At 0°K, hydrogen contains mainly para hydrogen which is more stable.
(b) At the temperature of liquefaction of air, the ratio of ortho and para hydrogen is 1:1.
(c) At the room temperature, the ratio of ortho to para hydrogen is 3:1.
(d) Even at very high temperatures, the ratio of ortho to para hydrogen can never be more than 3:1.
Thus, it has been possible to get pure para hydrogen by cooling ordinary hydrogen gas to a very low
temperature (close to 20 K) but it is never possible to get a sample of hydrogen containing more than 75%
of ortho hydrogen. i.e., Pure ortho hydrogen can not be obtained.

Isotopes of Hydrogen
Isotopes are the different forms of the same element which have the same atomic number but different mass

Isotopes of hydrogen

Name Symbol Atomic Mass Relative Nature radioactive or
number number abundance non-radioactive

Protium or Hydrogen 1 H or H 1 1 99.985% Non-radioactive
Deuterium 1 1
Tritium 1
2 H or D 2 0.015% Non-radioactive

3 H or T 3 10 −15 % Radioactive

Physical constants of H2 , D2 and T2

Property H2 D2 T2
Molecular mass 2.016 4.028 6.03
Melting point (K) 20.63
Boiling point (K) 13.8 18.7 25.0
Heat of fusion (kJ mol -1 ) 0.250
20.4 23.9
Heat of vaporisation (kJ mol -1 ) 1.393
0.117 0.197
Bond energy (kJ mol -1 ) 446.9
0.994 1.126

435.9 443.4


Water is the oxide of hydrogen. It is an important component of animal and vegetable matter. Water

constitutes about 65% of our body. It is the principal constituent of earth’s surface.

(1) Structure : Due to the presence of lone pairs, the Lone Pair of Electron
geometry of water is distorted and the H − O − H bond angle is
104.5°, which is less than the normal tetrahedral angle (109.5°) . ..

The geometry of the molecule is regarded as angular or bent. In :O:
water, each O − H bond is polar because of the high
electronegativity of oxygen (3.5) in comparison to that of H 104.5o H
hydrogen (2.1). The resultant dipole moment of water molecule is


In ice, each oxygen atom is tetrahedrally surrounded by four hydrogen atoms; two by covalent

bonds and two by hydrogen bonds. The resulting structure of ice is open structure having a number of

vacant spaces. Therefore, the density of ice is less than that of water and ice floats over water. It may be
noted that water has maximum density (1g cm−3 ) at 4°C.

(2) Heavy water : Chemically heavy water is deuterium oxide (D2O) . It was discovered by Urey. It
has been finding use in nuclear reactors as a moderator because it slows down the fast moving neutrons
and therefore, helps in controlling the nuclear fission process.

(3) Physical properties : Water is colourless, odourless and tasteless liquid at ordinary

Some physical constants of H2O and D2O at 298 K

Constant Ordinary water H2O Heavy water D2O

Molecular mass 18.015 20.028

Maximum density (g cm−3) 1.000 1.106

Melting point (K) 273.2 276.8
Boiling point (K) 373.2 374.4
Heat of fusion (kJ mol−1) at 273K 6.01 6.28

Heat of vaporisation (kJ mol −1) at 373K 40.66 41.61

Heat of formation (kJ mol−1) – 285.9 – 294.6

Ionisation constant 1.008 ×10−14 1.95 × 10−15

(4) Chemical properties : Water shows a versatile chemical behaviour. It behaves as an acid, a
base, an oxidant, a reductant and as ligand to metals.

(i) Dissociation of water : Water is quite stable and does not dissociate into its elements even at
high temperatures. Pure water has a small but measurable electrical conductivity and it dissociates as,

H 2O + H 2O ⇌ H 3O+ + OH − ; KW = 1.0 × 10−14 mol 2 L2 at 298K
Hydronium ion

(ii) Amphoteric nature : Water can act both as an acid and a base and is said to be amphoteric.
However, water is neutral towards litmus and its pH is 7.

(iii) Oxidising and reducing nature : Water can act both as an oxidising and a reducing agent in

its chemical reactions. e.g. 2Na + 2H 2O → 2NaOH + H 2 ; 2F2 + 2H 2O → 4HF + O2
Oxidi sin g agent Re ducing agent

(5) Hard and Soft water

Water which produces lather with soap solution readily is called soft water. e.g. distilled water, rain
water and demineralised water.

Water which does not produce lather with soap solution readily is called hard water. e.g. sea water,
river water, well water and tap water.

(i) Cause of hardness of water : The hardness of water is due to the presence of bicarbonates,
chlorides and sulphates of calcium and magnesium.

Hard water does not produce lather because the cations (Ca +2and Mg +2 ) present in hard water react

with soap to form insoluble precipitates, M +2 + 2C17 H 35COONa → (C17 H 35COO)2 M + 2Na + ,Where
M = Ca or Mg Sodium stearate(soap) Metal stearate(PPt.)
From hard water

Therefore, no lather is produced until all the calcium and magnesium ions are precipitated. This also
results into wastage of lot of soap.

(ii) Type of hardness of water : The hardness of water is of two types,

(a) Temporary hardness : This is due to the presence of bicarbonates of calcium and magnesium. It is

also called carbonate hardness.

(b) Permanent hardness : This is due to the presence of chlorides and sulphates of calcium and

magnesium. It is also called non-carbonate hardness.

(iii) Softening of water : The process of the removal of hardness from water is called softening of


(a) Removal of temporary hardness : It can be removed by the following methods,

By boiling : During boiling, the bicarbonates of Ca and Mg decompose into insoluble carbonates

and give CO2. The insoluble carbonates can be removed by filtration.

Ca(HCO3 )2 Heat → CaCO3 + CO2 + H 2O ; Mg(HCO3 )2 Heat → MgCO3 + CO2 + H 2O
Cal . bicarbonate PPt. Mag . bicarbonate PPt.

Clark’s method : This process is used on a commercial scale. In this process, calculated amount of

lime [Ca(OH)2 ]is added to temporary hard water.

Ca(HCO3 )2 + Ca(OH)2 → 2CaCO3 ↓ +2H 2O
Soluble Lime Insoluble

Mg(HCO3 )2 + Ca(OH 2 ) → MgCO3 ↓ +CaCO3 ↓ +2H 2O
Soluble Lime (Insoluble)

(b) Removal of permanent hardness : Permanent hardness can be removed by the following


By washing soda method : In this method, water is treated with a calculated amount of washing

soda (Na2CO3 ) which converts the chlorides and sulphates of Ca and Mg into their respective carbonates
which get precipitated.

CaCl 2 + Na2CO3 → CaCO3 + 2NaCl ; MgSO4 + Na2CO3 → MgCO3 + Na2SO4
ppt. ppt.

Permutit method : This is a modern method employed for the softening of hard water. hydrated
sodium aluminium silicate (Na2 Al2Si2O8 .xH 2O) is called permutit. These complex salts are also known as

The permutit as loosely packed in a big tank over a layer of coarse sand. Hard water is introduced
into the tank from the top. Water reaches the bottom of the tank and then slowly rises through the permutit
layer in the tank. The cations present in hard water are exchanged for sodium ions. Therefore this method
is also called ion exchange method.

Na2 Z+ Ca +2 → CaZ+ 2Na + ; Na2 Z + Mg +2 → MgZ + 2Na + , where
Cal (From hard Magnesium
Sodium (From hard zeolite water) zeolite
zeolite water)

Z = Al2Si2O8 . xH 2O

Hydrogen peroxide

Hydrogen peroxide (H 2O2 ) was discovered by French chemist Thenard.
(1) Preparation : It is prepared by

(i) Laboratory method : In laboratory, H2O2 is prepared by Merck’s process. It is prepared by
adding calculated amounts of sodium peroxide to ice cold dilute (20%) solution of H 2SO4 .

Na2O2 + H 2SO4 → Na2SO4 + H 2O2
(ii) Industrial method : On a commercial scale, H2O2 can be prepared by the electrolysis of 50%
H 2SO4 solution. In a cell, peroxy disulphuric acid is formed at the anode.

2H 2SO4 Elecrolysis → H 2S2O8 (aq.)+ H 2 (g)

Peroxy disulphuric

This is drawn off from the cell and hydrolysed with water to give H 2O2 .

H 2S2O8 + 2H 2O → 2H 2SO4 + H 2O2 The resulting solution is distilled under reduced pressure
when H2O2 gets distilled while H 2SO4 with high boiling point, remains undistilled.

(2) Physical properties : Pure H2O2 is a thick syrupy liquid with pale blue colour. It is more
viscous and dense than water. It is completely miscible with water, alcohol and ether in all proportions.

