SOME BASIC CONCEPTS OF
CHEMISTRY
4. How many significant figures are there in
(i) 0.00425
(Feb 17)
(ii) 20.0
(iii) 6.022 × 1023
(iv) 0.2500
(PUB)
(v) 3.10087
(vi) 8.2015
Ans. (i) 3 (ii) 3 (iii) 4 (iv) 4 (v) 6 (vi) 5
Weightage 8 marks
1 mark – 1Q (Q No. 1)
2 mark – 1Q (Q No. 11)
5 mark – 1Q (Q No. 27)
ONE MARK QUESTIONS
1. Following medicines are used in the
treatment of which disease?
(a) Cisplatin
(PUB)
(b) Taxol
(EQ)
(c) Drug AZT (Azidothymidine)
(EQ)
Ans: (a) Treatment of Cancer
(b) Treatment of Cancer
(c) Treatment of AIDS patients.
5. Give the relation between
(a) Degree Fahrenheit and degree Celsius.
(Feb 17)
(b) Degree Celsius and Kelvin
(EQ)
Ans.
(b) K = oC +273
2. Give the S.I. unit of
(i) Density (PUB, May–17, Feb -17,19, 20)
(ii) Temperature
(May 15)
(iii) Amount of substance
(Feb 18)
(vi) Electric current
(Feb 19)
(v) Luminous intensity
(vi) Volume
(Feb 20)
3
Ans: (i) kg/m
(ii) Kelvin
(iii) Mole
(iv) Ampere (v) Candela
(vi) m3
6. State law of Conservation of Mass
(EQ)
Ans: It states that “matter can neither be created nor
destroyed”.
7. State law of Definite Proportions
(May 15)
(iii) 0.00053
Ans: 5.3 × 10-4
(Feb 16)
(iv) 8008
Ans: 8.008 × 103
(v) 0.001023
Ans: 1.023 × 10-3
(vi) 232.508
Ans: 2.32508 × 102
(PUB)
Ans: The law states that a given compound always
contains exactly the same proportion of elements
by weight.
8. State law of ‘Multiple Proportions’.
(EQ)
Ans: When two elements can combine to form more
than one compound, the masses of one element
that combine with a fixed mass of the other
element, are in the ratio of small whole numbers.
3. Express the following in scientific notation
(i) 0.00016
(Feb 18)
–4
Ans: 1.6 × 10
(ii) 0.00085
Ans: 8.5 × 10–4
(a) º F = 9 / 5 × ( º C ) + 32
9. State ‘Gay Lussac’s Law’ of gaseous
volumes.
(EQ)
Ans: The volume of gases taking part in a chemical
reaction show simple whole number ratio to one
another when all the volumes are measured at
the same temperature and pressure.
(EQ)
10. State Avogadro’s law.
(PUB)
(May14, Feb-17,20)
Ans: It states that equal volumes of all gases contain
equal number of molecules at same temperature
and pressure.
(Feb 19)
1
I-PUC
Chemistry
1
HAND BOOK
2
11. Define atomic mass unit.
(May 14)
Ans: Atomic mass unit is defined as mass exactly
equal to one-twelfth the mass of one carbon-12
atom.
12. What is the value of Avogadro number?
(Feb -19)
23
Ans: 6.022 × 10
13. Define Mole.
(May-15,Feb – 16,17,19)
Ans: Amount of substance that contains Avogadro
number of particles [6.022 × 1023]
14. What is the mass of Avogadro number (1 mole)
of (a) CO2 Molecules (b) CH4 molecules (c) SO2
molecules (d) H2O molecules?
(EQ)
Ans: (a) 44g (b) 16g (c) 64g (d) 18g
I-PUC
Chemistry
15. How many moles are present in
(a) 32 g of methane
(b) 22g of CO2
(c) 6.4g of SO2
(d) 49g of H2SO4?
(EQ)
(EQ)
17. An organic compound of molecular mass 60
has the empirical formula CH2O. What is its
molecular formula?
Ans: Empirical formula mass = 12 + 2 + 16 = 30
60
Molar mass
=
=2
n =
30
Empirical formula mass
Molecular formula = (Empirical formula)n
Ans: Molarity
22. How molarity depends on temperature?
Ans: Increase in temperature decreases molarity.
23. Name the concentration term which is
independent of temperature.
(EQ)
1. Describe the classification of pure
substances.
(Mar 14)
Ans: Pure substances are classified as elements and
compounds.
An element consists of only one type of atoms.
Ex : Sodium, chlorine etc.
Compounds are formed by the combination of
two or more different elements in a fixed ratio by
mass
Ex: CO2, NaOH etc.
2. What is homogeneous mixture? Give one
example.
(Feb – 19)
Ans: The components completely mix with each other
and its composition is uniform throughout.
Ex: Air, Sea water etc.
= (CH2O)2 = C2H4O2
(PUB)
Ans: The reactant which is completely consumed and
limits the amount of product formed is called a
limiting reagent.
19. Define (a) Molarity
(b) Molality.
21. Among the molarity and molality, Which
one is temperature dependent? (Mar-16)
TWO MARK QUESTIONS
Ans: The mass of one mole of a substance in grams is
called its molar mass.
18. What is limiting reagent?
No. of moles of solute
Volume of the solution in litres
No.of moles of solute
Molality =
Mass of solvent in kg
Ans: Molarity =
Ans: Molality
Ans: (a) 2 mol (b) 0.5 mol (c) 0.1 mol (d) 0.5mol
16. Define molar mass.
20. Write the formula to calculate
(a) Molarity
(b) Molality
(Feb- 18, 20)
(Feb – 20)
Ans: Molarity - The number of moles of the solute in
1 litre of the solution
Molality - The number of moles of solute present
in 1kg of solvent.
3. What is heterogeneous mixture? Give one
example.
(May - 17)
Ans: The composition is not uniform throughout.
Ex: Mixtures of salt and sugar, grains and pulses
etc.
4. Classify the following into pure substance
and mixture. Copper(s), HCl(aq), NaCl(s)
and Gold in mercury.
(Feb 16)
Ans: Pure substance - Copper(s), NaCl(s)
Mixture - HCl(aq), Gold in mercury
3
I PUC - Chemistry
Ans: oF =
(PUB)
9
( º C) + 32 = 9 (37) + 32 = 98.6ºF
5
5
6. Express (i) 5L of milk in cubic meter
(ii) 25ºC in Kelvin
(PUB)
Ans: (i) 5 × 10-3m3
(ii) 298K
7. What is density? Write its SI unit. (Mar-16)
Ans: Density of a substance is amount of mass per
unit volume.
SI unit - kg/m3
12. Calculate the number of gold atoms present
in 98.5 g of gold (atomic mass of gold = 197
g)
(MQP)
Ans: 197g of gold contains → 6.022 × 1023 atoms
13. How many carbon atoms are present in 3 g?
(EQ)
Ans: Atomic mass of Carbon = 12u
12g of C contains --- 6.022 × 1023 atoms
8. Give any two postulates of Dalton’s atomic
theory.
(May - 14,15,18 Feb – 17,18)
Ans:
(i) Matter consists of indivisible atoms.
(ii) Atoms of same element have same mass
and shape
(iii) Atoms of different elements are different.
(iv) Atoms of different elements combine in
fixed ratio and form compounds.
(v) Atoms are neither created nor destroyed in
a chemical reaction.
3g of C contains ---
Cl
75.77 %
37
Cl
24.23 %
Ans: Average atomic mass
Molar mas
35
=
34.9689
36.9659
(75.77 × 34.9689) + ( 24.23 × 36.9659)
= 35.45u
75.77 + 24.23
10. Calculate the molar mass of glucose
(Feb – 17,18)
Ans: Molar mass of Glucose (C6 H12O6)
= (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
11. Calculate the mass of 220 cm3 of oxygen at
STP.
(EQ)
Ans: Molar mass of O2 = 32u
32g of O2
----- 22400cm3 at STP
?
---------220 cm3 at STP
32 × 220
=
= 0.314g
22400
?
6.022 × 1023 × 30
= 1.505 × 1023 atoms
12
14. How many H2O molecules are present in
90 g?
(EQ)
Ans: Molar mass of H2O = 18u
18g of H2Ocontains --- 6.022 × 1023molecules
90g of water contains --- ?
9. Calculate the average atomic mass of
chlorine using the following data.
(PUB)
Natural abundance
98.7 × 6.022 × 1023
197
= 3.01 × 1023 atoms
98.5 g of gold contains
6.022 × 1023 × 90
= 30.11 × 1023 molecules
18
15. How many atoms of Hydrogen are present
in 1 mole of water?
(PUB)
Ans: 1 mole of water (H2O) contains 2 moles of
hydrogen atoms i.e., 2 × 6.023 × 1023 atoms
= 12.046 ×1023 Hydrogen atoms.
16. Express 9.8g of sulphuric acid in mole(Atomic
mass of H=1, S= 32, O=16)
(PUB)
Ans: Molar mass of H2SO4 = 2(1) + 32 + 4(16) = 98
No. of moles =
Mass
9.8
=
= 0.1mol
Molar mass 98
17. Calculate number moles in 49g of H2SO4
(Atomic mass of H=1, S= 32, O=16)
(Feb – 17)
Ans: No. of moles =
Mass
49
=
= 0.5mol
Molar mass 98
I-PUC
Chemistry
5. Convert 37°C to °F.
HAND BOOK
4
18. Calculate the amount of water(g) produced
by the combustion of 16g of methane.(PUB)
Ans: CH4(g) + O2 (g) →CO2(g) + 2H2O(g)
1mol
1mol
1mol
2mol
5g of CaCO3 produces -------
56 × 5
= 2.8g of CaO
100
100g of CaCO3 produces ------ 44g of CO2
According to above balanced chemical equation
5g of CaCO3 produces -------
16g(1mol) of CH4 produces → 36g (2 mol) of H2O
on combustion
44 × 5
= 2.2g of CO2
100
19. Calculate the amount of carbon dioxide
produced by the combustion of 24g of methane.
