01. What is the oxidation number of fluorine in OF2? (1 mark)
(A) +2
(C) −2
(B) +1
(D) −1
Solution:
In OF2, oxidation state of O is +2 and that of F is –1.
Answer: (D)
02. Physical properties of Hydrogen are similar to….. (1 mark)
(A) metals
(C) insulator
(B) non-metals
(D) semiconductor
Solution:
Physical properties of Hydrogen are similar to non–metals.
Answer: (B)
03. Which scientist received the noble prize for separation of isotopes of hydrogen by
physical method? (1 mark)
(A) Michael Faraday
(C) Harold C. Urey
(B) J. D. Watson
(D) J. J. Thomson
Solution:
Harold C. Urey received the noble prize for separation of isotopes of hydrogen by physical method.
Answer: (C)
04. The atoms of protium are ……… times more than the atoms of dutarium. (1 mark)
(A) 5 × 103
(C) 5 × 10−3
(B) 2 × 103
(D) 2 × 10−4
Solution:
The atoms of protium are 5 × 103 times more than the atoms of dutarium.
Answer: (A)
05. Which catalyst is used in preparation of methanol form water gas? (1 mark)
(A) FeO
(C) Cu2O
(B) Ni
(D) CuO
Solution:
Cu2O catalyst is used in preparation of methanol form water gas.
Answer: (C)
06. Which element does not give interstitial hydride? (1 mark)
(A) Ga
(C) Fe
(B) Cu
(D) Ac
Solution:
Fe does not give interstitial hydride. (Elements of group 7, 8, 9 do not give hydride)
Answer: (C)
07. Which of the following hydride is electron rich? (1 mark)
(A) BH3
(C) CH4
(B) PH3
(D) AlH3
Solution:
(A) BH3 → electron deficient
(B) PH3 → electron rich
(C) CH4 → electron precise
(D) AlH3 → electron deficient
Answer: (B)
08. What is the level of Sodium in the red blood cells inn blood plasma? (1 mark)
(A) 10 m mol l−1
(B) 143 m mol l−1
(C) 105 m mol l−1
(D) 5 m mol l−1
Solution:
143 m mol l−1 of Sodium in the red blood cells inn blood plasma.
Answer: (B)
09. Give proportion of ATP in a person in resting condition at every 24 hours. (1 mark)
(A) 10 g
(C) 10 kg
(B) 15 g
(D) 15 kg
Solution:
15 kg of ATP is present in a person in resting condition at every 24 hours.
Answer: (D)
10. What would be the ratio of silica and alumina for good quality of cement? (1 mark)
(A) between 2 and 4
(C) between 2.5 and 4
(B) between 4.5 and 6.5
(D) between 4 and 6
Solution:
For good quality of cement the ratio of silica and alumina would be between 2.5 and 4.
Answer: (C)
11. What is the molality of 3 M H3PO4 solution whose density is 1.794 g/ml? (2 mark)
(A) 2.5
(C) 2.25
Solution:
m=
1000M
1000d − MM 2
m=
1000 × 3
1000 × 1.794 − 3 × 98
m=
3000
1794 − 294
(B) 2.45
(D) 2
m=
3000
1500
m = 2m
Answer: (D)
12. Atomic weight of CO2 is 44 u. Avogadro number is 6.022 × 1023. What is the atomic
mass unit of CO2 molecule in grams? (2 mark)
(A) 3.37 × 10−23 g
(C) 3.41 × 10−23 g
(B) 7.31 × 10−23 g
(D) 7.13 × 10−23 g
Solution:
The mass of 6.022 × 1023 CO2 molecules is 44 u.
44 × 1
∴ The mass of 1 CO2 molecules is
= 7.31 × 10−23g .