(3) Chemical properties
(i) Decomposition : Pure H2O2 is an unstable liquid and decomposes into water and O2 either
upon standing or upon heating, 2H 2O2 → 2H 2O + O2 ; ∆H = −196.0 kJ
(ii) Oxidising nature : It is a powerful oxidising agent. It acts as an oxidising agent in neutral, acidic
or in alkaline medium. e.g. 2KI + H 2O2 → 2KOH + I 2 [In neutral medium]

2FeSO4 + H 2SO4 + H 2O2 → Fe2 (SO4 )3 + 2H 2O [In acidic medium]

MnSO4 + H 2O2 + 2NaOH → MnO2 + Na2SO4 + 2H 2O [In alkaline medium]
(iii) Reducing nature : H2O2 has tendency to take up oxygen from strong oxidising agents and
thus, acts as a reducing agent, H 2O2 + O → H 2O + O2 . It can act as a reducing agent in acidic, basic or

From oxidising

even neutral medium.
In acidic medium, H 2O2 → 2H + + O2 + 2e −

In alkaline medium, H 2O2 + 2OH − → 2H 2O + O2 + 2e −
(iv) Bleaching action : H2O2 acts as a bleaching agent due to the release of nascent oxygen.

H 2O2 → H 2O + O

Thus, the bleaching action of H 2O2 is due to oxidation. It oxidises the colouring matter to a
colourless product, Colouring matter +O → Colour less matter

H 2O2 is used to bleach delicate materials like ivory, silk, wool, leather etc.

(4) Structure of H2O2 : H2O2 has non-planar structure in which two H-atoms are arranged in two
directions almost perpendicular to each other and to the axis joining the two oxygen atoms. The O – O
linkage is called peroxide linkage.

(5)Strength of H2O2 : The strength of H2O2 is expressed in terms of weight or volume,
(i) As weight percentage : The weight percentage of H2O2 gives the weight of H2O2 in 100 g of
solution. For example, a 40% solution by wt. means 40 g of H 2O2 are present in 100 g of solution.
(ii) As volume : The strength of H 2O2 is commonly expressed as volume. This refers to the volume
of oxygen which a solution of H 2O2 will give. For example, a “20 volume” of H 2O2 means that 1 litre of
this solution will give 20 litres of oxygen at NTP.

(6) Uses of H2O2
(i) It is used as an antichlor in bleaching because it can reduce chlorine.
Cl2 + H 2O2 → 2HCl + O2
(ii) It is used for restoring the colour of lead paintings.
(iii) It is used as an antiseptic for washing wounds, teeth and ears under the name perhydrol.

Alkali Metals and Their Compounds.

The group 1 of the periodic table contains six elements, namely lithium (Li), sodium (Na), potassium
(K), rubidium (Rb), caesium (Cs) and francium (Fr). All these elements are typical metals. These are
usually referred to as alkali metals since their hydroxides forms strong bases or alkalies.

Electronic configuration

Elements Discovery Electronic configuration ( ns1 )

3 Li Arfwedson (1817) 1s2 2s1 or [He]2 2s1
11 Na Davy (1807) 1s 2 2s 2 2p6 3s1 or [Ne]10 3s1
19 K Davy (1807) 1s 2 2s 2 2p6 3s 2 3p6 4s1 or [Ar]18 4s1
37 Rb 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4 p6 5s1 or [Kr]36 5s1
55 Cs Bunsen (1861) 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4 p6 4d10 5s 2 5 p6 6s1 or [Xe]54 6s1
87 Fr Bunsen (1860) 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4 p6 4d10 4 f 14 5s 2 5 p6 5d10 6s 2 6 p6 7s1 or [Rn]86 7s1
Percy (1939)

Note :  Francium is radioactive with longest lived isotope 223 Fr with half life period of only 21


(1) Because of similarity in electronic configuration, they exhibit similar properties. A regular
gradation in their properties with increase in at. no. is observed due to increasing size of atoms/ions and
the low binding energy of valency electrons.

(2) Of all the alkali metals, only sodium and potassium are found in abundance in nature. Francium occurs
only in minute quantities as a radioactive decay product.

Physical properties

(1) Physical state

(i) All are silvery white, soft and light solids. These can be cut with the help of knife. When freshly
cut, they have bright lustre which quickly tarnishes due to surface oxidation.

(ii) These form diamagnetic colourless ions since these ions do not have unpaired electrons, (i.e. M+
has ns0configuration). That is why alkali metal salts are colourless and diamagnetic.

(2) Atomic and ionic radii

(i) The alkali metals have largest atomic and ionic radii than their successive elements of other

groups belonging to same period.

(ii) The atomic and ionic radii of alkali metals, however, increases down the group due to progressive

addition of new energy shells.

No doubt the nuclear charge also increases on moving down the group but the influence of addition

of energy shell predominates

Li Na K Rb Cs Fr

Atomic radius (pm) 152 186 227 248 265 375

Ionic radius of M+ ions (pm) 60 95 133 148 169 –

(3) Density

(i) All are light metals, Li, Na and K have density less than water. Low values of density are because
these metals have high atomic volume due to larger atomic size. On moving down the group the atomic
size as well as atomic mass both increase but increase in atomic mass predominates over increase in
atomic size or atomic volume and therefore the ratio mass/volume i.e. density gradually increases down
the groups

(ii) The density increases gradually from Li to Cs, Li is lightest known metal among all.
Li = 0.534, Na = 0.972, K = 0.86, Rb = 1.53 and Cs = 1.87 g/ml at 200C.
(iii) K is lighter than Na because of its unusually large atomic size.

(iv) In solid state, they have body centred cubic lattice.

(4) Melting point and Boiling point

(i) All these elements possess low and in comparison to other group members.

Li Na K Rb Cs

Fr (K) 453.5 370.8 336.2 312.0 301.5 (K) 1620 1154.4 1038.5 961.0 978.0

(ii) The lattice energy of these atoms in metallic crystal lattice relatively low due to larger atomic size

and thus possess low and On moving down the group, the atomic size increases and binding

energy of their atoms in crystal lattice decreases which results lowering of m.pts.

(iii) Lattice energy decreases from Li to Cs and thus and also decrease from Li to Cs.

(5) Ionisation energy & electropositive or metallic character

(i) Due to unpaired lone electron in ns sub-shell as well as due to their larger size, the outermost electron is

far from the uncleus, the removal of electron is easier and these low values of ionisation energy.(I.E.)

(ii) Ionisation energy of these metal decreases from Li to Cs.

Ionisation energy Li Na K Rb Cs

IE1 520 495 418 403 376

IE2 7296 4563 3069 2650 2420

A jump in 2nd ionisation energy (huge difference) can be explained as,

Re movalof Re moval of
Li :1s 2 2s1 → Li + :1s 2 → Li 2+ : 1s1
2s electron 1s electron

Removal of 1s electrons from Li+ and that too from completely filled configuration requires much
more energy and a jump in 2nd ionisation is noticed

(iii) Lower are ionisation energy values, greater is the tendency to lose ns1 electron to change in M+
ion (i.e. M → M+ + e) and therefore stronger is electropositive character.

(iv) Electropositive character increases from Li to Cs.

Due to their strong electropositive character, they emit electrons even when exposed to light showing
photoelectric effect. This property is responsible for the use of Cs and K in photoelectric cell.

(6) Oxidation number and valency

(i) These elements easily form univalent + ve ion by losing solitary ns1 electron due to low ionisation
energy values.

(ii) Alkali metals are univalent in nature and form ionic compounds. Lithium salts are, however,

(iii) Further, the M+ ion acquires the stable noble gas configuration. It requires very high values of
energy to pull out another electron from next to outer shell of M+ ion and that is why their second
ionisation energy is very high. Consequently, under ordinary conditions, it is not possible for these metals
to form M2+ ion and thus they show +1 oxidation state.

(iv) Since the electronic configuration of M+ ions do not have unpaired electron and thus alkali metal
salts are diamagnetic and colouress. Only those alkali metal salts are coloured which have coloured anions
e.g. K2Cr2O7 is orange because of orange coloured Cr2O72- ion, KMnO4 is violet because of violet coloured
MnO41- ion.

(7) Hydration of Ions

(i) Hydration represents for the dissolution of a substance in water to get adsorb water molecule by
weak valency force. Hydration of ions is the exothermic process when ions on dissolution water get

(ii) The hydration is an exothermic process i.e energy is released during hydration.

(iii) The energy released when 1 mole of an ion in the gaseous state is dissolved in water to get it

hydrated is called hydration energy M(g) + + Aq → M+ ; ∆H = – energy.
( aq )

(iv) Smaller the cation, greater is the degree of hydration. Hydration energy, Li+ > Na+ > K+ > Rb+
> Cs+

(v) Li+ being smallest in size has maximum degree of hydration and that is why lithium salts are

mostly hydrated, LiCl. 2H2O Also lithuim ion being heavily hydrated, moves very slowly under the
influence of electric field and, therefore, is the poorest conductor current among alkali metals ions It may,

therefore, be concluded that it is the degree of hydration as well as the size of ion is responsible for the

current carried by an ion.

Relative ionic radii Cs+ > Rb+ > K+ > Na+ > Li+

Relative hydrated ionic radii Li+ > Na+ > K+ > Rb+ > Cs+

Relative conducting power Cs+ > Rb+ > K+ > Na + > Li+

(8) Electronegativities
(i) These metals are highly electropositive and thereby possess low values of electronegativities.

(ii) Electronegativity of alkali metals decreases down the group as the trend of numerical values of
electronegativity given below on Pauling scale suggests.

