 JEE Mains Previous Years Questions with Solutions

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1

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.
2

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.
3

AIEEE 2004

The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively
A
sp2 and sp2
B
sp3 and sp3
C
sp3 and sp2
D
sp2 and sp3

Explanation

Structure of  H3BO3  is Here Boron has 3 sigma bond. So it is sp2 hybridised.

And oxygen has two sigma bond and two lone pair. So it sp3 hybridised.
4

AIEEE 2004

The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species?
A
Bond length in NO+ is greater than in NO
B
Bond length is unpredictable
C
Bond length in NO+ in equal to that in NO
D
Bond length in NO is greater than in NO+

Explanation

Molecular orbital configuration of NO (15 electrons) is

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

$\therefore\,\,\,\,$ Nb = 10

Na = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Similarly Molecular orbital configuration of NO+ (14 electrons) is

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

$\therefore\,\,\,\,$ Nb = 10

Na = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Note :
(2) $\,\,\,\,$ Bond length $\propto$ ${1 \over {Bond\,\,order}}$

So, bond length in NO > NO+