CHEMISTRY
1. The spin only magnetic moment of Cr3+ is (atomic number of Cr is 24)
(1) 1.73 BM
(2) 2.83 BM
(3) 3.87 BM
(4) 4.90 BM
Solution
The electronic configuration of Cr (Z = 24): [Ar]3d44s2
Therefore, the electronic configuration of Cr3+ is [Ar]3d3
The number of unpaired electrons is 3, so the spin only magnetic moment is
  n(n  2)  3(3  2)  15  3.87 BM
Hence, the correct option is (3)
2. In which of the following reactions, polysubstitution takes place?
(1)
COCH3
+
CH3COCl
(2)
(3)
(4)
Solution
Friedel-Crafts alkylation reaction forms polysubstitution products. Due to activating effect of an alkyl
group connected to an aromatic ring, the monoalkylated reaction product is more reactive towards
electrophillic substitution than the original starting material.
Hence, the correct option is (2)
3. Write the period number, group number and block of the element having atomic number 42.
(1) 5, 5, d
(2) 5, 6, d
(3) 5, 2, d
(4) 5, 15, p
Solution
The element is Mo (atomic number 42). The electronic configuration is [Kr]5s14d5.
Hence, the correct option is (2)
HCl
4. In the reaction (CH 3 ) 2 CH  CH  CH 2 
 the product obtained is
(1) (CH3 )2 CCl  CH2CH3
(2) (CH 3 ) 2 CH  CH  CH 3
|
Cl
(3) a mixture of (1) and (2)
(4) none of these.
Solution
The reaction involved is
Hence, the correct option is (3)
5. The BCl3 is a planar molecule, whereas NCl3 is pyramidal because
(1) N−Cl bond is more covalent than B–Cl bond.
(2) B−Cl bond is more polar than N–Cl bond.
(3) Nitrogen atom is smaller than boron.
(4) BCl3 has no lone pair but NCl3 has one lone pair of electrons.
Solution
BCl3 has no lone pair as B has only three valance electrons thus its shape is trigonal planar ; while NCl3
has one lone pair of electron as N has 5 valence electrons , so its shape is pyramidal.
Cl
B
Cl
Cl
Hence, the correct option is (4)
6. In the following sequence of reactions, Z is
PCl5
(1) conc.H 2SO 4
Z 
 X 
Z
(2) H 2 O boil
(1)
(2)
(3)
(4)
CH 3CH 2 CH 2 OH
CH 3CHOHCH 3
CH3CH 2 CH 2 CH 2 OH
(CH 3 )3CCH 2 OH
Solution
The reaction is
Hence, the correct option is (2)
7. In the complex [Pt(py)4][PtCl4], the oxidation number of Pt atoms in the former and latter parts of the
complex are, respectively,
(1) 0 and 0
(2) +4 and +2
(3) +2 and +2
(4) 0 and +4
Solution
In both the complexes the oxidation number of Pt atom
[Pt (py)4]2+[PtCl4]2 is +2.
Hence, the correct option is (3)
8. During the preparation of arenediazonium salts, the excess of nitrous acid, if any, is destroyed by
adding
(1) aq. NaOH
(2) aq. Na2CO3
(3) aq. NH2CONH2
(4) aq. KI
Solution
Addition of aq. NH2CONH2 destroys excess HNO2 in the form of N2.
NH 2 CONH 2 (aq)  2HNO 2 (aq)  N 2 (g)  CO 2 (g)  3H 2 O(l)
Hence, the correct option is (3)
9. The correct statement regarding defects in solids is
(1) Frenkel defects are usually favoured by a very small difference in the sizes of a cation and anion.
(2) Frenkel defect is a dislocation defect.
(3) Greater the number of F-centres, lower is the intensity of the colour of the compound.
(4) Schottky defects have no effect on the physical properties of solids.
Solution
Frenkel defect is also called dislocation defect because smaller ions are dislocated from their lattice sites
into the interstitial sites.
In Frenkel defect , the ion instead of being in its expected location, is found in one of the interstices. This
defect is favoured by a large difference in size between the positive and negative ions.
Greater the number of F-centres, higher is the intensity of the colour of the compound.
Schottky defect involves the absence of cation or anion from the position which it is expected to occupy
in the periodic arrangement of ions. This affects the physical properties of solids.
Hence, the correct option is (2)
10. For a first-order reaction, the units of Arrhenius factor (A) will be
(1) J Kl sl
(2) s1
(3) mol L1 s1
(4) J Kl L1 s1
Solution
Arrhenius' equation gives the dependence of the rate constant k of a chemical reaction on the absolute
temperature T (in kelvin), where A is the pre-exponential factor (or simply the prefactor), Ea is the
activation energy, and R is the universal gas constant
k  Ae Ea / RT  A  ln k 
Ea
J mol1
 Units of A  s 1 
 s 1
1
1
RT
J K mol  K
Hence, the correct option is (2)
11. Select the structure of the major product in the following reaction.
i.) KMnO4, OH-, hot
?
ii) H3O
+
(1)cyclopentanol
(2) 1-cyclopentylcyclopentane
(3) cyclopentanone
(4) cyclopentene
Solution
i.) KMnO4, OH-, hot
?
ii) H3O+
Hence, the correct option is (3)
12. By how much would the oxidizing power of the (MnO4 | Mn 2 ) couple change if the H+ ions
concentration is decreased 100 times at 25°C?
(1) It will increase by 189 mV.
(2) It will decrease by 189 mV.
(3) It will increase by 19 mV.
(4) It will decrease by 19 mV.
Solution
The reaction involved is MnO4  5e  8H   Mn 2  4H 2 O
According to Nernst equation,
 [Mn 2  ] 
0.059
o
Ered  Ered

