1. Liquid Solutions
  2. Solid State
  3. Thermochemistry
  4. Electrochemistry
  5. Periodic Properties
  1. Chemical Bonding
  2. Concept of Acids & Bases 
  3. Transition Element & Coordination Compound




A solution is a homogeneous mixture of (two or more) substance, the composition of which may very between certain limits.


Ideal Solutions

The solutions which obey Raoult’s law at all composition of solute In solvent at all temperature are called ideal solution.


It can be obtained by mixing two components with  identical molecular  size, structure and polarity and they should have same intermolecular attraction. Two liquids A and B form an ideal solution when A–B molecular attraction will be almost same as A – B and B – B molecular attraction.


So, an ideal solutions is one.


  1. i) Which Obey Raoult’s law i.e.
  2. ii) ΔHmixing = 0 i.e. No heat is evolved or absorbed during mixing.

iii) ΔVmixing = 0 i.e. no expansioin or, contraction of volume occurs on mixing.


Colligative Properties


The properties of dilute or ideal solutions which depend on the solute particles or solute moles or solute concentration, but does not depend on the nature of solute is called colligative properties.


e.g. 50 molecules of NaCl and 100 molecules of urea have same colligative properties in the dilute solutions.


There are four kinds of colligative properties.


  1. i) Osmotic pressure
  2. ii) Lowering of vapour pressure

iii) Elevation in boiling point (Ebullioscopy)

  1. iv) Depression in freezing point (Cryoscopy).


Vont Hoff’s Factor (i)

For complete dissociation of

  1. a) NaCl, i = 2
  2. b) NaHCO3, i = 3
  3. c) Na2SO4, i = 3

Relation of ‘i’ with α

Case-1: Relation of “i” with α

Where α = degree of dissociation


Initially mole 1         0        0

After dissociation moles 1-α   2α α

∴ i = (1 – α) + 2 α + α = 1 + 2α

Similarly for partly dissociation of AxBy

i = 1  + (x + y –1)α


Case-2: Relation of “i” with α (degree of association)

at t = 0 1 0

After association moles 1-α α/n

∴ i =

Note: i) For dissociation i > 1

  1. ii) For association i < 1

iii) For non-electrolyte or, for ideal solution i = 1


Osmotic pressure

Excess pressure either withdrawn from solvant side or applied on the solution side in order to stop the flow of solvant, is called osmotic pressure.

πV = inST where π = Osmotic pressure

S = Solution constant

= R (molar gas constant)

T = Kelvin Temperature

⇒ π = iCST nf = final moles of solute

= i.n

n = initial moles

i = Vont Hoff’s factor


Iso-osmotic solution or, Isotonic

Two solutions having same osmotic pressure or, same concentration of solute is called isotonic solution.


For isotonic solution

π1 = π2

⇒ C1 = C2  ⇒  

Illustration-1: If a solution of mercuric cyanide of strength 3gm lit–1 has an osmotic pressure 0.3092 × 105 Nm–2 at 298 K, what is the apparent molecular weight and degree of dissociation of Hg(CN)2?

[Hg = 200, C = 12, N = 14]


Solution: Let the apparent molecular weight of Hg(CN)2 = m0 

Osmotic pressure = π =

0.3092 × 105 =

∴ where w = 3gm/lit

  = 3kg/m3

⇒ m0 = 240.2× 10–3 kg

For Hg(CN)2

∴ i = 1 + 2α

∴ i = = 1 + 2α

⇒ Solving α = 0.02457

∴ % α = 2.457%


Lowering Of Vapour Pressure

The pressure exerted by vapours over the liquid surface at equilibrium, is called vapour pressure of liquid.

Case-1: Addition of non-volatile solute in volatile solvant.

When non-volatile solute is added with volatile solvant, then fraction of surface is blocked due to which vapour is decreased i.e. vapour pressure decreases. That is why v.p. of solution becomes less than vapour pressure the pure solvant.

Let p0 = vapour pressure of pure solvant

Ps = vapour pressure solution

∴ p0 > ps

According to Raoult’s law

Ps = p0 xsolvant, where Xsolvant = mol-fraction of solvant

or, ps = p0 (1 – Xsolute) p0 – ps = lowering in vapour pressure

or, = relative lowering of vapour pressure

or Xsolute = mol-fraction of solute

(according to Oswald’s & Walker’s apparatus)

Illustration -2. The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 gm of salt per 100 gm of water at 100°C of 70%. Calculate the vapour pressure of solution.

Solution: Let initially 1 mole of Ca(NO3)2 is takne 

 Degree of dissociation of Ca(NO3)2 = = 0.7

Ionisation can be represented as 

Ca(NO3)2 Ca++ +

At start 1 0 0

At  equilibrium 1–0.7 0.7 2 × 0.7

Total moles at equilibrium = (1–0.7) + 0.7 + 2 × 0.7 = 2.4

Observed molar mass  =  

∴ No. of moles of solute  = = 0.102

n = 0.102

No. of moles of water  = N = = 5.55

Applying Raoult’s law,

∴ P = 746.3 mm Hg

Case-2: Addition of volatile solute in volatile solvant.

As we know

PA = where pA = partial vapour pressure of A

PB = = vapour pressure of A in solution or liquid phase.

Assuming that vapours of liquid is behaving like an ideal gas, then according to Dalton’s law of partial pressure 

pT = pA + pB

= …(1)


Mole fraction of componants in vapour phase

 x(mole fraction) =


Elevation In Boiling Point (Ebullioscopy)

Boiling point: The temperature at which vapour pressure of the liquid becomes equal to 1 atmospheric pressure, is called the boiling point of the liquid.

Let T0 = boiling pressure of the solvant

Ts = boiling point of the solution

Elevation in boiling point of solution = Ts – T0 = ΔTb

or, ΔTb  = Kb × i × Cm where i = Vont Hoff’s factor

= Kb × i × Cm = molality

iCm = final or, total molality

Kb = Ebullioscopic constant

ΔTb = Elevation in boiling point

W = Weight of solvant (in gm)

Thermodynamically, we know that

= where Tb = boiling pressure of the liquid

v = latent heat of vaporisation per gm

For H2O, Tb = 373 K, lv = 540 cal/gm

R = 2 cal mol–1 K–1

∴ Kb = 0.52 K mol kg–1


Illustration -3: River-water is found to contain 11.7% NaCl, 9.5% MgCl2 and 8.4% NaHCO3 by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl, 70% ionization of MgCl2 and 50% ionization of NaHCO3 (Kb for water = 0.52).


Solution: Let the weight of solution = 100 gm

∴ Weight of solvant = 100 – (11.7 + 9.5 + 8.4)

= 704 gm of H2O

Total moles of all solute after dissociation

in = iNaCl ⋅ nNaCl +

= (1 + α1) × + (1 + 2α2) × + (1 + 2α3) ×

= (1 + 0.9) × + (1 + 2× 0.7) × + (1 + 2 × 0.5) ×

Since, ΔTb = Kb × (in) × , where W = Wt. of solvant (in gm)

= 0.52 ×

= 5.94°C

∴ Boiling point of solution = 100 + 5.94°C = 10.5.94°C


Depression In Freezing Point or, Cryoscopy

Freezing point: The temperature at which the solid and liquid phase co-exists i.e. where the solid and the liquid have same vapour pressure i.e. the solid and the liquid phase be in equilibrium, is called freezing point. F.P. for water = 0°C.


Let T0 = freezing point solvant

Ts = freezing point of solution

∴ depression of freezing point  = T0 – Ts = ΔTf

ΔTf = Kf × iCm where Kf = Cryoscopic constant

× i ×

Where W = wt. of solvant (in gm)

w = wt. of the solute

mth = theoretical molecular weight


, where Tf = freezing point of liquid

f = latent heat of fusion

For H2O, = 1.86 K mol kg–1


Illustration -4: The freezing point of a solution of acetic acid (mol. fraction = 0.02) in benzene at 277.4K. Acetic acid exists partly as a dimer.

2A A2

Calculate the equilibrium constant for dimerisation. Freezing point of benzene is 278.4 K (Kf benzene = 5)

Solution: Let acetic acid → A

Benzene → B

 Mol-fraction of A = 0.02 (given)

∴ Mol-fraction of B = 1 – 0.02 = 0.98

∴ Molality of A in B = , where m = molar mass of B


i.e. Cm = 0.262 mol kg–1

As we know that

ΔTf = Kf × i × Cm

278.4 – 277.4 = 5 × i × 0.262

⇒ i = 0.76

Let α part of A forms dimer

2A A2

  1. 0

1-α α/2 initially moles moles after dimer in for mol.

∴ i = 1 – α + =  1 –

= 0.763

⇒ α = 0.47

So, [A] in molality = (1 – α) × Cm = (1 – 0.48) × 2.262

=0.52 × 0.262

[A2] in molality after dimer is formed

= × Cm =

= 0.06288 ∴ Kequilibrium

= 3.39 kg mol–1 



Problem-1. Calculate the molal elevation constant, Kb for water and the boiling point of 0.1 molal urea solution. Latent heat of vaporisation of water is 9.72 kcal mol–1 at 373.15K

Solution: , where Tb = boiling point of solvant

= 373.15 K

= Latent heat of vaporisation of water per gm

= 9.72 mol–1 = kcal per gm H2O

Kb =

Kb = 0.515 ≅  0.5 

Since elevation in boiling point,

ΔTb = Kb × i × molality, for urea, i = 1

ΔTb = 0.5 × 1 × 0.1 = 0.05

∴Boiling point of solution = boiling point of solvant + ΔTb

= 373.15 + 0.05 = 373.20 K


Problem – 2: A milimolar solution of pottassium ferricyanide is 70% dissociated at 27°C. Find out  the osmotic pressure of the solution. 

Solution: Given, concentration of solution  = C = M = 10–3M

K3[Fe(CN)6] ⎯→ 3K+ + [Fe(CN)6]–3

Initially moles   1 0 0

Moles after dissociation     1-α α Where α  = degree of dissociation = 70% = = 0.7

Total moles of after dissociation = (1–α) + 3α + α = 1+ 3α

∴ i =

i = 1+3α = 1+(3× 0.7) = 3.1

Since osmotic pressure 

π = iCRT

∴ π = 3.1 ×  10–3 × 0.0821 × 300 = 0.31 × 3 × 0.0821 atm

= 0.07635 atm

Problem – 3: 1kg of an aqueous solution of Sucrose is cooled and maintained at –4°C. How much ice will be separated out if the molality of the solution is 0.75?  Kf (H2O) = 1.86 Kg mol–1K.


Solution: Sucrose is a non–electrolyte,

Hence i = 1

Molecular weight of sucrose (C12H2O11) = m = 342 gm mol–1

Molality of the solution  = 0.75 m 

= 0.75 mol kg–1 solvant 

= 0.75 × 342 gm Sucrose  per kg solvant 

= 256.5 gm Sucrose per kg solvant

hence, weight of 1 molal solution = 1000 + 256.5 = 1256.5 gm 

Sucrose present in 1 kg solution = = 204.14 gm 

Weight of solvant (H2O) present in 1 kg solution = 1000– 204.14 

= 795.86 gm

Since depression in freezing point

ΔTf = Kf  × i × × , Where W = weight of solvant

w = weight of the solute  

or, 4 = 1.86 × 1 × ×

∴ W = 348.75 gm

i.e., weight of solvant required to maintain this solution at – 4°C is 

W = 348.75 

Hence rest weight of H2O will convert into ice.

Hence amount of ice formed = 795.86 – 348.75  = 447.11 gm

Problem – 4: A complex is represented as CoCl3.xNH3. Its 0.1 m solution in aqueous solution shows ΔTf = 0.558° Kf(H2O) = 1.86 mol–1 K  and assume 100% ionization and co–ordination number of Co(III) is six . What is the complex?

