1

### JEE Main 2018 (Online) 16th April Morning Slot

Which of the following conversions involves change in both shape and hybridisation ?
A
NH3 $\to$ NH4+
B
CH4 $\to$ C2H6
C
H2O $\to$ H3O+
D
BF3 $\to$ BF4$-$

## Explanation

BF3  $\to$  BF4$-$

2

### JEE Main 2019 (Online) 9th January Morning Slot

According to molecular orbital theory, which of the following is true with respect to Li2+ and Li2$-$ ?
A
$Li_2^ +$ is unstable and $Li_2^ -$ is stable
B
$Li_2^ +$ is stable and $Li_2^ -$ unstable
C
Both are stable
D
Both are unstable

## Explanation

$Li_2^ +$  (5 electrons) = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^1}}}$

$Li_2^ -$  (7 electrons) = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}$ $\,\sigma _{2{s^1}}^ * \,$

$\therefore$   Bond order of $Li_2^ +$ = ${{3 - 2} \over 2}$ = 0.5

Bond order of $Li_2^ -$ = ${{4 - 3} \over 2}$ = 0.5

As both $Li_2^ +$ and $Li_2^ -$ has non-zero bond order, so both are stable.
3

### JEE Main 2019 (Online) 9th January Evening Slot

In which of the following process, the bond order has increased and paramagnetic character has charged to diamagnetic ?
A
NO $\to$ NO+
B
N2 $\to$ N2+
C
O2 $\to$ O2+
D
O2 $\to$ O22$-$

## 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

And in NO one unpaired electron is present , so it is paramagnetic.

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

And in NO+ no unpaired electron is present , so it is diamagnetic.
4

### JEE Main 2019 (Online) 10th January Morning Slot

Two pi and half sigma bonds are present in :
A
O2
B
N$_2^ +$
C
O$_2^ +$
D
N2

## Explanation

Two pi and half sigma bonds are presents in molecule with bond order 2.5.

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

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

Bond order = ${1 \over 2}\left( {9 - 4} \right)$ = 2.5