1

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

The incorrect geometry is represented by :
A
BF3 - trigonal planar
B
H2O - bent
C
NF3 - trigonal planar
D
AsF5 - trigonal bipyramidal

## Explanation

NF3 has trigonal pyramidal geometry. Here central atom nitrogen has sp3 hybridization and one lone pair of electron. Because of lone pair there exist repulsion force between lone pair and bond pair of electrons, that is why bond angles are lower than tetrahedral bond angle.
2

### 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$-$

3

### 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.
4

### 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.