Solution: From question, molality of solution = Cm = 0.1 

Depression in freezing point = ΔTf = 0.558°C

Kf (H2O) = 1.86 kg  mol–1 K–1

As we know that 

ΔTf = Kf × i × Cm , Where i = Vont Hoff’s factor 

0.558 = 1.86 × 0.1 × i

∴ i = = 3

or, i 3 indicates that complex ionize to form three ions since co ordination  number is 6 hence x = 5 

i.e., CoCl3.5NH3 ⎯→ [Co(NH3)5Cl]++ + 2Cl

1 Cation 2 anions 

So, the complex is [Co(NH3)5Cl]Cl2

Problem-5. 1 g of monobasic acid when dissolved in 100 g of water lowers the freezing point by 0.168°C. 0.2 gm of the same acid when dissolved and titrated required 15.1 ml of N/10 alkali. Calculate degree of dissociation of the acid. ( Kf for water is 1.86)

Solution: 15.1 ml alkali is required by 0.2 gm monobasic acid

1000 ml (N) alkali is required by gm monobasic acid

= 132.45 gm monobasic acid

∴Equivalent wt. of monobasic acid, E = 132.45 = mol. wt.

ΔTƒ = Kƒ × i × Cm

ΔTƒ = Kƒ × i × ×

or, 0.168 = 1.86 × i × ×

i =

i = 1.196

For monobasic acid, HA     H+ + A

1–α       α        α

i = = 1 + α

∴1 + α = 1.196

∴α = 0.196 



  1. Solution of two volatile liquids A and B obey Raoult’s law. At a certain temperature, it is found that when the total pressure above a given solution is 400 mm Hg, the mol–fraction of A in the vapour is 0.45 and in the liquid is 0.65. What is the vapour pressure of the two pure liquids at the given temperature?


  1. A volume of 105 ml of pure water at 4°C saturated with NH3 gas, yielding a solution of density 0.9 gm / ml and containing NH3 30% by mass. Find the volume of NH3 solution and volume of NH3 gas resulting and volume of NH3 gas at 5°C and 775 mm of Hg which was used to saturate the water. 


  1. If 20 ml of ethanol (density  = 0.7893 gm / ml) is mixed with 40 ml (density = 0.9971 gms) at 25°C, the final solution has density of 0.9571 gm / ml. Calculate the percentage change in total volume of mixing. Also calculate the molality of alcohol in the final solution.


  1. Vapour pressure of C6H6 and C7H8 mixture at 50°C is givne by P (mm Hg) = 180
    XB + 90, where XB is the mole fraction of C6H6. A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C, what would be mole fraction of C6H6 in the vapour state?


  1. A curent of dry air was bubbled through a bulb containing 26.66 g of an organic substance in 200 gms of water, then through a bulb at the same temperature containing pure water and finally through a tube containing fused calcium chloride. The loss in weight of water bulb is 0.0870 gms and gains in weight of CaCl2 tube is 2.036 gm. Calculate the molecular weight of the organic substance in the solution.


  1. A dilute solution contains m mol of solute A in 1 kg of a solvent with molal elevation constant Kb; The solute dimerises in solution as 2A  A2. Show that equilibrium constant for this dimer formation is K = , where ΔTb is the elevation in boiling point for the given solution. (Assume molarity = molality).


  1. A 0.01 molal solution of Pt(NH3)4Cl4 in water had a freezing point depression of 0.0054°C. What is the formula for complex ? Kf(H2O) = 1.86 kg mol–1 K. Assume 100% ionization of the complex. 


  1. 10 gm of NH4Cl (mol. weight  = 53.5) when dissolved in 1000 gm of water lowered the freezing point by 0.637°C. Calculate the degree of hydrolysis of the salt if its degree of dissociation of 0.75. The molal depression constant of water is
    1.86 kg mol–1 K.


  1. Tritium, T (an isotope of H) combines with fluorine to form weak acid TF, which ionizes to give T+. Tritium is a radioactive and is a β-emitter. A freshely prepared aqueous solution of TF has pT (equivalent to pH) of 1.5 and freezes at – 0.372°C. If 600 ml of freshly prepared solution were allowed to stand for 24.8 years.

Calculate i) ionisation constant of TF

  1. ii) number of β-particles emitted

[Given Kf (H2O) = 1.86 kg mol–1 K, t ½ for tritium = 12.4 years] (ii) 4.55 × 102 β-particle]


  1. Find Ka, the ionization constant of tartaric acid if a 0.01 molal aqueous solutionof tartaric acid freezes at –0.205°C. Assume that only first ionization is of importance and that 0.1 m = 0.1M. Kf = 1.86 kg mol–1 K



  1. 276.9 mm Hg = 628.6 mm Hg
  2. 24 litres


  1. % change in volume = 3.05%

Molality = 8.604 m


  1. 0.93


  1. mB = 61.19

or mB =  63.59 (when is neglected in comparison to  


  1. The complex is  [Pt(NH3)4Cl2]Cl2


  1. h = 0.089
  2. (i) 7.299 × 10–3, (ii) 4.55 × 1022 β particles 
  3. Ka = 1.11 × 10–3



Solid: Substance having definite and orderly arrangement of atoms or molecules in 3-D space, is called solid. It has sharp (or very short range) melting point being definite and orderly arrangement of atoms or molecules.


Unit Cell: The smallest repeating unit of the space lattice, is called unit cell.


14-Bravais lattices


On the basis of arrangement of atoms, 7-systems of crystal lattices exist in 14-Bravais lattices. 7-crystal systems exist as 14-Bravais lattices as follow:

Sl. No. Crystal System Bravias lattices Parameters of Unit cell
Intercepts Crystal angle
1 Cubic Primitive,

Face centered 

Body centered

= 3 a = b = c α = β = γ  = 90°
2. Orthorhombic Primitive, 

Face centered 

Body centered

End centered

= 4 a ≠ b ≠ c α = β = γ  = 90°
3. Tetragonal Primitive

Body centered

=2 a = b ≠ c α = β = γ  = 90°
4. Rhombohedral Primitive = 1 a = b = c α = γ = 90

β ≠ 90°

5. Hexagonal Primitive = 1 a = b ≠ c α = β = 90°

γ = 120°

6. Monoclinic Primitive 

End-centered =2

a ≠ b ≠ c α = γ = 90°

β ≠ 90°

7. Triclinic Primitive = 1 a ≠ b ≠ c α ≠ β ≠ γ ≠ 90°


Calculation of rank or number of atoms in 1-unit cell of cubic system.


Sl. No. Types of atoms Contribution in 1-cubic unit cell
1. Corner atoms (nc) i.e. nc
2. Face atoms (nf) i.e.
3. Body atom (nb) Total contribution, i.e. nb


∴ Rank of atoms in 1-unit cell of cubic system

= + nb

1. Primitive unit cell: The same type atoms are arranged at all corners of the cubic unit cell.

Rank of atom = 1

P.F. =

= Where, r = radius atoms

            a = edge length of a cube = 2r

= = 0.52

 Void fraction + P.F. = 1 ∴ Void fraction = 1 – 0.52 = 0.48


2. Body centered cubic: In which all corners are arranged by the same atoms and the body centered atom may be same or different.

Length of body diagonal.

= BC = 4r = a

Here rank of atoms = 2

∴ P.F. = = 0.68 ∴ V.F. = 0.32


3. Face centered cubic (FCC) In cubic unit cell in which the same atoms are present at all corners and the atoms at the face-center may be same or different.

Here, rank of atoms = 4

Length of face diagonal = BC = 4r = √2a

∴ P.F. = =

P.F. = 0.74

V.F. = 0.26


  1. Hexagonal primitive unit cell: 12 atoms are arranged in the 12-corners of the unit cell. 2-atoms are present in the face centers, and 3-atoms remains present in mid-position of between of two faces.

Each corner atom would be common to six unit cells.

So, rank of atoms = 6

The height of the unit cell = C =

Length of unit cell = a = 2r

Since, volume of hexagonal unit cell

= its base area × height (C)

= area of 6-equilateral triangles × C

= × (2r)2 × 6 ×

P.F. = = 0.74

∴ void fraction = 0.26


Density of Crystal Lattice

Density of unit cell (ρ) =


Co-ordination no. (C.N.) and limiting radius ratio

C.N.: The no. of nearest atoms or neighbours is known as C.N.


Radius ratio =

When all anions and cations are closely packed, then it is known as ideal radius ratio.

Limiting radius ratio
= = x
Co-ordination number Shape Example
x < 0.155 2 Linear BeF2
0.155 ≤ x < 0.225 3 Triangular planar AlCl3
0.225 ≤ x < 0.414 4 Tetrahedron ZnS
0.414 ≤ x < 0.732 6 Octahedral NaCl
0.414 ≤ x < 0.732 4 Square planar [PtCl4]–2
0.732 ≤ x < 0.999 8 Body centered cubic CsCl


Tetrahedral and Octahedral Voids

  1. Tetrahedral void: is formed between three spheres on a closed-packed plane and a fourth sphere on the adjacent plane (either on the top or at the bottom) fitting in the cavity space between the three spheres. When the centres of four spheres are joined by line, the regular tetrahedron is obtained. So, this void is known as tetrahedral void.

In fcc unit cell, the point having co-ordinates from its each corner assuming origin (0, 0, 0) is the position of tetrahedral void. Since there are eight corners in fcc unit cell, hence no. of tetrahedral void = 8.


In hexagonal primitive unit cell, there are 12-corners, so it has 12-tetrahedral voids.


  1. Octahedral void: is formed with 3-spheres on a closed packed plane and 4-more spheres on an adjacent close-packed plane, so that the centers of the three spheres in one plane directly over the triangular valleys surrounding the central valley of the first plane and there is no sphere over the central valley. When the centers of the six spheres are joined by line, the regular octahedron is formed. So, void formed in regular octahedron is known as octahedral void.

In 3-D array, there is no octahedral void per sphere. The largest sphere that can fit into an octahedral void is 0.414r, where r = radius of the spheres of the closed packed array.

On calculation, it can be found that in fcc or hexagonal primitive unit cell


Effective no. of atoms = No. of octahedral voids.

i.e. No. of octahedral void = 4 (in fcc)

No. of octahedral void = 6 (in hexagonal primitive unit cell)

Various Structures of Ionic Solids

Sl. No. Crystal structure Brief description Co-ordination number No. of units per unit cell Examples
1. Rock salt (NaCl type) Cl ions occupy ccp and Na+ ions occupy all octahedral holes Na+ 6

Cl 6

4 Li, Na, K, Rb & halides, AgCl, AgBr, MgO, CaO, FeO
2. Zinc blende (Zns-type) S–2 ions occupy ccp and Zn++ ions occupy alternate tetrahedral voids Zn++ 4

S 4

4 BeS, HgS, CuI, CuBr, CuCl, AgI
3. Wurtzite (ZnS-type) S–2 ions in hcp and Zn++ ions occupy alternate tetrahedral voids Zn++ 4

S–2 4

4 ZnS, ZnO, CdS, BeO
4. Casium chloride
Cl ions in scp and Cs+ ion occupies body centre of the cube. Cs+ 8

Cl 8

1 CsBr, CsI, CsCN, CaS
5. Fluorite structure (CaF2-type) Ca++ ions in ccp F ions occupy all tetra-hedral holes/voids Ca++ 8

F 4

4 CaF2, SrF2, BaF2, BaCl2, SrCl2, CdF2, HgF2
6. Anti-fluorite (Li2O–type) O– – ions occupy ccp Li+ ions occupy all tetrahedral sites Li+ 4

O– – 8

4 K2O, Na2O, K2S, Na2S
  1. Prevoskite structure: The general formula of prevoskite is ABO3, where A+2 in bivalent cation, B+4 is tetravalent cation e.g. CaTiO3, BaTiO3. The bivalent ions are present in primitive cubic lattice, tetravalent cation occupies the centre of the body or octahedral void, whereas oxide ions occupy the all centres of six faces.


