Kendriya Vidyalaya Sangathan
Lucknow Region
Study Material
Subject : CHEMISTRY
Class - XII
:
Prepared under supervision of
Mrs. NOMITA WILSON
PRINCIPAL, KENDRIYA VIDYALAYA IIM,LUCKNOW
Astt. Commissioner KVS (LR)
RANVIR SINGH
Under Guidance of
Dr. Satyendra pal
Principal
Kendriya Vidyalaya No. 1 , AFS, Agra
Prepared By
S.NO / NAME OF THE TEACHER / KENDRIYA VIDYALAYA1 / Dr. ABHYAS YADAV / BARABANKI
2 / SANGITA RANI SINGH / GOMTI NAGAR, LUCKNOW
3 / R.C.TIWARI / B.K.T. , LUCKNOW
4 / B.K.SHARMA / ALIGANJ , LUCKNOW
5 / A.K.DIXIT / SITAPUR
6 / MANJULA DIXIT / I I M, LUCKNOW
7 / RAJ SHEKHAR / I I M, LUCKNOW
CONTENTS
ONE PAPER 70 MARKS
UNIT NO. / TITLE / MARKS
I / SOLID STATE / 4
II / SOLUTIONS / 5
III / ELECTROCHEMISTRY / 5
IV / CHEMICAL KINETICS / 5
V / SURFACE CHEMISTRY / 4
VI / GENERAL PRINCIPALES AND PROCESSES OF ISOLATION OF ELEMENTS / 3
VII / P-BLOCK ELEMENTS / 8
VIII / D & F BLOCK ELEMENTS / 5
IX / COORDINATION COMPOUNDS / 3
X / HALOALKANES & HALORENES / 4
XI / ALCOHALS, PHENOLS & ETHERS / 4
XII / ALDEHYDES, KETONES & CARBOXYLIC ACID / 6
XIII / ORGANIC COMPOUNDS CONTAINING NITROGEN / 4
XIV / BIOMOLECULES / 4
XV / POLYMERS / 3
XVI / CHEMISRY IN EVERDAY LIFE / 3
TOTAL / 70
Unit-I THE SOLID STATE
Density of unit cell = Mass of unit cell = z × M
Volume of unit cell Na × a3
z- Number of atoms per unit cell M – Atomic mass or formula mass for ionic solids
Na – Avogadro number a – Edge length
Radius ratio = r+ = Radius of the cation
r- Radius of the anion
If R is the radius of spheres in the close packed arrangement then
I. Radius of octahedral void = r = 0.414 R
II. Radius of tetrahedral void = r = 0.225 R
Packing fraction = Volume occupied by atoms in unit cell
Total volume of the unit cell
· Different types of solids (table)
Unit cell / Distance b/wnearest
neighbor (d) / Radius
Constituent
Atom / Coord
Ination
no. / Packing
fraction / Number of
Constituent
Atoms
Simple cubic / a / a/2 / 6 / 0.52 / 1/8×8 = 1
Face centered cubic (FCC) / a/ / a/2 / 12 / 0.74 / (1/8×8) + (1/2×6) = 4
Body centered cubic (BCC) / /2a / /4a / 8 / 0.68 / (1/8×8) + 1 = 2
Where, a = edge length of unit cell
Different Type of Solids
s.n. / Type ofSolids / Constituent
Particles / Bonding/
Attractive
Forces / Examples / Physical
Nature / Electrical
Conductivity / Melting
Point
1. / Molecular
Solids
(i)Non -polar
(ii)Polar
(iii)Hydrogen
bonded / Molecules / Dispersion or
London forces
Dipole-dipole
Interactions
Hydrogen
bonding / Ar, CCl4 ,
H2 , I2 , CO2
HCl , SO2
H2O(ice) / Soft
Soft
Hard / Insulator
Insulator
Insulator / Very low
Low
Low
2. / Ionic solids / Ions / Coulombic or
electrostatic / NaCl, MgO
Zns, CaF2 / Hard but
brittle / Insulators in
solid state but
conductors in
molten state
and in aqueous
solutions / High
3. / Metallic
solids / Positive ions
In a sea of
delocalized
electrons / Metallic
bonding / Fe, Cu, Ag, Mg / Hard but malleable
and ductile / Conductors in
solid state as well as in molten state / Fairly high
4. / Covalent or
Network solids / Atoms / Covalent
bonding / SiO2 (quartz),
SiC , C(diamond),
AIN, C(graphite) / Hard
Soft / Insulators
Conductor
(exception) / Very
high
Important Questions;
1. Name the non-stoichiometric point defect responsible for colour in alkali halides.
Ans: F-centres
2. What happens when a ferromagnetic substance is heated to high temperature?
Ans: Ferromagnetic substance changes to paramagnetic substance due to randomization of domains (spins) on heating.