(Feb – 19)
Ans: CH4(g) + O2 (g) → CO2(g) + 2H2O(g)
1mol(16g)
1mol(44g)
16g of methane produces --- 44g of CO2
Ans: C3H8 + 5O2
1mol (44g)
→
3CO2 + 4H2O
4mol(4 x 18g)
44g of C3H8 produces ------ 4 × 18g of H2O
120g of C3H8 produces -------
44 × 24
= 66g of CO2
16
4 × 18 × 120
= 196.36g of H2O
44
I-PUC
Chemistry
20. Calculate the amount of water produced in
grams by the combustion of 8g of methane.
(Feb – 18)
Ans: CH4(g) + O2 (g) → CO2(g) + 2H2O(g)
1mol(16g)
2mol(2 x 18)g
16g of methane produces ---(2 × 18)g of H2O
8g of methane produces --- ?
=
2 × 18 × 8
= 18g of H2O
16
21. What mass of calcium carbonate is to be
decomposed to obtain 4.4g of CO2 in the
following reaction (Molar mass of CaCO3 =
100)
(EQ)
Ans: CaCO3 → CaO + CO2
1 mole(100g)
1 mole(44g)
100g of CaCO3 produces ------ 44g of CO2
?
-------- 4.4g of CO2
100 × 4.4
= 10g
44
22. CaCO3 decomposes to give CaO and CO2
according to the equation CaCO3 → CaO
+ CO2. Calculate the mass of CaO and CO2
produced on complete decomposition of 5g
of CaCO3
(MQP)
Ans: CaCO3 →
1 mole(100g)
CaO + CO2
1 mol(56g)
1 mole(44g)
100g of CaCO3 produces ------ 56g of CaO
?
23. If 120g of propane(C3H8) is burnt in excess
of oxygen, how many grams of water are
produced?
(MQP)
24g of methane produces --- ?
=
?
?
24. Calculate the mole fraction of benzene and
CCl4 in a binary solution containing 0.3mol
of benzene and 0.4 mole of CCl4 (Feb -16)
Ans: n1 = 0.3 mol n2 = 0.4 mol.
X1= Mole fraction of benzene
=
n1
0.3
=
= 0.428
n1 + n 2 0.3 + 0.4
X2= Mole fraction of CCl4 = 1 – 0.428 = 0.571
25. Calculate the mole fraction of ethanol in a
solution containing 20g of ethanol and 100g
of water.
Ans: nethanol =
nwater =
mass
20
=
= 0.434mol.
mol.mass 46
mass
100 = 5.55mol.
=
mol.mass 18
Xethanol =
0.434
n ethanol
=
= 0.072
n ethanol + n water 0.434 + 5.55
26. Define gram molar volume. What is its value
at STP?
(PUB)
Ans: Volume occupied by one mole of any gas at STP.
(1 atm pressure and 273K temperature)
Its value is 22.4L at STP
5
I PUC - Chemistry
Volume of the solution = 250mL = 0.250L
27. A solution is prepared by adding 2g of substance
to 18g of water. Calculate the mass percent of the
solute.
(Feb – 19)
Ans: Mass % =
28. What is empirical formula? Give an example
for a compound whose empirical formula and
molecular formula are same.
(Feb – 17)
Ans: The simple formula which gives the ratio of number
of atoms of each element present in a molecule of the
compound.
Ex: H2O, CH4
Ans: Mass of Solution = 100g
Mass of solute (HNO3 ) = 69g
Mass of solvent (water) = 31g
Ans: Mol. Mass of methane(CH4)
Percentage of hydrogen =
4
× 100 = 25%
16
31. Calculate the percentage composition of
hydrogen and oxygen in water.
(Mar 18)
Ans: Mol. Mass of water (H2O)
= 2(1) + 1(16) = 18u
2
Percentage of hydrogen =
×100 = 11.11%
18
16
Percentage of oxygen =
× 100 = 88.88%
18
32. Calculate the molarity of NaOH in the solution
prepared by dissolving 4g in enough water to
form 250mL of the solution. (Given: Atomic
mass of Na- 23, O-16, H-1)
(May 15, 17)
Ans. Molecular mass of NaOH = 23 +16 + 1 = 40u
No. of moles of NaOH = mass/mol.mass
= 4/40 = 0.1 mol.
Mass of HNO3
M.M of HNO3
=
69
= 1.095 mol
63
Mass of the solution
Volume of the solution
100g
1.41 =
Volume of the soltion
Volume of solution = 100/1.41 = 70.92mL
30. What is the percentage composition of
carbon and hydrogen in methane? (PUB)
12
× 100 = 75%
16
Number of moles of HNO3 =
Density =
Molecular formula = (empirical formula)n
Percentage of carbon =
= 0.4 mol/L
33. Calculate the concentration of HNO3 in mol/L in
a sample which has density 1.41g/mL and mass
percent of 69% Mol. Mass of HNO3 = 63
(Feb – 18)
29. What is molecular formula? How it is
related to empirical formula?
(EQ)
Ans: The actual formula which gives the exact number
of atoms of each element present in a molecule
of the compound.
= 12 + 4(1) = 16u
0.1
0.250
=
= 70.92×10–3L
No. of moles of a solute
Volume of the solution in L
1.095
=
= 15.43mol/L
70.92 × 10−3
Molarity =
34. Calculate the molarity of NaOH in the solution,
prepared by dissolving 10g in enough water
to form 500mL of the solution. Molar mass of
NaOH is 40g/mol.
(Feb – 19)
Ans: No. of moles of NaOH = Mass/ Mol.Mass
= 10/40 = 0.25 moles
Volume of the solution = 500mL = 0.5L
Molarity =
=
No of moles of a solute
Volume of the solution in L
0.25
= 0.5 mol/L
0.5
I-PUC
Chemistry
Mass of solute
× 100
mass of solution
2
× 100 = 10%
=
2 + 18
No of moles of a solute
Volume of the solution in L
Molarity =
HAND BOOK
6
= 72 + 12 + 96 = 180
35. Calculate the molarity of a solution containing
2.3 moles of solute dissolved in 4.6L.
(Feb – 19,20)
Ans: Molarity =
72
× 100 = 40%
180
12
Percentage of hydrogen =
× 100 = 6.66%
180
96
Percentage of oxygen =
× 100 = 53.33%
180
Percentage of carbon =
2.3
No of moles of a solute
=
4.6
Volume of the solution in L
= 0.5 mol/L
36. Define mole fraction. Write equation to calculate
the mole fraction of solute in solution. (Mar - 14)
Ans: Mole Fraction is the ratio of numbers of moles
of the specific component to the total number of
moles in solution.
X solute =
2. Calculate the percentage of carbon,
hydrogen and oxygen in ethanol. (Mar 13)
Ans. Molecular mass of ethanol (C2H5OH)
No.of moles of solute
Total moles in solutions
=2(12) + 5(1) + 1(16) + 1 = 46
Percentage of carbon =
THREE MARK QUESTIONS
I-PUC
Chemistry
1. Calculate the percentage of carbon,
hydrogen and oxygen in glucose.
(EQ)
Ans. Molecular mass of glucose (C6H12O6)
24
× 100 = 52.1%
46
6
× 100 = 13 %
46
16
Percentage of oxygen =
× 100 = 34.7 %
46
Percentage of hydrogen =
= 6(12) + 12(1) + 6(16)
3. An organic compound contain 4.07% of hydrogen, 24.27% of carbon and 71.65% of chlorine. What
is its empirical formula?
[At. masses H = 1, C = 12, Cl = 35.5]. If two moles of this compound is dissolved in 5 litre of a solvent,
what is the molarity of the solution obtained?
(Feb- 17)
Ans.
Element
% composition
At. Mass
H
4.07
1
C
24.27
12
C1
71.65
35.5
% composition
At.mass
No. in previous column
Least no.
4.07
= 4.07
1
4.07
= 2.016
2.0183
24.27
= 2.0225
12
2.0225
=1.002
2.0183
71.65
=2.0183
35.5
2.0183
=1
2.0183
Empirical formula = CH2Cl
Molarity =
No. of moles of solute 2
= = 0.4 mol/kg
Volume of solution ( L ) 5
Simple whole
no. ratio
2
1
1
7
I PUC - Chemistry
4. An organic compound contains 4.05% hydrogen, 24.26% carbon and 71.67% chlorine. Its molecular
mass is 98.96. Find its empirical and molecular formula (Atomic mass of H = 1, C = 12, Cl = 35.45)
(Feb- 18,19,20)
Ans.
Element
% composition
At. Mass
% composition
At.mass
No. in previous column
Least no.
Simple whole
no. ratio
H
24.26
12
24.26
= 2.02
12
2.02
=1
2.02
1
C
4.05
1.008
4.05
= 4.05
1.008
4.05
=2
2.02
2
C1
71.67
35.45
71.67
= 2.02
35.45
2.02
=1
2.02
1
n=
I-PUC
Chemistry
Empirical formula = CH2Cl Empirical formula mass = 12 + 2 + 35.5 = 49.5
Molecular mass
98.96
=
=2
Empiricalformula mass 49.5
\ Molecular formula = (EF)n = (CH2Cl)2 = C2H4Cl2
5. Determine the empirical formula of an oxide of Iron which has 69.9% Iron and 30.1% dioxygen by
mass. [Atomic mass of Fe = 56, O = 16].
(Feb- 18, 20)
Ans.
Element % composition At. Mass
% composition
At.mass
No. in previous column
Least no.