Point defects in Crystal

Point defects in crystal arises, when any of the constituent particles has been either missing from the normal crystal site or has been dislocated to a position i.e. interstitial position. In point defect, there is no change in electrical neutrality of the crystal.


Difference between Schottky and of Frenkel defect

Schottky defect Frenkel’s defect
1. Arises when a pair of one cation and one anion are missing from the ionic crystal. 1. Arises when a cation leaves its lattice sites and inters an interstitial position.
2. Overall electrical neutrality of the crystal is maintained. 2. Overall electrical neutrality of crystal in preserved.
3. Density of the crystal is decreased 3. Density of the crystal doesn’t change.
4. This type of defects occurs in ionic solids with high C.N. and where cations and anions are of similar sizes. E.g. NaCl, CsCl 4. This type of defects occurs in ionic solids whose anions are large and cations are smaller in size. e.g. AgBr.

Whose crystal structure is open type with large interstices e.g. CaF2, ZnS.

Illustration -1: Which of these two, CdCl2 and NaCl, will produce Schottky defect, if added to a AgCl crystal?


Solution: CdCl2, in order to maintain the electrical neutrality of the solid crystal, one Cd++ ion replaces two Ag+ ions due to which cation vacancy is produced and density of solid crystal will decrease. Hence the fact.


Illustration-2: An element A (Atomic wt – 100) having bcc structure has unit cell edge length 400 pm. Calculate the density of A and number of unit cells and number of atoms in 10 gm of A.

Solution: a = 400 x 10-12m

= 400 x 10-12 x 100 cm  = 4 x 10-8cm

So, volume of unit cell edge length = a3 = (4 x 10-8)3

As we know that

density (ρ) =, here, n = no. of atoms per unit cell = 2

Mm = mass of 1 mole atom = 110 gm

NA = Avogadro’s no. = 6.023 x 1023

= =

= 5.188 gm/cm3

100 gm of A contains 6.023 × 1023 atoms of A

10 gm of A contains 6.023 ×1023 ×

= 6.023 x 1022 atoms of A

2 atoms remain present in 1-unit cell of bcc type 

6.0 23 ×1022 remains present in unit cells

= 3.0115 x 1022 unit cells


Illustration-3: In fcc arrangement of A & B atoms, where A atoms are at corners of the unit cell, B atoms at the face-centers, one of the atoms are missing from the corner in each unit cell then find the percentage of void space in the unit cell.

Solution: There areA atoms and 3 B atoms per unit cell

Also, 2 rA + 2 rB =

or, a =

Volume of unit cell = a3 =

Fraction of volume occupied per unit volume of the unit cell is given by

PF =


= =

= = 0.570

So, void space = 1 – packing fraction

= 1 – 0.570 = 0.430/unit volume of unit cell

percentage of void space = 43%


Illustration-4: KF has NaCl structure. What is the distance between K+ & in KF if density of KF is 2.48 gm/cm3.


Solution: Since KF has NaCl structure i.e. like Cl, F ions are present at all six faces and all corners of the cube. Similarly like Na+, all K+ ions occupy the middle position of 12 edges of the cube and center of the cube.

a = 2


As we know that

Density (ρ) = here, n = 4

Mm = molar mass of KF  =  (39 + 19) gm

= 58 gm

ρ =

or, a3 =

or, a = 5.37 × 10–8 cm

= 5.37Å

=  2.685Å 



Problem-1: A unit cell of sodium chloride has four formula units. The edge of length of the unit cell is 0.564 nm. What is the density of sodium chloride.


Solution: Edge-length of unit cell = a = 0.564 nm

or, a = 0.564 x 10-9 m

= 5.64 x 10-8 cm

Volume of unit cell = a3 = (5.64 x 10-8cm)3

As we know,

The density (ρ) of the unit cell

= where, n = no. of molecule in one unit cell

Mm = molar mass

NA = Avogadro’s No.

= = 2.16 gm/cm3


Problem-2: In a cubic crystal of CsCl (density = 3.97 gm/cm3) the eight corners are occupied by Cl ions with Cs+at the centre and visa-versa. Calculate the distance between the neighbouring Cs+ and Cl ions. What is the radius ratio of two ions?


Solution: As we know

Density (ρ) = here, n = 1

Mm = 132.9 + 35.5 = 168.4 gm

or, a3 =

or, a = 4.13 x 10-8= 4.13 Å

As it is BCC with Cs+ at center (radius r+) and Cl at corner (radius r)

So, body diagonal =

or, 2 r+ + 2r =

or, r+ + r = = 4.13 Å 

r+ + r = 3.57 Å

Now assume that two Cl touch each other

i.e. r + r = a

or, 2r = a = 4.13 Å

r = 2.065 Å





Problem-3: Metallic gold Crystallises in fcc lattice. The length of the Cubic unit cell
is a = 4.07 Å.

  1. What is the closest distance between gold atoms.
  2. How many “nearest neighbours” does each gold atom have at the distance calculated in (a).
  3. What is the density of gold.
  1. d) Prove that the packing fraction of gold is 0.74.

Solution: a) For fcc lattice nearest distance between two neighbours = 2r

As we know that

4r =

or, 2r =

= Å

= 2.88 Å


  1. If we consider a  face centred gold atom, it has four corner and eight other adjacent face centre atoms present at distance. Therefore there are 12 nearest neighbours. 
  1. c) density (ρ) = = = 4 g/cc
  1. PF =   0.74

Problem-4: The composition of a sample of wustite is Fe0.93O1.0 What percentage of iron is present in the form of Fe(III)?


Solution: Fe0.93O1.0 is a non-stoichiometric compound. It is a mixture of Fe+2 and Fe+3 ions

Let x atoms of Fe+++ ions are present in the compound

So, (0.93-x) atoms of Fe++ ions are present in the compound

In the compound,

Total positive charge = Total negative charge

x × (+3) +(0.93 – x) × (+2) = +2

or, 3 x + 0.93 × 2 – 2 x = +2

or, x + 2 x 0.93 = 2

x = 0.14

% of Fe+3 ions present = = 15.053

Problem-5: Pottassium Crystallizes in body centred cubic lattice  with a  unit cell length a = 5.2Å.

  1. What is the distance between nearest neighbours?
  2. What is the distance between next nearest neighbours?
  3. How many nearest neighbours does each K atom have?
  4. How many next-nearest neigbbours does each K atom have?
  1. e) What is the calculated density of crystalline Potassium? 


Solution: (a) The distance between nearest neighbours

= × length of the body diagonal =

= Å = 4.50 Å

  1. Distance between the next nearest neighbours

a = 5.20 Å

  1. Number of K atoms (nearest neighbours) present at a distance of from the body centred atom = 8
  1. d) Number of K atoms (next nearest neighbours) present at distance of edge length = 6
  2. e) density=where, n = no. of atoms in 1 unit cell of b.c.c. = 2

Mm = molar mass of K

= =  0.921 gm/cm3




  1. The unit cube length of LiCl (NaCl structure) is 5.14A°. Assuming anion-anion contact, Calculate the ionic radius of Cl ion.


  1. A metal crystallizes in two cubic phases, face centered cubic (fcc) and body centered cubic bcc whose unit cell length are 3.5 and 3.0A° respectively. Calculate the ratio of density of fcc and bcc. 


  1. The density of solid argon is 1.65gm per ml at – 233°C. If the Ar-atom is assumed to be a sphere of radius 1.54 x 10-8cm, what percentage of solid argon has apparently empty space [Ar = 40]


  1. Titanium has hexagonal close packing with cell edge length a = b = 295.3 pm, height = c = 472.9 pm. Calculate its density.


  1. If the density of crystalline CsCl is 3.988 g/cm3, (a) calculate the volume effectively occupied by a single CsCl ion pair in the crystal (b) calculate the smallest Cs-Cs inter nuclear distance.


  1. Compute the void space per unit volume of unit cell in the rock salt structure.


  1. The ZnS Structure is cubic. The unit cell be described as a face-centered sulphide ion sub-lattice with Zinc ions in the centers of alternating minicubes made by part on the main cube into eight equal parts.
  1. How many nearest neighbours does each Zn2+ have?
  2. How many nearest neigbours does each S2– have?
  3. What angle is made by the lines connecting any Zn++ to any two of its nearest neigbours?
  4. What minimumratio is needed to avoid anion-anion contact?

Closest cation-anion is assumed to touch.

  1. BaTiO3 crystallizes in the prevoskite structure. This structure may be described as a cubic lattice with barium ions occupying the corner of the unit cell, oxide ions occupying the face-centers and titanium ion occupying the center of the unit cell.
  1. If titanium is described as occupying holes in the BaO lattice, what types of hole does it occupy?
  2. What fraction of this type of hole does it occupy?
  3. Can you suggest a reason why it has certain holes of this type but not other holes of the same type?
  1. NH4Cl crystallizes in a body centered cubic lattice with a unit cell distance of
    387 pm. Calculate 

(a) the distance between oppositely charged ions in the lattice and

(b) the radius of theion if the radius of Cl ion is 181 pm.


10. Ice crystallizes in a hexagonal lattice. At low temperature at which the structure a = 4.13Å and b = 7.00Å (see figure). How many molecules are contained in a unit cell? [density of ice = 0.92 gm/cm3].



  1. 1.815 Å
  2. 1.259
  3. 62.1%


  1. 2.132 gm / cm3


  1. (a) 7.014 × 10–23 cm3 (b) 0.77 Å


  1. 0.206


  1. (a) 4, (b) 4, (c) 109° 28′


  1. (a) octahedral (b) 1/4
  2. (a) 335.15 pm, (b) 154.15 pm
  3. 4 molecules of H2O



Thermo-chemistry is the branch of physical chemistry which deals with the thermal or heat changes caused by chemical reactions.

  1. Heat of reaction

Heat of reaction is defined as the amount of heat absorbed or evolved at a given temperature when the reactant have combined to form the products as represented by a balanced chemical equations. It the heat of reaction is denoted by q, then the numerical value of q depends on the manner in which the reaction is performed. For the two methods of conducting chemical reactions is calorimeters,  we have:

Constant volume: w = 0 and qv= ΔE bomb calorimeter

Constant pressure: w = −ρdV and qp = ΔE + ρΔV: (open calorimeter)


⇒   constant pressure processes are much more common in chemistry therefore usually when ever we speak of heat of reaction it implies enthalpy change. 


  1. Standard states
State Standard State
Gas Ideal gas at 1 atm and the given temperature 
Liquid Pure liquid at 1 atm and the given temperature 
Solid Stable crystalline form at 1 atm and given temperature
  1. Factors influencing the heat of reaction 

(i) Physical state of substance:

Example – 1: H2(g) + O2 → H2O(g), ΔH = −57.8 kcal

H2(g) + O2 → H2O(), ΔH = −68.3 kcal

(ii) Allotropic forms of the elements: 

C(diamond) + O2(g) → CO2(g), ΔH = −94.3 kcal

C(graphite) + O2(g) → CO2(g), ΔH = − 97.6 kcal.

(iii) Temperature: Heat of reaction depends upon the temperature at which the reaction is carried out. This is due to the variation in the heat capacity of the system with temperature kirchoff’s has related heat of reaction (ΔH or ΔE) with temperature as following: 

H2 − H1 = ΔCp(T2 − T1)

E2 − E1 = ΔCv (T2 − T1)

ΔCp is the difference in the heat capacities (at constant pressure) of products and reactants. 