3. A compound is formed by two elements X and Y. Atoms of the element Y (anions) make ccp and those of the element X (cations) occupy all the octahedral voids. What is the formula of the compound?
Ans: Suppose no. of atoms Y in ccp = N
:. No. of octahedral voids = N
:. No. of atoms X = N
Ratio of X : Y = N : N =1: 1
Formula of compound is XY
4. Explain the following terms with suitable examples:
(i) Schottky defect (ii) F-centres
Ans (i) Stoichiometric defect when equal number of cations and anions are missing from the lattice. Cations and anions are of almost same size. It lowers the density of solid e.g. NaCl , KCl etc.
(ii)The electrons trapped at the anion vacancies are referred to as F-centres from German word Farbenzenter meaning colour centre e.g. yellow colour imparted to NaCl when heated in an atmosphere of sodium vapour.
5. Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm-3 Use this information to calculate Avogadro’s number. (At. Mass of iron = 56 g mol-1)
Ans : d = z x M
a3 x NA
For bcc lattice , z = 2
7.87 g cm-3 = 2 x 56 g mol-1
(286.65 x 10-10 cm)3 x NA
NA = 6.04 x 1023 mol-1
Unit-II SOLUTIONS
· Mass percent = Mass of component in solution x 100
Total mass of solution
· Mole fraction of a component = Number of moles of the component
Total number of moles of all the components
χB =
· Molarity (M) = Moles of solute
Volume of solution in litre
= WB x 1000
MB Volume of solution in ml
WB – mass of solute MB – molar mass of solute
· Molality (m) = Moles of solute
Mass of solvent in Kg
= WB x 1000
MB WA (in grams) WA – mass of solvent
· Henry’s law
P = KH x χ1
P – partial pressure KH – Henry’s constant χ1 - mole fraction
· Raoult’s law for volatile solute
PA = P°A χA
PA= partial pressure of component A in solution
P°A = vapour pressure of pure component
χA = mole fraction
P = PA + PB
= P°A χA + P°BχB
· Raoult’s law for non-volatile solute
P°A - PA = χB = nB = WB x MA
P°A nA + nB MB WA (dilute solution)
· Elevation in boiling point
ΔTb = Kb x m = Kb x WB x 1000
MB WA
· Depression in freezing point
ΔTf = Kf x m = Kf x WB x 1000
An ideal solution should have:
i. ΔVmix = 0
ii. ΔHmix = 0
iii. Obeys Raoult’s law over a wide range of concentration
· Non-ideal solution
i. ΔVmix ≠ 0
ii. ΔHmix ≠ 0
iii. Raoult’s law is not obeyed
Two types:
a) Positive Deviation : When partial pressure of each liquid and the resultant total pressure is greater than the pressure expected on the basis of the Raoult’s law
eg- Ether-acetone, Benzene-acetone , water-ethanol , n-hexane-ethanol etc.
ΔHmix = +ve
ΔVmix = +ve
PA > P°AχA ; PB > P°BχB
b) Negative deviation :When partial pressure of each liquid is less than the vapour pressure expected on basis of Raoult’s law.
Eg- Methanol-acetic acid , water-nitric acid , acetone-chloroform , chloroform-benzene , chloroform-ether etc.
ΔHmix = -ve
1. How is the molality of a solution different from its molarity?
Ans- Molarity is the number of moles of solute per litre of solution whereas molarity is the number of moles of solute in 1kg of the solvent.
2.State Henry’s law and mention two importants application.
Ans- Henry’s law : The pressure of gas over a solution is directly proportional to the mol fraction of the gas dissolve in the mol fraction of the gas dissolve in tje solution.
Applications :-
(i): To make soft drink bottle more rich in CO2 sealing is done at high pressure.
(ii):Deep sea divers use He and O2 mixture for respiration He is less soluble in blood as compared to N2 at high pressure and create the pain ful effect on the human body.
3. A 5 % sol (by mass) of Cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.
Ans- 5% glucose means 5gm in 100gm of sol
f =
For sugar 273.15 – 271 =
2.15 = kf × 0.154
For glucose f =
f = Kf × 0.292
Comparing the two
=
Freezing point of solution will be 273.15K – 4.08K
= 269.07K
UNIT-III - ELECTROCHEMISTRY
Important formulae
• R=V/I
Where R is Resistance, V is Voltage, I is Current.