23
6.022 × 10
Answer: (B)
13. The electron moving with the velocity of 600 m/s is measured with 0.005% accuracy
Calculate uncertainty in measured value of its position. (2 mark)
(A) 5.10 × 10−3 m
(C) 1.93 × 10−3 m
(B) 3.84 × 10−3 m
(D) 1.52 × 10−4 m
Solution:
0.005
= 3 × 10 −2 m / s
100
According to Heisenberg uncertainty principle;
h
∆ x i ∆p =
4π
h
∆ x i m∆v =
4π
h
∆x =
4π m∆v
Uncertainty in velcity of elctron =∆v = 600 ×
∆x =
6.626 × 10 −34
4 × 3.14 × 9.109 × 10 −31 × 3 × 10 −2
6.626 × 10 −1
4 × 3.14 × 9.109 × 3
∆ x = 1.930 × 10 −3 m
∆x =
Answer: (C)
14. A- In Rubidium atom, 37th electron enters in 5s instead of 4d.
R- Orbital’s minimum energy is decided by (n + l) rule. (2 mark)
(A) A is true but R is not true.
(B) A is not true but R is true.
(C) A and R both are right but R is not the explanation of A .
(D) A and R both are right and R is the explanation of A.
Solution:
In Rubidium atom, 37th electron enters in 5s instead of 4d orbitals is due to screening effect
experienced in 5s orbital is less than that of in 4d orbital so energy of 5s orbital is less than that of 4d
orbital.
The minimum energy guideline can be obtained for our understanding is decided by (n + l) rule.
Answer: (C)
15. If ionic radius of X+ and Y− is almost same (129 pm) then atomic radius of X and Y
are respectively……. (2 mark)
(A) 129 pm, 129 pm
(C) 156 pm, 156 pm
(B) 70 pm, 156 pm
(D) 156 pm, 70pm
Solution:
The ionic radius of X+ and Y− is almost same (129 pm).
The radius of cation is always smaller than its atom so the radius of X must be larger than that of X+
ie; 156pm.
The radius of anion is always larger than its atom so the radius of Y must be smaller than that of Y–
ie; 70pm.
Answer: (D)
16. What is the proportion of Ca2SiO4, Ca3SiO3 and Ca3Al2O6 in Portland cement
respectively? (1 mark)
(A) 51%, 26%, 11%
(C) 26%, 51%, 11%
(B) 11%, 26%, 51%
(D) 26%, 11%, 51%
Solution:
The proportion of Ca2SiO4, Ca3SiO3 and Ca3Al2O6 in Portland cement is 26%, 51%, 11%
respectively.
Answer: (C)
17. Which pump is working in human body? (1 mark)
(A) Ca-Mg
(C) Rb-Cs
(B) Na-K
(D) Ba-Sr
Solution:
Na-K pump is working in human body.
Answer: (B)
18. Caustic soda is used in ….. (1 mark)
(A) preparation of soap, paper and artificial silk
(B) purification of bauxite
(C) mercerization of cotton cloth
(D) all of the above
Solution:
Caustic soda is used in preparation of soap, paper and artificial silk, purification of bauxite and
mercerization of cotton cloth.
Answer: (D)
19. Which salt cannot be obtained by Solvay ammonia method? (1 mark)
(A) Na2CO3
(C) KHCO3
(B) NaHCO3
(D) NH4HCO3
Solution:
KHCO3 salt cannot be obtained by Solvay ammonia method because it is very highly soluble in water
at normal temperature.
Answer: (C)
20. Which alloy is used in formation of spring? (1 mark)
(A) Be-Mg
(C) Cu-Mg
(B) Cu-Be
(D) Be-Al
Solution:
Cu-Be alloy is used in formation of spring.
Answer: (B)
21. What is the correct order of ionic radius of S, Cl, K and Ca? (2 mark)
(A) S−2 > Cl− > K+ > Ca+2
(C) Cl− > S−2 > K+ > Ca+2
(B) K+ > Ca+2 > S−2 > Cl−
(D) S−2 < Cl− < K+ < Ca+2
Solution:
S−2, Cl−, K+, Ca+2are isoelctronic ions. In isoelctronic ions, higher the nuclear charge smaller the
ionic size.