log 

 8
5
 [MnO4 ][H ] 
After decrease in concentration, let [H+] = x/100 = x/102, therefore,
o
Ered  Ered

 [Mn 2 ]  1016 
0.059
log 

8 
5
 [MnO4 ][X] 
0.059
log1016  0.189 V = 189 mV
5
This Ered decreases by 0.189 V. The tendency of the half-cell to get reduced is its oxidizing power. Hence,
the oxidizing power decreases by 0.189 V.
Hence, the correct option is (2)

13. What volume of 0.1 M H2SO4 will be required to produce 17.0 g of H2S by the following reaction?
5H 2SO 4  10NaI  4Na 2SO 4  5I 2  H 2S  4H 2 O
(1) 30.0 L
(2) 50.0 L
(3) 25.0 L
(4) 5.0 L
Solution
Moles of H 2S 
Also,
17
17 1

 mol
MH2S 34 2
1 mol H 2S  5 mol of H 2SO4
(from stoichiometry of reaction)
2.5
1
5
 25.0 L
or
mol of H 2S  mol of H 2SO4  M  VL  VL 
0.1
2
2
Hence, the correct option is (3)
14. Proteins contains
(1) exclusively D-form of amino acids.
(2) exclusively L-form of amino acids.
(3) 50% of each of D-form and L-form of amino acids.
(4) 33% of D-form and 67% of L-form of amino acids.
Solution
Proteins are condensation polymers which contains L-amino acids as monomeric units. All the proteins in
the human body are made up of L-isomers of amino acids.
Hence, the correct option is (2)
15. The quantum numbers of 1/2and 1/2 for the electron spin represent
(1) rotation of the electron in clockwise and anticlockwise direction, respectively
(2) rotation of the electron in anticlockwise and clockwise direction, respectively
(3) magnetic moment of the electron pointing up and down, respectively
(4) two quantum mechanical spin states that have no classical analog.
Solution
An electron in its motion in an orbit may rotate or spin about its own axis and spin may be clockwise and
anticlockwise. Assuming this spin to be quantized there are two possible values of s, that is, s = +1/2 and
s = 1/2, depending upon whether the electron spins are in one direction or in the other.
Hence, the correct option is (4)
16. The product formed in the following reaction is
(1)
(2)
(3)
(4)
Solution
CH2NH2 is a strong electrophile and readily releases N2.
..
CH 2 N 2  CH 2  N 2
Hence, the correct option is (4)
17. Which statement below does not occur during the formation of an addition polymer?
(1) Free radicals initiate the process.
(2) Certain double bonds in monomers are replaced with single bonds.
(3) Propagation involves a reaction between two free radicals.
(4) Termination occurs when the free radicals are used up.
Solution
In the chain propagation step, a free radical attacks another monomer to produce yet another free radical
and the process continues. The reaction between two free radicals leads to termination of reacton.
Hence, the correct option is (3)
18. If n = 6, the correct sequence of filling of electrons will be
(1) ns  np  (n  1)d  (n  2)f
(2) ns  (n  2)f  (n  1)d  np
(3) ns  (n  1)d  (n  2)f  np
(4) ns  (n  2)f  np  (n  1)d
Solution
According to Aufbau principle, the correct sequence for filling up orbitals when n = 6 is 6s, 4f, 5d, 6p.
Hence, the correct option is (2)
19. In the following reaction, compound A in acidic medium will have the structure
(1)
(2)
(3)
(4)
Solution
The product formed on nucleophilic substitution reaction with ammonia will be different in acidic and
basic medium. In acidic medium, H+ will be in higher concentration and it will combine with base to form
ammonium ion.
Nucleophilic substitution reaction of substrate with ammonia