• Conductivity or conductance = 1/R = 1/V
The unit of conductance is Ohm-1 = Siemens = S.
• Specific Conductance = 1/ Specific resistance
K = 1 / p
• ^m = Specific Conductance/ Molarity
K / M
• Kohlrausch law - ^m = V+ ^+ + v-^-
• Faraday’s First Law of Electrolysis
m = Z x Q
m = Z x I x t
Where is mass of substance deposited, I is current in ampere, t is time in seconds,
Q = I x t
Z = Eq. Wt./ 96500
Where z is electrochemical equivalent. Unit of electrochemical equivalent is gram/coulomb
Faraday is charge on 1 mole of electrons.
Charge on 1 electron = 1.602 x 10 -19 C
Charge on 1 mole of electrons = 1.602 x 10 -19 x 6.023 x 10 23
= 96487 C
= 96500 C
• Faraday’s Second Law of Electrolysis :
• W1/E1 = W2/E2 = W3/E3
• Where W1 is mass of substance 1 deposited and E1 is its equivalent weight,
• W2 is mass of substance 2 deposited and E2 is its equivalent weight,
• W3 is mass of substance 3 deposited and E3 deposited and E3 is its equivalent weight.
• Nernst Equation
Eo oxidation = - Eo reduction
Eo cell = Eo cathode – Eo anode
Eo cell = Eo anode – Eo cathode
Mn+ + ne- M
E = Eo – 2.303 RT / nF log [M]/[Mn+]
where E = reduction electrode potential,
Eo = Standard Reduction electrode Potential,
n = No. of electrons
T = Temperature in Kelvin, F = 96500 C
Change in Gibb,s free energy ( ∆G ) = - n F E0 Cell
∆G0 = - 2.303RTlogKc
Important Questions
• Why is the equilibrium constant K related to only Eocell and not Ecell ?
Ans: It is because Ecell at equilibrium is 0 volt .
• Write the nernst Equation for the electrode reaction
M n+ + ne- → M(s)
Ans : Emn+/m = Eomn+/m +2.303RT/nF log[Mn+]
• Predict the product of Electrolysis in each of the Following
a) An aq Sol. Of AgNO3 with silver electrodes.
b) An aq. Sol. Of AgNO3 with platinum electrodes.
Ans : At Cathode : Ag+ (aq) + e- Ag(s)
At anode Ag(s) Ag+ (aq)+ e-
AgNO3(aq) Ag+(aq) + NO3-(aq); H2O(l) H+(aq) + OH- (aq)
b) At cathode 2Ag+(aq) + 2e- 2Ag(s)
At anode 2OH-(aq) O2(g) + 2H+(aq) + 4e-
● Can you store copper sulphate in zinc pot ?
Ans: No , because Zn is more reactive than Cu .
●If a current of 0.5 ampere flows through a metallic wire for 2 hours , then how many electrons flow through the wire ?
Ans : Q = it = 5 Х 2 Х 60 Х 60 = 3600 C
No. of electrons flow = 3600 Х 6.022 Х 1023
9600
= 2.246 Х 1022
UNIT-IV Chemical Kinetics
Important formulae
• Rate Of reaction : it is defined as the change in concentration of reactant (or product) in a particular time interval. Unit of rate of reaction is mol L-1 s-1. If time is in minutes, then units is mol L-1 min-1 and so on. R = K ﴾ Conc ﴿n
• Rate Law or rate equation : It is the expression which relates the rate of reaction with concentration of the reactants. The constant of proportionality ‘k’ is known as rate constant.
R = K [ A ]p [ B ]q Where , p and q are not stoichiometric coefficient but they are order of the reaction
• Rate constant : when concentration of both reactants are unity, then the rate of reaction is known as rate constant. It is also called specific reaction rate.
Unit of K = ﴾ Conc ﴿ 1- n time -1
But in gaseous reaction , unit of K = atm 1-n time-1
• Molecularity : Total number of atoms, ions or molecules of the reactants involved in the reaction is termed as its molecularity. It is always in whole number. It is never more than three. It cannot be zero.
Examples: Unimolecular reaction
NH4NO2 N2 + 2H2O
Bimolecular reaction
2HI(g) H2(g)+I
Trimolecular reaction
2NO(g) + O2 (g) 2NO2(g)
• Order of reaction : The sum of the exponents of the concentration of reactants in the rate law is termed as order of the reaction. It can be in fraction. It can be zero also.