 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

Which of the following molecules/ions does not contain unpaired electrons?
A
$O_2^{2−}$
B
B2
C
$N_2^+$
D
O2

Explanation

(A) Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ *$

So O $_2^{2 - }$ has no unpaired electrons.

(B) Molecular orbital configuration of B2 (10 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$

Here in B2, 2 unpaired electrons present.

(C) Moleculer orbital configuration of $N_2^{ + }$ (13 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Here in $N_2^{ + }$, 1 unpaired electron present,

(D) Molecular orbital configuration of O2 (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ *$

So O$_2$ has 2 unpaired electrons.
2

AIEEE 2005

Lattice energy of an ionic compounds depends upon
A
Charge on the ion only
B
Size of the ion only
C
Packing of ions only
D
Charge on the ion and size of the ion

Explanation

Electrostatic force of attraction between cation and anion is called Lattice energy.

Lattice energy (F) = $K{{{q_1}{q_2}} \over {{r^2}}}$

q1 and q2 are the charges of cations and anion. So Lattice energy depens on the charges of ions.

r = distance between center of ions

And, r = r+ + r- where r+ = radius of cation and r- = radius of anion

So, more the size of ions less the value of Lattice energy.
3

AIEEE 2004

The maximum number of 90° angles between bond pair of electrons is observed in
A
dsp3 hybridization
B
sp3d2 hybridization
C
dsp2 hybridization
D
sp3d hybridization

Explanation Here eight 90o angles between bond pair and bond pair. Those angles are $\angle$1M2, $\angle$2M3, $\angle$3M4, $\angle$4M1, $\angle$5M1, $\angle$5M2, $\angle$5M3, $\angle$5M4. Here twelve 90o angles between bond pair and bond pair. Those angles are $\angle$1M2, $\angle$2M3, $\angle$3M4, $\angle$4M1, $\angle$5M1, $\angle$5M2, $\angle$5M3, $\angle$5M4, $\angle$6M1, $\angle$6M2, $\angle$6M3, $\angle$6M4. Here four 90o angles between bond pair and bond pair. Those angles are $\angle$1M2, $\angle$2M3, $\angle$3M4, $\angle$4M1. Here six 90o angles between bond pair and bond pair. Those angles are $\angle$1M3, $\angle$1M4, $\angle$1M5, $\angle$2M3, $\angle$2M4, $\angle$2M5.
4

AIEEE 2004

Which one of the following has the regular tetrahedral structure?
(Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)
A
XeF4
B
[Ni(CN)4]2-
C
$BF_4^-$
D
SF4

Explanation

Regular Tetrahedral structure is possible in sp3 hybridization where central atom has 4 bond pair and no lone pair.

(a)   XeF4  is sp3d2 hybridised and structure is square planar. (b)   [Ni(CN)4]$-$2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni+2 is $=$ [Ar]3d8 But because of CN$-$ ion which is a strong field ligand , it can perform pairing of electron. And the structure of dsp2 hybridization is square planar. (C) $\,\,\,\,$ BF$_4^ -$, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral. (d) SF4 is sp3d hybridised and structure is see-saw.