Answer: (A)
22. xKMnO 4 + yNH 3 
→ KNO3 + MnO 2 + KOH + H 2O
What is x and y in the above reaction? (2 mark)
(A) 4, 6
(B) 3, 8
(C) 8, 6
(D) 8, 3
Solution:
The balanced reaction equation is as under;
8KMnO 4 + 3NH3 
→ 3KNO3 + 8MnO2 + 5KOH + 2H 2O
Answer: (D)
23. Which of the following reaction is redox? (2 mark)
(A) 3Cu + 2Au (NO 3 )3 
→ 3Cu (NO3 ) 2 + 2Au
(B) Fe + CdSO 4 
→ FeSO 4 + Cd
(C) 3Ag + Au (NO3 )3 
→ 3AgNO3 + Au
(D) All of the above are true
Solution:
In all above options change in oxidation state takes place so all are redox reactions.
Answer: (D)
24. How many gram H2O2 is there in 600 ml solution of 10 volume H2O2? (2 mark)
(A) 18.2 g
(C) 1.82 g
(B) 20.6 g
(D) 2.06 g
Solution:
Volume strength = Molarity×11.2
Molarity=
Volume strength
11.2
10
11.2
Molarity=0.8928M
Molarity=
Molarity=
0.8928=
w2 =
w2 ×1000
M 2 ×V
w2 × 1000
34×600
0.8928 × 34×600
1000
w2 =18.21 g H 2 O 2 .
Answer: (A)
25. Find out mole fraction and molality of 10% w/w NaOH. (3 mark)
(A) 0.0476, 2.77 m
(C) 0.045, 2.9 m
Solution:
The concentration of NaOH solution is 10%w/w.
10 g NaOH is dissolved in 100 g solution.
∴10 g NaOH is dissolved in 90 g water.
X2 =
X2 =
n2
n2 + n1
w2
M2
w2 w1
+
M 2 M1
(B) 0.045, 2.5 m
(D) 0.51, 3.5 m
X2 =
10
40
10 90
+
40 18
X2 =
0.25
0.25 + 5.00
X2 =
0.25
5.25
X 2 = 0.0476
m=
X 2 ×1000
X1 × M1
0.0476 × 1000
0.9524 × 18
m = 2.777 m
m=
Answer: (A)
26. Match column I and Column II. (4 mark)
Column I
Column II
Cl
a. 1-Chloro-4-methylpentan-2-ol
NH 2
1.
O
OH
b. 4-Chloro-4-methylpentan-2-ol
OH
2.
O
c. 2-Methyl-5-chloropentan-4-ol
Cl
3.
OH
d. 2-Hydroxypentanoic acid
Cl
4.
OH
e. 2-Chloro-5-aminopentane
f. 4-Chloropentanamide
(A) 1→f; 2→d; 3→b; 4→a
(B) 1→e; 2→b; 3→d; 4→a
(C) 1→f; 2→a; 3→b; 4→c
(D) 1→e; 2→c; 3→d; 4→a
Solution:
5
4
5
4
3
2
1
CH3 − CH − CH 2 − CH 2 − C − NH 2
|
||
1.
Cl
O
4-Chloropentanamide
3
2
1
CH3 − CH 2 − CH 2 − CH − C − OH
|
||
2.
OH O
2-Hydroxy pentanoic acid
CH3
|
5
3.
4
3
2
1
3
2
1
CH3 − C− CH 2 − CH − CH3
|
|
Cl
OH
4-chloro-4-methyl pentan-2-ol
5
4
CH3 − CH − CH 2 − CH − CH 2 − Cl
|
|
4.
CH3
OH
1-Chloro-4-methyl pentan-2-ol
Answer: (A)
27. What is used as antacid in medicine? (1 mark)
(A) Magnesium carbonate
(C) Calcium hydroxide
(B) Milk of magnesia
(D) Barium oxide
Solution:
Milk of magnesia is used as antacid in medicine.
