Chemistry | 30th January 2024 _ Shift-1

SECTION – A
1. Given below are two statements :
Statements I : The gas liberated on warming a salt with dil H2SO4, turns a piece of paper dipped in lead acetate
into black, it is a confirmatory test for sulphide ion.
Statements II : In statement-I the colour of paper turns black because of formation of lead sulphite.
In the light of the above statements, choose the most appropriate from the options given below :
(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Ans. 3
Na 2S  H 2SO 4 
 Na 2SO 4  H 2S

(CH3COO) 2 Pb  H 2S 
 PbS  2CH3COOH
Black lead sulphide

O
C CHO
Cl H2
2. Pd BaSO4
This reduction reaction is known as :
(1) Rosenmund reduction (2) Wolff-Kishner reduction
(3) Stephen reduction (4) Etard reduction
Ans. 1
Rosenmund reduction
O
C–Cl CHO
a) H2
Pd-BaSO4
Benzoyl chloride Benzaldehyde
Wolff Kishner
O
H H
C–R C
b) N2H4 R
KOH, 

Stephen reduction
CN CHO
SnCl2
(c) H2O/

Etard reduction
CH3 CHO
(d)
+ CrO2Cl2

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3. Sugar which does not give reddish brown precipitate with Fehling's reagent is :
(1) Sucrose (2) Lactose (3) Glucose (4) Maltose
Ans. 1
Sucrose is non reducing sugar and does not contain free aldehyde group thus does not give reddish brown ppt
with fehling reagent.

4. Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : There is a considerable increase in covalent radius from N to P. However from As to Bi only a
small increase in covalent radius is observed.
Reason (R) : Covalent and ionic radii in a particular oxidation state increases down the group.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is false but (R) is true
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are true and (R) is the correct explanation of (A)
Ans. 2
Statement – I Factual data
According to NCERT
Statement – II True

5. Which of the following molecule/species is most stable?




(1) (2) (3) (4)

An.s 1
Stability order Aromatic > Non Aromatic > Anti aromatic
+ +
(1) (2) (3) (4)
2 e– 4 e– Carbene Non
Huckel rule Anti aromatic
aromatic aromatic

6. Diamagnetic Lanthanoid ions are :

(1) Nd3+ & Eu3+ (2) La3+ & Ce4+ (3) Nd3+ & Ce4+ (4) Lu3+ & Eu3+
Ans. 2
Ce [Xe] 4f15d16s2
Ce4+  Diamagnetic
La  [Xe] 4f 05d16s2
La3+  Diamagnetic

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7. Aluminium chloride in acidified aqueous solution forms an ion having geometry :
(1) Octahedral (2) Square planar (3) Tetrahedral (4) Trigonal bipyramidal
Ans. 1
AlCl3 in acidified aqueous solution forms octahedral geometry [Al(H2O)6]3+

8. Given below are two statements :

Statement (I) : The orbitals having same energy are called as degenerate orbitals.
Statement (II) : In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Statement I is true but Statement II is false
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Ans. 1
For single electron species the energy depends on principal quantum number(n) only so statement –II is false.

9. Example of vinylic halide is :

X
CH2X X
x
(1) (2) (3) (4)
Ans. 1
Vinylic carbon

X CH2X
A) B)
Vinylic Halide Allylic carbon

X
Alylic carbon
X
C) D)
Aryl halide

10. Structure of 4-Methylpent-2-enal is :

CH3 O O
(1) H2C C C CH2 C H (2) CH3 CH2 C CH C H
H H CH3
O O
(3) CH3 CH2 CH C C H (4) CH3 CH CH CH C H
CH3 CH3
Ans. 4
5 4 3 2 1
H3C–CH–CH=CH–CHO
CH3
4-Methyl pent-2-enal

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11. Match List – I with List – II.
List – I List – II
Molecule Shape
(A) BrF5 (I) T-shape
(B) H2O (II) See saw
(C) ClF3 (III) Bent
(D) SF4 (IV) Square pyramidal
Choose the correct answer from the options given below :
(1) (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (2) (A) – (II), (B) – (I), (C) – (III), (D) – (IV)
(3) (A) – (III), (B) – (IV), (C) – (I), (D) – (II) (4) (A) – (IV), (B) – (III), (C) – (I), (D) – (II)
Ans. 4