Compound A in acidic medium H+ will combine with the  NH
2 and form  N H 3 .Therefore, the final
(Basic)
product formed is
Hence, the correct option is (1)
20. The correct order about of C–O bond length in the following is
(I) CO
(II) CO2
(III) CO32 
(1) II  I  III
(2) III  II  I
(3) I  II  III
(4) I  III  II
Solution
The order is as follows:
Hence, the correct option is (3)
21. Which of the following statement is incorrect?
(1) The system of constant entropy and constant volume will attain the equilibrium in a state of minimum
energy
(2) The entropy of the universe is increasing
(3) The process would be spontaneous when (ΔS)U,V  0, (ΔU)S,V  0
(4) The process would be spontaneous when (ΔS)U,V  0, (ΔU)S,V  0
Solution
For spontaneous process
G  U  pV  T S
At constant U and V
U  0 and V  0
Therefore,
G  T S
For spontaneity G  0 , therefore S  0
At constant S and V
S  0 and V  0
Therefore,
G  U
For spontaneity G  0 , therefore U  0
Hence, the correct option is (3)
22. At temperature T, a compound AB2(g) dissociates according to the reaction
2AB2 (g)
2AB(g)  B2 (g)
with degree of dissociation , which is small compared with unity. The expression for Kp in terms of 
and the total pressure pT is
p 3
(1) T
2
(2)
(3)
pT 2
3
pT 2
2
(4) None of these
Solution
For the given equilibrium, the equilibrium concentration is
2AB2(g)
2AB(g) + B2(g)
c
Equilibrium concentration
c(1  )
c
2
Therefore,
Kp 
( pB2 )( pAB ) 2
( pAB2 ) 2
Substituting for pAB, pAB2 and pB2 in terms of mole fraction and pT , we get
pT
 3  pT
(c / 2)  (c )2
Kp 


[c(1   )]2
[c(1   / 2)]
 
2(1   ) 2 1  
2

3
  pT

1 therefore K p 
As  is small compared with unity, so 1   1and1 
2
2
Hence, the correct option is (1)
23. Select the incorrect statement among the following:
(1) NF3 is a weaker base than NH3.
(2) AlCl3 is largely covalent while AlF3 is largely ionic.
(3) B2H62NH3 is ionic.
(4) NO has one unpaired electron in bonding molecular orbital.
Solution
MO configuration of NO:  1s 2  *1s 2  2s 2  *2s 2 2px2  2px2  2p y2  * 2p1y
NO has one unpaired electron in antibonding  molecular orbital.
Hence, the correct option is (4)
24. How many enantiomer pairs are obtained by monochlorination of 2,3-dimethylbutane?
(1) One
(2) Four
(3) Two
(4) Three
Solution
On monochlorination of 2,3-dimethylbutane, only one enantiomer pair is obtained.
*
Cl
1-chloro-2,3-dimethylbutane
monochlorination
2,3-dimethylbutane
Cl
2-chloro-2,3-dimethylbutane
Hence, the correct option is (1)
25. A 1.0 M solution of acetic acid has a pH of 2.37. What percentage of the acetic acid is ionized in the
solution?
(1) 0.51%
(2) 0.43%
(3) 0.59%
(4) 0.72%
Solution
pH   log[H  ]
[H  ]  antilog(2.37)
 4.27  10 3 M
Percentage ionization 
4.27  103
 100%  0.43%
1.0
Hence, the correct option is (2)
26. How many valence electrons are there in nitrogen atom in case of a nitrene?
(1) 5
(2) 6
(3) 7
(4) 8
Solution
The nitrogen atom in nitrenes has a sextet of electrons. Nitrenes are the nitrogen analogues of carbenes.
These are electron-deficient monovalent nitrogen species. The nitrogen atom in nitrenes has a sextet of
electrons. As in the case of carbenes, the nitrenes exist in singlet and triplet states.
..
R  N:
..
R  N:
Singlet nitrene
Triplet nitrene
Hence, the correct option is (2)
27. How many mL of 0.05 M KMnO4 is required to oxidize 2.0 g of FeSO4 in dilute solution?
(1) 52.63 mL
(2) 52.96 mL
(1) 67.63 mL
(2) 70.96 mL
Solution
The redox reactions are
MnO4  8H  5e  Mn 2  4H2O
Fe2  Fe3  e
Therefore, the overall reaction is
MnO4  8H   5Fe2  Mn 2  4H 2O  5Fe3
Now, by the relation M1V1 = M2V2, we have
2
5  0.05  V 
 1000
152
Solving, we get V = 52.63 mL.
Hence, the correct option is (1)
28. The colligative properties of a solution depend on
(1) nature of solute particles present in it.
(2) nature of solvent used.
(3) number of solute particles present in it
(4) number of moles of solvent only.
Solution
The colligative properties of a solution depend on the number of solute particles irrespective of nature.
Hence, the correct option is (3)
29. Which of the following patterns is correct for [Fe(H 2 O) 6 ]3+ and [Fe(CN) 6 ]3  respectively?
(1)
(2)
(3)
(4)
Solution
Since CNis a stronger ligand than H2Otherefore, CFT spilitting in [Fe(CN)6]3will be more than
[Fe(H2O)6]3+.
Hence, the correct option is (1)
30. Four particles have velocities 1, 0, 2, and 3 m s1. The root mean square velocity of the particles is (in
m s1)
(1) 3.5
(2) 3.5
(3) 1.5
(4) 14 / 3
Solution
urms 
Hence, the correct option is (2).
12  02  22  32
 3.5
4