Simple whole no.
ratio
Fe
69.9
56
69.9
= 1.24
56
1.24
=1
1.24
1×2=2
O
30.1
16
30.1
= 1.88
16
1.88
= 1.5
1.24
1.5 × 2 = 3
Empirical formula = Fe2O3
HAND BOOK
8
6. An unknown compound was found to contain 43.4% sodium, 11.3% carbon and 45.3% oxygen. The
molar mass of the compound is 110 g/mol. Find out its empirical formula and molecular formula.
(EQ)
Ans.
Element
% composition
At. Mass
Na
43.4
23
C
11.3
12
O
45.3
16
Simple whole
no. ratio
% composition
At.mass
No. in previous column
Least no.
43.4
= 1.886
23
1.88
= 2.004
0.941
2
45.3
= 2.83
16
2.83
= 3.007
0.941
3
11.3
= 0.941
12
0.941
=1
0.941
1
I-PUC
Chemistry
Empirical formula = Na2CO3
Empirical formula Mass = (23 × 2) + 12 + (16 × 3) = 46 + 12 + 48 = 106
110
Molar mass 134
=
=
==11.03
26 = 11
106
EF mass
106
\ Molecular formula = (EF)n = (Na2CO3)1 = Na2CO3
7. A compound gave on analysis, the following percentage composition, K = 26.27% (atomic mass of
K = 39), Cr = 35.36% (atomic mass of Cr = 52) and rest is oxygen. (atomic mass of oxygen = 16).
Determine the empirical formula of the compound.
(EQ)
Ans.
Element
% composition
At. Mass
K
26.57
39
Cr
35.36
52
O
38.07
16
∴ Emperical formula is K2Cr2O7.
% c omp osition
A t .ma ss
No. in previous coloum
Least no.
Simple whole
no. ratio
35.36
= 0.68
52
0.68
=1
0.68
1×2=2
26.57
= 0.68
39
38.07
= 2.379
16
0.68
=1
0.68
1×2=2
2.379
= 3.5
0.68
3.5 × 2 = 7
9
I PUC - Chemistry
8. An organic compound found to contain 39.9% carbon, 6.7% hydrogen and the rest is oxygen. If the
molecular mass of the compound is 60, determine the molecular formula of the compound.
(EQ)
Ans.
% composition
At. Mass
C
39.9
12
H
6.7
1
O
53.4
16
% c omp osition
A t .ma ss
Simple whole
no. ratio
39.9
= 3.33
12
3.33
=1
3.33
1
53.4
= 3.33
16
3.33
=1
3.33
1
6.7
= 6.7
1
Empirical formula = CH2O
Empirical formula mass = 12 + 2 + 16 = 30
molecular mass
60
n=
=
= 2
Empirical formula mass
30
No. in previous column
Least no.
2
6.7
= 2.012
3.33
Molecular formula = (EF)n = (CH2O)2 = C2H4O2
9. A Hydrocarbon contains 80%C. If the molar mass of the compound is 30u, derive empirical formula
and molecular formula.
(EQ)
Ans.
Element
% composition
At. Mass
C
80
12
H
20
1
% c omp osition
A t .ma ss
80
= 6.66
12
20
= 20
1
Empirical formula = CH3
Empirical formula mass = 12 + 3 = 15
molecular mass
30
n=
=
= 2
Empirical formula mass 15
Molecular formula = (EF)n = (CH3)2 = C2H6
No. in previous column
Least no.
Simple whole
no. ratio
20
=3
6.66
3
6.66
=1
6.66
1
I-PUC
Chemistry
Element
2
STRUCTURE OF ATOM
(iii) IR region
Weightage 10 marks
9. Write the wave length range of visible
region.
(EQ)
5 mark – 2Q (Q No. 28 and 29)
Ans: 400 nm – 750 nm
ONE MARK QUESTIONS
10. Give the shape of s,p,d and f orbitals
(Mar 16)
Ans: s orbital - Spherical shape
1. Name the fundamental particles of an atom
(PUB)
Ans: Electrons, Protons and Neutrons.
Ans:
p orbital - Dumb-bell shape
(ii) Protons
(EQ)
d orbital - Double dumb-bell shape
f orbital - Complex shape
(i) Electrons - J.J. Thomson
(ii) Protons - Goldstein
(iii) Neutrons - James Chadwick
11. Name the quantum number that specifies (i) Size of the orbital (ii) Shape of the orbital
(iii) Orientation of atomic orbital (iv) Spin
of the electron.
(PUB)
3. What is the relative charge of an electron
(Feb 15)
Ans: -1
Ans:
4. Name the fundamental particle of an atom
that has highest value of e/m ratio (PUB)
Ans: Electron
Principal quantum number
Azimuthal quantum number
Magnetic quantum number
Spin quantum number.
12. Draw the shape of (i) px orbital. (Feb -15)
(ii) py orbital (iii) pz orbital (iv) dxy orbital
(v) dyz orbital (vi) dzx orbital (v) dx2-y2 orbital
(v) dz2 orbital.
(EQ)
5. Calculate the number of neutrons in 15P
(EQ)
31
Ans: A = 31, Z = 15
Ans.
Number of neutrons = A - Z = 31 - 15 = 16
6. How do isotopes of an element differ from
one another?
(Mar 16,20)
Ans: Isotopes of an element possess different mass
number.
7. Name the spectral lines which lies in
(i) UV region
(ii) Visible region
(iii) IR region
(Feb 20)
Ans: (i) Lyman series
(i)
(ii)
(iii)
(iv)
(ii) Balmer series
(iii) Paschen series
8. In which region (i) Lyman, (ii) Balmer, and
(iii) Panchen series appear in hydrogen
spectrum?
(Feb – 16,17)
Ans: (i) UV region (ii) Visible region
1
I-PUC
Chemistry
2. Who discovered (i) Electrons
(iii) Neutrons
HAND BOOK
2
13. State Aufbau principle.
(Feb – 17)
Ans. In the ground state the orbitals are filled in the
order of increasing energy.
14. State Pauli’s exclusion principle.
(EQ)
Ans. It states that “No two electrons of the same atom
can have all the four quantum numbers same”.
15. Give the possible values of l for (a) n=1
(b) n= 2 (c ) n = 3 (d) n= 4
(EQ)
Ans: (a) l = 0 (b) l= 0,1 (c ) l = 0,1,2 (d) l = 0,1,2,3
16. Name the orbital when (a) n = 3 and l = 2
(b) n = 4 and l = 0
(c) n = 2 and l =1
(d) n = 5 and l =3
Ans: (a) 3d
(b) 4s
(c) 2p
(d) 5f
I-PUC
Chemistry
17. Write the electronic configuration of
(i) Cr (EQ) (ii) Cu (EQ) (iii) Cu+ (Mar 16)
(iv) Zn (PUB)
Ans:
(i)
(ii)
(iii)
(iv)
Cr(Z = 25) - [Ar] 4s1 3d5
Cu(Z = 29) - [Ar] 4s1 3d10
Cu+(Z = 29) - [Ar] 4s0 3d10
Zn(Z = 30) - [Ar] 4s23d10
18. What is the atomic number of the element
whose outermost electrons are represented
by (i) 3s1 (ii)3p5 ?
(PUB)
Ans: (i) 11 (ii) 17
TWO MARK QUESTIONS
1. The atomic number and atomic mass of iron
are 26 and 56 respectively. Find the number
of protons and neutrons in its atom. (EQ)
Ans: A = 56, Z = 26
No. of protons = Z = 26
No. of neutrons = A - Z = 56 - 26 = 30
Ans: Atoms with same atomic number but different
mass number are known as Isotopes.
Ex: 6C and 6C
14
3. What are isobars? Give one example.
7. Write any two limitations of Rutherford’s
model of an atom.
Ans: (i) It fails to explain the stability of an atom.
(ii) It fails to explain electronic structure of
atoms.
8. Define Wavelength and frequency of
electromagnetic radiation. How these
two are related with velocity? (Feb – 15)
Ans, Wavelength (λ): It is the distance between any
two successive crests or troughs of a wave.
Frequency (υ): It is the number of waves passing
through a point in unit time.
Wavelength and frequency are related to velocity
as C = υ λ
9. Define Wave number. How it is related to
wavelength?
(EQ)
Ans: It is defined as the number of wavelengths per
unit length.
1
λ
10. Calculate the frequency of yellow radiation
having wavelength 5800Aº.
[1Aº = 10-10m, C = 3 × 108 ms-1]
8
(EQ)
Ans: Isobars are the atoms with same mass number
but different atomic number
Ex: 6C14 and 7N14
6. Explain Rutherford’s model of an atom
Ans. (i) The positive charge and most of the mass of
the atom is concentrated in extremely small
region called nucleus.
(ii) The nucleus is surrounded by electrons that
move around the nucleus with very high
speed in a circular path called orbits.
(iii) Electrons and the nucleus are held together
by electrostatic forces of attraction.
υ=
2. What are isotopes? Give one example.
(PUB)
12
5. What are the observations made by
Rutherford in α- particle experiment? (PUB)
Ans: (i) Most of the α- particles passed through the
gold foil without any deflection.
(ii) Some deflected through small angles and
very few α- particles almost bounce back.
(Feb -18)
Ans: υ = C , υ = 3 × 10 −10 = 5.172 × 1014 s −1
λ
5800 × 10
3
I PUC - Chemistry
11. The wavelength of red light from beacon
light is 680 nm. Calculate the frequency
of the light.
(Feb -17)
Ans. Given:
16. Explain photoelectric effect.
Ans: The phenomenon of emission of electrons from
metal surfaces exposed to light energy of suitable
frequency is known as photoelectric effect.
= 680nm= 680 × 10–9m C = 3 x 108 ms-1
C
λ
,=
= 4.411 × 1014 s–1
12. The Vividh Bharati station of All India
Radio, Delhi, broadcasts on a frequency
of 1,368 kHz (kilo hertz). Calculate the
wavelength of the electromagnetic radiation
emitted by transmitter.