ΔCv is the difference in the heat capacities (at constant volume) of products and reactants. 

(iv) The condition of constant volume or constant pressure: 

The heat of reaction depends on the fact that whether the reaction is carried out at constant volume (ΔE) or at constant pressure (ΔH). The two are, however related as follows:

ΔH = ΔE + Δn RT

Δn = Gaseous product moles – Gaseous reactant moles

  1. Calorimetry

An object undergoing a temperature change without a chemical reaction or change of state, absorbs or discharges an amount of heat equal to its heat capacity times of the temperature change. 

Heat exchange = (heat capacity) × (temperature change)

Ilustration -1: It 1.00 kcal of heat is added to 1.2 C of oxygen in a cylinder at constant pressure of 1.00 atm, the volume increases to 1.5 C. calculate ΔE for this process. 


Solution: ΔH = ΔE + PΔV

1.00 = ΔE + × 2 × 10−3

ΔE = 0.993 kcal


Illustration-2: A sample of solid naphthalene C10H8, weighing 0.600 g is burnt to CO2(g) and H2O(l) in a constant volume calorimeter at T = 298K. In this experiment the observed temperature rise of the calorimeter and its contents is 2.27°C. In a separate experiment, the total heat capacity of the calorimeter was found to be 2556 cal/deg. What is ΔE for the combustion of one mole of naphthalene? What is for this reaction? Also calculate the enthalpy of formation of naphthalene. 

(H2O, ) = −68.32 kcal/mol

(CO2, g) = −94.05 kcal/mol


Solution: C10H8(s) +12O2(g) → 10CO2(g) + 4H2O(g)

Heat released = 2.27 × 2556 = 5.802 kcal 

ΔE = − = − 1237.76 kcal/mol

Also, ΔH = ΔE + Δn RT

For the reaction: 

C10H8(s) + 12O2(g) → 10CO2(g) + ΔH2O()

Δn = 10 – 12 = 2

∴  ΔH = −1237.76 − 2 × 2 × 10−3 × 298 = −1238.952 kcal/mole.

Also – 1238.952 = 10 ΔHf(CO2)(g) + 4ΔHf(H2O)() − ΔHfC10H8(s)

−1238.952 = 10 × (−94.05) + 4(−68.32) − ΔHF(C10H8)

∴ HF(C10H8) = 25.172 kcal/mol.


  1. Heat of formation

The change in enthalpy that takes place when one mole of the compound is formed from its elements. 

H2(g) + ½ O2(g) → H2O(), ΔH = −68 38 kj/mol


  1. Heat of combustion

The change is enthalpy when one 


  1. Heat of neutralization

The quantity of heat evolved when one gram equivalent of an acid is completely neutralized by one gram equivalent of a base in dilute solution

HCl + NaOH → NaCl + H2O() ΔH = −13.7 kcal

(aq) (aq)   (aq)


The heat of neutralization of strong acid and a strong base is 13.7 kcal. on the basis of electrolytic dissociation theory, it has been clearly explained that this heat of neutralization is merely the heat of formation of water from H+ of an acid and OH of a base. 

H+(aq) + OH(aq) → H2O() ΔH = −13.7 kcal.

The heat of neutralization in case of a weak acid or a weak base is some what less than 13.7 kcal because some energy is used 4p in dissociating the weak electrolyte. The difference in the values gives the dissociation energy of the weak acid or a weak base. 


  1. Hess’s Law: The enthalpy change in a chemical or physical process is same whether the process is carried out in one step or in several steps. 


Illustration – 3: The standard enthalpies of formation at 298 K for CCl4(g), H2O(g), CO2(g) and HCl(g) are −106.7, −241.8, −393.7 and −92.5 kJ mole−1, respectively. Calculate for the reaction.

CCl4(g) + 2H2O(g) → CO2(g) + 4HCl(g)


Solution: ΔH = ΔHF(CO2,g) + 4ΔHF(HCl, g) − ΔHF(CCl4, g) − 2ΔHF(H2O, g)

= (−393.7 − 4 × 92.5 + 106.7 + 2 × 241.8) KJ mole-1

= 173.4 KJ mol-1.

  1. Born-Haber Cycle

This cycle devised by Born and Haber in 1919 relates the lattice energy of a crystal to other thermo-chemical data. The elements in their standard state are first converted to gaseous atoms, then to ions and finally packed into the crystal lattice 

Na(s) + Cl2(g) → NaCl(s), ΔHf

ΔH1 ↓ ΔH2 ↑ ΔH5 ΔH


Cl(g) Cl(g)


Na(g) Na+(g)


H1 = Sublimation energy ΔH2 = , ΔH3 = I.E. ΔH4 = E.A.

ΔH5 = Lattice energy

ΔHf = ΔH1 + ΔH2 + ΔH3+ ΔH4 + ΔH5


Illustration -4: Use the Born-haber cycle and the following data to calculate the electron affinity of chlorine 

ΔHf (RbCl) = −102.9 kcal/mol

ionization energy of Rb = 95 kcal/mol

ΔHsub(Rb) = +20.5 kcal/mol

B.E. (Cl2) = 54 kcal/mol

Cattice energy of RbCl = −166 kcal/mol



Rb(s) + Cl2(g) RbCl (s)

  20.5 ↓ 27 ↑ −166

Cl(g) Cl(g)

Rb(g) Rb+(g)

ΔH = −E.A. = −102.9 − (20.5 + 27 + 95 − 166) = −79 kcal/mol


  1. Bond Dissociation energy

For a diatomic molecule, the band dissociation energy is same as bond energy, but for polyatomic molecule having various types of same bonds, bond energy is the average amount of energy to break one mole bonds of that type in gaseous molecule.


Example: H − OH(g) → H(g) + OH(g), ΔH1 = 498 KJ

O − H (g) → H(g) + O(g), ΔHg = 430 KJ

ΔH0 − H = = 464 KJ mol−1

Enthalpy of reaction = [sum of bond energies of reactants] – [sum of bound energies of products]

Illustration -5: Using the bond enthalpy data given below, compute the enthalpy of formation of gaseous isoprene 

CH2 = C(CH3)CH = CH2

B.E. (C − H) = 413.4 KJ mole-1

B.E. (C − C) = 347.7 KJ mol-1

B.E. (C = C) = 615.1 KJ mol-1

Enthalpy of sublimation of C(graphite) = 718.4 KJ mol-1

Enthalpy of formation of H(g) = 218.0 KJ mol−1


Solution: We have to compute the enthalpy change of the following reaction

5C(graphite) + 4H2(g) → CH2 = C (CH3) CH = CH2 (g)

ΔH = [5ΔHC(s) → C(g) + 4ΔHH-H] − [2ΔHC = C + 8ΔHC-H + 2ΔHC−C]

= [5 + 718.4 + 4 × 2 × 218] – [2 × 615.1 + 8 × 413.4 + 2 × 347.7] 

ΔH = 103.2 KJ mol−1



  1. The heat of combustion of ethane gas is 368 kcal/mol assuming that 60% of the heat is useful, how many m3 of ethane measured at STP must be burnt to supply enough heat to convert 50 kg of water at 10°C to steam at 100°C. specific heat of water is 1 cal/g. Heat of vaporization of H2O is 540 cal/g.   
  2. Calculate the heat of neutralization from the following data. 200 ML of 1M HCl is mixed with 400 ML of 0.5 M NaoH. The temperature rise in calorimeter was found to be 4.4°C. Water equivalent of calorimeter is 12g and specific heat is 1 cal/ml/degree for solution. 


  1. Using bond energy data, calculate heat of formation of isoprene

5C(s) + 4H2(g) → H2C = C — C = CH2

| |


Given C — H = 98.8 Kcal, C — C = 83 Kcal

H — H = 104 Kcal, C = C = 107 Kcal

And C(s) → C(g) = 171 Kcal  


  1. Calculate the resonance energy of C6H6 using Kekule formula for C6H6 from the following data

(1) for C6H6 = −358.5 KJ mol−1

(2) Heat of atomization of C = 716.8 KJ mol-1

(3) Bond energy of C — H, C — C and C = C and H — H are 490, 340, 620, 436.9 KJ mol-1 respectively.  


  1. One mole of water at 373 K is converted into steam at a pressure of one atm, 40.68 KJ of heat is absorbed. The molar volume of water and steam are 18 ML and 30600 ML. Calculate ΔE for process (1 atm = 1.01 × 105 NM-2)  
  2. Using the data given below, calculate the bond enthalpy of C — C and C — H bonds.

(ethane) = −1556.5 KJ mol-1

(propane) = −2117.5 KJ mol-1

ΔHC(s) → C(g) = 719.7 KJ mole-1

Bond enthalpy of H — H = 435.1 KJ mol-1

(H2O, ) = −284.5 KJ mol−1

(CO2, g) = −393.3 KJ mol-1


  1. When 12.0 g of carbon (graphite) reacted with oxygen to form CO and CO2 at 25°C and constant pressure, 313.8 KJ of heat was released and no carbon. Calculate the mass of oxygen consumed. 

Given: (CO, g) = −110.5 KJ mol-1

(CO2, g) = −393.5 KJ mol-1 

  1. A couple sitting in a warm room on a winter day takes kg of cheese sand witches (an energy intake of 8130 Kj of both). Supposing that none of energy is stored in body, what mass of water would they need to perspire in order to maintain their original temperature. The enthalpy of vaporization of water is 40.65 KJ mol-1.


  1. The heat evolved on combustion of 1.00 g of starch, (C6H10O5)x into CO2(g) and H2O() is 4.18 kcal. Calculate the standard enthalpy of formation of 1.00 g of starch. Given H2O() = − 68.32 kcal mol-1, CO2(g) = -94.05 kcal mol-1.
  2. All sample of a mixture of methane gas and oxygen, measured at 25°C and 740 ton, was allowed to react at constant pressure is a calorimeter which together with its contents had a heat capacity of 1260 cal/k. The complete combustion of the methane to CO2 and H2O caused a temperature rise in the calorimeter of 0.667 k. What was the mole percent of CH4 in the original mixture? 

ΔH combustion of CH4(g) = -210.8 kcal/mol




  1. 3.196 m3
  2. −13.464 Kcal
  3. 20.6 Kcal
  4. –150.0 kJ mol–1
  5. 37.601 kJ
  6. B.E. (C — C) = 414.0 KJ mol-1, B.E. (C — H) = 344.3 KJ mol-1)
  7. 27.5 Kg
  8. 3.6 Kg
  9. 9.768 Kcal
  10. 10%



Faraday’s Law Of Electrolysis

  1. First Law:

W ∝ It

W = ZIt, where Z =  E.C.E.

I = current in amp.

t = time in sec.


  1. Second Law of Electrolysis

For constant value of Q or, It

W ∝ E where W = wt.

E = Equivalent weight

or, W ∝ E ∝ Z


Combining I & II Law Of Electrolysis

W = where A = Atomic weight

n = no. of moles of transferred electrons

F = Faraday’s

or, = 96,500 c

∙ 2nd law of electrolysis follows


law of equivalents i.e. same equivalent of the reactant reacts and the same equivalents of 

the products produce.


Illustration -1: A current of 15 amperes is employed to plate nickel in a NiSO4 bath. Both Ni and H2 are formed at cathode. The current efficiency with respect to the formation of Ni is 60%. (a) How many g. of Ni are plated out in one hour? (b) What is the thickness of the plating of the cathode consisting of a square sheet of metal of 4 cm. side which is coated on both faces? The density of nickel is 8.9 g/cc. (c) What volume  of H2 (STP) is formed per hour?


