CBSE XI | Chemistry

Sample Paper – 3 Solution

CBSE
Class XI Chemistry
Sample Paper – 3 Solution

Section A
1. HI>HBr>HCl>HF

2. BeCl2: Linear
SiCl4: Tetrahedral
OR

The electron pairs involved in the bond formation are known as bond pairs or shared
pairs.

3. IUPAC name of allyl alcohol: Prop-2-en-1-ol

4. The amount of oxygen required by bacteria to breakdown the organic matter present in
a certain volume of a sample of water is called biochemical oxygen demand.
OR
Carboxyhaemolobin is the compound formed when CO combines with blood.

5. Sodium (Na) – Yellow

Poassium (K) - Violet

Section B

6. Metallic character decreases and non metallic character increases in moving from left to
right in a period. This is due to increase in ionization enthalpy and electron gain
enthalpy.

7. Increasing order of size: Al3+ < Mg2+ < Na+ < F- < O2-
This is an isoelectronic series i.e. the number electrons are the same in all the elements.
Thus, as the effective nuclear charge decreases, electrons are held away from the
nucleus and thus size increases.

8. Given:
Velocity of electron = 2.07×107 m/s
Mass of electron = 9.1×10-31 kg
We know,

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

h

mv
6.63  1034

9.1  1031  2.05  0.5  107
 3.55  1011 m

9. Given:
Pressure P = 5 bar
Molar mass of nitrogen M = 28 g/mol
Density of nitrogen,
PM

RT
Density of gaseousoxideis,

0.987  5  28 0.987  5   x 

273  0.0821 273  0.0821
0.987  5  28  273  0.0821
X 
273  0.0821  0..987  5
 70

The molar mass of the oxide is 70 g/mol

10. 2H2O+ 2 F2 → 4HF+ O2
In this reaction H2O is getting oxidized to O2 and F2 is getting reduced to F- ion.
Therefore, F2 is the oxidizing agent and H2O is reducing agent.
OR
(a) PbS(g) + H2O2(aq) → PbSO4(s) + 4H2O(l)

(b) CO(g) + 2 H2(g) 

Cobalt
catalyst
 CH 3OH

11. Given:
No. Nuclei No. of protons No. of neutrons
56
1 26
Fe 26 30

88
2 38
Sr 38 50

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

12. Molar mass of methanol (CH3OH) = 32 g/mol

= 0.032 kg/mol
Molarity of solution =
0.793

0.032

 24.78mol / l
We have,
M1V1 = M2V2
24.78 × V1 = 0.25 × 2.5
V1 = 25.22 ml
The required volume is 25.22 ml
OR
Given:
M= 3 mol/lit
Mass of NaCl in 1 litre solution = 3×58.5
= 175.5 g
Mass of water in solution = 1000 × 1.25
= 1250 g
Mass of water in solution = 1250- 175.5
= 1074.5 g

Molarity
Number of moles of solute

Massof solvent in kg
3

1.074

 2.79m

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

Section C

13.
We know,

En 

 2.18  1018 Z2 
2
n

For He ,
n  1, n  2

E1 

 2.18  1018 22 
2
1
  8.72  1018 J

rn 
 0.0529  n2
Z
Here,n  1, Z  2

rn 
0.0529 12
2
 0.02645nm
Energy is 8.72 ×10-18 J

Radius of the orbit is 0.02645 nm

OR

Let us find out the atomic number of element.

Let the number of protons = x
x  31.7
Number of neutrons  x 
100

  x  0.317x 
Massno.of element  No.of protons  No.of neutrons
81  x  x  0.317x
81  2.317x
x  35
No.of protons  35
No. of neutrons  81  35
 46
Atomic numer of element is 35

The element with atomic number 35 is bromine (Br).

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

14. The balanced chemical equation is

2CO  O2  2CO2
2mol 1mol
2x22.4L 22.4L

Volume of oxygen required to convert 2 x 22.4 L of CO at N.T.P. = 22.4 L

22.4
Volume of oxygen required to convert 5.2 L of CO at N.T.P. = x 5.2  2.6 L
2 x 22.4

OR

(i) H2+ is more stable than H2- as it contains no electron in antibonding MO while latter
contains an electron in antibonding MO making it less st
(ii) PCl5 contains axial and equatorial bonds. Axial bonds are longer than equatorial
bonds as they face more repulsion from equatorial bonds. Hence axial bonds are
weaker than equatorial bonds.
(iii) NaI is more covalent due to high polarizability of iodide ion due to its bigger size
than chloride ion.
15.
(i) This is due to the reason that the molecules which undergo evaporation are high
energy molecules and therefore, the kinetic energy of the remaining molecules
becomes less. Since the remaining molecules have lower average kinetic energy,
their temperature becomes low.
(ii) This is due to surface tension of liquids. Due to surface tension, the molecules of a
liquid, try to make surface area to be minimum and for a given volume, sphere has
the minimum surface area. Therefore the falling liquid drops are spherical.
(iii) Intermolecular forces are stronger in acetone than in ether. Thus the vapour
pressure of acetone is less than ether.

