HBSE Class 11 Sample Paper 2024 Answers Chemistry

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Chemistry Sample Paper (2023-24) CHE-856 Class: 11th

1. (a) 2 (1)
2. (b) 6 unpaired e- (1)
3. (c) F<Cl>Br>I (1)
4. (d) O22- (1)
5. (b) 18 gm/mol (1)
6. (c) q=0 (1)
7. (a) +2 (1)
8. (a) electron deficient species (1)
9. (c) < U° (1)
10. (c) (1)
11. (a) 0 (1)
12. (a) planar (1)
13. (b) 6.022 x1021 molecules (1)
14. (a) 1 (1)
15. (d) Assertion(A) is false but Reason(R) is true. (1)
16. (a) Assertion(A) and Reason(R) are correct and (R) is correct explanation of
Assertion(A) (1)
17. (a) Assertion(A) and Reason(R) are correct and (R) is the correct explanation
of Assertion(A) (1)
18. (d) Assertion(A) is wrong statement Reason(R) is correct statement. (1)
SECTION-B
19. (a) n=5 (1)
(b) 16e- (1)
20. (a) as the volume increases, pressure decreases so equilibrium move in
forward direction where number of moles increases. (1)
1
(b) KC(Reverse) = = 120.48 (1)
8.3𝑥15−3

21. 16H++Cr2O72- + 6e- 2Cr3++7H2O
[C2H4O+H2O C2H4O2 + 2e- + 2H+]3
Cr2O72- + 16H+ + 3C2H4O + 3H2O 2Cr3+ + 7H2O + 3 C2H4O2 + 6H+
Cr2O72- + 3 C2H4O2 +10H+ 2Cr3+ + 3 C2H4O2 + 4H2O (2)
22.

(2)
23. Buffer solution is the solution which resist change in the pH value of
solution when small amount of acid or base is added to it.
Buffer solutions are of two types
(a) Acidic buffer:- It contains equimolar quantities of weak acid and its salt
with strong base.
Example acetic acid and sodium acetate
(b) Basic Buffer:- It contains equimolar quantities of weak base and its salt
with strong acid
Example ammonium hydroxide and ammonium chloride (2)
COCH3
Anhy
24. + CH3COCl + HCl
AlCl3
Mechanism
(i) Generation of Electrophile

(ii) Attack of electrophile to form intermediate carbocation.

(iii) Loss of proton from carbocation

(2)
AlCl4- + H+ AlCl3 + HCl
OR
Newman's Representation
H H
H H H

H H HH HH
H (1)
Staggerred Eclipsed
Staggerred Conformation is more stable due to less repulsions. (1)
25. (a) S<P<O<N (1)
(b) P<S<N<O (1)
OR
(a) 3s23p4
Group 16th group
Block p-block
Period 3rd period (1)
(b) 4s2 3d2
Group 4th group
Block d-block
Period 4th period (1)
SECTION-C
26. (a) 1 mole C2H6 contains 2 moles of carbon, 3 mole C2H6 contains 3x2=6
moles carbon. (1)
(b) 1 mole C2H6 contains 6 mole hydrogen, 3 mole C2H6 contains 6x3=18
mole hydrogen. (1)

(c) 1 mole C2H6 = 6.022 x 1023 molecules
3 mole C2H6 = 3x6.022 x 1023 molecules
= 18.066 x 1023 molecules (1)
27. (a) C non metal (1)
(a) A alkali metal (1)
(b) B alkaline earth metal (1)
28. G = H – T S (1)
= 40.63x1000 – 300 x 108.8
= 7990 J/mol (1)
G = +ve  reaction is not feasible (1)
OR
1
SO2(g) + O2  SO3(g)
2

H =  fH° products -  fH° reactants (1)
= fH° SO3 - fH° SO2
= - 395.6 – (-296.6) (1)
= - 395.6 + 296.6
= - 99 KJ (1)
29. (a) AgCl  Cl
108+35.5 35.5
143.5 35.5
35.5 Amt.of AgCl formed
% of Cl = x x 100
143.5 Wt.of Organic Compound
35.5 0.5740
% of Cl = x x 100 (2)
143.5 0.3780

= 37.57%
(b) Paper chromatography is a type of partition chromatography which is
based upon the differences in the tendencies of substance to distribute
between two phases. (1)
OR

(c) The process of displacement of  electrons along the saturated carbon
chain due to the presence of a polar covalent bond at one end of the chain
is called inductive effect (I effect). It is of two types.
(i) + I effect  substituent has less e- attracting power than H.
(ii) - I effect  substituent has more e- attracting power than H.
+ I effect example - CH3, - C2H5
- I effect example – NO2, - CN, -X (1½)
Electromeric effect is the complete transfer of shared pair of  e- to one of
the atom joined by multiple bond. It is of two types
(i) + E effect
(ii) - E effect

(i) + E effect is when  e-s are transferred to atom to which attacking
reagent get attached example

(ii) - E effect is when  bond e- all transferred to atom other than the one
to which reagent get attached.
Example

(1½)
30. (a) NaOH  Na+ + OH-
0.01 M 0.01M
KW = [H+] [OH-] = 10-14
10−14
[H+] = x 100 = 10-12 (1)
0,01

pH = - log [H+]