Li Na K Rb Cs Fr

Electronegativity 0.98 0.93 0.82 0.82 0.79 –

Note :  Fr being radioactive elements and thus studies on physical properties of this element

are limited.

(9) Specific heat : It decreases from Li to Cs.

Li Na K Rb Cs


Specific heat (Cal/g) 0.941 0.293 0.17 0.08 0.049

(10) Conduction power : All are good conductors of heat & electricity, because of loosely held
valence electrons.

(11) Standard oxidation potential and reducting properties

(i) Since alkali metals easily lose ns1 electron and thus they have high values of oxidation potential

M + aq → M+ +e
( aq )

(ii) The standard oxidation potentials of a alkali metals (in volts) are listed below,

Li Na K Rb Cs

+3.05 +2.71 +2.93 +2.99 +2.99

(iii) More is oxidation potential, more is the tendency to get oxidized and thus more powerful is
reducing nature in aqueous medium. That is why alkali metals liberate H2 from H2O and HCl.

2H 2O + 2M → 2MOH + H 2 ; 2HCl + 2M → 2MCl + H 2

(iv) However, an examination of ionisation energy for alkali metals reveals that Li should have the

minimum tendency to lose electron and thus its reducing nature should be minimum. The greatest

reducing nature of Li in aq. medium is accounted due to the maximum hydration energy of Li+ ion. For

Li(s) → Li(g) ; ∆H1 = Heat of sublimation, ∆Hs

Li(g) → Li + + e; ∆H2 = IE1

Li → Li+ + ∆H3 = – Heat of hydration, ∆Hh
( aq );

Li(s) + H 2O → Li + (aq) + e; ∆H = ∆H1 + ∆H 2 + ∆H 3 = ∆H s + IE1 − ∆H h

Similarly, for sodium, Na(s) + H2O → Na + + e; ∆H = ∆H5 + IE1 − ∆Hh
( sq )

∆Hh for Li > ∆Hh for Na. Therefore, large negative ∆H values are observed in case of Li and this
explains for more possibility of Li to get itself oxidized or have reducing nature.

(12) Characteristic flame colours : The alkali metals and their salts give characteristic colour to
Bunsen flame. The flame energy causes and excitation of the outermost electron which on reverting back
to its initial position gives out the absorbed energy as visible light. These colour differ from each other Li –
crimson, Na–Golden yellow, K – Pale violet , Rb and Cs –violet. These different colours are due to
different ionisation energy of alkali metals. The energy released is minimum in the case of Li+ and
increases in the order.

Energy released : Li+ < Na+ < K+ < Rb+ < Cs+

λ released : Li+ > Na+ > K+ > Rb+ > Cs+

Frequency released : Li+ < Na+ < K+ < Rb+ < Cs+

Chemical properties

(1) Occurrence : Alkali metals are very reactive and thus found in combined state Some important

ores of alkali metals are given ahead.

(i) Lithuim : Triphylite, Petalite, lepidolite, Spodumene [LiAl(SiO3)3], Amblygonite [Li(Al F)PO4]
(ii) Sodium : Chile salt petre (NaNO3), Sodium chloride (NaCl), Sodium sulphate (Na2SO4), Borax
(Na2B4O710H2O), Glauber salt (Na2 SO4.10H2O)
(iii) Potassium : Sylime (KCl), carnallite (KCl.MgCl2.6H2O)and Felspar (K2O.Al2O3.6SiO2)
(iv) Rubidium : Lithuim ores Lepidolite, triphylite contains 0.7 to 3% Rb2 O
(v) Caesium : Lepidolite, Pollucite contains 0.2 to 7% Cs2O
(2) Extraction of alkali metals : Alkali metals cannot be extracted by the usual methods for the

extraction of metals due to following reasons.

(i) Alkali metals are strong reducing agents, hence cannot be extracted by reduction of their oxides or

other compounds.

(ii) Being highly electropositive in nature, it is not possible to apply the method of displacing them

from their salt solutions by any other element.

(iii) The aqueous solutions of their salts cannot be used for extraction by electrolytic method because

hydrogen ion is discharged at cathode instead of an alkali metal ions as the discharge potentials of alkali

metals are high. However, by using Hg as cathode, alkali metal can be deposited. The alkali metal readily

combines with Hg to form an amalgam from which its recovery difficult. The only successful method,

therefore, is the electrolysis of their fused salts, usually chlorides. Generally, another metal chloride is

added to lower their fussion temperature.

Fused NaCl : NaCl fusion → Na + + Cl – Electrolysis : Anode : 2Cl − → Cl2 + 2e
of fused salt : Cathode : 2Na + + 2e → 2Na

(3) Alloys Formation
(i) The alkali metals form alloys among themselves as well as with other metals.
(ii) Alkali metals also get dissolved in mercury to form amalgam with evolution of heat and the
amalgamation is highly exothermic .

(4) Formation of oxides and hydroxides

(i) These are most reactive metals and have strong affinity for O2 quickly tranish in air due to the
formation of a film of their oxides on the surface. These are, therefore, kept under kerosene or paraffin oil to
protect them from air,

M + O2 → M 2O → M 2O2
Oxide Peroxide

(ii) When burnt air (O2), lithium forms lithium oxide (Li2O) sodium forms sodium peroxide (Na2O2)

and other alkali metals form super oxide (Mo2 i.e. KO2,RbO2 or CsO2)

2Li + 1 O2 → Li2O ; 2Na + O2 → Na2 O2 ; K + O2 → KO2
2 Lithuim oxide Potassium super oxide

The reactivity of alkali metals towards oxygen to form different oxides is due to strong positive

field around each alkali metal cation. Li + being smallest, possesses strong positive field and thus combines
with small anion O2- to form stable Li2O compound. The Na+ and K+ being relatively larger thus exert less
strong positive field around them and thus reacts with larger oxygen anion i.e, O22−and O21− to form stable

The monoxide, peroxides and superoxides have O2 and O22− ,O21− ions respectively. The structures of

each are, [ •• ]2− [ •• •• ]2– •• •• ]1−
O•• ; O– O•• , [ O• O••
• • • • •
• • •

×× •• ••

The O2–1 ion has a three electron covalent bond and has one electron unpaired . It is therefore
superoxides are paramagnetic and coloured KO2 is light yellow and paramagnetic substance.

(iii) The oxides of alkali metals and metal itself give strongly alkaline solution in water with evolution

of heat

M + H2O → MOH + 1 H2; ∆H = −ve

Li2O + H 2O → 2LiOH; ∆H = −ve

Na2O2 + 2H 2O → 2NaOH + H 2O2(l) ; ∆H = −ve

2KO2 + 2H 2O → 2KOH + H 2O2(l) + O2(g) ; ∆H = −ve

The peroxides and superoxides act as strong oxidising agents due to formation of H2O2

(iv) The reactivity of alkali metals towards air and water increases from Li to Cs that is why lithium
decomposes H2O very slowly at 25oC whereas Na does so vigorously, K reacts producing a flame and Rb,
Cs do so explosively.

M + H2O → MOH + 1 H2

(v) The basic character of oxides and hydroxides of alkali metals increases from Li to Cs. This is due
to the increase in ionic character of alkali metal hydroxides down the group which leads to complete
dissociation and leads to increase in concentration of OH- ions.

(5) Hydrides
(i) These metal combines H to give white crystalline ionic hydrides of the general of the formula MH.
(ii) The tendency to form their hydrides, basic character and stability decreases from Li to Cs since

the electropositive character decreases from Cs to Li.

2M+ H2 → 2MH ; Reactivity towards H2 is Cs < Rb < K < Na < Li
(iii) The metal hydrides react with water to give MOH & H2 ; MH + H2 O → MOH + H2
(iv) The ionic nature of hydrides increases from Li to Cs because of the fact that hydrogen is present
in the these hydrides as H- and the smaller cation will produce more polarisation of anion (according to
Fajan rule) and will develop more covalent character.

(v) The electrolysis of fused hydrides give H2 at anode. NaH fused Contains Na + and H −i.e.,

At cathode: Na+ +e → Na ; At anode: H− → 1 H 2 +e

(vi) Alkali metals also form hydrides like NaBH4, LiAlH4 which are good reducing agent.

(6) Carbonates and Bicarbonates

(i) The carbonates (M2CO3) & bicarbonates (MHCO3) are highly stable to heat, where M stands for
alkali metals.

(ii) The stability of these salts increases with the increasing electropositive character from Li to Cs. It
is therefore Li2 CO3 decompose on heating, Li2CO3→ Li2O+CO2

(iii) Bicarbonates are decomposed at relatively low temperature,

2MHCO3 3000C → M 2CO3 + H 2O + CO2

(iv) Both carbonates and bicarbonates are soluble in water to give alkaline solution due to hydrolysis
of carbonate ions or bicarbonate ions.

(7) Halides

(i) Alkali metals combine directly with halogens to form ionic halide M + X − .
(ii) The ease with which the alkali metals form halides increases from Li to Cs due to increasing
electropositive character from Li to Cs.