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

16.
(i) Combustion of methanol
3
CH OH (l) + O (g)  CO (g) + 2 H O(l) ;ΔH= -726 kJ/mol (Eq-1)
3 2 2 2 2
(ii) Enthalpy of formation of CO
2
C (graphite) + O (g)  CO (g) ;ΔH= -393 kJ/mol (Eq-2)
2 2
(iii) Enthalpy of formation of H O
2
1
H (g) + O (g)  H O (l) ;ΔH= -286 kJ/mol (Eq-3)
2 2 2 2
Required reaction :

1
C(graphite) + 2H (g) + O  CH OH(l) ;ΔH= ? (Eq-4)
2 2 2 3
(Eq-2) + (2 x Eq-3) - (Eq-1) gives the required enthalpy

for formation of methanol

ΔH = (-572-393)+726
-1
= -239 kJ mol

17. For AgBr, Ksp = 5.0 x 10-13

Precipitation of AgBr will occur when ionic product [Ag+] [Br-] becomes larger than Ksp.
AgNO3 Ag + + NO3-

[Ag + ]=0.05M
Theconcentration of Br- required tostart precipitation.
K sp 5.0x10-13
[Br- ]= +
= =1.0x10-11
[Ag ] 0.05
Now, [Br- ] = [KBr]=1.0x10-11
Molar mass of KBr=120
Therefore, theamount of KBr required=1.0x10-11 x120
= 1.20x10-9 g

OR

For the reaction CH4(g) +H2O(g) CO(g) + 3H2 (g)

(p CO )(pH 2 )3
(i) K p =
(pCH4 )(pH2O )
(ii) On increasing pressure, the reaction equilibrium will shift in the backward
direction.
There is no effect of catalyst in equilibrium composition; however the equilibrium
will be attained faster.

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

18. Reduction half reaction:

Step 1:

MnO4-+ 5e-  Mn2+

Step 2: MnO4-+ 8H++ 5e-  Mn2+ + 4H2O

Oxidation half reaction:

Step 1:

SO2  HSO4-+ 2e-

Step 2: SO2 + 2H2O  HSO4-+ 3H++ 2e-

Multiply by required coefficient and add the two equations

MnO4-+ 8H++ 5e-  Mn2+ + 4 H2O) x2

(SO2 + 2H2O  HSO4-+ 3H++ 2e-) x5

Final reaction: 2MnO4- + 5SO2 + 2H2O + H+  2Mn2+ +5 HSO4-

OR

P+5 HNO3 
 H3PO4 + 5 NO2 +H2O
O = 15 O =15
H=5 H=5
Oxygen and Hydrogen atoms are balanced.

19.
(a) H2O is covalent hydride whereas NaH is ionic or saline hydride.
(b) Group 7 to group 9 elements do not form hydrides. This region of periodic table from
group 7 to 9 is called as hydride gap.
(c) 1 L of H2O2 gives 15 L of O2 at NTP.

20.
(a) 2-Ethyl-3methylpentan-1-ol
(b) 1-Chloropropan-2-one
(c) 2,4,6-Tribromophenol

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

21. A reagent which can accept an electron pair in a reaction is called an electrophile.
Examples: H+, Cl+,NO2 +, R3C+.

22.
(a) Lithium and magnesium follow diagonal relationship and so lithium like magnesium
forms nitride while other alkali metals do not.

(b) Size of O 2- ion is smaller than SO42-. Since a bigger anions stabilizes bigger cation
more than a smaller cation stabilizes a bigger anion, lattice enthalpy of BaO is
smaller than BaSO4. BaO is soluble as hydration energy is more than lattice energy
but BaSO4 (as hydration energy is less than lattice energy) is insoluble in water.

23.
(a) The inductive effect is least in C2-C3 bond because the magnitude of inductive
effect decreases as the number of intervening bonds increases.

(b)
(i) Metamerism

(ii)Functional group isomerism

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

24.
(a)
(i) When quicklime is heated with silica it gives calcium silicate.

CaO  SiO2 

Heat
 CaSiO3
Calcium Silica Calcium
oxide silicate

(ii)When calcium nitrate is heated it forms CaO, NO2 and O2

2Ca(NO3 )2  2CaO  4NO2  O2

(b) A cation is highly polarizing if its charge/radius ratio is high.

Li+ ion has the highest polarizing power among the alkali metal ions because it has
the highest charge/radius ratio.
Section D
25.