Answer: (B)
28. What is the hybridization of middle carbon atom of C3H8? (1 mark)
(A) sp
(C) sp3
(B) sp2
(D) (B) & (C)
Solution:
C3H8 is a saturated hydrocarbon propane so all the carbon atoms including the middle one has sp3
hybridization.
Answer: (C)
29. What is the IUPAC name of (CH3)2C(CI3)2? (1 mark)
(A) 1, 1, 1, 3, 3, 3-Hexaiodopropane
(B) 1, 1, 1-Triiodopropane
(C) 1, 1, 1, 3, 3, 3-Hexaiodo-2, 2-dimethylpropane
(D) 1, 1, 1-Triiodo-2, 2-dimethylpropane
Solution:
CH3
|
3
2
1
CI3 − C − CI3
|
CH 3
1,1,1,3,3,3-Hexaiodo-2,2-dimethyl propane
Answer: (C)
30. The percentage of carbon and hydrogen in an organic substance are 68.85 and 4.92
respectively. Find its empirical formula. (3 mark)
(A) C3H3O
(C) C7H3O2
(B) C4H3O
(D) C7H6O2
Solution:
% of O = 100 – (% of C + % of H)
% of O = 100 – (68.85 + 4.92)
% of O = 100 – 73.77
% of O = 26.23
Element
A
%
No. of atoms
Relative no of atoms
Integer
C
12
68.85
68.85
= 5.7375
12
5.7375
= 3.4955
1.6394
7
H
1
4.92
4.92
= 4.92
1
4.92
=3
1.6394
6
O
16
26.23
26.23
= 1.6394
16
1.6394
=1
1.6394
2
The empirical formula of compound is C7H6O2.
Answer: (D)
31. Calculate the wavelength of electron of hydrogen atom emitted during a transition
from n = 4 state to n = 2. (3 mark)
(A) 4878 Ao
(C) 4634 Ao
(B) 5272 Ao
(D) 5084 Ao
Solution:
ν=
1 1
= 3.28 × 1015  2 − 2 
λ
 ni n f 
c
3 ×1010
λ
3 ×1010
λ
3 ×1010
λ
3 × 1010
λ
λ=
1 1
= 3.28 ×1015  2 − 2 
2 4 
1 1 
= 3.28 ×1015  − 
 4 16 
 4 − 1
= 3.28 ×1015 
 16 
= 3.28 × 1015
3
16
16 × 3 × 1010
3.28 × 1015 × 3
λ = 4.878 × 10 −5 cm
o
λ = 4878 A
Answer: (A)
32. Match P and Q. (3 mark)
Column P
1. Order of first ionization enthalpy
2. Order of electron gain enthalpy
3. Ascending order of electronegativity
(A) 1→a; 2→c; 3→b
(C) 1→c; 2→a; 3→b
Column Q
a. Li>Be<B < C < N
b. Be < B < C < O
c. Be > B<C<N>O
(B) 1→a; 2→b; 3→c
(D) 1→c; 2→b; 3→a
Solution:
1. Order of first ionization enthalpy
2. Order of electron gain enthalpy
→
→
c. Be > B<C<N>O
a. Li>Be<B < C < N
3. Ascending order of electronegativity
→
b. Be < B < C < O
Answer: (C)
33. Which of the following pair does not show tautomerism? (4 mark)
(A) Eth-1-en-1-ol and Ethanal
(B) Propan-2-one and Prop-1-en-2-ol
(C) Pentan-2, 4-dione and 4–Hydroxypent-3-en-2-one
(D) 3-Methylpentan-2-ol and 3-Methylpentan-3-ol
Solution:
2
1
(A) CH 2 = CH − OH , CH3 − CHO
Ethanal
Eth-1-en-1-ol
1
2
3
1
1
2
3
4
1
2
2
3
CH3 − C− CH3 CH 2 = C − CH3
||
|
,
(B)
O
OH
Propan-2-one Prop-1-en-2-ol
1
5
3
2
4
5
CH 3 − C − CH 2 − C − CH3 CH3 − C − CH = C − CH3
||
|
||
||
(C)
,
O
OH
O
O