(March – 16) (Feb -17)
c
Ans: c = υλ λ =
υ
3 × 108
=
= 294.11 m
1020 × 103
13. The FM station of All India Radio,
Hassan, broadcast on a frequency of
1020kilohertz. Calculate the wavelength of
the electromagnetic radiation emitted by
transmitter.
(PUB)
Ans: υ = 1368 × 103 Hz
c = 3.0 × 108 ms–1
c
λ=
υ
λ=
3.0 × 108
= 219.3m
1369 × 103
14. Calculate the energy of one mole of photon
of radiation whose frequency is 5 × 1014 Hz.
(Given h = 6.626 × 10-34 Js).
(Feb – 16)
Ans: E = hυ
E = 6.626 × 10–34 × 5 × 1014
= 33.13 × 10–20 J/ photon
\ Energy of 1 mole of photons
= 33.13 × 10–20 × 6.022 × 1023 = 199.51 kJmol–1
15. Calculate the energy of radiation with wave
length 500nm. (h = 6.626 × 10–34Js) (Feb-20)
c
c
Ans: E = hυ = h ∵ υ = 
λ
λ
=
6.626 × 10−34 Js × 3 × 108 ms −1
= 3.98 × 10−19 J
500 × 10−9 m
Each metal has a minimum energy needed for an
electron to be emitted. This is known as the work
function, W(hυ0).
The excess energy is the kinetic energy of the
emitted electron.
hυ = hυ0 + 1/2 mV2
17. What is photo electric effect? Does the effect
support particle nature or wave nature of
light?
(PUB)
Ans: The process of emission of electrons from the
surface of metal when they are exposed to light
of suitable frequency is called photoelectric
effect.
It support particle nature.
18. Write Rydberg equation and explain the
terms.
(EQ)
 1
1
1
= RH  2 − 2 
λ
 n1 n2 
Ans:
where:
RH : Rydberg constant
n1 : Quantum number of initial state
n2 : Quantum number of final state
19. Calculate the wave number of the spectral
line of shortest wavelength appearing in
the Balmer series of Hydrogen spectrum.
[Given R = 1.097 × 107m–1]
(PUB)
Ans: n1 = 2 (Balmer series), n2 = ∞
1
υ = R
2
n1
−
1

n22 
1

υ = 1.097 × 107  2 − 0 = 2.742 × 106 m −1
2

I-PUC
Chemistry
υ =
HAND BOOK
4
20. What is the wavelength of light emitted
when the electron in an hydrogen atom
undergoes transition from an energy
level with n=4 to an energy level with
n=2?
(May – 14)
Ans: n1 = 2, n2 = 4
1
1
υ = R  2 − 2
n2 
n1
1
1
υ = 1.097 × 10  2 − 2  = 0.204 × 107 m −1
2
4


1
= 0.204 × 107 m −1
υ=
λ
1
λ =
= 4.901 × 10−7 m
0.204 × 107
7
I-PUC
Chemistry
21. Calculate the wave number of the first
spectral line of Lyman series of hydrogen
spectrum (R = 10.97 × 106 m-1)
(PUB)
Ans: n1 = 1 (Lyman series), n2 = 2 (1st line)
 1
1
1 1
υ = R  2 − 2  = 10.97 × 106  2 − 2 
1 2 
 n1 n2 
 1
= 10.97 × 106  1 − 
 4
3
= 10.97 × 106 × = 8.2275 × 106m–1
4
22. Calculate the wave number for the longest
wavelength transition in the Balmer series
of hydrogen given RH = 1.09×107m–1.
(May – 17)
Ans: n1 = 2 (Balmer series),
n2 = 3 (Longest wavelength)
 1
1
υ = R 2 − 2
 n1 n2 
 1 1
 1 1
= 1.09 × 107  2 − 2  = 1.09 × 107  − 
2 3 
 4 9
= 1.09 × 107
(9 − 4) = 1.51 × 106 m −1
36
23. Calculate the energy associated with 5th
orbit of Hydrogen atom.
(PUB)
Ans: n = 5, Z = 1
 Z2 
E n = −2.18 × 10−18  2 
n 
 12 
E n = −2.18 × 10−18  2  = –8.72 × 10–20J
5 
24. Calculate the energy associated with the
first orbit of He+
(EQ)
Ans : n = 1 Z = 2
 Z2 
E n = −2.18 × 10−18  2 
n 
 22 
E n = −2.18 × 10−18  2  = – 8.72 × 10–18 J
1 
25. Calculate the radius of 3rd orbit in
(a) H atom (b) He+ ion (c) Li2+ ion
 n2 
Ans: rn = 52.9   pm
Z
(EQ)
 32 
(i) rn= 52.9   pm = 476pm
 1
 32 
(ii) rn = 52.9   pm = 238.05pm
 2
 32 
(iii) rn = 52.9   pm = 158.7pm
 3
26. Mention any two merits of Bohr’s theory.
(PUB)
Ans: a. Bohr’s theory has given satisfactory
explanation for the emission spectrum of
hydrogen.
b. The values of Rydberg constant, wavelength
of spectral line and radius of atoms / ions
containing one electron, calculated using
Bohr’s theory was in good agreement with
experimental results.
27. Mention any two limitation of Bohr’s theory.
(EQ)
Ans: a. Bohr’s theory could not explain the
emission spectrum in the case of atoms
having more than one electron.
b. It fails to account for the finer details of
hydrogen spectrum.
5
I PUC - Chemistry
28. Write de-Broglie equation and explain the
terms involved in it.
(Mar 16)
h
h
OR λ =
mv
P
33. Name the set of d-orbitals having the
electron density along the axis.
λ = wave length
m = mass of the particle
v = velocity of the particle
p = momemtum
Ans. d x 2 − y 2 and d z 2
29. What will be the wavelength of a ball of mass
0.1 kg moving with a velocity of 10 ms-1?
Given h = 6.626 × 10-34 Js.
(EQ)
Ans: λ =
h
6.626 × 10−34
=
= 6.626 × 10−34 m
mV
0.1 × 10
30. State Heisenberg’s uncertainty principle.
Give its mathematical form. (Feb – 17,20)
Ans: It states that it is impossible to determine
simultaneously, the exact position and
momentum (or velocity) of an electron.
Mathematical formula, ∆∆xx,, ∆∆pp ≥≥
hh
44ππ
31. A microscope using suitable photons is
employed to locate an electron in an atom
within a distance of 0.1 A0. What is the
uncertainty involved in the measurement of
its velocity? (h = 6.626 × 10-34 Js).
(PUB)
Ans: Dx = 0.1Aº = 0.1 × 10–10 m
m = 9.11 × 10
∆x m ∆v =
=
–31
m Dv = ?
h
h
, ∆v =
4π
∆x m4π
6.626 × 10 −34
0.1 × 10 −10 × 9.11 × 10 −31 × 4 × 3.14
= 5.79 × 106 ms −1
32. Mention any two differences between orbit
and orbital
(Mar 16)
Ans.
Orbit
It is a definite circular
path around the nucleus
where electrons are
suppose to move.
Orbital
It is a three dimentional
region
arround
the
nucleus where there is
maximum probability of
finding the electrons.
34. Write the value of n, l and m for an electron
in 2p orbital.
(Feb – 18)
Ans: n = 2, l = 1,
m = -1, 0, +1
35. How many total number of orbitals are
associated with Principal quantum number
n= 3?
(Mar16)
Ans: n = 3
Subshells - s,p and d
Total number of orbitals = 1 + 3 + 5 = 9
36. Calculate the maximum number of electrons
present in third main energy level. (PUB)
Ans: n = 3
Subshells - s,p and d
Total no.of orbitals = 1 + 3 + 5 = 9
Total no. of electrons = 2+ 6 + 10 = 18
37. Give the possible values of l and m when
n = 2.
(Mar16)
Ans:
n
2
subshell
s
p
l
0
1
m
0
–1, 0, +1
38. Using s, p and d notations, describe the
orbitals with the following quantum
numbers (i) n = 2, l = 1 (ii) n = 4, l = 0
(iii) n = 3, l = 2.
(May – 14,Feb – 18)(1 mark each)
Ans: (i) n = 2, l = 1 ⇒ 2p
(ii) n = 4, l = 0 ⇒ 4s
(iii) n = 3, l = 2 ⇒ 3d
39. State Hund’s rule of maximum multiplicity.
Give one example.
(Feb – 15, 19)
Ans. “Among the orbitals of same energy, electron do
not start pairing until they contain one electron
each or singly occupied.
Ex : Nitrogen ( Z =7)
Electronic Configuration → 1s 2 2s2 2p1x 2p1y 2p1z
I-PUC
Chemistry
Ans: λ =
Orbit can accomodate Oribital can accomodate a
a maximum of 2n2 maximum of 2 electrons.
electrons.
HAND BOOK
6
40. Write the electronic configuration of copper
(Z = 29) and give the value of n and l for the
valence shell electron.
(May – 14)
2
2
6
2
6
1
10
Ans. 29Cu: 1s 2s 2p 3s 3p 4s 3d
n = 4, l = 0 for 4s
THREE MARK QUESTIONS
1. Write the Rydberg’s equation to calculate
the wave number of a spectral line in
hydrogen spectrum. What is the value of
n1 for a spectral line appearing in visible
region of hydrogen spectrum? Name the
series.
(May 14)
1
1
− 2
2
n2 
n1
Ans. υ = R 
n1 = 2 (for visible region) Balmer series
I-PUC
Chemistry
2. Give any three postulates of Bohr’s model of
an atom.
(Feb – 14,15,17,19,May -17,20)
Ans. Postulates of Bohr’s atomic model:
1. Electrons revolve around the nucleus in a
definite circular path called orbits.
2. As long as electrons revolve in a particular
orbit it does not emit energy.