(iii) Lithium halides however have more covalent nature. Smaller is the cation, more is deformation
of anion and thus more is covalent nature in compound. Also among lithium halides, lithium iodide has
maximum covalent nature because of larger anion which is easily deformed by a cation (The Fajan’s rule)
Thus covalent character in lithium halides is, LiI > LiBr > LiCl > LiF

(iv) These are readily soluble in water. However, lithium fluoride is sparingly soluble. The low
solubility of LiF is due to higher forces of attractions among smaller Li+ and smaller F- ions (high lattice

(v) Halides having ionic nature have high and good conductor of current. The melting points of
halides shows the order, NaF > NaCl > NaBr > Nal

(vi) Halides of potassium, rubidium and caesium have a property of combining with extra halogen
atoms forming polyhalides.

KI + I2 →KI3 ; In KI3(aq) the ions K+ and I–3 are present

(8) Solubility in liquid NH3

(i) These metals dissolve in liquid NH3 to produce blue coloured solution, which conducts electricity
to an appreciable degree.

(ii) With increasing concentration of ammonia, blue colour starts changing to that of metallic copper

after which dissolution of alkali metals in NH3 ceases.

(iii) The metal atom is converted into ammoniated metal in i.e. M+ (NH3) and the electron set free
combines with NH3 molecule to produce ammonia solvated electron.

Na + (x + y) → NH 3[Na(NH 3 )x ]+ + [e(NH 3 )y ]−
Ammoniated cation Ammoniated electron

(iv) It is the ammoniated electron which is responsible for blue colour, paramagnetic nature and
reducing power of alkali metals in ammonia solution. However, the increased conductance nature of these
metals in ammonia is due to presence of ammoniated cation and ammonia solvated electron.

(v) The stability of metal-ammonia solution decreases from Li to Cs.

(vi) The blue solution on standing or on heating slowly liberates hydrogen, 2M + 2NH3 →2MNH2 +
H2 . Sodamide (NaNH2) is a waxy solid, used in preparation of number of sodium compounds.

(9) Nitrates : Nitrates of alkali metals (MNO3) are soluble in water and decompose on heating.
LiNO3 decomposes to give NO2 and O2 and rest all give nitrites and oxygen.

2MNO3 → 2MNO2 + O2 (except Li) ; 4 LiNO3 →2Li2O + 4NO2 + O2

(10) Sulphates

(i) Alkali metals’ sulphate have the formula M2SO4 .
(ii) Except Li2SO4, rest all are soluble water.

(iii) These sulphates on fusing with carbon form sulphides, M2SO4 + 4C → M2S + 4CO
(iv) The sulphates of alkali metals (except Li) form double salts with the sulphate of the trivalent

metals like Fe, Al, Cr etc. The double sulphates crystallize with large number of water molecules as alum. e.g.

K2SO4 . Al2 (SO4)3. 24 H2O.

(11) Reaction with non-metals

(i) These have high affinity for non-metals. Except carbon and nitrogen, they directly react with
hydrogen, halogens, sulphur, phosphorus etc. to form corresponding compounds on heating.

2Na + H2 3000C → 2NaH ; 2K + H2→2KH

2Na + Cl2 → 2NaCl ; 2K + Cl2 → 2KCl

2Na + S → Na2S ; 2K + S → K2S

3Na + P → Na3P ; 3K + P → K3P

(ii) Li reacts, however directly with carbon and nitrogen to form carbides and nitrides.

2Li + 2C → LiC2 ; 6Li + 2N2 → 2 Li3N
(iii) The nitrides of these metals on reaction with water give NH3. M3N + 3H2O → 3MOH + NH3

(12) Reaction with acidic hydrogen : Alkali metals react with acids and other compounds

containing acidic hydrogen (i.e, H atom attached on F,O, N and triply bonded carbon atom, for example,

HF, H2O, ROH, RNH2, CH≡CH) to liberate H2 .

M + H2O → MOH + 1 H2 ; M + HX → MX + 1 H2
2 2

M + ROH → ROH + 1 H 2 ; M + RNH 2 → RNHNa + 1 H2
2 2

(13) Complex ion formation : A metal shows complex formation only when it possesses the
following characteristics, (i) Small size (ii) High nuclear charge (iii) Presence of empty orbitals in order to
accept electron pair ligand. Only Lithium in alkali metals due to small size forms a few complex ions Rest
all alkali metals do not possess the tendency to form complex ion.

Anomalous behaviour of Lithium

Anomalous behaviour of lithium is due to extremely low size of lithium its cation On account of small
size and high nuclear charge, lithium exerts the greatest polarizing effect out of all alkali metals on negative
ion. Consequently lithium ion possesses remarkable tendency towards solvation and develops covalent
character in its compounds. Li differs from other alkali metals in the following respects,

(1) It is comparatively harder than other alkali metals.

(2) It can be melted in dry air without losing its brilliance.

(3) Unlike other alkali metals, lithium is reactive among all. It can be noticed by the following

(i) It is not affected by air. (ii) It decomposes water very slowly to liberate H2. (iii) It hardly reacts with
bromine while other alkali metals react violently.

(4) Lithium is the only alkali metal which directly reacts with N2.

(5) Lithium when heated in NH3 forms imide, Li2 NH while other metals form amides, MNH2.

(6) When burnt in air,, lithium form Li2O sodium form Na2O and Na2O2 other alkali metals form
monoxide, peroxide and superoxide.

(7) Li2O is less basic and less soluble in water than other alkali metals.
(8) LiOH is weaker base than NaOH or KOH and decomposes on heating.

2LiOH ∆ → Li2O + H 2O

(9) LiHCO3 is liquid while other metal bicarbonates are solid.

(10) Only Li2CO3 decomposes on heating Li2CO3 heat → Li2O + CO2 . Na2CO3, K2CO3 etc. do
not decompose on heating.

(11) LiNO3 and other alkali metal nitrates give different products on heating

4LiNO3 = 2Li2O + 4NO2 + O2 ; 2NaNO3 = 2NaNO2 + O2

(12) LiCl and LiNO3 are soluble in alcohol and other organic solvents. These salts of other alkali
metals are, however, insoluble in organic solvents.

(13) LiCl is deliquescent while NaCl, KBr etc. are not. Lithium chloride crystals contain two
molecules of water of crystallisation (LiCl. 2H2O). Crystals of NaCl KBr, KI etc do not conation water of

(14) Li2SO4 does not form alums like other alkali metals.

(15) Li reacts with water slowly at room temperature Na reacts vigorously Reaction with K. Rb and
Cs is violent.

(16) Li reacts with Br2 slowly. Reaction of other alkali metals with Br2 is fast.

(17) Li2 CO3 Li2C2O4, LiF , Li3PO4 are the only alkali metal salts which are insoluble or sparingly soluble
in water.

Diagonal Relationship of Li with Mg

Due to its small size lithium differs from other alkali metals but resembles with Mg as its size is closer
to Mg Its resemblance with Mg is known as diagonal relationship. Generally the periodic properties show
either increasing or decreasing trend along the group and vice versa along the period which brought the
diagonally situated elements to closer values. Following are the characteristic to be noted.

Period Group I Group II

2 Li Be

3 Na Mg

(1) Both Li and Mg are harder and higher than the other metals of their groups.

(2) Due to covalent nature, chlorides of both Li and Mg are deliquescent and soluble in alcohol and
pyridine while chlorides of other alkali metals are not so.

(3) Fluorides, phosphates of li and Mg are sparingly soluble in water whereas those of other alkali
metals are soluble in water.

(4) Carbonates of Li and Mg decompose on heating and liberate CO2 Carbonates of other alkali
metals are stable towards heat and decomposed only on fusion.

Li2CO3 → Li2O + CO2 ; Mg CO3 → MgO + CO2
(5) Hydroxides and nitrates of both Li and Mg decompose on heating to give oxide. Hydroxides of
both Li and Mg are weak alkali.

4 LiNO3 → 2Li2O + 4NO2 + O2 ; 2Mg(NO3)2 →2MgO + 4NO2 + O2

2LiOH → Li2O + H2O ; Mg(OH)2 → MgO + H2O
Hydroxides of other alkali metals are stable towards heat while their nitrates give O2 and nitrite.

2KNO3 → 2KNO2 + O2
(6) Both Li and Mg combine directly with N2 to give nitrides Li3 N and Mg3 N2. Other alkali metals
combine at high temperature, 6Li + N2 → 2Li3N; 3Mg + N2 → Mg3 N2. Both the nitrides are decomposed
by water to give NH3

Li3N + 3H2O → 3LiOH + NH3 ; Mg3N2 + 6H2o → 3Mg(OH)2+ 2NH3
(7) Bicarbonates of Li and Mg are more soluble in water than carbonates whereas carbonates
whereas carbonates of alkali metals are more soluble.

(8) Both Li and Mg combine with carbon on heating.

2Li + 2C → Li2C2 ; Mg + 2C → Mg C2
(9) The periodic properties of Li and Mg are quite comparable

Li Mg

Electronegativity 1.0 1.2

Atomic radii 1.23 1.36

Ionic radii 0.60(Li+) 0.65(Mg+2)

Atomic volume 12.97 c.c 13.97 c.c

(10) Both have high polarizing power. Polarizing Power = Ionic charge / (ionic radius)2.