4-Hydroxy pent-3-en-2-one
Pentan-2,4-dione
3
CH 3
|
5
4
CH3 − CH − CH − CH 2 − CH3 1
3
3
2
4
|
|
(D)
, CH3 − CH 2 − C− CH 2 − CH 3
OH CH 3
|
OH
3-Methyl pentan-2-ol
3-Methyl pentan-3-ol
Answer: (D)
34. (P) As2S3 + (Q) NO 3− + (R) H 2O + (S) H + 
→ (x) NO + (y) SO 4−2 + (z) H 3AsO 4
What is the value of P, Q, R and S in above reaction? (4 mark)
(A) 6, 14, 18, 5
(B) 4, 14, 6, 5
(C) 6, 28, 4, 10
(D) 3, 28, 4, 10
Solution:
As2 S3 + N O3− 
→ N O + S O−42 + H3 As O4
+3
-2
+5
oxidation
+2
+6
+5
24 units
oxidation
4 units
As2 S3 + N O 3− 
→ N O + 3SO 4−2 + 2H 3 As O 4
+6
-6
+5
reduction
+2
3 units
+18
+10
3As2S3 + 28NO3− + 10H + 
→ 28NO + 9SO 4−2 + 6H 3AsO 4
3As2S3 + 28NO3− + 10H + + 4H 2O 
→ 28NO + 9SO 4−2 + 6H3AsO4
Answer: (D)
35. Which is the correct order of the stability of carbonium ion? (1 mark)
+
R −C−R
|
R
(P)
+
H−C−R
|
H
(R)
(A) Q > R > S > P
(C) R > S > Q > P
+
H−C−H
|
H
(Q)
+
R −C−R
|
H
(S)
(B) P > S > R > Q
(D) S > R > P > Q
Solution:
P is 3o carbonium ion which is stabilized by 3 +I effect and 9 hyperconjugations.
Q is methyl carbonium ion which has no inductive effect or hyperconjugation.
R is 1o carbonium ion which is stabilized by 1 +I effect and 3 hyperconjugations.
S is 2o carbonium ion which is stabilized by 2 +I effect and 6 hyperconjugations.
∴The stability order of these carbonium ions will be P > S > R > Q.
Answer: (B)
36. Atoms or group of atoms having electron attraction capacity more than that of
hydrogen is called ………….. (1 mark)
(A) + E effect
(B) − E effect
(C) +I effect
(D) − I effect
Solution:
Atoms or group of atoms having electron attraction capacity more than that of hydrogen is called – I
effect.
Answer: (D)
37. What is the volume of 1.5 N H2O2 solution at STP? (2 mark)
(A) 4.8 l
(C) 2.4 l
(B) 8.4 l
(D) 4.2 l
Solution:
Volume strength = Normality×5.6
= 1.5× 5.6
= 8.4 l
Answer: (B)
38. Match A and B. (2 mark)
Column A
1. Ionic hydride
2. Metallic hydride
3. Molecular hydride
4. Interstitial hydride
Column B
a. LaH2
b. LiH
c. TiH1.5-1.8
d. HF
(A) 1→a; 2, 4→b; 3→ c
(B) 1→b; 2→a; 4→ c; 3→d
(C) 1→c; 2→b; 3→d; 4→a
(D) 1→d; 2→c; 3→d; 4→a
Solution:
1. Ionic hydride
→ b. LiH
2. Metallic hydride
→ a. LaH2
3. Molecular hydride
→ d. HF
4. Interstitial hydride
→ c. TiH1.5-1.8
Answer: (B)
39. What are the names of chief minerals of Barium and Strontium respectively? (2 mark)
(A) Crinite, Silastine
(C) Witherite, Silastine
(B) Witherite, Criserite
(D) Crinite, Criserite
Solution:
The of chief minerals of Barium and Strontium are Witherite and Silastine respectively.