Solution: Total charge passed in one hour = (15 × 60 × 60) C

= 54000 C

But current efficiency with respect to Ni is 60%

Therefore, charge used up in deposition of Ni = 54000 ×   C

= 32400 C

And remaining charge will be used up to evolve H2

i.e. equal to = (54000 – 32400) C

= 21600 C

Ni+2 + 2e Ni 

Wt. Of Ni (plated) = gm =  9.85 gm

Now, volume of square sheet of Ni metal required to coated on
both faces =

so volume of square sheet of Ni metal required to coated on one face

which should be equal to = (area of square) × thickness of plate

= 16  cm3   × thickness of plate

thickness of plate = = 0.035 cm

Now, 2H+ + 2e H2

Volume of H2 evolved at STP =

= 2.5 litre


Illustration-2: Per di-sulphuric acid (H2S2O8) can be prepared by electrolytic oxidation of H2SO4 as 2H2SO4 H2S2O8 + 2H+ + 2e. O2 and H2 are byproducts. In such an electrolysis 0.87 g of H2 and 3.36 g of O2 were generated at STP. Calculate the total quantity of current passed through the solution to carry out electrolysis. Also report the weight of H2S2O8 formed.


Solution: At anode: 2H2SO4 H2S2O8 +  2H+ + 2e

2H2O O2 + 4H+ + 4e

At cathode: 2H+ + 2e H2

In electrolysis of H2SO4, firstly H2SO4 oxidises at anode untill all H2SO4 converts into H2S2O8 and then O2 is evolved by the electrolysis of H2O.

simultaneously at that time H2 is evolved at cathode which continue during whole electrolysis process.

So, total charge required to produce 0.87 gm H2 = C

= 83955 C

Now, quantity of charge required to produce 3.36 gm
O2 = C = 40530 C

Thus, quantity of charge used up in the formation of H2S2O8  
= (83955 – 40530) =  43425 C

Therefore, weight of H2S2O8 formed =

=  43.65 gm


Illustration-3: A current of 40 microamperes is passed through a solution of AgNO3 for 32 minutes using Pt electrodes. An uniform single atom thick layer of Ag is deposited covering 43% cathode surface. What is the total surface area of cathode, if each Ag atom  covers 5.4 × 10-16 cm2?


Solution: Total charge passed = (4 ×10–6 × 32 × 60) C =  0.0768 C

Ag+ + e Ag

So, weight of Ag deposited at cathode surface = gm

= 8.6 × 10-5 gm

Now, total no. of Ag atoms deposited =

= 4.8 × 1017

So, surface area of cathode which is covered by Ag atoms 

= (4.8 × 1017 × 5.4 × 10–16) = 259.2 cm2

Total surface area = cm2

= 602.8 cm2


Electrochemical Cells: In which chemical energy converts into electrical energy.

Single Electrode potential: Due to electrical double layer formation at the phase boundary of metal and its solution, a potential is developed. The potential difference between the metal and the metal ion in which electrode is dipped, is called electrode potential and denoted as Eel.


In standard states i.e. 1 atm pressure, 298 K temperature and at 1 M concentration it is called standard electrode potential ()


IUPAC Cell Representation:


Anodic oxidation part || Cathodic reduction part


Zn | Zn++ (C1) || Cu++ (C2) /Cu ⎯→ Daniel cell


Where: double slash (||) indicates salt bridge.


A slash (|) indicates phase difference of the electrolyte.


Rules: i) Concentration term of electrolyte is written in bracket.

  1. ii) For “ous” and ìc” ion of the same metal, a single slash is written with Pt.

iii) For gaseous cell 1 atm in bracket is written and Pt is used.

e.g. Pt,

Cell reaction


Oxidation half cell: Zn ⎯⎯→ Zn++ + 2e

Reduction half cell: Cu++ + 2e ⎯⎯⎯→ Cu


Zn + Cu++  ⎯⎯⎯→ Zn++ + Cu

E.M.F. of a Cell:

or, Ecell is the potential difference between the two half cell. Since the potential difference is the driving force for electrons.

= (only oxidation potential)


Relation between EMF & free energy

Or, ΔG = maximum amount of electrical work

ΔG = –nF Ecell

At standard states

ΔG0 = –nF

From Nernst equation

Ecell =

Where Q = reaction quotient

n = no. of moles of transferred electrons

  • If Ecell = positive, reaction occurs spontaneously in the forward direction.
  • If Ecell = negative, reaction occurs spontaneously in the reverse direction.
  • If cell reaction is reversed, the sign of Ecell changes.

Relation Between And K Equilibrium Constant


At equilibrium ΔG = 0

∴ Ecell  = 0

So, from Nernst equation

Ecell = logQ

At equilibrium, Q becomes  K  Where K = equilibrium constant

Ecell = 0

∴ 0 = log K 

or, logK =


If two half reactions having potentials and are combined to give a third-half reaction having potential , then 

= +

– n3F = – n1 – n2


Relation between enthalpy of reaction and Ecell.

ΔH = nF

where = temperature coefficient with EMF

= Rate of change of EMF with temperature

ΔS = , where ΔS = entropy change.

Concentration Cell:

Cells in which two electrodes of the same materials are dipped into solution of its ions having different concentrations are known as concentration cells or, cells having

e.g. Ag(s) | Ag+(Ca) || Ag+(Cc) | Ag

Anode reaction Ag(s) ⎯→ + 1e


Cathode reaction + 1e ⎯→ Ag(s)



= 0

∴ From Nernst equation,

or, Ecell = , where n = 1 Where Ca = concentration of Ag+ at anode

            CC = concentration at cathode

For Ecell = +ve, Cc > Ca

Likewise the emf of the cell consisting of two hydrogen electrodes operating at different pressures P1 and P2 (P1 > P2) and dipping into solution of HCl is:

Ecell = log

Where P2 > P1


Illustration -4: Calculate the e.m.f. of the cell.

Ka for CH3COOH = 1.8 × 10–5

Kb  for NH4OH = 1.8 × 10–5


Solution: Since CH3COOH CH3COO + H+

∴ [H+] =

= 1.342 × 10–3 mol lit–1

Similarly [OH] = =

= 0.424 × 10–3 mol lit–1

∴ [H+] = = 2.359× 10–11 mol lit–1

Now the cell is

 Ecell = log

= log

= – 0.4575 volt


Illustration-5: For the galvanic cell

Calculate the emf generated and assign correct polarity to each electrode for the spontaneous process after taking an account of the cell reaction at 25°C. Given Ksp (AgCl) = 2.8× 10–10

Ksp (AgBr) = 3.3 × 10–13

Solution: The galvanic cell is 

[Ag+] =

= 1.4 × 10–9

[Ag+] = = = 3.3 × 10–11

Now the given cell is converted into the following cell

Ag | Ag+ (1.4 × 10–9) || Ag+ (3.3 × 10–11) | Ag

Ecell = log

So, Ecell = log

= – 0.037 volt

To get Ecell positive, the polarity of the cells should be reversed i.e. cell is

Ag | AgBr(s) KBr (0.01 M) || AgCl, KCl  (0.2 M) | Ag and Ecell = + 0.037 V



Problem-1: Calculate the potential of an indicator electrode versus the standard hydrogen electrode which originally 0.1M  MnO4 and 0.8M H+ and which has been treated with 90% of the Fe2+ necessary to reduce all the MnO4 to Mn2+.

MnO4 + 8H+ + 5e ⎯→ Mn2+ + 4H2O E° = 1.51 V


Solution: Let us consider Galvanic cell is 

H+ (1M)  | H2(1atm), Pt || MnO4 (H+) | Mn+2, Pt 

Anode half cell : 2H+ (1M) ⎯→ H2 (1atm) + 2e

Cathode half cell:  MnO4 + 8H+ + 5e ⎯→  Mn+2 + 4H2O

Initial Conc.: 0.1 0.8   0 0

Alter Complete  

reaction with Fe+2

(0.01) (0.08) (0.09)  

So, electrode potential of indicator electrode 

= 1.51 –

= 1.51 –

= 1.51 – (5.36 × 109) = 1.51 – 0.1149

= 1.395 V

Thus, potential of an indicator electrode versus the SHE is 1.395 V because  ESHE = 0   


Problem-2: The EMF of the following cell is –0.46 V

Pt,H2 (1 atm) | (0.4 M), (6.4 × 10-3 M) || Zn+2 (0.3 M) | Zn if = – 0.76 V, Calculate Ka of i.e. for the equilibrium 

H+ +


Solution: Anode half cell: H2 2H+ + 2e

Cathode half cell: Zn+2 + 2e Zn


Net cell reaction Zn2+ + H2 2H+ + Zn

– 0.46 = – 0.76 –

0.3 = –

= –10.1523

= 7.04 × 10–11

[H+] = 4.6 × 10–6 M

Now,         H+         +      

(0.4 M) (4.6 ×10–6M) (6.4 × 10–3 M)

Ka =

= = 7.36 × 10–8


Problem-3: The standard reduction potential at 25°C for the reaction
2H2O + 2e → H2 + 2OH is –0.8277 V. Calculate the equilibrium constant for the reaction 2H2O → H3O+ + OH at 25°C.


Solution: Consider an electrode H as

2H+ + 2e → H2 E0 = 0.0 V

Given electrode is

2H2O + 2e → H2 + 2OH E0 = –0.8277 V

∴Cell reaction are 

At anode: H2 + 2OH → 2H2O + 2e

At cathode: 2H+ + 2e → H2


Net cell reaction is 2H+ + 2OH → 2H2O

Ecell  =

= (0.8277 + 0) – = 0.8277 – 0.591

At equilibrium, Ecell  = 0

∴0.8277 = 0.0591

= 14.0051

log Kw = –14.0051

Kw = 9.88 ×10–15

Thus for 2H2O H3O+ + OH

Kw = [H3O+]  [OH]

Kw = 9.88 × 10–15


Problem-4: How many moles of iron metal will be produced by passage of 4A of current through 1L of 0.1 M Fe3+ solution for 1 hour? Assume only iron Fe+3 + 3e Fe is reduced.


Solution: Total charge passed = (4× 1 × 60 × 60) C= 14400 C = 0.149 F

First of all ferric ion charges to ferrous and  than  ferrous changes to Fe.

Fe+3 + e ⎯→ Fe+2

Fe+2 + 2e  ⎯→ Fe

So, charge used up in formation of Fe(II) = 0.1 F

Therefore, remaining charge left = (0.149 – 0.1) F   =  0.049 F

Which is used in conversion of Fe+2 to Fe 

So, number of mole of Fe produced = 0.0245

Problem-5: The standard reduction potential for the half-cell

+ 2H+(aq) + e ⎯→ NO2(g) + H2O is 0.78 volt

  1. i) Calculate the reduction potential in 8 M H+
  2. ii) What will be the reduction potential of the half cell in a neutral solution? Assume all other spacies to be at unit concentration.


Solution: The half cell is

(aq) = 2H+ (aq) + e ⎯→ NO2(g) + H2=O

Given log


= 0.78 –

= 0.78 –

= 0.78 + 0.0592 log 64 = 0.78 + 0.0592 log26 = 0.88969 volt

In neutral solution

[H+] = 10–7 M

∴ Eel = log =

= 0.78 – = 0.78 –

= 0.78 – 0.8288 = – 0.0488 volt




  1. Calculate emf of silver-silver chloride electrode immersed in 1 M kcal at 25°C. Given that Ksp of AgCl = 1.8 × 10–10, = 0.799 volt.


  1. Two weak acid solutions and each with same concentration and having pKa values 3 and 5 are placed in contact with hydrogenelectrode (1 atm, 25°) and are interconnected through a salt-bridge. Find e.m.f. of cell.


  1. Estimate the cell potential of a Daniel cell having 1.0 M Zn++ and originally having 1.0 M Cu+2 after sufficient ammonia has been added to the cathode compartment to make the NH3 concentration 2.0 M. Given and are 0.76 and –0.36 volt respectively. Also equilibrium constant for the [Cu(NH3)4]+2 formation is 1 × 1012.