3. Only these orbits are permitted in which
angular momentum of an electron is an
integral multiple of h
2π
mvr =
nh
2π
{m → mass of e–, v → velocity, r → radius of
orbit, n = 1, 2, 3, 4}
3. State (n+l ) rule. Which one of the two
orbitals has lower energy: 4s, 3d. (Feb – 15)
Ans. It states that the orbitals having lower (n+l) value
has lower energy.
Among 4s and 3d.
Orbital n + l
4s
4+0=4
3d
3+2=5
∴ 4s has lower energy than 3d.
4.
For the element with atomic number 24.
(i) Write the electronic configuration.
(ii) Write the value of n and l for its electron
in the valence shell.
(iii) How many unpaired electrons are
present in it?
(PUB)
2
2
6
2
6
1
5
Ans. (i) 1s 2s 2sp 3s 3p 4s 3d
(ii) n = 4, l = 0 for 4s (valence shell)
(iii) Six or 6
FOUR MARK QUESTIONS
1. Explain the significance of four quantum
numbers.
(Feb – 14,15,17,18,19)
Ans. 1. Principal Quantum Number (n): It
specifies the size and energy of orbital.
2. Azimuthal Quantum Number (l): It
indicates the three dimensional shape of the
orbital.
3. Magnetic Quantum number (m): It
indicates the orientation of atomic orbital
in space.
4. Spin Quantum Number (s): It indicates
spin of an electron about its axis. Its value
is +1/2.
CLASSIFICATION OF ELEMENTS AND
PERIODICITY IN PROPERTIES
9. Write
the general outer electronic
configuration of s-, p-(Feb 19), d- (Feb 20),
f- block elements, halogens, and Inert gases/
noble gases.
(1Mark each – EQ)
1-2
Ans: s block - ns
Weightage 4 marks
1 mark – 1Q (Q No. 4)
3 mark – 1Q (Q No. 19)
ONE MARK QUESTIONS
p- block – ns2 np1-6
1. State Mendeleev’s periodic law.
(May14)
Ans: “The properties of the elements are a periodic
function of their atomic weights”.
d block - (n-1) d1-10ns1-2
2. Which important property did, Mendeleev
use to classify the elements in his periodic
table?
Ans: Atomic mass
Inert gases - ns2np6
f block - (n-2) f 1-14 (n-1) d0–1 ns2.
Halogens – ns2np5
10. Give reason (1 mark each)
(i) A cation is smaller than its parent
atom.
(ii) An anion is larger than its parent atom
(Feb 16)
(iii) Nitrogen has a higher Ionisation
enthalpy than oxygen/ Oxygen has
lower ionisation enthalpy than nitrogen
(May 15,17, Feb 17)
(iv) Ionisation enthalpy of inert gases is
very high.
(EQ)
(v) Noble gases have large positive electron
gain enthalpies.
(EQ)
(vi) Group 17 elements (halogens) have
very high negative electron gain
enthalpies.
(vii) Second ionisation enthalpy is higher
than first ionisation enthalpy
(EQ)
(viii) Electron gain enthalpy of chlorine is
higher than fluorine.
(EQ)
Ans: (i) Because it has lesser number electrons.
Electrons are strongly attracted by the
nucleus and pulled inwards. As a result, the
size decreases
(ii) Because the addition of one or more
electrons would result in increased
repulsion among the electrons and a
decrease in effective nuclear charge.
(iii) Nitrogen has half-filled p – orbital
(iv) Outermost shell is completely filled.
3. On what parameter do the elements are
classified in the modern periodic table?
(PUB)
Ans: Atomic number.
4. State Modern Periodic law.
(May 14,Feb. 19)
Ans: The physical and chemical properties of the
elements are the periodic functions of their
atomic numbers.
5. Write the IUPAC name of an element with
atomic number (i) 104 (Feb 19) (ii) 107(Feb
16) (iii) 108 (May 15) (iv) 111 (May 15)
Ans: (i) Unnilquadium (ii) Unnilseptium
(iii) Unniloctium (iv) Unununium
6. What are representative elements? (EQ)
Ans: s & p - block elements together are called
representative elements.
7. Among N,Cu,Ca and U identify the element
which belongs to (i) p- block (ii) d- block
(iii) Actinoid (iv) s block
(1Mark each – EQ, PUB)
Ans: (i) N (ii) Cu (iii) U (iv) Ca
8. Which group elements in the periodic
table are called (i) Halogens (ii) noble
gases?
(PUB)
Ans: (i) Group - 17 (ii) Group - 18
1
I-PUC
Chemistry
3
HAND BOOK
2
(v) Because the electron has to enter the next
higher principal quantum level leading to a
very unstable electronic configuration.
(vi) Because they can attain stable noble gas
electronic configurations by picking up an
electron.
(vii) Due to increase in effective nuclear charge
(viii) Because added electron should go to
smaller orbit in Fluorine (n = 2) which
suffers electron – electron repulsion.
11. Name a species (or ion) that will be
isoelectronic with Mg2+ ion.
(May 14)
+
3+
–
Ans. Na or Al or F .
12. Arrange the following species in the
increasing order of ionic radii.
(PUB)
2+
+
3+
2–
–
Ans. Mg , Na , Al , O , F
Al3+ < Mg2+ < Na+ < F– < O2–.
I-PUC
Chemistry
13. Which of the following species will have the
smallest size? Mg, Mg2+, Al, Al3+. (May 14)
Ans. Al3+
14. Among O2- and F- ions which one has smaller
in size?
(Feb 18)
Ans. F– ion.
15. Select isoelectronic species among Na+,Cl–,
F– and Li+
(Feb 20)
+
Ans: Na and F
16. How electronegativity vary down the group
in the periodic table ?
(EQ)
Ans: Electronegativity decreases down the group in
the periodic table.
17. Which is the least electro negative element
in periodic table?
(EQ)
Ans: Francium
18. Which one has highest electron gain
enthalpy among P,S,Cl and F
Ans: Chlorine
19. Name the halogen which has highest
electron gain enthalpy.
Ans: Chlorine
20. Which is the most electro negative element?
Ans: Fluorine
TWO MARK QUESTIONS
1. What is the basic difference in approach
between the Mendeleev’s Periodic Law and
the Modern Periodic Law?
(PUB)
Ans. Mendeleev’s Periodic Law states that the
properties of the elements are periodic functions
of their atomic weights.
Modern Periodic Law states that the properties
of the elements are periodic functions of their
atomic numbers.
2. Define atomic radius? How does it vary
along a period?
(Mar 15)
Ans. Atomic radius is the distance between the centre
of the nucleus and the outermost shell of an
atom.
Across a period: It decreases.
3. How does the atomic radius vary in the
periodic table?
(EQ)
Ans. Increases down the group and decreases across
the period.
4. Which of the following species will have the
largest and the smallest size? Mg, Mg2+, Al,
Al3+
Ans: Largest size – Mg
Smallest size – Al3+
5. What are Iso-electronic species? Arrange
the following in the increasing order of their
ionic radius : N-3, Mg+2, Na+ and O-2. (May
14, Feb 17, 18)
Ans. Atoms / ions containing same number of
electrons are called isoelectronic species.
Mg+2, Na+ O-2. N-3
6. Arrange the following ions in the order of
increasing ionic radii and also indicate the
number of electrons present around each
ion: K+, Ca2+ , S2–
(EQ)
Ans. Increasing order of ionic radii
Ca2+ < K+ < S2–
K+, Ca2+, S2–
No. of electrons 18, 18 and 18 respectively
7. Define ionisation enthalpy of an element.
How does it very in a periodic table?
(Feb 18,19,20 May15)
Ans. The ionisation enthalpy of an element is defined
as the energy required to remove the outermost
electron from an isolated neutral gaseous atom.
It increases across the period and decreases
down the group.
8. What is screening effect? How does it
influence the Ionisation enthalpy?
Ans: The nucleus reduces its force of attraction on the
valence electrons due to the presence of electrons
in the inner-shell.
Increase in screening effect decreases ionisation
enthalpy.
9. Define electronegativity. Which is the most
electronegative element?
(Feb 17, 19, May 15)
Ans: The ability of an atom to attract shared pair of
electrons towards itself.
The most electronegative element is fluorine.
10. How electronegativity vary in a periodic
table?
(PUB)
Ans : It increases across the period and decreases
down the group.
11. What is electron gain enthalpy? How does it
vary in a periodic table?
(Feb 20)
Ans : Electron gain enthalpy of an element is defined as
the energy required or liberated when an electron
is added to outermost orbit of an isolated neutral
gaseous atom.
It increases across the period and decreases
down the group.
THREE MARK QUESTIONS
1. Mention three defects of Mendeleev’s
periodic table.
Ans:
(i) Some elements are not arranged in the
increasing order of their atomic masses.
3
Co is placed before Ni, Te is placed before
I etc.
(ii) Position of hydrogen is not clear because it
shows properties similar to metals as well
as non metals.
(iii) The position of isotopes of elements is not
celar.
2. Write a brief note on s, p and d block
elements.
(PUB)
Ans: s-block elements:- The elements of group-1
and group-2 of modern periodic table are called
s-block elements. The group - 1 elements are
called alkali metals and group-2 elements are
called alkaline earth metals the general electronic
configuration is ns1–2
p-block elements:- The elements of group-13
to group-18 of modern periodic table are
called p-block elements. The general electronic
configuration is ns2 np1–6
d-block elements:- The elements of group-3
to group-12 of modern periodic table are called
d-block elements. They are also called transition
elements due to their gradual change in
properties. The general electronic configuration
is (n-1)d1-10 ns1-2.
2. Among Chlorine and Flourine which is
having
(i) More negative value of electron gain
enthalpy.
(ii) More ionisation energy.
(iii) Smaller size.
(PUB)
Ans: (i) Chlorine → More negative value of electron
gain enthalpy.
(ii) Fluorine → More ionisation energy
(iii) Fluorine → Smaller size.