Answer: (C)
40. Which of the following is true for alkaline earth metal? (2 mark)
(A) they make M−2 ion.
(C) they make M+2 ion.
(B) they make M+ ion.
(D) they have highest electronegativity.
Solution:
Alkaline earth metals have ns2 general electron configuration so they can lose two electrons to
achieve nearest inert gas configuration and make M+2 ion.
Answer: (C)
41. Which of the following pair shows tautomerism? (2 mark)
(A) Ethanal and vinyl alcohol
(C) Diethylether and methyl propyl ether
(B) Neopentane and n-pantane
(D) Propanone and propanal
Solution:
Tautomerism is observed when there exists presence of Hydrogen atom adjacent to carbonyl group
which is possible just in option A only which is clear from their structures.
CH3 − CHO CH 2 = CH − OH
(A)
,
Vinyl alcohol
Ethanal
CH 3
|
CH
−
C
− CH 3 , CH3 − CH2 − CH2 − CH2 − CH3
3
(B)
n-Pentane
|
CH 3
Neo Pentane
(C)
CH3 − CH 2 − O − CH 2 − CH 3 CH3 − O − CH 2 − CH 2 − CH 3
,
Diethyl ether
Methyl propyl ether
(D)
CH3 − CO − CH3 CH3 − CH 2 − CHO
,
Propanone
Propanal
Answer: (A)
42. Which of the following IUPAC name is not true? (2 mark)
(A) CH3CH2CH2COOCH2CH3 →Ethylbutanoate
(B) (CH3)2CHCH2CHO →3-Methyl butenal
H3C
CH3
(C)
3,5-Dimethylcyclopent-1-ene
(D) CH3CH(OH)CH(CH3)2 →3-Methyl-2-butanol
Solution:
(A) CH3CH2CH2COOCH2CH3 →Ethylbutanoate
3
4
2
1
CH3 − CH − CH 2 − CHO
|
(B)
CH3
3-Methyl butanal
H3C
CH3
(C)
3,5-Dimethylcyclopent-1-ene
1
3
2
4
CH3 − CH − CH − CH3
|
|
(D)
OH CH3
3-Methyl-2-butanol
Answer: (B)
43. P4 + NO3− 
→ PO −4 3 + NO 2 This reaction occurs in acidic medium. Which of the
following is its balance reaction? (3 mark)
(A) P4 + 10NO3− + 8H + 
→ 4PO 4−3 + 10NO 2 + 4H 2O
(B) P4 + 20NO 3− 
→ 4PO −43 + 20NO 2
(C) P4 + 20NO 3− + 8H + 
→ 4PO 4−3 + 20NO 2 + 4H 2O
(D) P4 + 20NO 3− + 4H + 
→ 4PO 4−3 + 20NO 2 + 2H 2 O
Solution:
P4 + N O3− 
→ P O4−3 + N O2
+5
+5
0
oxidation
+4
20 units
P4 + N O 3− 
→ 4 P O −4 3 + N O 2
0
+20
+5
reduction
+4
1 unit
P4 + 20NO3− 
→ 4PO −43 + 20NO 2
P4 + 20NO3− + 8H + 
→ 4PO −43 + 20NO 2
P4 + 20NO3− + 8H + 
→ 4PO−43 + 20NO 2 + 4H 2O
Answer: (C)
44. Which of the following pair has same number of σ and π bonds? (3 mark)
(A) Propanol and Propanone
(B) Ethanoic acid and Methyl methanoate
(C) Diethylketone and Methyl propyl ester (D) Propyl amine and Methyl ethyl amine
Solution:
CH3 − C − CH3
CH3 − CH 2 − CH 2 − OH
||
(A)
,
O
Propanol
Propanone
CH3 − C − OH H − C − O − CH3
||
||
(B)
,
O
O
Ethanoic acid Methyl methanoate
CH3 − CH 2 − C − CH 2 − CH3 CH3 − CH 2 − CH 2 − C − O − CH3
||
||
,
(C)
O
O
Diethyl ketone
Methyl propyl ester
(D)
CH3 − CH 2 − CH 2 − NH 2 CH 3 − CH 2 − NH − CH 3
,
Propyl amine
Methyl ethyl amine
Answer: (B)
45. With which reference there is no importance of Dalton’s law? (1 mark)
(A) isotope
(B) isobar
(C) isoosmotic
(D) isothermic
Solution:
Dalton suggested that the mass of all the atoms of the given element is same. So after the discovery
of isotopes the fact that atoms of the same elements can have different masses also due to presence of
different number of neutrons in the nuclei.