= - log 10-12
= 12 (1)
(b) SO42- (1)
31. (a) N2 + 3H2  2NH3
ng = 2 - 4 = -2
H = U + ng RT
= - 87.425 x 1000+(-2) x 8.314 x 298 (2)
= - 87425 – 4955.14
= - 77514.7J
OR
(a) H = U when ng=0
eg. H2 + I2  2HI
ng = 2-2=0
 H = U + 0RT
 H = U
H = U + PV
when V = 0 then H = U (2)
OR
any other relevant example.
(b) (i) ng = 1-3 = -2 H = U - 2RT H<U
(ii) ng = 1 – 1 = 0 H = U+O
(iii) ng = 1 – 0 = 1 H = U+RT H>U
H < U in (i) (1)
H > U in (iii) (1)
32. (a)
O O
2 3
H 1 4
5 6

3-keto-2-methylhex – 4 – en – 1 – al (1)

(b) pent - 3 – en -1 - yne
HCC-CH=CH-CH3

(1)
O
||
(c) (i) H – C – OH Methonic acid
O
||
(ii) CH3 - C – OH Ethanoic acid
O
||
(iii) CH3-CH2-C-OH Propanoic acid
O
||
(iv) CH3-CH2-CH2-C-OH Butanoic acid (2)
OR
(i)
5 3
CN
1
4
2
3-methyl pentane nitrile (1)
(ii) 1 2 3 4
CH3 – CH – CH – CH3
| |
OH Br
3-Bromobutan-2-o1 (1)
SECTION-E
2
33. (a) V = 45 x = 0.9 (1)
100
40
m=
1000

h
x = (1)
4mv
6.626𝑥10−34
=
4x3.14x40x10−3x0.9

= 1.46x10-33m (1)
(b) Orbit Orbital
(a) It I well defined It is region in space
circular path around nucleus
around which where the probability
e- revolve to find e- is maximum
(b) Represent planar three dimensional motion of electron
motion

OR / any other relevant difference. (2)
OR
A h PB P𝐵
(a) = = (1)
B PA h PA

P
PB= A
2

A P 1
= A = (1)
𝐵 2.P𝐴 2

5𝑥10−8 1
= B = 10x10-8m
B 2
= 10-7m (1)

(b) Emission Spectrum Absorption Spectrum

(i) It is obtained when It is obtained when white
radiation emitted by white light is passed
the excited substance through solution and
are analysed with transmitted light is
spectroscope analysed through
spectroscope

(ii)
Emission spectrum Consist of dark lines
consist of bright lines in otherwise
coloured lines separated continuous spectrum
by dark spaces. (2)
34. (a) P(Ground State) ls2 2s2 2p6 3s2 3px1 3py1 3pz1
1
P(Excited State) ls2 2s2 2p6 3px1 3py1 3pz1 3dz2

(1)
5 orbital hybridise and form 5 new orbitals of same energy and same
shape.

5 sp3d hybrid orbitals adopt trigonal bipyramidal arrangement which
overlap with 3pz orbital to form 5  bonds

(1)

Axial bonds are longer than equatorial bonds due to greater repulsions
from other bonds in axial position. (1)
(b) BeH2 is linear molecule. Therefore the resultant dipole moment of two
Be-H bonds get cancelled giving zero dipole moment.

H – Be – H =0 (2)
OR
(a) O2 = ls2 *ls2 2s2 *2s2 2pz2 2px2=2py2 *2px1 = *2py1
O+2 = ls2 *ls2 2s2 *2s2 2pz2 2px2=2py2 *2px1 = *2py0
O-2 = ls2 *ls2 2s2 *2s2 2pz2 2px2=2py2 *2px2 = *2py1 (1)
8−4 4
Bond order O2 = = =2
2 2
8−3 5
O+2 = = = 2.5
2 2
8−5 3
O-2 = = = 1.5 (1)
2 2

Higher is the bond order, more is the stability.
O2+ > O2 > O2- (1)
(b) CO2 has zero dipole moment so CO2 is linear as two C=O bond moments
get cancelled where as H2O molecule has resultant dipole moment. Two
O-H bonds are arranged in angular shape and the bond moment of two
O-H bonds give resultant dipole moment. (2)
35. (a)
Ni
(i) CH2=CH2+H2 CH3-CH3 (1)
523- 573K

(ii) 2CH3CH2Cl + Hg2F2 2CH3CH2F + Hg2Cl2 (1)

Dry
(iii) CH3Br + 2Na + CH3Br CH3-CH3 (1)
ether

HBr
(b) H2C=CH-CH2-CH2-CH2-CH3 CH2 - (CH2)4 –CH3
peroxide |
Br
1 – bromohexane (2)
OR
hv 2Na
(a) C2H6+Cl2 CH3-CH2 CH3CH2CH2CH3 (1)
| dry
C1 ether Butane
Red Hot
(b) 3CHCH (1)
Fe tube

(c) Two isomers of C3H6O are
CH3-CH2-CHO and CH3-C-CH3 (1)
||
O
CH3-C= O O =C-CH3 (1)
| |
H CH3
CH3-CH=C-CH3 (1)
|
CH3
2-methyl but-2-ene