I-PUC
Chemistry
I PUC - Chemistry
CHEMICAL BONDING AND
MOLECULAR STRUCTURE
(ii) Bond angle: It is defined as the angle
between the orbitals containing bonding
electron pairs around the central atom in a
molecule.
(iii) Bond ethalpy: It is defined as the amount
of energy required to break one mole of
bonds of a particular type between two
atoms in gaseous state.
(iv) Bond order: It is the number of covalent
bonds between two atoms in a molecule.
Weightage 11 marks
2 mark – 1Q (Q No. 13)
3 mark – 3Q (Q No. 20,21,22)
ONE MARK QUESTIONS
1. Write the Lewis dot structure of (i) O2(Mar14)
(ii) N2(Mar16) (iii) CO32- (Mar14, May 15)
(iv) CH4(Feb 18) (v) O3(Feb 19) (vi) CO2
(Mar16,Feb 18) (vii) CO(Feb 18) (viii) C2H2
(Feb 18) (ix) NO2- (EQ)
Ans:
(i)
(iii)
(v)
(vii)
8. Define dipole moment of a polar bond.
(Mar 16, Feb 17, 20)
Ans. The product of the magnitude of the charge on
any one of atoms and distance between them.
(ii)
7. Between H2O and H2S which is more polar?
Why?
(PUB)
Ans: H2O, because oxygen is more electronegative
(iv)
8. Write resonance structures of ozone.
(PUB,Feb 19)
Ans:
(vi)
(viii)
(ix)
2. State octet rule.
(EQ)
Ans. Tendancy of an atom to have eight electrons in
its valence shell.
9. Mention the shape of (i) BCl3, (ii) BeCl2
(iii) CH4 (iv) PCl5 (v) SF6 molecule. (EQ)
Ans.
3. How many valence electrons are present
around (i) P in PCl5 (ii) S in SF6.
(PUB)
Ans: (i) 10 (ii) 12
4. Define lattice energy.
(Mar 14)
Ans. The energy required to completely separate one
mole of ionic solid into its constituent ions in
gaseous state.
(i)
(ii)
(iii)
(iv)
(v)
BCl3 →Trigonal planar.
BeCl2→ Linear
CH4 → Tetrahedral
PCl5→ Trigonal bipyramidal
SF6→ Octahedral
11. Predict the shape of the molecule having
(i) 2 lone pairs and 4 bond pairs for it central
atom.
(PUB)
7. Define the terms: (i) Bond length (ii) Bond
angle (iii) Bond ethalpy (iv) Bond order.
(May 15)
Ans. (i) Bond length: It is defined as the equilibrium
distance between the nuclei of two bonded
atoms in a molecule.
(ii) 1 lone pairs and 3 bond pairs for it central
atom
(EQ)
(iii) 2 lone pairs and 2 bond pairs for it central
atom
(EQ)
1
I-PUC
Chemistry
4
HAND BOOK
2
Ans: (i) Square planar
(ii) Trigonal pyramidal
(iii) Bent
12. What type of orbital overlapping results in
the formation of (i) sigma bond (ii) pi bond?
(PUB)
Ans: (i) Axial overlapping
(ii) Side ways overlapping.
13. What is hybridisation?
(Mar 14)
Ans: It is the process of intermixing of the orbitals
of slightly different energies resulting in the
formation of new set of orbitals of equivalent
energies and shape.
I-PUC
Chemistry
14. Why is a sigma bond stronger than a pi bond?
(EQ)
Ans: Because axial overlapping is more effective
than lateral overlapping.
16. Why He2 (Helium) molecule is not stable?
(Mar 14,Feb 20)
Ans: Bond order of Helium molecule is zero. Hence
bond does not exist between helium atoms.
17. Name the type of hydrogen bonding in
o- nitrophenol.
(FEB 19)
Ans: Intramolecular hydrogen bonding
18. Between O2 and O2 which one has higher
bond order?
(PUB)
Ans: O2
TWO MARK QUESTIONS
1. Mention two conditions for an atom to form
ionic bond.
(PUB)
Ans: (i) One of the atoms must be a metal and the
other must be a non-metal.
(ii) Ionization potential of one atom must be
small.
(iii) Electron affinity of second atom must be
high.
2. Mention the factors affecting ionic bonding.
(Mar 16)
Ans: Ionization energy, electronegativity, and lattice
energy.
3. Mention any two limitations of octet rule.
(Mar 14)
Ans. (a) Octet rule cannot explain the relative
stability of the molecules.
(b) Octet rule cannot explain the shape of
molecules.
4. What are non-polar molecules? Give one
example.
(EQ)
Ans. When covalent bond is formed between two
identical atoms, the shared pair of electrons are
equally shared by two atoms in a molecules.
Ex: H2, O2, N2 etc.
5. What are polar molecules? Give one example.
(EQ)
Ans. When covalent bond is formed between two
different atoms, shared pair of electron gets
displaced towards more electronegative atom.
Ex: HCl, H2O, etc.
6. Show that BeF2 molecule has zero dipole
moment.
(Feb 17)
Ans. The dipole moment of BeF2 is zero because the
bond dipoles are equal and in opposite directions.
Hence cancel the effect of each other.
7. Draw the shapes of BMO and ABMO formed
by combination of 1s and 1s atomic orbitals.
(Feb 19)
+
.
.
.
+
Ans.
+
+
σ1s(BMO)
1s
1s
.
+
1s
−
.
+
1s
+. . −
σ*1s(AMBO)
8. Write any two conditions for hybridization.
(EQ)
Ans. (a) The orbitals present in the valence shell of
an atom are hybridised.
(b) The orbitals under going hybridisation
shoud have almost equal energy.
3
I PUC - Chemistry
Covalent bond formed by sideways overlapping
of bonding orbitals is called pi bond.
11. Distinguish between a sigma and a Pi bond.
(Feb 17,19,20, May 19)
Ans.
Sigma bond
Pi bond
It is formed by the axial It is formed by the
overlaping of bonding side ways (lateral)
orbitals
overlaping of bonding
orbital.
σ bond is stronger.
π bond is weaker.
12. Mention two conditions for the linear
combination of atomic orbitals.
(EQ)
Ans.
* The combining atomic orbitals must have
comparable energies.
* The extent of overlapping must be large.
13. Give two difference between bonding and
antibonding molecular orbitals.
(EQ)
Bonding molecular
Antibonding molecular
orbital
orbital
It is formed by the addition It is formed by the
of atomic orbitals.
subtraction of atomic
orbitals.
Energy of BMO is lower Energy of ABMO is higher
than energy of combining than energy of combining
orbitals.
orbitals.
14. Write the electronic configuration of diatomic
helium molecule? Calculate its bond order.
(May 15)
2
2
Ans. Electronic configuration of He : (σ1s) (σ *1s)
1
1
[Nb − Na ] = [2 − 2] = 0
2
2
Bond order = 0
Bond order =
THREE MARK QUESTIONS
1. Explain main postulates of VSEPR thory.
(Feb 17,18,19,20,May 14)
Ans. Postulates of VSEPR theory:
(a) The geometry of a molecule depends on
the number of valence shell electron pairs
around the central atom.
(b) Electron pairs tend to repel one another
because electron clouds have similar
charge.
(c) The repulsive interaction of electron pair
decreases in the order.
Lone pair – lone pair > lone pair – bond pair >
bond pair – bond pair.
2. Account for the shape of H2O molecule based
on VSEPR theory.
(May 15)
2
2
2
1
1
Ans. 8O: 1s 2s 2px 2py 2pz
Oxygen atom has 2 lone pair & 2 bond pair of
electrons. Due to greater replusion of lone pair
- lone pair compare to lone pair - bond pair, the
tetrahedral bond angle is distorted to 104.5°.
Water has bent or V shape.
O
H
104.5°
H
3. Account for the shape of NH3 molecule based
on VSEPR theory.
(May 16)
2
2
1
1
1
Ans. 7N: 1s 2s 2px 2py 2pz
Nitrogen has 1 lone pair & 3 bond pair of
electrons. Due to greater replusion of lone pair bond pair compare to bond pair - bond pair the
tetrahedral bond angle is distorted to 1070
Ammonia has Trigonal pyrimidal shape.
..
N
H
107°
H
H
I-PUC
Chemistry
10. What are sigma and pi bonds?
(EQ)
Ans. Covalent bond formed by axial overlapping of
bonding orbitals is called sigma bond.
34. Give one example for a molecule having
(i) Intermolecular hydrogen bonding
(ii) Intramolecular hydrogen bonding (EQ)
Ans: (i) p – nitrophenol or HF
(ii) o - nitrophenol
..
..
9. What is the shape and hybridization of
(i) NH3 (ii) H2O?
(Mar 14)
Ans. (i) Shape: Trigonal pyramidal,
Hybridisation: Sp3
(ii) Shape: Bent, Hybridisation: Sp3
HAND BOOK
4
4. Mention the postulates Valence bond theory.
(Feb 18,19)
Ans: (i) A bond is formed when atomic valence
orbitals overlap with each other
(ii) Overlapping orbitals contain a pair of
electrons.
(iii) Electron density concentrates between
bonded atoms.
(iv) The strength of the bond depends on the
degree of overlapping.
I-PUC
Chemistry
5. Describe any three important salient features
of hybridisation.
(Mar 14)
Ans. (i) The hybrid orbitals have equivalent in
energy and identical shape.
(ii) Both half filled and completely filled
orbitals take part in the hybridisation.
(iii) The no. of hybrid orbitals is equal to the no.
of orbitals taking part in hybridisation.
(iv) Hybrid orbitals form more stable bonds.
6. Explain the formation of Methane molecule
on the basis of hybridisation.
(Feb 18,19)
2
2
1
1
Ans. 6 C − 1s 2s 2p 2py (Groundstate)
x
6
C − 1s2 2s1 2p1x 2p1y 2p1z (Excitedstate)
In an excited state, carbon undergoes sp3
hybridisation to give four sp3 hybrid orbitals
which are directed towards four corners of
tetrahedron. They overlap with 1s atomic orbitals
of hydrogen atom to form four C-H σ bonds.
