CLASS XII CBSE
CHEMISTRY
Set | A – 31 | B – 51 |
31.
conc.HI
B(mix)  (CH3)3C—O—CH3
HI
anhydrous


A
(mix)
(A) A and B are identical mixture of CH3I and (CH3)3C—OH
(B) A and B are identical mixture of CH3OH and (CH3)3C—I
(C) A is mixture of CH3I and (CH3)3C—OH B is a mixture of CH3OH and (CH3)3C—I
(D) B is mixture of CH3I and (CH3)3C—OH A is a mixture of CH3OH and (CH3)3C—I
Set | A – 32 | B – 52|
32.
O
O
(A)
1. KBrO, 
COCH3 

 A
2. H 
3. 
CH3
O
COOH
(B)
CH3
(C)
CH3
COOH (D)
COOH
OH
CH3
OH
Set | A – 33 | B – 53 |
33.
End product of the following sequence of reactions is

CH3MgBr
CO 2 / H3 O
HgSO4 / H2 SO4
Ag2 O
CH  CH 

 
 
 


O
||
(A) CH 3  C  COOH
O
||
(C) CH 3  C  CHO
(B) CH2(COOH)2
O
||
(D) H  C  CH 2COOH
1
CLASS XII CBSE
Set | A – 34 | B – 54|
34.
In two separate experiments equal quantities of an alkyl halide, C4H9Cl were treated
at the same temperature with equal volume of 0.1 molar and 0.2 molar solutions of
NaOH respectively. In the two experiments, the times taken for the reaction of
exactly 50% of the alkyl halide were the same. The most likely structure of halide is:
(A) CH3CH2CH2CH2Cl
(C) (CH3)2 CHCH2Cl
(B) CH3CH(Cl) CH2 CH3
(D) (CH3)3 CCl
Set | A – 35 | B – 55|
35.
When nitrobenzene is treated with 𝐵𝑟2 in presence of 𝐹𝑒𝐵𝑟3 , the major product
formed is m-bromonitrbenzene. Statements which are related to the formation of
m-isomer are
(A) the electron density on meta carbon is less than that on ortho and para positions
(B) the intermediate carbonium ion formed after initial attack of 𝐵𝑟 + at the meta
position is least destabilized
(C) loss of aromaticity when 𝐵𝑟 + attacks at the ortho and para positions and not at
meta position
(D) easier loss of 𝐻 + to regain aromaticity from the meta position than from ortho
and para positions
2
CLASS XII CBSE
Set | A – 36 | B – 56|
36.
Nylon-66 is a polyamide of
(A) Vinylchloride and formaldehyde
(C) Adipic acid and hexamethylene diamine
(B) Adipic acid and methyl amine
(D) Formaldehyde and malamine
Set | A – 37 | B – 57|
37.
The pKa of acetylsalicylic acid (aspirin) is 3.5 . The pH of gastric juice in human
stomach is about 2-3 and pH in the small intestine is about 8. Aspirin will be.
(A) Unionized in the small intestine and in the stomach
(B) Completely ionized in the stomach and almost unionized in the small intestine.
(C) Ionized in the stomach and almost unionized in the small intestine
(D) Ionized in the small intestine and almost unionized in the stomach
Set | A – 38 | B – 58|
38.
The ease of alkaline hydrolysis is more for
COOCH3
COOCH3
COOCH3
(C)
(B)
(A)
NO2
COOCH3
(D)
OCH3
Cl
3
CLASS XII CBSE
Set | A – 39 | B – 59|
39.
By how much would the oxidizing power of the MnO 4 / Mn 2  couple change if the H+
ions concentration is decreased 100 times?
(A) increases by 189 mV
(C) will increase by 19 mV
(B) decreases by 189 mV
(D) will decrease by 19 mV
Set | A – 40 | B – 60|
40.
A solution of sodium sulphate in water is electrolysed using inert electrodes. The
products at cathode and anode are respectively.
(A) H2, O2
(B) O2, H2
(C) O2, Na
(D) O2, SO2
Set | A & B – 41 |
41.
The molar freezing point constant for water in 1.86 K. molarity–1. If 34.2 g of cane
sugar (C12H22O11) are dissolved in 1000g of water, the solution will freeze at