(11) Lithium and Mg Form only monooxide on heating in oxygen.

4Li + O2 → 2 Li2O ; 2Mg + O2 → 2 MgO
(12) Li2SO4 Like MgSO4 does not form alums.
(13) The bicarbonates of Li and Mg do not exist in solid state, they exist in solution only.

(14) Alkyls of Li and Mg (R.Li and R. MgX) are soluble in organic solvent.

(15) Lithium chloride and MgCl2 both are deliquescent and separate out from their aqueous solutions
as hydrated crystals, LiCl. 2H2O and MgCl2 . 2H2O.

Sodium and its compounds

(1) Ores of sodium : NaCl (common salt), NaNO3 (chile salt petre), Na2SO4 .10H 2O (Glauber's
salt), borax (sodium tetraborate or sodium borate, Na2 B4 O7 .10H 2O) .

(2) Extraction of sodium : It is manufactured by the electrolysis of fused sodium chloride in the
presence of CaCl2 and KF using graphite anode and iron cathode. This process is called Down process.

NaCl ⇌ Na + + Cl − .
At cathode : Na + + e − → Na ; At anode : Cl − → Cl + e − ; Cl + Cl → Cl2 ↑

Note :  Sodium cannot be extracted from aqueous NaCl because E0 (–0.83V) is more
H2O / H2

than E 0 Na + / Na (–2.71V).

 Anode and cathode are separated by means of a wire gauze to prevent the reaction between Na
and Cl2 .

(3) Compound of sodium
Sodium hydroxide (Caustic soda), NaOH

(i) Preparation

(a) Gossage process : Na 2soCluOtio3n)+ Ca(OH)2 → 2NaOH ↓ +CaCO3


(b) Electrolytic method : Caustic soda is manufactured by the electrolysis of a concentrated solution
of NaCl.

At anode: Cl − discharged; At cathode: Na + discharged

(c) Castner - Kellener cell (Mercury cathode process) : NaOH obtained by electrolysis of aq.
solution of brine. The cell comprises of rectangular iron tank divided into three compartments.

Outer compartment – Brine solution is electrolysed ; Central compartment – 2% NaOH solution
and H 2

(ii) Properties : White crystalline solid, highly soluble in water, It is only sparingly soluble in alcohol.

(a) Reaction with salt : FeCl3 + 3NaOH → Fe(OH)3 ↓ + 3NaCl
(Insoluble hydroxide)

HgCl2 + 2NaOH → 2NaCl + Hg(OH) 2 → H 2O + HgO ↓

unstable yellow

AgNO3 + 2NaOH → 2NaNO3 + 2AgOH → Ag 2O ↓ +H 2O

Note :  Zn, Al, Sb, Pb, Sn and As forms insoluble hydroxide which dissolve in excess of NaOH

(amphoteric hydroxide).

 NH 4 Cl + NaOH heat → NaCl + NH 3 ↑ +H 2O
(b) Reaction with halogens : X 2 + 2NaOH (cold) → NaX + NaXO + H 2O

sod. hypohalite

3X 2 + 6NaOH (hot) → 5NaX + NaXO3 + 3H 2O ; (X = Cl, Br, I)
(Sod. halate)

(c) Reaction with metals: Weakly electropositive metals like Zn, Al and Sn etc.

Zn + 2NaOH → Na2 ZnO2 + H 2 ↑
(d) Reaction with sand, SiO2 : 2NaOH + SiO2 → Na2SiO3 + H 2O

Sod. silicate (glass)

(e) Reaction with CO: NaOH + CO 150−200oC → SHoCd.OfoOrmNatae
5−10 atm

Note :  NaOH breaks down the proteins of the skin flesh to a pasty mass, therefore it is

commonly known as caustic soda.
(iii) Uses : In the manufacturing of sodium metal, soap, rayon, paper, dyes and drugs. For
mercuring cotton to make cloth unshrinkable and reagent in lab.

Sodium carbonate or washing soda, Na 2CO3

(i) Preparation : Solvay process : In this process, brine (NaCl) , NH3 and CO2 are the raw

NH 3 + CO2 + H 2O → NH 4 HCO3
NH 4 HCO3 + NaCl 30oC → NaHCO3 ↓ + NH 4 Cl
2NaHCO3 250oC → Na2CO3 + H 2O + CO2

2NH 4 Cl + Csala(kOedH)2 → CaCl 2 + 2H 2O + 2NH 3


Note :  CaCl2 so formed in the above reaction is a by product of solvay process.

(ii) Properties : (a) Na(2dCecOah3y.d1r0atHe)2O dryair → N(Ma2oCnoOhy3d.Hrat2eO) + 9H 2O

Na2CO3 . H 2O ∆ → Na 2 CO3 → It does not decompose on further
heating even to redness ( 853o C)

(b) It is soluble in water with considerable evolution of heat.

Na2CO3 + H 2O → H 2CO3 + 2Na + + 2OH −
Weak acid

(c) It is readily decomposed by acids with the evolution of CO2 gas.

(d) Na2CO3 + H 2O + CO2 → 2NaHCO3

(iii) Uses : In textile and petroleum refining, Manufacturing of glass, NaOH soap powders etc.
Sodium peroxide (Na2O2)

(i) Preparation : It is manufactured by heating sodium metal on aluminium trays in air (free
from CO2 )

2Na + O2 (air) ∆ → Na2O2
(ii) Properties : (a) When pure it is colourless. The faint yellow colour of commercial product is
due to presence of small amount of superoxide (NaO2 ).

(b) On coming with moist air it become white due to formation of NaOH and Na2CO3 .

2Na2O2 + 2H 2O → 4 NaOH + O2 ; 2NaOH + CO2 → Na2CO3 + H 2O

(c) It is powerful oxidising agent. It oxidises Cr (III) hydroxide to sodium chromate, Mn (II) to sodium
manganate and sulphides to sulphates.

(iii) Uses : As a bleaching agent and it is used for the purification of air in confined spaces such as
submarines because it can combine with CO2 to give Na2CO3 and oxygen,
2CO2 + 2Na2O2 → 2Na2CO3 + O2 .

Alkaline Earth Metals and Their Compounds .

The group 2 of the periodic table consists of six metallic elements. These are beryllium (Be),
magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba) and radium (Ra). These (except Be) are
known as alkaline earth metals as their oxides are alkaline and occur in earth crust.

Electronic configuration

Element Electronic configurations ( ns 2 )

4 Be 1s 2 2s 2 or [He] 2s 2

12 Mg 1s 2 2s 2 2p6 3s 2 or [Ne]3s 2

20 Ca 1s 2 2s 2 2p6 3s 2 3p6 4s 2 or [Ar]4s 2

38 Sr 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4 p6 5s 2 or [Kr]5s 2

56 Ba 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4 p6 4d10 5s 2 5 p6 6s 2 or [Xe]6s 2

88 Ra 1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4 p6 4d10 4 f 14 5s 2 5 p6 5d10 6s 2 6 p6 7s 2 or [Rn]7s 2

Note :  Radium was discovered in the ore pitch blende by madam Curie. It is radioactive in

Physical properties

(1) Physical state : All are greyish-white, light, malleable and ductile metals with metallic lustre.
Their hardness progressively decrease with increase in atomic number. Althought these are fairly soft but
relatively harder than alkali metals.

(2) Atomic and ionic radii

(i) The atomic and ionic radii of alkaline earth metals also increase down the group due to

progressive addition of new energy shells like alkali metals.

Be Mg Ca Sr Ba Ra

Atomic radius (pm) 112 160 197 215 222 –

Ionic radius of M2+ ion (pm) 31 65 99 113 135 140

(ii) The atomic radii of alkaline earth metals are however smaller than their corresponding alkali metal of

the same period. This is due to the fact that alkaline earth metals possess a higher nuclear charge than

alkali metals which more effectively pulls the orbit electrons towards the nucleus causing a decrease in


(3) Density

(i) Density decreases slightly upto Ca after which it increases. The decrease in density from Be to Ca
might be due to less packing of atoms in solid lattice of Mg and Ca.

Be Mg Ca Sr Ba Ra

1.84 1.74 1.55 2.54 3.75 6.00

(ii) The alkaline earth metals are more denser, heavier and harder than alkali metal. The higher
density of alkaline earth metals is due to their smaller atomic size and strong intermetallic bonds which
provide a more close packing in crystal lattice as compared to alkali metals.

(4) Melting point and Boiling point
(i) Melting points and boiling points of alkaline earth metals do not show any regular trend.

Be Mg Ca Sr Ba Ra (K) 1560 920 1112 1041 1000 973 (K) 2770 1378 1767 1654 1413 –

(ii) The values are, however, more than alkali metals. This might due to close packing of atoms in

crystal lattice in alkaline earth metals.

(5) Ionisation energy and electropositive or metallic character

(i) Since the atomic size decreases along the period and the nuclear charge increases and thus the

electrons are more tightly held towards nucleus. It is therefore alkaline earth metals have higher ionisation

energy in comparison to alkali metals but lower ionisation energies in comparison to p-block elements.

(ii) The ionisation energy of alkaline earth metals decreases from Be to Ba.

