### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2011

The hybridization of orbitals of N atom in $NO_3^-$, $NO_2^+$ and $NH_4^+$ are respectively :
A
sp , sp2, sp3
B
sp2, sp , sp3
C
sp , sp3, sp2
D
sp2, sp3, sp

## Explanation

2

### AIEEE 2011

Among the following the maximum covalent character is shown by the compound :
A
SnCl2
B
AlCl3
C
MgCl2
D
FeCl2

## Explanation

Charge of cation/Size of cation is called polarising power.

$\therefore$ (i) Polarising power $\propto$ charge of cation

(ii) Polarising power $\propto$
1
size of cation

Here the AlCl3 is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size. So it has more covalent character.
3

### AIEEE 2009

Using MO theory, predict which of the following species has the shortest bond length?
A
$O_2^+$
B
$O_2^-$
C
$O_2^{2-}$
D
$O_2^{2+}$

## Explanation

Note :

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

(2) $\,$ Bond order $= {1 \over 2}$ [Nb $-$ Na]

Nb = No of electrons in bonding molecular orbital

Na $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is

Here Na = Anti bonding electron $=$ 4 and Nb = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -

Here Na = 10

and Nb = 10

In O atom 8 electrons present, so in O2, 8 $\times$ 2 = 16 electrons present.

Then in $O_2^ +$ no of electrons = 15

in $O_2^ -$ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O2 (16 electrons) is

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

$\therefore\,\,\,\,$Na = 6

Nb = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ +$ (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

Molecular orbital configuration of O$_2^ {2+ }$ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^o}^ *$

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

Na = 4

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

Molecular orbital configuration of $O_2^ -$ (17 electrons) is

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

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

Na = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

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

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

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

Na = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond length $\propto$ ${1 \over {Bond\,\,order}}$

So $O_2^{2+}$ has the shortest bond length.
4

### AIEEE 2008

The bond dissociation energy of B - F in BF3 is 646 jJ mol-1 whereas that of C - F in CF4 is 515 kj mol-1. The correct reason for higher B - F bond dissociation energy as compared to that of C - F is
A
stronger $\sigma$ bond between B and F in BF3 as compared to that between C and F in CF4
B
significant $p\pi - p\pi$ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4
C
lower degree of $p\pi - p\pi$ interaction between B and F in BF3 than that between C and F in CF4
D
smaller size of B - atom as compared to that of C- atom.

## Explanation

In BF3, B is sp2 hybridised and by Back Bonding method strong p$\pi$-p$\pi$ bond is created between filled p-orbital of F and vacant p-orbital of B which leads to shortening of B–F bond length which results in higher bond dissociation energy of the B–F bond. However in CF4, C does not have any vacant p-orbitals to undergo $\pi$-bonding.