F F
Br
BrF5 F [Square pyramidal]
F
F

O
H2O H 
Bent H
F

Cl F
ClF3 
T-shape
F

F
F
SF4  S
F [Sea–saw]
F


12. The final product A, formed in the following multistep reaction sequence is :
(i) Mg, ether
Br then CO2, H+
(ii) NH3, 
A
(iii) Br2, NaOH
O
NH2
(1) NH2 (2)

O O

(3) Br (4) OH

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Ans. 2
MgBr COOH
Br
Mg, ether CO2, H+

NH3, 
NH2
CONH2
Br2
NaOH
(Hoffmann Bromamide
Aniline reaction)

13. In the given reactions, identify the reagent A and reagent B.

CH3
"A"+(CH3CO)2O
[Intermediate] H3O+
273-283 K

"B"+CS2 CHO
[Intermediate]
H3O+
(1) A-CrO3 B-CrO3
(2) A-CrO3 B-CrO2Cl2
(3) A-CrO2Cl2 B-CrO2Cl2
(4) A-CrO2Cl2 B-CrO3
Ans. 2
CH3 CH(OCOCH3) CHO
CrO3+ H3O+
(CH3CO)2O 
(Benzaldehyde)

OCrCl2OH
CH3 CH CHO
OCrCl2OH
2CrO2Cl2 H3O+
CS2 Hydrolysis
Brown chromium (Benzaldehyde)
complex

14. Given below are two statements : one is labelled as Assertion (A) and the other is labelled Reason (R).
Assertion (A) : CH2 CH CH2 Cl is an example of allyl halide.
Reason (R) : Allyl halides are the compounds in which the halogen atom is attached to sp 2 hybridized carbon
atom.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) (A) is false but (R) is true
(4) Both (A) and (R) are true and (R) is the correct explanation of (A)
Ans. 1
C = C–C–Cl is an example of allyl halide.
Allyl halide – Halogen atom is bonded to sp3 hybridised carbon atom adjacent to C = C to an allyic carbon.

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15. What happens to freezing point of benzene when small quantity of naphthalene is added to benzene?
(1) Increases (2) Remains unchanged
(3) First decreases and then increases (4) Decreases
Sol. 4
When small amount of naphthalene is added to benzene, depression in freezing point takes place and freezing
point of benzene decreases.

16. Match List I with List II :

List I List II
(Species) (Electronic distribution)
+2
A. Cr I. 3d8
B. Mn+ II. 3d34s1
C. Ni+2 III. 3d4
D. V+ IV. 3d54s1
Choose the correct answer from the options given below :
(1) (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (2) (A) – (III), (B) – (IV), (C) – (I), (D) – (II)
(3) (A) – (IV), (B) – (III), (C) – (I), (D) – (II) (4) (A) – (II), (B) – (I), (C) – (IV), (D) – (III)
Ans. 2
Cr  [Ar]3d54s1, Cr2+  [Ar]3d4
Mn  [Ar]3d54s2, Mn+  [Ar]3d54s1
Ni [Ar]3d84s2, Ni2+  [Ar]3d8
V[Ar]3d34s2, V+  [Ar] 3d34s1

17. Compound A formed in the following reaction reacts with B gives the product C. Find out A and B.
B
CH3 C CH + Na A CH3 C C CH2 CH2 + NaBr
(C)
CH3
– +
(1) A CH3 C CNa, B CH3 CH2 CH2 Br
(2) A CH3 CH CH2, B CH3 CH2 CH2 Br
(3) A CH3 CH2 CH3, B CH3 C CH
– +
(4) A CH3 C CNa, B CH3 CH2 CH3
Ans. 1

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 CH3–C  CH + Na CH3–C  C–Na+
(A)
C–C–C–Br
(B)
C–C  C–C–C–C + NaBr

18. Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B).
A  B
NH2 N2Cl Scarlet red dye
NaOH
OH
(1) A = HNO3/H2SO4 B=

NH2
(2) A = NaNO2 + HCl, 0 – 5°C; B=

OH

(3) A = NaNO2 + HCl, 0 – 5°C; B=

OH
(4) A = NaNO2 + HCl, 0 – 5°C; B=

Ans. 4
NH2 N2+Cl– OH
(B)
NaNO2 + HCl Scarlet red dye
0–5ºC NaOH
(A)

19. The Lassiagne's extract is boiled with dil HNO3 before testing for halogens because,
(1) AgCN is soluble in HNO3 (2) Silver halides are soluble in HNO3
(3) Ag2S is soluble in HNO3 (4) Na2S and NaCN are decomposed by HNO3
Ans. 4
If Nitrogen or Sulphur is present in the compound the sodium Fusion extract is boiled with concentrate HNO 3
to decomposed sulphide or cyanide of sodium during lassiagne’s test.