Be Mg Ca Sr Ba Ra

First ionisation energy (k J mol-1) 899 737 590 549 503 509

Second ionisation energy (kJ mol-1) 1757 1450 1146 1064 965 979

(iii) The higher values of second ionisation energy is due to the fact that removal of one electron

from the valence shell, the remaining electrons are more tightly held in which nucleus of cation and thus

more energy is required to pull one more electron from monovalent cation.

(iv) No doubt first ionisation energy of alkaline earth metals are higher than alkali metals but a closer

look on 2nd ionisation energy of alkali metals and alkaline earth metals reveals that 2nd ionisation energy

of alkali metals are more

Li Be

1st ionisation energy (kJ mol-1) 520 899

2nd ionisation energy (kJ mol-1) 7296 1757

This may be explained as, Li : 1s2, 2s1 removal of 2s → Li +: 1s2 removal of1s → Li2+ : 1s1
electron electron

Be : 1s2 , 2s2 removal of 2s → Be+ : 1s2, 2s1 removal of 2s → Be2+ : 1s2
electron electron

The removal of 2nd electron from alkali metals takes place from 1s sub shell which are more closer to

nucleus and exert more nuclear charge to hold up 1 s electron core, whereas removal of 2nd electron from

alkaline earth metals takes from 2s sub shell. More closer are shells to the nucleus, more tightly are held

electrons with nucleus and thus more energy is required to remove the electron.

(v) All these possess strong electropositive character which increases from Be to Ba.

(vi) These have less electropositive character than alkali metals as the later have low values of

ionisation energy.

(6) Oxidation number and valency

(i) The IE1 of the these metals are much lower than IE1 and thus it appears that these metals should
form univalent ion rather than divalent ions but in actual practice, all these give bivalent ions. This is due
to the fact that M2+ ion possesses a higher degree of hydration or M2+ ions are extensively hydrated to
form [M(H2O)x]2+, a hydrated ion. This involves a large amount of energy evolution which counter
balances the higher value of second ionisation energy.

M →M2+ , ∆H = IE1 + E2

M2+ + xH2O → [M(H2O)x]2+; ∆H = – hydration energy.

(ii) The tendency of these metals to exist as divalent cation can thus be accounted as,

(a) Divalent cation of these metals possess noble gas or stable configuration.

(b) The formation of divalent cation lattice leads to evolution of energy due to strong lattice structure
of divalent cation which easily compensates for the higher values of second ionisation energy of these

(c) The higher heats of hydration of divalent cation which accounts for the existence of the divalent

ions of these metals in solution state.

(7) Hydration of ions

(i) The hydration energies of alkaline earth metals divalent cation are much more than the hydration

energy of monovalent cation.

Mg+ Mg2+

Hydration energy or Heat of hydration (kJ mol-1) 353 1906

The abnormally higher values of heat of hydration for divalent cations of alkaline earth metals are
responsible for their divalent nature. MgCl2 formation occurs with more amount of heat evolution and thus
MgCl2 is more stable.

(ii) The hydration energies of M2+ ion decreases with increase in ionic radii.

Be2+ Mg2+ Ca2+ Sr2+ Ba2+

Heat of hydration kJ mol-1 2382 1906 1651 1484 1275

(iii) Heat of hydration are larger than alkali metals ions and thus alkaline earth metals compounds
are more extensively hydrated than those of alkali metals e.g MgCl2 and CaCl2 exists as Mg Cl2 .6H2O and
CaCl2. 6H2O which NaCl and KCl do not form such hydrates.

(iv) The ionic mobility, therefore, increases from Ba2+ to Ba2+, as the size of hydrated ion decreases.

(8) Electronegativities
(i) The electronegativities of alkaline earth metals are also small but are higher than alkali metals.
(ii) Electronegativity decreases from Be to Ba as shown below,

Be Mg Ca Sr Ba

Electronegativity 1.57 1.31 1.00 0.95 0.89

(9) Conduction power : Good conductor of heat and electricity.

(10) Standard oxidation potential and reducing properties
(i) The standard oxidation potential (in volts) are,

Be Mg Ca Sr Ba

1.69 2.35 2.87 2.89 2.90

(ii) All these metals possess tendency to lose two electrons to give M2+ ion and are used as reducing

(iii) The reducing character increases from Be to Ba, however, these are less powerful reducing agent
than alkali metals.

(iv) Beryllium having relatively lower oxidation potential and thus does not liberate H2 from acids.

(11) Characteristic flame colours
(i) The characteristic flame colour shown are : Ca-brick red; Sr –crimson ; Ba-apple green and Ra-
(ii) Alkaline earth metals except Be and Mg produce characteristic colour to flame due to easy
excitation of electrons to higher energy levels.
(iii) Be and Mg atoms due to their small size, bind their electrons more strongly (because of higher
effective nuclear charge) Hence these requires high excitation energy and are not excited by the energy of
flame with the result that no flame colour is shown by them.
Chemical properties
(1) Occurrence : These are found mainly in combined state such as oxides, carbonates and
sulphates Mg and Ca are found in abundance in nature. Be is not very abundant, Sr and Ba are less
abundant. Ra is rare element. Some important ores of alkaline earth metals are given below,
(i) Baryllium : Beryl (3BeO.Al2O3.6SiO2); Phenacite (Be2SiO4)
(ii) Magnesium : Magnesite (MgCO3); Dolomite (CaCO3. MgCO3); Epsomite(MgSO4. 7H2O);
Carnallite (MgCl2.KCl. 6H2O); Asbestos [CaMg3(SiO3)4]
(iii) Calcium : Limestone (CaCO3); Gypsum : (CaSO4.2H2O), Anhydrite (CaSO4); Fluorapatite
[(3Ca3(PO4)2.CaF2)] Phosphorite rock [Ca3(PO4)2]
(iv) Barium : Barytes (BaSO4) ; witherite (BaCO3)

(v) Radium : Pitch blende (U3O8); (Ra in traces); other radium rich minerals are carnotite [K2UO2)]
(VO4)2 8H2O and antamite [Ca(UO2)2]

(2) Extraction of alkaline earth metals
(i) Be and Mg are obtained by reducing their oxides carbon,

BeO + C → Be + CO ; MgO + C → Mg + CO
(ii) The extraction of alkaline earth metals can also be made by the reduction of their oxides by
alkali metals or by electrolysing their fused salts.

(3) Alloy formation : These dissolve in mercury and form amalgams.

(4) Formation of oxides and hydroxides

(i) The elements (except Ba and Ra) when burnt in air give oxides of ionic nature M2+O2- which are
crystalline in nature . Ba and Ra however give peroxide. The tendency to form higher oxides increases
from Be to Ra.

2M + O2 → 2MO (M is Be, Mg or Ca )

2M + O2 → MO2 (M is Ba or Sr)

(ii) Their less reactivity than the alkali metals is evident by the fact that they are slowly oxidized on

exposure to air, However the reactivity of these metals towards oxygen increases on moving down the


(iii) The oxides of these metals are very stable due to high lattice energy.

(iv) The oxides of the metal (except BeO and MgO) dissolve in water to form basic hydroxides and
evolve a large amount of heat. BeO and MgO possess high lattice energy and thus insoluble in water.

(v) BeO dissolves both in acid and alkalies to give salts i.e. BeO possesses amphoteric nature.

BeO + 2NaOH → Na2BeO2 + H2O ; BeO + 2HCl → BeCl2 + H2O

Sod. beryllate Beryllium chloride

(vi)The basic nature of oxides of alkaline earth metals increases from Be to Ra as the electropositive
Character increases from Be to Ra.

(vii)The tendency of these metal to react with water increases with increase in electropositive
character i.e. Be to Ra.

(viii) Reaction of Be with water is not certain, magnesium reacts only with hot water, while other
metals react with cold water but slowly and less energetically than alkali metals.

(ix) The inertness of Be and Mg towards water is due to the formation of protective , thin layer of
hydroxide on the surface of the metals.

(x) The basic nature of hydroxides increase from Be to Ra. It is because of increase in ionic radius
down the group which results in a decrease in strength of M –O bond in M –(OH)2 to show more
dissociation of hydroxides and greater basic character.

(xi) The solubility of hydroxides of alkaline earth metals is relatively less than their corresponding
alkali metal hydroxides Furthermore, the solubility of hydroxides of alkaline earth metals increases from Be
to Ba. Be (OH)2 and Mg (OH)2 are almost insoluble, Ca (OH)2 (often called lime water) is sparingly soluble
whereas Sr(OH) 2 and Ba (OH)2 (often called baryta water ) are more soluble.

The trend of the solubility of these hydroxides depends on the values of lattice energy and hydration
energy of these hydroxides. The magnitude of hydration energy remains almost same whereas lattice

energy decreases appreciably down the group leading to more –Ve values for ∆H solution down the group.

∆H = ∆H + ∆Hsolution
lattice energy hydration energy

More negative is ∆H solution more is solubility of compounds.

(xii) The basic character of oxides and hydroxides of alkaline earth metals is lesser than their

corresponding alkali metal oxides and hydroxides.

