Class 11 Sample Paper 2022 Solution Chemistry Term 2

ANSWER
1. I-D,II-C Or C III-D IV-A

2. V-A,VI-D, VII-B,VIII-C Or A

3. B

4. D or D

5. B

6. D or C

7. A or C

8. D or A

9. A

10. C

11. D

12. D

13.C

14.B or A

15.B

16.D

17.

a- Sawhorse: Newman :

b- Ethene decolourise the red liquid Br 2 due to double bond and Ethane does not
decolorize Br 2
OR
(a) Kolbe’s Electrolytic Process: Sodium salt of carboxylic acid is electrolysed to form
hydrocarbon containing even number of carbon atom. Hydrogen gas is released at
cathode.
(b) Wurtz Reaction: Alkyl halid on reaction with Na in presence of Dry ether produce
hydrocarbon containing double number of carbon atom as are present in parent alkyl
halide.
18.(i)Cosmic rays<X-Rays<Amber colour<microwaves<radiation from FM radio.(ii)1
19.(i)

(a) ClF3 (b) XeF4

(ii) Due to zero bond order
Or
(i) O2 is paramagnetic as it contains unpaired electron in pie antibonding molecular
orbital.

(ii) Due to hydrogen bonding in H2O which results large association of water
molecules together but in H2S there is no hydrogen bonding.
20. Δx=[h/(4x3.14)]/Δv=2.11x10-14cm Or [1/λ]= 109737 [1/n12 - 1/n22] n1=1, n2=3,
λ=1.025x10-5cm.

21. ΔH=2x(-94.05)+ (-68.32)-(-310.62)=54.20kcal
22. Na + C+N + Heat→NaCN
6CN-+ Fe2+→ [Fe(CN)6]4-
3[Fe(CN)6]4- + 4Fe3+→Fe4[Fe(CN)6]3.x H2O( Prussian Blue)

23. 2MnO4-(aq.) +6 I-(aq.)+ 4H2O→ 2MnO2(s) + 3I2(s) +8OH- ( Basic medium).
24. (A)Hydrogenation of Alkene: When alkene are subjected to react with hydrogen
gas in the presence of the catalyst such Ni, then alkane are produced.
CH2=CH2 + H2 + CATALYST→ C2H6

B) Dehalogenation Reaction: When vicinal dihalides are treated with zinc metal, lose
a molecule of ZnX2 to form alkene e.g CH2Br-CH2Br + Zn→CH2=CH2 + ZnBr2
25. (A) Addition of HCl gas ,cause forward reaction as HCl will react with NH3 to
form white fumes of NH4Cl and hence according to Le Chatelier principle reactions
will move in forward direction.
(B) Addition of NH3 will result backward direction for the equilibrium reaction as it
will result more consumption of the Ammonia gas added at equilibrium.
26. (a) CaH2 <BeH2 <TiH2 order of increasing electrical conductance
(b)LiH < NaH <CsH order of increasing ionic character
(c) H-H <D-D <F-F order of increasing bond dissociation enthalpy.
OR
(a)I <Br<Cl<F order of increasing electronegativity (b)Al3+<Mg2+<Na+<F-order of
increasing ionic size. (c)B<Be<C<O<N order of increasing ionization potential.
27. (a) CH2=CH2,CH3CH=CH2,CH3CH=CHCH3 order of increasing rate of reaction
with HBr. (b)ClCH2COOH <Cl2CHCOOH< Cl3CCOOH increasing order of acidic
strength
(c) Acetophenone <1-MethylBenzene<Anisole, increasing rate of Electrophilic
Aromatic chlorination substitution reaction.

OR
‘A’ is Hex-3-ene and

‘A’+ H2→Propane

‘A’+ KMnO4+ Heat →Propanoic Acid

28. SO2(g) + NO2(g). SO3(g) + NO(g)

t=0 1 1 1 1

t=teq 1-x 1-x 1+x
1-x 1+x
1-x

Therefore K=16=[1+x]2/[1-x]2
Or 4=[1+x]/[1-x] Therefore x=0.6 Hence [SO2]=1-0.6=0.4 moles/Litre

29. CH3CH=CH2 , Propene will not shows geometrical isomerism
CH3CH=CHCH3 , But-2-ene will shows geometrical isomerism
CHCl=CHCl , 1,2 Dicholoroethene will shows geometrical isomerism.

30. (a)2Mg + O2 →2MgO
(b)CaO + SiO2 →CaSiO3
(c) 2Ca(NO3)2 →2CaO+ 4NO2 + O2
31. (a) PCl5 has sp3d hybridation. As the axial bond pairs suffer more repulsive
interaction from the equatorial bond pairs, therefore axial bonds have been found to be
slightly longer and hence slightly weaker than the equatorial bonds, which makes PCl5
more reactive
(b) 2Ca + O2 + Heat → 2CaO , 3Ca + N2 +Heat→Ca3N2
Ca3N2 + 6H2O→3Ca(OH)2 + 2NH3(A)

Ca(OH)2 + CO2→CaCO3[B-White] + H2O

OR
(i)-(a) BCl3, due to inert pair effect.
(b) As Zn,Cd and Hg do not have partially filled ‘d’ orbital in their ground state as
well as in common oxidation states.
(ii) X-Al, A- Al(OH)3, B-Na[Al(OH)4], C-AlCl3, D-Al2O3

32. [A] = CH3CH=C(CH3)2 [B] = CH3CHBr-CBr(CH3)2 [C] = CH3CH(OH)-
C(OH)(CH3)2
OR
(i) (a) (CH3)3C+>(CH3)2HC+>(CH3)H2C+>CH3+
(b) Toluene > Benzene > m-Dinitrobenzene
(ii) CH3CH=CH2 + HBr→CH3CH(Br)-CH3 (Major)+ CH3CH2-CH2Br(Minor)
33. (a) Empirical formula is the shortest formula of the compound without considering
its bonding aspect. It may or may not be same as Molecular formula of a substance.
(b) Empirical formula : Na 2SH20O14 , Molecular Formula : Na2SO4.10H2O
OR
(a) The standard enthalpy change for the formation of one mole of a compound from
its element in their most stable states of aggregation.
(b) 2Xequ (ii) + 3Xequ (iii)- equ (i) give ΔH=2825.77 kJ for the reaction C2H6(g) →
2C(g) + 6H(g)
Let ‘x’ is the C-C bond energy then ‘x’ + 6 BE(C-H)=2825.77 and hence BE
(C-C)=329.77 kJ