  1. For the following cell:

Ag(s) | Ag+ (satd. AgI(aq) || Ag+ (0.10 M) | Ag(s)

Ecell = 0.417  calcualte Ksp for AgI

  1. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298 K, if the emf of the cell

Ag(s) | Ag+ (satd Ag2CrO4 solution) || Ag+ (0.1 M) | Ag(s) is

0.164 V at 298 


  1. A cell contains two H-electrodes. The negative electrode is in contact with a solution of 10–6 MH+ ions. The emf of the cell is 0.118 volt at 25°C. Calculate [H+] at positive electrodes.


  1. The reduction potential diagram for Cu in acid solution

Find x. Does Cu+ disproportionate in solution?


  1. For the reaction 

H2(g) + 2AgCl(s) + 2H2O(l) ⎯→ 2Ag(s) + 2H3O+ + 2Cl at 25°C

The standard free energy of formation of AgCl(s), H2O(l), H3O+, Cl are –110,
–237, –200 –α, (–168 + α) KJ / mol (where α is not known). Calculate the cell voltage if this reaction is run at 25°C and 0.8 atm in a cell in which [H3O+] and [Cl] are 0.006 M and 0.02 M respectively.


  1. A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4 at 25°C. The measured potential difference between the rod and the SHE is 0.589V, the rod being positive. Calculate the solubility product of silver oxalate. Given  = + 0.8V
  2. When AgCl is dissolved in a large excess of NH3, practically all silver can be assumed to exist in the form of a single species . Compute the values of m and n using the following two cells.

Ag | 0.379 × 10-3M AgCl,  1M NH3 || 37.9 ×10-3M AgCl, 1M NH3 | Ag

Ecell = 0.1185V at 298K

Ag | 3.4 ×10-3M AgCl,  1M NH3 || 3.4 × 10-2M AgCl, 0.1M NH3 | Ag

Ecell = 0.1263V at 298K



  1. 0.222 volt 2. 0.0592
  2. 0.71 volt 4. Ksp (AgI) = 7.73 × 10–17)


  1. Ksp = 2.287 × 10–12 mol3 lit] 6. 10–4 M]


  1. x = + 0.325 volt 8. 0.446 V
  2. 9.7 × 10–12 M3 10. m = 1, n = 2



On the basis of electronic configuration, elements may be divided into 4-blocks

  1. s-block elements (normal elements):

General  electronic configuration nsx where x = 1 to 2

Since maximum e = 2, so, maximum groups = 2

i.e. Group IA and Group II A

alkali metal alkaline earth metal

General characteristics

  1. i) It has no variable and oxidation states.
  2. ii) Doesn’t form complex, generally

iii) Forms colourless compounds

  1. iv) Compounds are diamagnetic.
  1. p-block elements (Representative Elements)

General electronic configuration: ns2 npx where x = 1 to 6

Contains six groups  i.e. Groups III A to zero

i.e. Group 3 to 18.

General characteristics

  1. i) Most of elements are metalloids and non-metals and some are metals.
  2. ii) They show variable valency and oxidation state due to vacant d-orbital and inert pair effect.
  3. d-block elements

Transition elements

or, Transition metals (T.M)

General Electronic configuration = ns2 (n – 1)dx np6 where x = 1 to 9 T.E.

It contains 10 group i.e. III B to VII B, VIII, IB & IIB


General characteristics

  1. i) It shows variable valency and oxidation states.
  2. ii) It shows complex formation and forms coloured compounds.

iii) Generally forms paramagnetic compounds

  1. iv) M. shows catalytic properties.
  2. f-block elements

Inner-transition elements

or Inner T.M.(Transition metals)

General electronic configuration = ns2 (n–2) fx (n–1)d0 to 1 np6

Here x = 1 to 14

Transuranic element: Elements from atomic no. (Z) = 93 to onwards

Atomic Radius

iCovalent radius – One half of the distance between the nuclei (internuclear distance ) of two covalently bonded like atoms in a homodiatomic molecule is called the covalent radius of that atom. The covalent bond must be single covalent bond. The covalent radius (rA) of atom A in a molecule A2 may be given as –

i.e. the distance between nuclei of two single covalently bonded atoms in a homodiatomic molecule is equal to the sum of covalent radii of both the atoms.

dA-A = rA+rA

In a heterodiatomic molecule AB where the electronegativity of atoms A and B are different, the experimental values of internuclear distance dA-B is less than the theoretical values. According to Schomaker and Stevenson (1941) – 

DA-B = rA + rB – 0.09 Δx

Where Δx is the difference of electronegativities of the atoms of A and B.

According to  Pauling – If the electronegativities of the two atoms A and B are xA and xB respectively then

DA-B = rA + rB – (C1xA – C2xB)

C1 and C2 are the Stevenson’s coefficients for atoms A and B respectively.


Ionic Radius

A neutral atom changes to a cation by the loss of one or more electrons and to an anion by the gain of one or more electrons. The number of charge on cation and anion is equal to the number of electrons lost or gained respectively. The ionic radii of the ions present in an ionic crystal may be calculated from the internuclear distance between the two ions

Illustration -1: The radii of Ne is greater than the radii of fluorine 

Solution: In fluorine, the radii means the atomic or covalent radii which is actually half the intermolecular distance between  2 atoms whereas in neon the radii means the  Vanderwaals radii as neon is not a diatomic  molecule. Vanderwaals radii is actually half the distance between adjacent molecule. So Vanderwaal’s radii being larger than atomic radii, neon has got a larger radii than F2

Illustration-2: Lithium show similarity with magnesium although they are not in the same period or groups 

Solution: On  moving across a period the charge on the  ions increases and the size decreases, causing the polarising power to increase. On moving down a group the  size increases and polarising power decreases. On moving diagonally i.e., from Li to Mg these two effects partly cancel each other and so there is no marked change in properties. 

Ionisation potential or Ionisation Energy

  1. i) The amount of energy required to remove the most loosely bound electron of the outermost shell (i.e. the outermost electron) from one mole of an isolated gaseous atom of an element in its ground state to produce a cation is known as ionisation energy of that element.
  2. ii) Because ionisation energy is generally expressed in electron volts, so it is also known as ionisation potential.

iii) Energy required for the removal of first, second and third electron from the gaseous atom is called first, second and third ionisation energy respectively.

A(g)+I1 → A(g)+ +

A+(g)+I2 → A(g)2+ +

A(g)2+ + I3 → A(g)3+ +

  1. iv) The order of first, second and third ionisation energies may be given as 


This is because second and third electron is removed from monopositive and dipositive cations respectively. Effective nuclear charge increase with the increase of positive charge. So the attraction between the nucleus and the outermost electron increases and more energy is required for the removal of electron.




  1. ii) Ionisation potential α Effective nuclear charge (Zeff)
  2. iv) Ionisation potential α


Illustration – 3: The ionisation energy of potassium is 415 kJ / mole  while that of Cu is 750 kJ / mole 


Solution: The configuration of K is  [Ar] 4s1 while that of Cu is [Ar] 3d10 4s1. So the answer lies in the presence of 10 d electrons. As the  d electrons have got very poor screening effect so the nuclear charge is not properly  screened. Therefore the effective nuclear charge is high and so the outermost electron is tightly held and high energy is needed to remove the electron.


Illustration – 4: The ionisation energy of  the  coinage metals fall in the order Cu > Ag < Au.


Solution: In all the 3 cases an s-electron in the unpaired state is to be removed. In the case of Cu a 4s electron is to be removed which is closer to the nucleus than the  5s electron of Ag. So I.P. decreases from Cu to Ag. However from Ag to Au the 14 f electrons are added which provide very poor shielding  effect. The nuclear charge is  thus enhanced  and therefore the outer electron of Au  is more tightly held and so the IP is high.

Electron affinity 

  1. i) The amount of energy released when an electron is added to the outermost shell of one mole of an isolated gaseous atom in its lower energy state.
  2. ii) Electron affinity just defined is actually first electron affinity since it corresponds to the addition of one electron only. In the process of adding further electron, the second electron will be added to gaseous anion against the electrostatic repulsion between the electron being added and the gaseous anion. Sometimes energy instead of being released is supplied for the addition of an electron  to an anion.

Factors affecting the magnitude of electron affinity 

  1. i) Electron affinity α
  2. ii) Electron affinity α Effective nuclear charge (Zeff)

iii) Electron affinity α

  1. iv) Stability of half  filled and completely filled orbitals – The stability of half filled and completely filled degenerate orbitals of  a sub shell is comparatively more, so it is difficult to add electron in such orbitals and lesser energy is released on addition of electron hence the electron affinity value will decrease.


Illustration – 5: Electron affinity of SF5 is among the highest known but that of SF6 is quite modest.


Solution: Sulfur in SF6 is saturated in terms of maximum covalency and maximum co-ordination number and so it has very little tendency to attract electron. In SF5 which is actually a free radical so it has got a very strong tendency to attract an electron to fulfill its covalency and co-ordination number. There fore the electron affinity of SF5 is high.


  1. i) It may be defined as the tendency of an atom to attract shared pair of electrons towards itself in a covalently bound molecules.
  2. ii) The numerical value of the electronegativity of an element depends upon its ionisation potential and electron affinity values. Higher ionisation potential and higher electron affinity values implies higher electronegativity value.



Some arbitrary scales for the quantitative measurement of electronegativities are as under

  1. i) Pauling’s scale – Pauling related  the resonance energy(Δ)AB of a molecule AB with the electronegativities of the atoms A and B. If xA and xB are the electronegativities of atoms A and B respectivey then

0.208 = xA – xB if xA > xB

or 0.043 ΔAB = 23.06 (xA – x-B)2

ΔAB = EA-B(experimental) – EA-B(theoretical)  where EA-B is the energy of A-B bond. In a purely covalent molecule, AB, the experimental and theoretical values of bond energy A-B are equal.

So ΔAB = 0

Or 0=23.06 (xA – xB)2

Or xA = xB

In an ionic molecule AB, EA-B(experimental) is more than EA-B(Theoretical).

Pauling assumed the electronegativity value of fluorine 4 and calculated the electronegativity values of other elements from this value.

iii) Alred Rochow’s electronegativity 

The electronegativity of an element is the electrostatic force of attraction between the electron present on the circumference of the outermost shell of this atom and the atomic nucleus. If the distance between the circumference of outermost shell and the nucleus is r and the effective nuclear charge Zeff then 


  1. i) Atomic radius —As the atomic radius of the element increases the electronegativity value decrease.

iii) Effective nuclear charge

The electronegativity value decreases as the effective nuclear charge on the atomic nucleus increases.

Electronegativity α Effective nuclear charge (Zeff)

iii) Oxidation state of the atom —

The electronegativity value increases as the oxidation state (i.e. the number of positive charge) of the atom increases.

  1. iv) Hybridisation state of an atom in a molecule —if  the s- character in the hybridisation state of the atom increases because  s-electrons are comparatively near to the nucleus. For example the electronegativity values of C-atom in various hybridisation states are as under —

Hybridisation states sp3 sp2 sp

s-Character 25% 33.33% 50%

Electronegativity 2.48 2.75 3.25


s-character is increasing

So the electronegativity value is increasing 


Problem 1: The first I.P. of nitrogen is greater than oxygen while the reverse is true for their second I.P. values.

Solution: Nitrogen has half filled configuration (p3) while oxygen has p4 configuration. So due to half filled stability the I.P. of nitrogen is more, while after the removal of the 1st electron oxygen attains p3 configuration i.e. half filled stability.

Problem 2: On moving down the group the higher oxidation state becomes less stable 

Solution: On moving down the group due to inert pair effect higher oxidation state becomes less stable.