(xiii) Aqueous solution of lime water [Ca(OH)2] or baryta water [Ba(OH)]2 are used to qualitative
identification and quantative estimation of carbon dioxide, as both of them gives white precipitate with

CO2 due to formation of insoluble CaCO3 or BaCO3

Ca(OH)2 + CO2 → CaCO3 + H2O ; Ba(OH)2 + CO2 → BaCO3 + H2O

(white ppt) (white ppt)

Note :  SO2 also give white ppt of CaSO3 and BaSO3 on passing through lime water or baryta

water However on passing CO2 in excess, the white turbidity of insoluble carbonates
dissolve to give a clear solution again due to the formation of soluble bicarbonates,

CaCO3 → H2O + CO2 → Ca(HCO3)2

(5) Hydrides

(i) Except Be, all alkaline earth metals form hydrides (MH2) on heating directly with H2 . M+ H2 →

(ii) BeH2 is prepared by the action of Li Al H4 On BeCl2 ; 2BeCl2 + LiAlH4 → 2BeH2 + LiCl +

(iii) BeH2 and MgH2 are covalent while other hydrides are ionic.

(iv) The ionic hydrides of Ca, Sr, Ba liberate H2 at anode and metal at cathode.
fusion Ca2+ + 2H–

Anode : 2H- →H2 + 2e Cathode : Ca2+ + 2e →Ca

(v) The stability of hydrides decreases from Be to Ba.

(vi) The hydrides having higher reactivity for water, dissolves readily and produce hydrogen gas.

CaH2(s) + 2H2O →Ca(OH) 2 + 2H2↑

(6) Carbonates and Bicarbonates

(i) All these metal carbonates (MCO3) are insoluble in neutral medium but soluble in acid medium .
These are precipitated by the addition of alkali metal or ammonium carbonate solution to the solution of
these metals.

(NH4)2 CO3 + CaCl2 →2NH4Cl + CaCO3 ; Na2CO3 + BaCl2 →2NaCl + BaCO3
(ii) Alkaline earth metal carbonates are obtained as white precipitates when calculated amount of
carbon dioxide is passed through the solution of the alkaline metal hydroxides.

M(OH)2 (aq) + CO2 (g) →MCO3(s) + H2O(l)

and sodium or ammonium carbonate is added to the solution of the alkaline earth metal salt such as

CaCl2 (aq) + Na2CO3 (aq) → CaCO3(s) +2 NaCl(aq)
(iii) Solubility of carbonates of these metals also decreases downward in the group due to the
decrease of hydration energy as the lattice energy remains almost unchanged as in case of sulphates.
(vi) The carbonates of these metals decompose on heating to give the oxides, the temperature of
decomposition increasing from Be to Ba. Beryllium carbonate is unstable.

MCO3 Heat → MO + CO2

(7) Halides

(i) The alkaline earth metals combine directly with halogens at appropriate temperatures forming
halides, MX2. These halides can also be prepared by the action of halogen acids (HX) on metals, metal
oxides, hydroxides and carbonates.

M + 2HX → MX2 + H2 ; MO + 2HX →MX2 + H2O
M(OH)2 + 2HX→ MX2 +2H2O ; MCO3 + 2HX → MX2 + CO2 + H2O
Beryllium chloride is however, conveniently obtained from oxide

BeO + C + Cl2 870−1070K → BeCl 2 + CO.

(ii) BeCl2 is essentially covalent, the chlorides MgCl2, CaCl2 , SrCl2 and BaCl2 are ionic; the ionic
character increases as the size of the metal ion increases. The evidence is provided by the following facts,

(a) Beryllium chloride is relatively low melting and volatile whereas BaCl2 has high melting and

(b) Beryllium chloride is soluble in organic solvents.

(iii) The halides of the members of this group are soluble in water and produce neutral solutions from

which the hydrates such : MgCl2 6H2O, CaCl2.6H2O.BaCl2 2H2O can be crystallised. The tendency to form
hydrated halides decreases with increasing size of the metal ions.

(iv) BeCl2 is readily hydrolysed with water to form acid solution, BeCl2 + 2H2O →Be (OH)2 + 2HCl.
(v) The fluorides are relatively less soluble than the chlorides due to high lattice energies. Except

BeCl2 and MgCl2 the chlorides of alkaline earth metals impart characteristic colours to flame.

CaCl2 SrCl2 BaCl2

Brick red colour Crimson colour Grassy green colour

Structure of BeCl2 In the solid phase polymeric chain structure with three centre 2 electron bonding
with Be-Cl-Be bridged structure is shown below,

202 PM Cl Be Cl Be Cl
Be 98 82 Cl
263 pm

In the vapour phase it tends to form a chloro-bridged dimer which dissociates into the linear
triatomic monomer at high temperature at nearly 1200 K.

(8) Solubility in liquid ammonia : Like alkali metals, alkaline earth metals also dissolve in liquid
ammonia to form coloured solutions When such a solution is evaporated, hexammoniate, M(NH3)6 is

(9) Nitrides
(i) All the alkaline earth metals directs combine with N2 give nitrides, M3N2.
(ii) The ease of formation of nitrides however decreases from Be to Ba.

(iii) These nitrides are hydrolysed water to liberate NH3, M3N2 + 6H2O →3M(OH)2 + 2NH3

(10) Sulphates
(i) All these form sulphate of the type M SO4 by the action of H2 SO4 on metals, their oxides,
carbonates or hydroxides.

M + H2SO4 → MSO4 + H2 ; MO + H2SO4 → MSO4 + H2O

MCO3+ H2SO4 → MSO4 + H2O+CO2 ; M(OH)2 + H2SO4 → MSO4 + 2H2O
(ii) The solubility of sulphates in water decreases on moving down the group BeSO4 and MgSO4 are
fairly soluble in water while BaSO4 is completely insoluble. This is due to increases in lattice energy of
sulphates down the group which predominates over hydration energy.

(ii) Sulphate are quite stable to heat however reduced to sulphide on heating with carbon.

MSO4 + 2C →MS+2CO2
(11) Action with carbon : Alkaline metals (except Be, Mg) when heated with carbon form carbides
of the type MC2 These carbides are also called acetylides as on hydrolysis they evolve acetylene.

MC2 + 2H2O→M(OH) 2 + C2H2

(12) Action with sulphur and phosphorus : Alkaline earth metals directly combine with sulphur
and phosphorus when heated to form sulphides of the type MS and phosphides of the type M3P2

M + S → MS ; 3M + 2P → M3P2
Sulphides on hydrolysis liberate H2S while phosphides on hydrolysis evolve phosphine.

MS + dil. acid → H2S ; M3P2 + dil. acid → PH3
Sulphides are phosphorescent and are decomposed by water

2MS + 2H2O→ M(OH) 2 + M(HS)2

(13) Nitrates : Nitrates of these metals are soluble in water On heating they decompose into their
corresponding oxides with evolution of a mixture of nitrogen dioxide and oxygen.

M(NO3 )2 → MO + 2NO2 +  1 O2
 2

(14) Formation of complexes

(i) Tendency to show complex ion formation depends upon smaller size, high nuclear charge and
vacant orbitals to accept electron. Since alkaline metals too do not possess these characteristics and thus
are unable to form complex ion.

(ii) However, Be2+ on account of smaller size forms many complex such as (BeF3)1-, (BeF4)2-.
Anomalous behaviour of Beryllium

Beryllium differs from rest of the alkaline earth metals on account of its small atomic size, high
electronegativity Be2+ exerts high polarizing effect on anions and thus produces covalent nature in its
compounds. Following are some noteworthy difference of Be from other alkaline earth metals,

(1) Be is lightest alkaline earth metal.
(2) Be possesses higher and than other group members.
(3) BeO is amphoteric in nature whereas oxides of other group members are strong base.
(4) It is not easily effected by dry air and does not decompose water at ordinary temperature.
(5) BeSO4 is soluble in water.
(6) Be and Mg carbonates are not precipitated by (NH4)2 CO3 in presence of NH4 Cl.
(7) Be and Mg salts do not impart colour to flame.
(8) Be does not form peroxide like other alkaline earth metals.
(9) It does not evolve hydrogen so readily from acids as other alkaline earth metals do so.
(10) It has strong tendency to form complex compounds.
(11) Be3N2 is volatile whereas nitrides of other alkaline earth metals are non-volatile.
(12) It’s salts can never have more than four molecules of water of crystallization as it has only four
available orbitals in its valence shell.
(13) Berylium carbide reacts water to give methane whereas magnesium carbide and calcium
carbide give propyne and acetylene respectively.

Be2C+4H2O→2Be(OH)2 + CH4 ; Mg2C3 + 4H2O → 2Mg(OH)2 + C3H6

CaC2 + 2H2O → Ca(OH)2 + C2H4
Diagonal relationship of Be with Al
Due to its small size Be differs from other earth alkaline earth metals but resembles in many of its
properties with Al on account of diagonal relationship.
(1) Be2+ and Al3+ have almost same and smaller size and thus favour for covalent bonding.
(2) Both these form covalent compounds having low m. pt and soluble in organic solvent.
(3) Both have same value of electronegativity (i.e.1.5).
(4) The standard O.P of these elements are quite close to each other ; Be2+=1.69 volts and Al3+=
1.70 volts.
(5) Both become passive on treating with conc. HNO3 in cold.
(6) Both form many stable complexes e.g. (BeF3)-, (AlH4)-.
(7) Like BeO, Al2O3 is amphoteric in nature. Also both are high m. pt. solids.