(A) –1.86°C
(B) 1.86°C
(C) –3.92°C
4
(D) 2.42°C
CLASS XII CBSE
Set | A & B – 42 |
42.
The vapour pressures of ethanol and methanol are 44.5mm and 88.7mm Hg
respectively at the same temperature. An ideal solution is formed by mixing 60gm
of ethanol and 40gm of methanol. The mole fraction of methanol in the vapour
phase is.
(A) 0.66
(B) 0.55
(C) 0.11
(D) 0.33
Set | A & B– 43 |
43.
The incorrect statement is
(A) Calamine and siderite are carbonates
(B) Argentite and cuprites are oxides
(C) Zinc blende and iron-pyrite are sulphides
(D) Malachite and azurite are ores of Cu
Set | A & B – 44 |
44.
Xenon crystallizes in face centre cubic lattice and the edge of the unit cell is 620 pm,
then the radius of xenon-atom is.
(A) 438.5pm
(B) 219.25pm
(C) 536.94pm
5
(D) 265.5pm
CLASS XII CBSE
Set | A & B – 45 |
45.
Fraction of total volume occupied by atoms in a simple cube is
𝜋
(A) 2
(B)
√3𝜋
8
(C)
√2𝜋
6
𝜋
(D) 6
Set | A & B – 46 |
46.
Carbonate ion(𝐶𝑂32− ) acts as a
(A) tridentate ligand
(C) monodentate ligand
Solution:
(B) bidentate ligand
(D) tetradentate ligand
(B)
Set | A & B – 47 |
47.
Which of the following compound will not react with ethanolic KCN?
(A) ethyl chloride
(B) acetyl chloride
(C) chlorobenzene
(D) benzaldehyde
Solution:(c) As benzaldehyde will react as nucleophilic addition reaction and Ethyl chloride (𝐶2 𝐻5 𝐶𝑙) and
acetyl chloride (𝐶𝐻3 𝐶𝑂𝐶𝑙)react with alc. KCN by nucleophilic substitution reaction.
Set | A & B – 48 |
48.
Which of the following will react with water?
(A) CHCl3
(B) Cl3 CCHO
(C) CCl4
(D) ClCH2 CH2 Cl
Solution: (b) Choral (𝐶𝐶𝑙3 𝐶𝐻𝑂) on reaction with water forms choral hydrate [𝐶𝐶𝑙3 𝐶𝐻𝐶(𝑂𝐻)2 ] which is quite
stable because of intramolecular hydrogen bonding.
Set | A & B – 49 |
6
CLASS XII CBSE
49.
In a Cannizzaro reaction, the intermediate that will be the best hydride donor is
Solution: (d) −𝑁𝑂2 group is an electron withdrawing group and its presence facilities the release of hydride
ion.
Set | A & B – 50 |
50.
The half-cell reaction for the corrosion
1
2H + + O2 + 2e− → H2 O, E 0 = 1.23 V
2
Fe2+ + 2e− → Fe(s) ; E 0 = −0.44 V
Find the ΔG0 (in kJ) for the overall reaction
(A) -76
(B) -322
(C) -161
Solution:(b) Applying Δ𝐺 0 = −𝑛𝐹𝐸 0
𝐹𝑒(𝑠) → 𝐹𝑒 2+ + 2𝑒 − , Δ𝐺10
2𝐻 + + 2𝑒 − + 1⁄2 𝑂2 → 𝐻2 𝑂(𝑙) ; Δ𝐺20
1
𝐹𝑒(𝑠) + 2𝐻 + + 𝑂2 → 𝐹𝑒 2+ + 𝐻2 𝑂; Δ𝐺30
2
Applying Δ𝐺10 + Δ𝐺20 = Δ𝐺30
Δ𝐺30 = (−2𝐹 × 0.44) + (−2𝐹 × 1.23)
= −(2 × 96500 × 0.44 + 2 × 96500 × 1.23) = −322310𝐽
∴ Δ𝐺30 = −322 𝐾𝐽
Set | A – 51 | B – 31|
51.
The reaction of
with HBr gives
Solution: (b) We can represent the mechanism as follows:
7
(D) -152
CLASS XII CBSE
Set | A – 52 | B – 32|
52.
The geometry of Ni(CO)4 and Ni(PPh3 )2 Cl2 are
(A) both square planar
(B) tetrahedral and square planar respectively
(C) both tetrahedral
(D) square planar and tetrahedral respectively
Solution(C) In 𝑁𝑖(𝐶𝑂)4 , oxidation state of 𝑁𝑖 is zero. Its
configuration is 3𝑑 8 4𝑠 2
In 𝑁𝑖(𝐶𝑂)4 , the unpaired electrons in 3d and 4s pair up and 3d orbitals are filled up and there is no electron
in 4𝑠 − orbital. Then 𝑠𝑝3 hybridization occurs. 𝑁𝑖(𝐶𝑂)4 :
𝑁𝑖(𝑃𝑃ℎ3 )2 ]𝐶𝑙2 O.S. of Ni
In
is +2, i.e. 𝑁𝑖 2+ so its electronic configuration is 3𝑑 8 with 2 unpaired electrons (𝑃𝑃ℎ3 )𝐶𝑙2 cannot pair up