Problem 3: The first I.P. of Na is the same as electron affinity of Na+ ion

Solution: First I.P. means energy needed to remove an electron while electron affinity is energy released when an electron is added. As the energy needed is equal to the energy released. So I.P. is same as electron affinity.

Problem 4: The second I.P. for alkali metals shows a jump while the third I.P. for alkaline earth metal shows a jump.

Solution: As alkaline earth metals have ns2 configuration, so after removal of 2 electrons, it attains the inert gas configuration. So there is a jump in I.P.

Problem 5: Mn2O7 is an acidic oxide 

Solution: In Mn2O7 manganese is in the highest possible oxidation, so electron deficient and hence tendency to accept electron is very high. So acidic in nature.


  1. NaOH behaves as a base while Zn(OH)2 is amphoteric, why? 
  2. The first IP of Be is less than that of boron. Explain  
  3. Electron affinity of chlorine is higher than fluorine, why? 
  4. Among fluorine – fluorine bond and chlorine – chlorine bond, which is more stronger and why?
  5. The successive I.P values of B are 8.3, 23, 38 and 260 eV. Why is the 4th value so high?
  6. Of all metals, gold has got a relatively high electron affinity. Explain
  7. The third ionisation potential of nitrogen less than that of chlorine. Explain 
  1. The electronegativities of B, Al, Ga are 2.0, 1.5, 1.6 respectively. The trend is not  regular, why?
  2. Electron affinity of oxygen is negative but a number of stable ionic oxides are known, why?
  1. A chemist has one mole of X atoms He finds that on absorption of 410 kJ, half of the X atoms transfer one electron to the other half. If all the resulting X′ ions are subsequently converted to X+ ions, an additional 735 kJ must be added. Find out the ionisation potential and electron affinity of X.


The formation of a molecule from its constituent atoms must be energetically favourable leading to minimum energy i.e., the molecule formed must be more stable than the separated atoms.

3 types of bonding generally exists (i) ionic (ii) covalent (iii) coordinate covalent 

Ionic Bond: Transfer of electrons from one atom to the other whereby each atom may attain octet in their outermost shell. 

Criteria for ionic bond: (I) One of the species should have low I.P. ii) The other should have electron affinity 

The first property is generally exhibited by alkali and alkaline earth metals while the second property in shown by non metals.

Eg.s of ionic bond: KCl, Na2CO3, Al2O3 etc.

Covalency: This involves mutual sharing of electron by virtue of which they can attain their inert gas configuration.

E.g. CO2, O2, HCl etc.

Coordinate covalent: This is a special type of covalent bond where the sharing pair is donated by one species.

Eg. SO2 , , SO3 etc.

Hybridisation: Process of mixing of dissimilar orbitals of similar energy / comparable energy to give rise to an equal number of  orbitals with mixed character. The hybrid orbitals undergo better overlap and form stronger bonds than pure orbitals in conformity with the most stable geometry for a molecule.

Main points of hybridisation;

  1. i) Only orbitals of same energy belonging to the same atom or ion undergoes hybridisation.
  2. ii) of hybridised orbitals  = total number of sigma bonds + lone pair/s 

iii) Hybridised bond in stronger than the non – hybridised bond.

VSEPR Theory

This theory stands form the general principle that valence shell electron occupy essentially localised orbitals. Mutual interaction among the electron orient the orbitals in space to an equilibrium position where repulsion becomes minimum. The extent of repulsive interactive follows the order.

Lone pair —  lone pair > lone pair — bond pair  > bond pair — bond pair 

Illustration-1: Bond angle of NH3 is 107° while that of H2O is 104° 

Solution: NH3 contains 1 lone pair while H2O has two lone pair. The lp – b.p. and lp – lp repulsion tend to reduce the bond angles. The  bond pair – bond pair tends to open the angles. But as this interaction is weaker than the lone pair  bond pair repulsion. So the ultimate bond angle is smaller. ∠of NH3 = 107° and ∠H2O is 104°. H2O  has a smaller bond ∠ because there are 2 lone pair which reduces it further.

Illustration -2: The bond angle of NH3 is 107°  while that that of NF3 is 102°.

Solution: Both NH3 and NF3 have a lone pair of electrons. But  fluorine being highly electronegative, the bond pair electrons are drawn more towards F in NF3, whereas in NH3 it is drawn towards N. So in NF3 the bond pairs being displaced away from the central atom has very little tendency to open up the angle. But in NH3 this opening up is more as the bond pair electrons are closer to each other. So bond ∠ of NF3 is less than NH3.

Dipole moment:  The degree of polarity is measured by dipole moment(μ)

μ  = e × d [ e  = electronic charge 

d = dipole length]

Unit of dipole moment is Debye 

1D = 10–18 = 3.3 × 10–30 Cm

Vectorical resultant of bond moment (B1 & B2)

μ =

If a molecule is symmetrical the molecular dipole  = 0

% ionic character  =

Hydrogen Bonding

An atom of hydrogen linked covalently to a strongly electronegative atom can establish an extra weak attachment to another electronegative atom in the same or different molecules. This attachment is called a hydrogen bond. To distinguish from a normal covalent bond, a hydrogen bond is represented by a broken line eg X – HY where
X & Y are two electronagative atoms. The strength of hydrogen bond is quite low about 

2-10 kcal mol–1 or 8.4–42 kJ mol–1 as compared to a covalent bond strength 50–100 kcal mol–1 or 209 – 419 kJ mol–1

Conditions for Hydrogen Bonding

  1. Hydrogen should be linked to a highly electronegative element.
  2. The size of the electronegative element must be small.

These two criterias are fulfilled by F, O, and N in the periodic table. Greater the electronegativity and smaller the size, the stronger is the hydrogen bond which is evident from the relative order of energies of hydrogen bonds.



Problem 1: CO2 has got dipole moment of zero why?

Solutions: The structure of CO2 is. This is a highly symmetrical structure with a plane of symmetry passing through the carbon. The bond dipole of C–O is directed towards oxygen as it is the negative end. Here two equal dipoles acting in opposite direction cancel each other and therefore the dipole moment is

Problem 2. Dipole moment of BF3 is zero.

Solution : BF3 is a planar molecule with the bond angles being 120°.

In each B-F bond the bond dipole is projected towards Fluorine. Now the direction of the resultant bond dipoles of these two bonds is shown in the diagram as R. If R is equal to the dipole of the 3rd B-F bond then R and B-F

dipole will neutralize each other as equal dipoles acting in opposite direction cancel each other. Now
let’s see the magnitude of R. As dipole moment is a vector quantity, so the resultant of two dipoles can
be obtained from the law of
vector addition. Suppose the
bond dipole is μ1.

R = a & b are the individual vectors

= θ = angle between vectors

= cos120° =

= = μ1 = μB–F

R is equal in magnitude to B-F bond dipole. Similar is the case with the other two resultants. So the net dipole of BF3 is zero.

Problem 3: o-hydroxy benzaldehyde is more volatile than p-hydroxy benzaldehyde.


Solution: More volatility means it has got lower boiling point. Now p-hydroxy benzaldehyde remains associated through intermolecular hydrogen bonding. But in o-hydroxy benzaldehyde only intramolecular H-bonding  formation takes places, as a result of which there is no association. So p-hydroxy benzaldehyde which remains as an associated species has got higher boiling point and so less volatile.

Problem 4: O-nitrophenol is less soluble in water than p-nitrophenol.

Solution: A substance is said to be soluble in water if it is capable of forming H-bonding with water molecule. In o-nitrophenol due to chelation the –OH group is not available to form hydrogen bond with water hence it is sparingly soluble in water. On the other hand the –OH group is available in p-nitrophenol to form H-bond with water and hence it is more soluble compared to the o-isomer. 

Problem 5: Ethyl alcohol (C2H5OH) has got a higher boiling point than dimethyl ether (CH3-O-CH3) although the molecular weight of both are same.

Solution: Though ethyl alcohol and dimethyl ether have the same molecular weight but in ethyl alcohol the hydrogen of the O-H groups forms intermolecular  hydrogen bonding with the OH group in another molecule. But in case of ether the hydrogen is linked to C is not so electronegative to encourage the hydrogen to from hydrogen bonding.

C2H5 C2H5


—–O ⎯ H ———-O ⎯ H ——

Due to intermolecular H-bonding, ethyl alcohol remains in the associated form and therefore boils at a higher temperature compared to dimethyl ether.


  1. PCl5 exists in the solid state in the form of [PCl4]+ [PCl6] yet it is a non-conductor of electricity. Explain why?
  2. Why is ice lighter than water?
  3. Suggest a reason for the large difference between the boiling points of butanol and butanal, although they have almost the same solubility in water.
  4. F2O has got a bond angle of 103° but Cl2O has got a bond angle of 110°. Why?
  5. The melting point of AgCl is only 455°C while that  of KCl is 776°C though the crystal radii of Ag+ and K+ ions are almost the same. Why?
  6. The geometry of trimethyl amine and trisilyl amine are not the same. Explain why?
  7. The bond angle of NH3 is 109° while that of PH3 is 93° . Explain why? 
  8. LiClO4 forms a stable trihydrate whereas the other alkali metal perchlorates do not why?
  9. Lithium nitrate decomposes to lithium oxide nitrogen dioxide and O2 while sodium nitrate decomposes to sodium nitrite. Explain why?
  10. Sulfur form SF6 but not SH6, Why?


  1. Bronsted – Lowry Concept

Proton donor is called acid

Proton acceptor is called base.

i.e. BH+ is conjugate acid of the base B and vice-versa. Stronger the acid, weaker is its conjugate base and vice-versa.


  1. Lewis Concept

Long pair acceptor is Lewis acid and lone pair donar is Lewis base.

Strength of Lewis acid ∝ nuclear charge

∝ no. of electronegative atoms

Order of strength of Lewis acid

⇒ Fe+3 > Fe+2

⇒ Li+ > Na+ > K+

Illustration –  1:  Arrange according to increasing Lewis acid character, 

BF3,BCl3,BBr3, BI3

Solution:  On the basis of the relative electronegativities of the halogens, one might expect the B atom in BF3 to be most electron deficient and hence most acidic. However, the compounds involve extensive π-interaction from a filled p orbital of the halogen and an empty p-orbital on B. The small F-atom forms the most stable p-p π-bond by efficient match of energy and size of the orbitals. Formation of an adduct with a base converts the geometry of bonds around boron from planar to pyramidal, thus rupturing the π-bonds. The Lewis acidity therefore increases with decrease in the extent of π-conjugation, as the halogen atom increases in size giving poorer overlap, i.e. from BF3 to BI3. Hence the order is 

BI3< BF3 < BCl3 < BBr3 .

Illustration –  2:  NF3 is least basic NCl3.

Solution: Due to greater electronegativity of F the lone pair of electrons on N is pulled more towards F making it less available for donation.

Illustration- 3:

Among maleic acid and fumaric acid which is a stronger  acid and  why.

Solution: Maleic acid is nothing but cis-butene dioic acid while fumaric acid is the trans isomer. The conjugate base of maleic acid  i.e., the maleate ion is stabilized due to intramolecular hydrogen bonding which is not possible in the fumarate ion. Now more than stable the conjugate base is the more more the acidity. Therefore  maleic acid is stronger  than fumaric acid.


Higher the electron-density available for protonation, higher the basic characters (strength). As the no. of alkyl group increases on N-atom, the electron density on N-atom increases for protonation so, basic strength increases up to secondary amine, but in 3° amine the base strength decreases

  1. i) due to sterric effect
  2. ii) due to decrease in solvation of cation via hydrogen bonding.

As the no. of alkyl group or, size of alkyl group increases, the degree of solvation of cation decreases, i.e. uptake of proton of decreases; so basic character decreases.