Given:
CCl4(l) → CCl4(l (g) ΔH= 30.5 kJ/mol ……………(1)
C(s)+ Cl2(g) → CCl4(g) ΔH= -135.5 kJ/mol……………(2)
C(s) → C(g) ΔH= 715.0 kJ/mol……………(3)
Cl2(g) → 2Cl(g) ΔH= 242 kJ/mol……………(4)
CCl4(g) → C(g) + 4Cl(g) ΔH=?
By solving,
Equation(3) +2(equation4) – equation (1) –equation(2)

Substituting the values,

ΔH= [715.0 + 2(242) -30.5-(-135.5)] kJ/mol
= 1304 kJ/mol
Bond enthalpy of C-C in CCl4 (average value)
1304

4
 326 kJmol 1
Enthalpy of change for the process is 1304 kJ/mol
Bond enthalpy is 326 kJ/mol

OR
(a) Bond energy is the amount of energy required to dissociate one mole of bonds
present between the atoms in the gaseous phase. As molecules dissociate
completely into atoms in the gaseous phase therefore bond energy of a diatomic
molecule is called enthalpy of atomization.

(b) C(s) +2H2(g) → CH4(g) ΔrH = -74.8 kJ

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

C(s) → C(g) ΔrH0 = +719.6kJ

H2(g) → 2H(g) ΔrH0 = +435kJ

C(s) + 2H2(g) → C(g) + 4 H (g)

C(s) - 2H2(g) - CH4(g)
0 = C(g) + 4 H (g) - CH4(g)
ΔrH = 719.6 +2(435.4) + 74.8
ΔrH = +1665.2 kJ
This gives the enthalpy of dissociation of four moles C-H bons.

Hence bond energy for C-H bond

1665.2

4

 416.3kJ / mol

26.
(a) This is because boric acid does not act as proton donor rather it accepts a lone pair
of electrons from OH- ions. There by acting as monobasic lewis acid.
(b) PbO2 and SnO2 both are in +4 oxidation state. But due to sronger ionetr par effect
Pb2+ ion is more stable than Sn2+ ion.
(c) In other way Pb2+ ions is more easily reduced to Pb2+ ions. Thus PbO2 acts as a
stronger oxidizing agent than SnO2.
(d) The molecules of CO2 are held together by weak Van der Waals forces of attraction
which can be easily overcome by collisions of the molecules at room temperature.
Onsequently CO2 is a gas.
While silicon atoms forms four single covalent bonds with 0-atm which are
tetrahedrally arranged and form a three- dimensional structure. Thus SiO2 is a high
melting solid.
(e) SiF62- is known but SiCl62- is not known, because interaction between lone pair of
chloride ion and Si4+ ion is not strong. Also six large chloride ions cannot be
accommodated around Si4+ due to limitation of its size.
(f) Borazine is called inorganic benzene, as its structure is similar to that of benzene.
Its formula is B3N3H6. The compound is isoelectric and isostructural with benzene.

OR

Metal X on treatment with sodium hydroxide gives white precipitate which dissolves in
excess of NaOH to give soluble complex (B), therefore, the metal X is Al.

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

Al  3NaOH  Al OH 3   3Na 

X

Al  OH 3  NaOH  2Na   Al  OH 4 
 Excess  Sodium tetrahydroaluminate

Al  OH 3  3HCl aq   AlCl3  3H2O

A C 
Al  OH 3 
Δ
 Al2O3  3H2O
A  D

27.
(a)
(i) Benzene to p-Nitrobromobenzene

(ii) Ethyl chloride to ethene

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

(b) Mechanism of addition of HBr to propene

(c) Friedel- Crafts alkylation- It is the reaction of benzene with alkyl halide in presence
of anhydrous aluminium chloride. The reaction results in the formation of alkyl
benzene.

OR

(a)
CH3 -CH2 -CH2 -CH2 -CH2 -CH3 CH  CH

n-hexane Ethyne
3
sp hybridised sp hybridized
carbon carbon
s-character 25% s-character 50%

Since s-orbital are closer to the nucleus, hence due to more s character in ethyne (sp
hybridized) the hybridized orbital is nearest to this carbon atom in comparison to
sp2 hybridised carbon. This leads to the movement of C-H bond pair more towards sp
hybridized carbon, leading to the development of partial positive charge on the
hydrogen attached to sp hydridised carbon and eventually helps in release of proton
(H+). Thus, ethyne is more acidic than n-hexane.

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 CBSE XI | Chemistry
Sample Paper – 3 Solution

(b)
(i) Wurtz reaction: Alkyl halides on treatment with sodium metal in dry ether
medium give higher alkanes. This is called Wurtz reaction and is used for the
preparation of alkanes with even number of carbon atoms.

Ether
CH3Br + 2Na + BrCH3   CH3 -CH3 + 2NaBr
Bromomethane Ethane

(ii) Acidic dehydration: Alcohols on heating with conc. H2SO4 at 443 K form
alkenes with elimination of one water molecule. Since a water molecule is lost
in the presence of acid, the reaction is called acidic dehydration of alcohols.

conc. H SO
CH3 -CH2OH 
2 4CH =CH + H O
2 2 2

(c)

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