electrons in 3d-orbitals , thus in [𝑁𝑖(𝑃𝑃ℎ3 )2 𝐶𝑙2 , 𝑠𝑝3 hybridization occurs
Because of 𝑠𝑝3 hybridization, it is tetrahedral
Set | A – 53 | B – 33|
53.
Innitroprussideion, the iron and NO exist as FeII and NO+ rather than FeIII and NO.
These forms can be differentiated by
(A) estimating the concentration of iron
(B) measuring the concentration of CN
(C) measuring the solid state magnetic moment
(D) thermally decomposing the compound
Solution: (c) Magnetic moment 𝜇 is given by = √𝑛(𝑛 + 2)𝐵. 𝑀.
Where n is the number of unpaired electrons
Number of unpaired electrons in various species are
𝐹𝑒 2+ : 𝐼𝑡 𝑖𝑠 3𝑑 6 𝑖. 𝑒 4unpaired electrons
𝐹𝑒 3+ : 𝐼𝑡 𝑖𝑠 3𝑑 5 𝑖. 𝑒. 5 unpaired electrons
they (𝑖. 𝑒 .+ 𝑁𝑂, 𝑁𝑂) have different number of unpaired electrons so they can be differentiated by the
measurement of the solid state magnetic moment of the nitroprusside ion
8
CLASS XII CBSE
Set | A – 54 | B – 34|
54.
Saturated solution of KNO3 is used to make the ‘salt-bridge’ because
(A) velocity of K + is greater than that of NO−
3
+
(B) velocity ofNO−
is
greater
than
that
of
K
3
(C) velocities of both K + and NO−
3 are nearly the same
(D) KNO3 is highly soluble in water
Solution:(c) The ionic mobility’s of 𝐾 + 𝑎𝑛𝑑 𝑁𝑂3− ions are nearly the same. It helps to keep the cathode and
anode half-cells natural at all times
Comprehension Type 55, 56 & 57
An unsaturated hydrocarbon A (C7H12) absorbs two molecule of hydrogen when
hydrogenated. On oxidation it gives one molecule each of acetic acid, and acetoacetic acid
and on reduction give 2-methylhexane
Set | A – 55 | B – 35|
55.
The structure of A:
(A) CH3—CH2—
(C) CH3—
CH2—CH=CH2
(B) CH3—CH=
CH2—CH=CH2
CH2—CH=CH—CH 3
(D) CH3—CH=
CH=CH—CH3
Sol : (C)
Set | A – 56 | B – 36|
56.
The compound A on heating produces a more stable compound:
(A) CH3—CH2—
—CH=CH2
(B) CH3—CH=
(C)
(D)
Sol : (D)
9
CH=CH—CH3
CLASS XII CBSE
Set | A – 57 | B – 37|
57.
The A on Birch Reduction with Na/Lq.NH3 gives:
(A) CH3—CH2—
(C) CH3—
CH2—CH=CH2
(B) CH3—CH=
CH2—CH=CH—CH 3
CH2—CH2—CH 3
(D) CH3—CH2—
—CH2—CH2 —CH3
Sol : (C)
Comprehension Type 58, 59 & 60
Wurtz reacton involves the condensation of two molecules of alkyl halides in the presence
of sodium and dry ether
𝐷𝑟𝑦 𝐸𝑡ℎ𝑒𝑟
𝑅 − 𝑋 + 2𝑁𝑎 + 𝑋 − 𝑅 →
𝑅 − 𝑅 + 2𝑁𝑎𝑋
In this reaction small amount of alkene is also formed as by-product.
𝑁𝑎/𝐷𝑟𝑦 𝑒𝑡ℎ𝑒𝑟
𝐶𝐻3 𝐶𝐻2 𝐵𝑟 + 𝐶𝐻3 𝐶𝐻2 𝐵𝑟 →
𝐶𝐻3 − 𝐶𝐻2 − 𝐶𝐻2 − 𝐶𝐻3 + 𝐶𝐻2 = 𝐶𝐻
⏟ 3 − 𝐶𝐻3
𝐵𝑦−𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
Tertiary alkyl halides do not give Wurtz reaction. Frankland reaction is similar but has
similar but has certain advantages over Wurtz reaction. It is useful in the synthesis of
symmetrical alkanes. Frankland reaction is shown by primary, secondary as well as tertiary
alkyl halide.
Answer the following questions
Set | A – 58 | B – 38|
58.
Which of the following alkanes is not obtained from Wurtz reaction?
(A) Methane
(B) Ethane
(C) Propane
(D) Butane
Sol : (A)
Set | A – 59 | B – 39|
59.
A mixture of ethyl iodide and methyl iodide is subjected to the Wurtz reaction. The
products which cannot be formed:
(A) ethane
(B) butane
(C) propane
(D)2-methylpropane
Sol : (D)
Set | A – 60 | B – 40|
60.
The intermediate compound(s) formed in frankland reaction is/are:
Sol (C)
(A) RZnI2
(B) RZn
(C) RZnI
10
(D) R2ZnI