Answer: (A)
46. There is a difference of 25K in boiling point of two liquids. How much difference is
there in Fahrenheit? (1 mark)
(A) 25
(C) 45
(B) 77
(D) 54
Solution:
K1 = oC1 + 298
K2 = oC2 + 298
K2 − K1 = (oC2 + 298) − (oC1 + 298)
25 = oC2 −o C1
o
F1 =
9 o
( C1 + 32)
5
o
F2 =
9 o
( C 2 + 32)
5
o
F2 − o F1 =
9 o
9
( C 2 + 32) − ( o C1 + 32)
5
5
9
= ( o C 2 + 32 − o C1 − 32)
5
9
= ( o C 2 − o C1 )
5
9
= (25)
5
=45o F
Answer: (C)
47. What is the value of one atom of 12C obtained by mass spectrometer? (1 mark)
(A) 12 g
(B) 1.992648 × 10−23 g
(C) 1/12 g
(D) 6.022 × 1023 g
Solution:
The value of one atom of 12C obtained by mass spectrometer is 1.992648 × 10−23 g.
Answer: (B)
48. How many grams of sugar should be dissolved in 60 g water so that 25% w/w solution
will be obtained? (1 mark)
(A) 15 g
(C) 25 g
Solution:
(B) 20 g
(D) 40 g
The concentration of sugar solution is 25%w/w.
In 100 g solution 25 g sugar is dissolved.
∴ In 75 g water 25 g sugar is dissolved.
∴ In 60 g water
60×25
= 20g sugar is dissolved.
75
Answer: (B)
49. What is true for 1735 Cl and 37
17 Cl ? (1 mark)
(A) they both have same number of electrons.
(B) they both have same number of nucleons.
(C) they both have same number of neutrons.
(D) all of the above
Solution:
They have 17 number of electrons.
They have 35 and 37 number of nucleons respectively.
They have 18 and 20 number of neutrons respectively.
Answer: (A)
50. Which of the following has 10 electrons, 8 protons and 8 neutrons? (1 mark)
(A) N−3
(B) N−2
(C) O−2
(D) O−1
Solution:
N−3 has 10 electrons, 7 protons and 7 neutrons.
N−2 has 9 electrons, 8 protons and 8 neutrons.
O−2 has 10 electrons, 8 protons and 8 neutrons.
O−1 has 9 electrons, 8 protons and 8 neutrons.
Answer: (C)
51. According to Bohr’s principle, which of the following is quantized for
electron? (1 mark)
(A) angular momentum
(C) acceleration
(B) angular velocity
(D) velocity
Solution:
According to Bohr’s theory, angular momentum of electron is quantized for electron as per following
formula;
h
2π
mvr = n
Answer: (A)
52. The wavelength of electron having energy E is 4000 Ao. What is the wavelength of
electron having energy 0.5E? (1 mark)
(A) 2000 Ao
(C) 6000 Ao
(B) 8000 Ao
(D) 5000 Ao
Solution:
E1 =
E2 =
hc
λ1
hc
λ2
hc
E 2 λ2 λ1
=
=
E1 hc λ2
λ1
0.5E 4000
=
E
λ2
∴ 0.5 =
∴ λ2 =
4000
λ2
4000
0.5
∴λ2 = 8000o A
Answer: (B)
53. Which of the following quantum number set is not possible? (1 mark)
(A) 7, 0, 0 +1/2
(B) 4, 3, +2, −1/2
(C) 5, 2, −2, +1/2
(D) 6, 1, −2, −1/2
Solution:
n can assume any positive integer number which is valid in all options.
l can assume 0 to n–1 integer number, which is valid in all options.
m can assume –l to +l integer values including zero, which is violated in D option in which value of
m is larger than the value of l.