20. Choose the correct statements from the following :

(A) Ethane-1, 2-diamine is a chelating ligand.
(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.
(C) Cyanide ion is used as ligand for leaching of silver.
(D) Phosphine act as a ligand in Wilkinson catalyst.
(E) The stability constants of Ca2+ and Mg2+ are similar with EDTA complexes.
Choose the correct answer from the options given below :
(1) (B), (C), (E) only (2) (C), (D), (E) only (3) (A), (B), (C) only (4) (A), (D), (E) only
Ans. 3
(A) (en) is chelating ligand (B) Fact (C) Fact

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 SECTION – B
21. The rate of First order reaction is 0.04 mol L–1 s–1 at 10 minutes and 0.03 mol L–1 s–1 at 20 minutes after
initiatation. Half life of the reaction is ______ minutes. (Given log2 = 0.3010, log3 = 0.4771)
Ans. 24
r1=K[A1]
0.04 = K [A1]
r2=K[A2]
0.03=K[A2]
A1 0.04 4
 
A2 0.03 3
let A1 be the intial and A2 be the final concentration after a given gap of 10 min
2.303 A 
then K  log  1 
t  A2 
2.303 4
K log  2.855  10 –2 min –1
10 3
0.693 0.693
t1/2    24.08
k 2.855 10 –2
= 24 min.

22. The pH at which Mg(OH)2 [Ksp] = 1×10–11] begins to precipitate from a solution containing 0.10 M Mg2+ ions
is ______.
Ans. 9
[Mg2+] [OH–]2 = Ksp
[0.1] [OH–]2 = 1×10–11
[OH–]=10–5 M
POH = –log(10–5)=5
pH=14–POH
pH = 14–5=9

B
30
V
(dm3)
20

10 C
23. A
0
10 20 30
P(kPa)
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by
tracing the path A  B  C  A as shown in the diagram above. The total work done in the process is
________ J.

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Ans. 200
W = Area of ABC
1
  30 – 10  30 – 10 
2
1
  20  20
2
400
  200J
2

24. If IUPAC name of an element is "Unununnium" then the element belongs to nth group of Periodic table. The
value of n is _______.
Ans. 11
Atomic number  111
Group No.  11th

25. The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule is _____.
Ans. 8
Two molecular orbital 2s and *2s
six molecular orbital 2pz, *2pz 2px, *2px, 2py, *2py
Total = 2 + 6 = 8

26. On a thin layer chromatographic plate, an organic compound moved by 3.5 cm, while the solvent moved by 5
cm. The retardation factor of the organic compound is _______ × 10 –1.
Ans. 7
Distance travelled by solute
Rf (retardation factor) =
Distance travelled by solvent

3.5
  0.7  7  10 –1
5

27. The compound formed by the reaction of ethanal with semicarbazine contains _______ number of nitrogen
atoms.
Ans. 3
H+
CH3–CHO + H2N–NH CONH2 CH3–CH=N–NH CONH2
Ethanal Semicarbazide (Semicarbazone derivative )
+
H2O
28. 0.05 cm thick coating of silver is deposited on a plate of 0.05 m2 area. The number of silver atoms deposited on
plate are _______ × 1023. (At mass Ag = 108, d = 7.9 g cm–3)
Ans. 11
Volume of Ag deposited = 0.05×104 cm2×0.05 cm
= 25 cm3
Mass of Ag deposited = V×d
Vd
Number of Ag atom =  NA
108
25  7.9  6.023  1023

108
= 1.1×1024
= 11×1023

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29. 2MnO4  bI  cH 2O  xI 2  yMnO2  zOH
If the above equation is balanced with integer coefficients, the value of z is ________.
Ans. 8
7 –1 o 4
2Mno 4–  bI –  cH 2 O  xI 2  y MnO 2  zOH –
nf=3 nf = 1
2MnO4–  6I –  4H 2O 
 3I 2  2MnO 2  8OH –

30. The mass of sodium acetate (CH 3COONa) required to prepare 250 mL of 0.35 M aqueous solution is ________
g. (Molar mass of CH3COONa is 82.02 g mol–1)
Ans. 7
0.35 M  1000 ml Solution contains 0.35 mole of CH3COONa
0.35
then 250 ml solution. contains  250 mole of CH3COONa
1000
= 0.0875 moles
Mass of CH3COONa = 0.0875×82.02
= 7.17 gm
= 7 gm

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Revised