Geometry: Tetrahedral
Hybridisation: Sp3
Bond angle: 109.5°
7. Explain sp2 hybrdisation by taking BCl3 as an
example.
(Feb 17,18, May 15)
Ans.
5
5
B − 1s2 2s2 2p1x (Groundstate)
B − 1s2 2s1 2p1x 2p1y (Excitedstate)
In an excited state, boron undergoes sp2
hybridisation to give three sp2 hybrid orbitals
which are oriented in trigonal planar arrangement
and overlaps with half filled orbitals of chlorine
atom along the axis to form three B-Cl σ - bonds.
8. Explain sp2 hybridisation in ethene.
(Mar 16, May 16,Feb 20)
Ans:
In ethene, both the carbon atoms are sp2
hybridized and have one unpaired electron in
unhybridized p orbital.
One of the sp2 hybrid orbital of both the carbon
atoms overlap each other to form C-C σ bond.
Two sp2 hybrid orbitals on each carbon atom
form sigma bonds with hydrogen atoms.
The unhybridised orbital on each carbon atom
overlap each other side ways to form one C-C
p -bond.
5
I PUC - Chemistry
9. Explain sp hybridisation in ethyne.
(EQ)
Ans:
8. Explain sp hybrdisation by taking BeCl2 as an
example.
(Feb 19)
2
2
Ans. 4 Be − 1s 2s (Groundstate)
4
Be − 1s2 2s1 2p1x (Excitedstate)
In an excited state, bereyllium undergoes sp
hybridisation to give two sp hybrid orbitals
which are oriented in opposite direction and
overlaps with half filled orbitals of chlorine atom
along the axis to form two Be-Cl σ - bonds.
Cl - Be - Cl
One of the sp hybrid orbital of both the carbon
atoms overlap each other to form C-C σ bond.
Other sp hybrid orbital on each carbon atom form
sigma with hydrogen atoms. Two unhybridized
orbitals on each carbon atom overlap each other
side ways to form two C-C p-bonds.
15
P − 1s2 2s2 2p6 3s1 3p1x 3p1y 3p1z 3d1 (Excitedstate)
In an excited state, phosphrous undergoes sp3d
hybridisation to give five sp3d hybrid orbitals
which are directed towards the five corners of
trigonal bipyramidal. They overlap with half
filled orbitals of chlorine atom to form five P-Cl
σ bonds.
Geometry: Trigonal bipyramidal
Hybridisation: sp3d
Cl
Cl
P
6. With respect to the formation of Ethene
molecule, mention (i) hybridization of carbon.
(ii) number of sigma bonds in the molecule
(iii) number of pi bonds in the molecule. (EQ)
Ans.
(i) Hybridisation of carbon in ethene →sp2
(ii) No. of bonds → 5
(iii) No. of pi bonds → 1
7. With respect to the formation of Ethyne
molecule, mention (i) hybridization of carbon.
(ii) number of sigma bonds in the molecule
(iii) number of pi bonds in the molecule.
(May 15)
Ans. (i) Hybridisation of carbon in ethyne →sp
(ii) No. of bonds → 3
(iii) No. of pi bonds → 2
Cl
Cl
Cl
10. Explain the formation of SF6 molecule on the
basis of hybridisation.
(EQ)
2
2
6
2
2
1
1
Ans. 16 S − 1s 2s 2p 3s 3px 3py 3pz (Groundstate)
16
S − 1s2 2s2 2p6 3s1 3p1x 3p1y 3p1z 3d2 (Excitedstate)
In an excited state, sulphur undergoes sp3d2
hybridisation to give six sp3d2 hybrid orbitals
which are directed towards the six corners of
regular octahedron. They overlap with half
filled orbitals of fluorine atom to form six S-F σ
bonds.
Geometry: octahedral
Hybridisation: sp3d2
I-PUC
Chemistry
In ethyne, both the carbon atoms are sp
hybridized and have two unpaired electrons in
unhybridized p orbitals.
9. Explain the formation of PCl5 molecule on the
basis of hybridisation.
(EQ)
Ans.
2
2
6
2
1
1
1
15 P − 1s 2s 2p 3s 3px 3py 3pz (Groundstate)
HAND BOOK
6
F
F
(ii) Bond order =
F
S
F
F
F
11. Describe LCAO method for the formation
molecular orbitals of Hydrogen molecule.
Write the energy level diagram for these
orbitals.
(EQ)
Ans. Hydrogen molecule has two hydrogen atoms A
and B. Their atomic orbitals may be represented
by wave functions ψA and ψB .
Molecular orbitals are formed by addition and
subtraction of 1s atomic orbitals.
Ψ Mo = Ψ A ± Ψ B
(iii) Diamagnetic
1
11
N b − N a ] = ([48 −−24)]==12
[
2
22
14. For C2 molecule;
(i) Write the electronic configuration.
(ii) Calculate its bond order.
(iii) State its magnetic property.
(May 15,Feb 17)
Ans. C2 molecule contain 12 electrons
Electronic configuration of C2 molecule
(
)(
σ1s 2 2 σ*1s
(σ1s)
2p )= (ππ2p
2p ))
) (σ2s) (σ 2s) ((ππ2p
2
Bond order =
Diamagnetic
2
2
*
2 2
xx
1
1
N b − N a ] = [8 − 4 ] = 2
[
2
2
2 2
yy
I-PUC
Chemistry
15. Write the molecular orbital electronic
configuration of oxygen molecule and
calculate the bond order.
(Mar 14, 16,May 16,Feb 17,20)
Ans. O2 molecule contains 16 electrons.
Electronic configuration of O2 molecule:
( σ1s ) ( σ *1s ) ( σ2s ) ( σ *2s ) ( σ2pz ) ( π2px )
2
2
2
( π2p ) ( π *2p ) ( π *2p )
The molecular orbitals σ & σ * formed as
σ = ψA + ψB ,
(BMO)
*
σ = ψA − ψB
(ABMO)
12. For H2 molecule;
(i) Write the electronic configuration.
(ii) Calculate its bond order.
(iii) State its magnetic property.
(May 15, Feb 19, Mar 16)
Ans. (i) Electronic configuration H2 molecule (σ 1s)2
1
(ii) Bond order = 1 [ N b − N a ]== 1( 2
8−−04) ]= =1 2
22 [
2
(iii) Diamagnetic
13. For Li2 molecule;
(i) Write the electronic configuration.
(ii) Calculate its bond order.
(iii) State its magnetic property.
(Feb 18)
Ans. (i) Electronic configuration of Li2 molecule
(σ1s)2 (σ*1s)2 (σ2s)2
OR
y
2
x
1
1
y
2
2
2
KK ( σ2s ) ( σ *2s ) ( σ2pz ) ( π2px ) ( π2py )
2
2
2
( π *2px ) ( π *2py )
1
Bond order =
2
2
1
[ N b − Na ] == 810−−46==22
It is paramagnetic.
2
22
16. What is hydrogen bonding? Mention the
types?
(EQ)
Ans. Attractive force which binds hydrogen atom
of one molecule with the electronegative atom
(F, O, or N) of another molecule or in the same
molecule is called hydrogen bonding.
Types:
1. Intermolecular Hydrogen Bonding
Eg: p-nitrophenol
2. Intramolecular Hydrogen Bonding
Eg: o-nitrophenol
5
STATES OF MATTER
7. State Charles’ law.
(May 14,Feb 20)
Ans. At constant pressure the volume of a given
mass of gas is directly proportional to absolute
temperature.
Weightage 8 marks
1 mark – 1Q (Q No. 2, Part - A)
2 mark – 1Q (Q No. 12 Part - B)
5 mark – 1Q (Q No. 30 Part - D)
8. Give the mathematical form of Charles’ law.
(May 14,Mar 15)
Ans. V ∝ T at constant pressure.
ONE MARK QUESTIONS
1. Mention the type of intermolecular
attractions that exists between
i) Non-polar molecules.
ii) HCl Molecules
(EQ)
Ans:
i) London forces or Dispersion forces.
ii) Dipole-Dipole forces.
I-PUC
Chemistry
9. Represent Charles’ law graphically
(p Vs V) (EQ)
Ans.
2. Name the type of Intermolecular forces
broken in the process of evaporation of
water.
(May 15)
Ans. Hydrogen bonds
3. State Boyle’s law.
(Feb 17,18)
Ans. At constant temperature the volume of a given
mass of gas is inversely proportional to pressure.
10. What is meant by absolute zero? (May 14)
Ans. The lowest hypothetical or imaginary
temperature at which gases are supposed to
occupy zero volume is called Absolute zero. I
4. Give the mathematical form of Boyle’s law.
(Feb 17,18)
ts values is - 273.15° C
1
at constant tamperature
v
5. “At constant temperature, the pressure of a
fixed amount of gas varies inversely with its
volume” Which law is this?
(EQ)
Ans. Boyle’s law
Ans: p ∝
11. What will be the volume of 10cm3 of oxygen
gas at –273.15°C ?
(EQ)
Ans. Zero.
12. State Gay-Lussac’s law.
(EQ)
Ans. At constant volume pressure of a given mass
of gas is directly proportional to absolute
temperature.
6. Represent Boyle’s law graphically
(p Vs V) (May 15)
Ans:
13. What is an ideal gas?
(Mar 16,Feb 17)
Ans: A gas which obeys Boyle’s law, Charles’ law and
Avogadro law strictly is called an ideal gas.
14. Write ideal gas equation for one mole of a
gas.
(Mar 16, Feb 18)
Ans. PV = RT
1
HAND BOOK
2
15. Write the expression for combined gas
equation.
(Feb 19)
P1 V1 P2 V2
=
Ans.