Illustration -4 . Explain the order of basic character of the following structures.

Solution: In IV the lone pair is delocalised i.e., involved in resonance and therefore least basic. In II  and I the lone pair is localised on nitrogen but in II due to presence of oxygen the lone pair is some what pulled towards oxygen whereas in I there is no such factor resulting in the high basicity.

Illustration – 5: Among N,N dimethyl aniline and N,N,2,6 – tetramethyl aniline which  one is a stronger base and why.


In N,N-2,6 – tetramethyl aniline the methyl groups on nitrogen and the methyl groups at the ortho position are very close to each other resulting in a steric crowding. Now to avoid steric crowding the C—N bond rotates and becomes perpendicular to the benzene ring. In this process the lone pair on nitrogen becomes perpendicular to the p-orbitals of benzene ring thereby inhibiting resonance. But in N,N dimethyl aniline there is no steric hindrance, so the lone pair is in the same plane as the benzene ring and undergoes resonance. Therefore the lone pair on  the tetramethyl derivative is more available and hence it is more basic.


Problem 1: Write down the conjugate base of the following 

(i) HS (ii) H2PO3 (iii) H2O , (iii) [Fe(H2O)6]3+

Solution: To form a conjugate base means removal of a proton 

So answer is 

(i) S– – (ii) HPO3– –

(iii) OH (iv) [Fe(H2O)5OH]2+

Problem 2: Which of the following oxide is most acidic 

Ag2O, V2O5, CO, N2O5

Solution: The oxide with the highest positive oxidation state on the element other than O should be most acidic. Oxidation states  of V in V2O5 and N in N2O5 are both +5. But the electronegativity of N is higher, making N2O5 the most acidic oxide.

Problem 3:  Which one is the strongest base  towards a proton,  ,

Solution: Bond energy (N—H>P—H) consideration suggest that will be the stronger base. This is also consistent with the relative strengths of the respective conjugate acids NH3<PH3.

Problem 4:  Arrange according to increasing Lewis acid character, 

B(n–Bu)3,  B(t–Bu)3

Solution: The highly branched tertiary butyl group involve appreciable back – strain (B-strain) when the boron atom changes to pyramidal environment on adduct formation. This destabilizes the adduct. Hence the order is 

B(t–Bu)3 < B(n–Bu)3

Problem 5:  Arrange according to increasing Lewis acid character, 

SiF4, SiCl4, SiBr4, Sil4

Solution: The order in this case is the reverse of that for BX3. π-conjugation from the halogen p-orbital to the Si-d orbital is not as intense as in the case of BX3 and the order of acidity follows the increase in electron withdrawing power of the halogen from I to F. Hence the order is 

SiI4 < SiBr4 < SiCl4 < SiF4


Transition Elements

The elements lying in the middle of the periodic table between group-2 and group-13 are known as d-block elements. These d-block elements are called transition elements because they exhibit transitional behaviour between s-block and p-block elements

Transition elements may be defined as the elements whose atoms or simple ions have partially filled d-orbitals. Zinc, Cadmium and mercury are not considered as transition elements because their atom as well as ion does not have partially filledd orbitals 

General Electronic Configuration : 

(n-1) d1-9 ns1-2

(n-1) stands for the penultimate shell and the d-orbital may have one to nine electrons and the s-orbital of the ultimate shell (n) may two (or in some cases one) electrons.

Physicochemical Properties 

  1. Metallic Character:
  2. Oxidation States: 
  3. Ionization Energy : 
  4. Complex formation (Complexation): 
  5. Coloured Complexes:  
  6. Catalytic properties: 
  7. Magnetic properties: 

Further paramagnetism is expressed in terms of magnetic moment, which is related to the no. of unpaired electrons as follows:


n – no. of unpaired electrons

B.M. – Bohr mageton, a unit of magnetic moment 

More the magnetic moment, more will be the paramagnetic character.   

  1. Formation of Alloys: 
  2. Interstitial Compound: 

Inner Transition Elements 

The f- block elements are known as inner transition  elements because they involve the filling of inner sub-shells (4f or 5f).

  1. Lanthanides: It consists of elements that follow lanthanum and involve the filling of 4f subshell.
  2. Actinides: It consists of elements that follow Actinium and involve the filling of 5f  subshell.
  3. Lanthanide Contraction: The steady decrease in the size of lanthanide ions (M 3+) with the increase in atomic no. is called lanthanide contraction.

Causes: As we move down the group from left to right in a lanthanide series, the atomic no. increases and for every proton in the nucleus the extra electron goes to 4f orbital. The 4f orbital is too diffuse to shield the nucleus effectively, thus their is a gradual increase in the effective nuclear charge experienced by the outer electrons. Consequently, the attraction of the nucleus for the electrons in the outermost shell increases with the increase of atomic no., thus size decreases.                    

Illustration- 1: Write down the IUPAC name of the complex  (1) K4[Fe(CN)6].

Solution: Firstly the +ve part should be named followed by the negative part in which the name of the ligand should be given in alphabetical order with the metal in the negative part ending in – ate (oxidation state in parethesis) Thus the name is Potassiumhexacyano – C-ferrate (II). Hyphen C is shown to indicate CN bonded via carbon

Illustration – 2: Now let take complex where there is an organic ligand like ethylene diamine. [Co(NH3)2Cl(en)2]Cl2

Solution: The approach is same as the earlier one with the exception that in case of  -en which is actually ethylene diammine the term  bis –  comes to indicate two – en groups instead of – bi. The name is Diamminechlorobis (ethylene diamine) cobalt(III) chloride 

Illustration – 3: Write IUPAC name of [Pt(Py)4][PtCl4].

Solution: Here both the positive and negative part has the same metal. Procedure is same earlier are the IUPAC name. Tetrapyridineplatinum(II) tetrachloroplatinate(II).

Illustration – 4: Write the IUPAC name of [Fe(NH3)4O2C2O4]Cl

Solution: In this charge on the complex part is +1. The ligand oxalato has a charge of –2, so iron should be in  +3 state meaning O2 to be neutral. Now had O2 been superoxo (O2) or peroxo (O2– – ) the negative charge of the ligands should have been –3 and –4 respectively. In that case Iron has to be +4 and +5 which is not possible. So O2 will behave as a neutral ligand and IUPAC name is Tetraammineoxalatodioxygeniron (III) chloride.

Bonding in Co-Ordination Compounds 

Pauling proposed a simple valence bond theory to explain bonding in Co-ordination Compounds According to this theory:

  1. Central metals loses a requisite no. of electrons to form the ion.
  2. The cation orbitals hybridize to form a new set of equivalent hybridized orbitals with definite directional properties.
  3. Each ligand contains a lone – pair of electrons. A covalent bond is formed by the overlap of a vacant hybridized metal orbital and a filled orbital of the ligand.
  4. If the complex have unpaired electron then it will be paramagnetic if not then it will be diamagnetic.

To understand the valence bond concept let us take some examples of co-ordination compounds.

  1. [Cr(NH3)6]3+, the central metal is in +3 oxidation state

Orbitals of Cr3+ion

d2sp3 hybridized

orbitals of Cr3+

d2sp3hybrid orbitals

of [Cr(NH3)6]3+

shape is octahedral


2. [Ni(CN4)]2-, Ni  is in +2 oxidation state


Atomic orbitals of Ni


dsp2 hybridized

orbitals of Ni+2


hybridized orbitals

of [Ni(CN)4]2-

shape is square planar

Illustration-  5: [FeF6]4– is an outer orbital complex while [Fe(CN)6]4– is an inner orbital complex 

Solution:  The oxidation state of Fe in both complexes is +2. i.e, the configuration is 4s°3d6.

Now F is a very weak ligand compared to CN. Now a strong field ligand is one which can pair up the electrons while a weak field ligand is one which cannot pair up. So in presence of CN′ the electrons in the d – orbitals of Fe2+ will be paired up. This will leave two  vacant 3d orbitals  which will then undergo hybridisation  to form d2sp3. As the 3d orbital has participated with 4s and 4p orbitals for hybridisation it is called  inner orbital complex. While had the outer d orbital ie 4 d participated with 4s and 4p, then the complex should have been called outer orbital complex. This happens in [FeF6]4–. In presence of  F which is a weak field ligand, electrons are not paired up and therefore there is no vacant 3d orbitals and so it is the outer 4d that participates.

Moreover in [Fe(CN)6]4– there is no unpaired electrons and hence diamagnetic while in [FeF6]4– there are unpaired electron, hence paramagnetic.

Illustration – 6: [Ni(CN)4]2– is  square planar whereas [Ni(Cl)4]2– is tetrahedral.  

Solution : CN being  a strong ligand pairs up the electron of the 3d level. So one inner d orbital is available along with 4s and 4p and the mode of hybridisation is dsp2 and hence square planar. Cl being a weak lignad cannot pair the electrons and so the mode of hybridisation is sp3 and hence geometry is tetrahedral.


Write the IUPAC name of the following (1 to 9) :

  1. (NH4)3[ZrF7]
  2. [FeCl2(H2O)4]+
  3. [Ag(NH3)2]4 [Fe(CN)6]
  4. [Cr(en)4Cl2]2[PdCl4]
  5. [Al(H2O)2(OH)4]2
  6. [Co(en)2(CN)2]ClO3
  7. [Rh(NH3)4Cl2]+
  8. [Pt(NH3)4Cl2]2+
  9. [Pt (NH3)3Br]NO2
  10. A,B and C are three complexes or chromium (III) with the empirical formula H12O6Cl3Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H2SO4, whereas complexes B and C lose 6.75% and 13.5% of their original weight, respectively, on treatment with concentrated H2SO4 identify A, B and C.


  1. Ammonium heptafluorozirconate (IV)
  2. Tetraaquadichloroiron(III) ion.
  3. Diamminesilver (I) hexacyanoferrate (II)
  4. Dichlorobis(ethylenediamine) chromium (III) tetrachloropalladate (II)
  5. Tetrahydroxodiaquaaluminate (III) ion
  6. Dicyanobis(ethylenediamine) cobalt (III) chlorate
  7. Dichlorotetraamminerhodium (III) ion
  8. Dichloro tetraammineplatinum (IV) ion
  9. Bromotriammineplatinum (II) nitrite
  1. Conc. H2SO4 is a dehydrating agent. It removes the water of crystallization from a complex. However it fails to absorb water molecules from the non ionizable part of the co-ordination compound.

Hence A: [Cr[H2O)6]Cl3

B: [Cr(H2O)5Cl] Cl2.H2O

C: [Cr(H2O)4Cl2]Cl.2H2O



Write the IUPAC name

  1. Na3[Ag(S2O3)2]
  2. K2[OsCl5N]
  3. [Zn(NCS)4]2–
  4. [Cr(en)3]Cl3
  5. [Co(NH3)5NO2](NO3)2
  6. Na2[Fe(CN)5NO]
  7. [Ir(Ph3P)2(CO)Cl]
  8. Na4[Fe(CN)5NOS]
  9. [Pt(NH3)4Cl2][PtCl4]
  10. [Co(NH3)5CO3]2[CuCl4]


  1. Sodiumbisthiosulfatoargentate (I)
  2. Potassiumpentachloromitridosmate (VI)
  3. Tetrakisthiocyanato(N)zincate (II)
  4. Trisethylenediaminechromium(III)chloride
  5. Pentamminenitrocobalt(III)nitrate
  6. Sodiumpentacyanonitrosylferrate(III)
  7. Carbonylchloribis(triphenulphosphine)iridium(I)
  8. Sodiumpentacyanonitrosylsulfidoferrate(III)
  9. Tetramminedichloroplatinum(IV)tetrachloroplatinate(II)
  10. Pentaamminecarbonatocobalt(III)tetrachlorocuprate(II)