Answer: (D)
54. Energy of 1 mole photon is….. (1 mark)
2e4π 2 z 2 m
(A) E =
n2h2
(B) E = hν
(C) E = cp
(D) E = Nhc/λ
Solution:
Energy of one mole photon is called Einstein which is given by following formula;
E = Nhc/λ
Answer: (D)
55. What is the ratio of energy of photons having wavelength 2000Ao and 4000Ao
respectively? (1 mark)
(A) 1:2
(C) 2.5:1
Solution:
E1 =
E2 =
hc
λ1
hc
λ2
hc
E1 λ1 λ2
=
=
E 2 hc λ1
λ2
E1 λ2
=
E 2 λ1
(B) 1:2.5
(D) 2:1
E1 4000 2
=
=
E 2 2000 1
E1 : E2 = 2 : 1
Answer: (D)
56. Which of the following graph suggest that the property of an atom depends on its
atomic number? (1 mark)
(A) ν2→Z
(B) ν →Z
(C) ν2→A
(D) ν →A
Solution:
ν →Z graph suggest that the property of an atom depends on its atomic number.
Answer: (B)
57. Uub means…. (1 mark)
(A) element having atomic number 111
(C) element having atomic number 113
(B) element having atomic number 112
(D) element having atomic number 114
Solution:
Uub means element having atomic number 112.
Answer: (B)
58. Chancourtois classified elements on the basis of……..(1 mark)
(A) atomic number
(C) atomic volume
(B) oxidation number
(D) atomic mass
Solution:
Chancourtois classified elements on the basis of atomic mass.
Answer: (D)
59. How many elements are there in p block? (1 mark)
(A) 31
(C) 28
(B) 30
(D) 29
Solution:
Total 30 elements are there in p–block.
Answer: (B)
60. Which of the following is true for ionization enthalpy? (1 mark)
(A) Na > Mg > Al > Si
(C) Si > Mg > Al > Na
(B) Na > Al > Si > Mg
(D) Si > Al > Mg > Na
Solution:
The correct order of ionization enthalpy is Si > Mg > Al > Na.
Answer: (C)
61. For which element, the highest shielding effect for outermost electron is
observed? (1 mark)
(A) element of group 13 and period 2
(C) element of group 13 and period 4
(B) element of group 13 and period 3
(D) element of group 13 and period 5
Solution:
After completion of d–orbitals electron start filling in p–orbitals from 4th period ie; elements of 13th
group. As we go down in the group shielding effect gradually increases. So it will be observed
maximum for elements of 13th group and 5th period in the given options.
Answer: (D)
62. What is the order of electron gain enthalpy for F, Cl, Br and I? (1 mark)
(A) F < Cl < Br < I
(C) I < Br < F < Cl
(B) F < Br < I < Cl
(D) Br < I < Cl < F
Solution:
The correct order of electron gain enthalpy is I < Br < F < Cl.
Answer: (C)
63. What is the oxidation number of underlined carbon in CH3COOC2H5? (1 mark)
(A) +2
(C) 0
(B) +4
(D) +3
Solution:
Calculate oxidation state taking COO–.
Answer: (D)
64. Choose correct answer. (T for true and F for false) (1 mark)
(a) reducing agent is reduced.
(b) oxidizing agent is oxidized.
(c) oxidizing agent is reduced.
(d) oxidation reduction does not occur in redox reaction.
(A) FTFT
(B) FFTT
(C) FFTF
(D) TTFT
Solution:
(a) reducing agent is oxidized.
(b) oxidizing agent is reduced.
(c) oxidizing agent is reduced.
(d) oxidation reduction occurs in redox reaction.
Answer: (C)