T1
T2
16. Between CO and CO2, which diffuses faster?
(EQ)
Ans. CO (Because of its lower molar mass compare to
CO2)
17. What is aqueous tension
(May 14)
Ans. Pressure exerted by the saturated water vapour is
called aqueous tension.
I-PUC
Chemistry
18. Define the following terms(1 Mark each)
(i) Boyle temperature.
(ii) Critical temperature
(iii) Critical volume
(iv) Critical pressure
(Mar 15, Feb 17, 18, 19)
Ans:
(i) The temperature at which a real gas obeys
ideal gas law over an appreciable range of
pressure is called Boyle temperature
(ii) The temperature above which gas can not
be liquified by applying pressure is called
critical temperature
(iii) Volume of one mole of the gas at critical
temperature is called critical volume.
(iv) The pressure required to liquefy the
gas at critical temperature is called as
the critical pressure.
19. What is the value of critical temperature of
CO2?
(Feb 19)
Ans: 30.98°C
20. What is the effect of increase in temperature
on
(i) Vapour pressure
(ii) Surface tension
(iii) Viscosity
(EQ)
Ans. (i) Increase (ii) decrease (iii) decrease
21. Give reason: Liquid drops are spherical in
shape.
(Feb 20)
Ans: Because they tend to acquire minimum surface
area to reduce surface tension.
TWO MARK QUESTIONS
1. Explain with an example London forces.
(EQ)
Ans. The force of attraction between two temporary
dipoles.
Ex: Interaction between Noble gas atoms.
2. Explain with an example dipole-dipole
interaction between the molecules of a
compound.
(EQ)
Ans. It is interaction between polar molecules.
Ex: Interaction between HCl molecules.
3. Explain with an example dipole-induced
dipole interaction between the molecules of
a compound.
(Mar 15)
Ans. A non-polar molecule may be polarised by the
presence of a polar molecule near it, thereby
making it an induced dipole.
Ex: Noble Gas get polarized in the presence of
polar molecule.
4. A balloon is filled with hydrogen at room
temperature. It will burst if pressure exceeds
0.2 bar. If at 1 bar pressure the gas occupies
2.27 L volume, upto what volume can the
balloon be expanded?
(May 17)
Ans. According to Boyle’s law P1V1 = P2V2
P1 = 1 bar, V1 = 2.27 L
P2 = 0.2 bar V2 = ?
V2 =
P1 V1 1bar ×2.27L
=
= 11.35L
P2
0.2bar
5. A balloon is filled with 2L of air at 300K
temperature. What will be the volume of the
balloon at 320K temperature?
(EQ)
Ans. V1 = 2L, T1 = 300K, V2 = ? T2 = 320K
According to Charles’ law
V1 V2
=
T1 T2
V2 =
V1 T2 2L × 320
=
= 2.13L
T1
300
6. On a ship sailing in Pacific ocean where
temperature is 300K, a ballon is filled with
3L of air. What will be the volume of the
ballon when the ship reaches Indian ocean,
where the temperature is 310K? (May 15)
Ans. V1 = 3L, T1 = 300K, V2 = ? T2 = 310K
3
I PUC - Chemistry
proportionality constant called universal gas
constant. On rearranging we get pV = n RT
According to Charles’ law
V1 V2
=
T1 T2
V1 T2 3L
2L × 310K
320
V2 = 3.1L
=
= 2.13L
300
T1
300K
7. On a ship sailing in Pacific Ocean where
temperature is 23.4C, a balloon is filled
with 2L air. What will be the volume of the
balloon when ship reaches Indian ocean
where temperature is 26.1C.
(Feb 17)
Ans. V1 = 2L, T1 = 273+23.4 = 296.4 K, V2 = ?
T2 = 273+26.1 = 299.1 K
V = 22.7L = 22.7 × 10–3 m3
V1 V2
=
T1 T2
2 × 299.1
V2 =
= 2.018L
296.4
pV = nRT
R=
8. At 25°C and 760 mm of Hg pressure a gas
occupies 600 mL volume. What will be its
pressure at a height where temperature
is 10°C and volume of the gas is 640 mL.
Solution
(EQ)
Ans: p1 = 760 mm Hg, p2 = ?
V1= 600 mL V2 = 640 mL
T1 = 25 + 273 = 298 K, T2 = 10 + 273 = 283 K
According to Combined gas law
105 Pa × 22.7 × 10−3 m 3
1 mol × 273.15K
= 8.314 Pa m3 K–1 mol–1
= 8.314 J K–1 mol–1
12. Calculate the volume occupied by 8.8g
of CO2 at 31.1ºC and 1 bar pressure R =
0.0831L bar mol–1 K–1.
(May 14, Mar 16)
Ans: T = 273 + 31.1 = 304.1K, p = 1 bar
Number of moles (n) =
p1V1 T2
×
T1 V2
pV = nRT
760 × 600 × 283
=
= 676. 6 mm Hg
298 × 640
9. Derive ideal gas equation.
(Feb 17)
1
Ans: At constant T and n; V ∝ (Boyle’s Law)
p
At constant p and n; V ∝ T (Charles’ Law)
At constant p and T ; V ∝ n (Avogadro Law)
Thus,
=
pV
nT
W 8.8
=
M 44
= 0.2 mol
p1V1 p2V2
=
T1
T2
nT
RnT
V∝
⇒V =
p
p
11. Calculate the value of R in SI unit.
(Mar 16, Feb 17)
Ans: Consider one mole of an ideal gas under STP
conditions (273.15 K and 1 bar pressure) It
occupies a volume of 22.7L.
P = 1 bar = 105 Pa,
According to Charles’ law
p2 =
p = Pressure, V = Volume, n = Number of moles,
R = Gas constant, T = Temperature
where
R
is
V=
0.2 × 0.0831 × 304.1
= 5.054L
1
13. Calculate the pressure exerted by 4 mole of
a gas occupying a volume of 1.5m3 at 100°C.
Given R = 8.314 J/K/mol.
(EQ)
Ans. n = 4 mol, V = 1.5 m3, T = 273 + 100 = 373 K
pV = nRT
p=
nRT 4 × 8.314 × 373
=
= 8269.65Pa
V
1.5
I-PUC
Chemistry
V2 =
10. Give the ideal gas equation for ‘n’ moles of a
gas and explain the terms.
(Feb 19,20)
Ans: pV = nRT
HAND BOOK
4
14. Calculate the temperature of 6.0 mole of a
gas occupying a volume of 4dm3 at 2 bar
presure. (R = 0.0825 L atm mol-1K-1)
(May 15)
Ans. n = 6 mol, V = 4dm3, T = ? p = 2 bar
19. Draw Z Vs p graph for (i) Ideal gas (ii) H2
(iii) N2 (iv) CH4 (v) CO2
(EQ)
Ans.
pV = nRT
T=
pV
2× 4
=
= 16.16K
nR 6 × 0.0825
15. Derive the relationship between Density and
molecular mass of a gaseous substance.
(May 15)
Ans. Ideal gas equation is given by
pV = n RT
Replacing n by
m
,weget
M
m
RT
,weget
M
m RT
Rearranging M =
V p
dRT
M=
p
(Where d is density and M = Molecular mass of
gas)
pV =
I-PUC
Chemistry
16. Mention two conditions at which real gases
approach ideal behaviour.
(EQ)
Ans. Low pressure and High temperature.
17. Write the expression for compressibility
factor Z for one mole of a gas. What
happens to the compressibility factor Z
for CO2 at very high pressure at ordinary
temperature?
(Mar 16)
Ans. Compressibility factor (Z) is given by Z =
Z > 1 under high pressure.
pV
nRT
18. What is compressibility factor (Z)? What is
its value for an ideal gas? (Feb 18, May 15)
pV
Ans. Compressibility factor (Z) =
nRT
Z = 1 for an ideal gas.
20. Two gases A and B have critical temperatures
as 250K and 125K respectively. Which one
of these can be liquified easily and why?
(EQ)
Ans. Gas A : Higher the critical temperature, greater is
the intermolecular force of attraction.
21. Critical temperature of CO2 and CH4 are
31.1°C and -81.9°C respectively. Which
is one of this has stronger intermolecular
forces and why?
(EQ)
Ans. CO2 because, gas with higher critical temperature
will have higher intermolecular forces. Hence
liquified first.
22. What is meant by normal boiling point and
standard boiling point?
(May 14)
Ans. Normal boiling point: Temperature at which
the vapour pressure of a liquid becomes equal
to 1 atmosphere is called normal boiling point.
Normal boiling point of water is 100°C.
Standard boiling point: Temperature at which
the vapour pressure of the liquid becomes equal
to 1 bar is called standard boiling point. Standard
boiling point of water is 99.6°C.
23. Define surface tension. Give its SI unit. (EQ)
Ans. If we imagine a line of unit length on the surface
of the liquid, the force acting perpendicular to it
is called surface tension.
SI unit of Surface tension is Nm-1.
5
I PUC - Chemistry
SI unit of Viscosity / Co-efficent of Viscosity is
Pas.
25 . Why do real gas deviate from ideal
behaviour?
(Feb 20)
Ans. Because 1) they have intermolecular forces of
attraction, 2) Gas molecules have considerable
volume.
THREE MARK QUESTIONS
1. Write the postulates of ‘kinetic theory of
gases’.
(May 14, Feb 18, 17,20,)
Ans. 1. Gases are made up of a large number of
small particles called molecules.
2. Volume occupied by the gas molecules is
negligible as compared to the total volume
of the gas.
3. There is no loss of kinetic energy during
collision.
4. The average kinetic energy of the gas
molecules is directly proportional to
absolute temperature.
I-PUC
Chemistry
24. What is Viscosity? Give its SI unit.
(May 14, Feb 17,20,)
Ans. The resistance possesed by liquid for its flow
which arises due to the internal friction between
layers of fluid is called Viscosity.