Al2O3 + 2NaOH → 2NaAlO2 + H2O ; Al2O3 + 6HCl → 2AlCl3 + 3H2O
(8) Be and Al both react with NaOH to liberate H2 forming beryllates and alluminates.

Be + 2NaOH→Na2BeO2+H2 ; 2Al + 6NaOH→ 2Na3AlO3 + 3H2
(9) Be2 C and Al4C3 both give CH4 on treating with water.

Be2C+ 2H2O→CH4 + 2BeO ; Al4C3 + 6H2O→3CH4 + 2Al2O3
(10) Both occur together in nature in beryl ore, 3BeO. Al2O3. 6SiO2.
(11) Unlike other alkaline earths but like aluminium, beryllium is not easily attacked by air (Also Mg
is not attacked by air)
(12) Both Be and Al react very slowly with dil. HCl to liberate H2.
(13) Both Be and Al form polymeric covalent hydrides while hydrides of other alkaline earth are
(14) Both BeCl2 and AlCl3 are prepared is similar way.

BeO+ C+ Cl2 →BeCl2 + CO ; Al2O3 + 3C +3Cl2 → 2AlCl3 + 3CO
(15) Both BeCl2 and AlCl3 are soluble in organic solvents and act as catalyst in Friedel –Crafts
(16) Both Be (OH)2 and Al (OH) 3 are amphoteric whereas hydroxides of other alkaline earths are strong
(17) The salts of Be and Al are extensively hydrated.
(18) BeCl2 and AlCl3 both have a bridged polymeric structure.
(19) Be and Al both form fluoro complex ions [BeF4]2- and [AlF6]3- in solution state whereas other
members of 2nd group do not form such complexes.

Difference between alkali metals and alkaline earth metals

Properties Alkaline earth metals Alkali metals
Electronic configuration Two electrons are present in the valency
shell. The configuration is ns2 One electron is present in the
Valency valency shell. The configuration is
Electropositive nature Bivalent ns1
Less electropositive

More electropositive

Carbonates Insoluble in water. Decompose Soluble in water. Do not decompose
Hydroxides On heating on heating (Li2CO3 is an exception).
Bicarbonates Strong bases, highly soluble and
Action of carbon Weak bases, less soluble and decompose stable towards heat.
on heating These are known is solid state.
Action of nitrogen Do not directly combine with carbon.
These are not known in free state.
Do not directly combine with
Exist only in solution nitrogen.
Directly combine with carbon and form

Directly combine with nitrogen and form

Nitrates Decompose on heating evolving a mixture Decompose on heating evolving Only
Hydration of compounds of NO2 and oxygen oxygen

The compounds are extensively hydrated. The compounds are less hydrated
MgCl2. 6H2O,CaCl2, 6H2O, BaCl2, 2H2O NaCl, KCl, RbCl form non-hydrated
are hydrated chlorides chlorides.

Solubility of salts Sulphates, phosphates, fluorides, Sulphates, phosphates, fluorides,
chromates, oxalates, etc. are soluble
chromates, oxalates etc. are insoluble in in water


Physical properties Comparatively harder. High melting Soft. Low melting points,

points. Diamagnetic Paramagnetic

Magnesium and its compounds

(1) Ores of magnesium : Magnesite (MgCO3 ), Dolomite (MgCO3 .CaCO3 ), Epsomite (epsom
salt) (MgSO4 .7H 2O) Carnallite (MgCl2 . KCl.6H 2O) Asbestos (CaMg3 (SiO3 )4 ), Talc
(Mg2 (Si2O5 )2 . Mg(OH)2 ) .

(2) Extraction of magnesium : It is prepared by the electrolysis of fused magnesium chloride
which is obtained from carnallite and magnesite.

(3) Compounds of magnesium
(i) Magnesia (MgO) : It is used as magnesia cement. It is a mixture of MgO and MgCl2. It is also
called Sorel's cement.
(ii) Magnesium hydroxide : It aqueous suspension is used in Medicine as an antacid. Its medicinal
name is milk of magnesia.
(iii) Magnesium sulphate or Epsom salt (MgSO4 .7H 2O) : It is isomorphous with ZnSO4 .7H 2O.
It is used as a purgative in medicine, as a mordant in dyeing and as a stimulant to increase the secretion of
(iv) Magnesium chloride (MgCl2.6H 2O): It is a deliquescent solid. Hydrated salt on heating in air
undergoes partial hydrolysis. MgCl2.6H 2O Heat → Mg(OH)Cl + HCl + 5H 2O .

Calcium and its compounds
(1) Ores of calcium : Lime stone or marble or chalk (CaCO3 ), Gypsum (CaSO4 . 2H 2O),
Dolomite (CaCO3 . MgCO3 ), Fluorspar (CaF2 ), phosphorite Ca3 (PO4 )2 . Calcium phosphate is a
constituent of bones and teeth.
(2) Manufacture : It is manufactured by the electrolysis of a molten mixture of calcium chloride
containing some calcium fluoride. Calcium chloride in turn is obtained as a by product of the solvay

(3) Compounds of calcium
(i) Calcium oxide or Quick lime or Burnt lime (CaO) : It's aqueous suspension is known as slaked

CaO + H 2O hissingsound → Casl(aOkeHd )2 + Heat,


When exposed to oxy-hydrogen flame, it starts emitting light called lime light.

Note :  CaO is used as basic flux, for removing hardness of water, as a drying agent (for

NH3 gas) for preparing mortar (CaO+ sand +water).
(ii) Calcium chloride (CaCl2.6H 2O) : Fused CaCl2 is a good dessicant (drying agent). It can't be
used to dry alcohol or ammonia as it forms additional products with them.
(iii) Calcium carbonate (CaCO3) : Ca(OH)2 + CO2 → CaCO3 + H 2O .

Note :  It is insoluble in water but dissolves in the presence of CO2 due to the formation of

calcium bicarbonate. CaCO3 + H 2O + CO2 → Ca(HCO3 )2
 It is a constituent of protective shells of marine animals.

s and p block elements

(iv) Gypsum (CaSO4 . 2H 2O) : On partially dehydrates to produce plaster of paris.

CGaypSsOum4 . 2H 2O 120oC → PClaasSteOr o4f . 1 H2O + 1 1 H2O
2 2

Plaster of paris : CaSO4 . 1 H 2O SeHtt2iOng→ CoarSthOor4h.o2mHbi2cO Hardening → MConaoSclOin4ic.(2gHyp2suOm)
Plaster of paris

Gypsum 200oC → C(adSeaOd4b(uarnnht pyldasrtoeru)s)

Gypsum when heated to about 200o C is converted into anhydrous calcium sulphate. The anhydrous form
(anhydrite) is known as dead burnt plaster because it does not set like plaster of paris when moistened with water.

(v) Calcium Hydroxide Ca(OH)2 (slaked lime)
CaO + H 2O → Ca(OH)2 ; Ca(OH)2 + CO2 →CaCO3 + Ca(HCO3 )2

Suspension of Ca(OH)2 in water is called milk of lime.
Ca(OH)2 + Cl2 → CaOCl2 + H 2O

Boron Family.

Group 13 of long form of periodic table (previously reported as group III A according to Mendeleefs periodic
table) includes boron (B) ; aluminium (Al) , gallium (Ga), indium (In) and thallium (Tl) Boron is the first member of
group 13 of the periodic table and is the only non-metal of this group. The all other members are metals. The non-
metallic nature of boron is due its small size and high ionisation energy. The members of this family are collectively
known as boron family and sometimes as aluminium family.

Electronic configuration

Element Electronic configuration ( ns 2 np1 )

5B 1s 2 ,2s 2 2p1 or [He] 2s 2 2p1
13 Al 1s 2 ,2s 2 2p6 ,3s 2 3p1 or [Ne] 3s 2 3p1
31 Ga 1s 2 ,2s 2 2p6 ,3s 2 3p6 3d10 ,4s 2 4 p1 or [Ar]3d10 4s 2 4 p1
49 In 1s 2 ,2s 2 2p6 ,3s 2 3p6 3d10 ,4s 2 4 p6 4d10 ,5s 2 5p1 or [Kr]4d10 5s 2 5p1
81 Tl 1s 2 ,2s 2 2p6 ,3s 2 3p6 3d10 ,4s 2 4 p6 4d10 4 f 14 ,5s 2 5p6 5d10 ,6s 2 6p1 or [Xe]4 f 14 5d10 6s 2 6 p1

Physical properties

(1) A regular increasing trend in density down the group is due to increase in size.
(2) Melting points do not vary regularly and decrease from B to Ga and then increase.
(3) Boron has very high because it exist as giant covalent polymer in both solid and liquid state.
(4) Low of Ga (29.80C) is due to the fact that consists of only Ga2 molecule; it exist as liquid upto
20000C and hence used in high temperature thermometry.

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