1

### AIPMT 2015 Cancelled Paper

Which of the following options represents the correct bond order ?
A
O2$-$  > O2 < O2+
B
O2$-$  <  O2  >  O2+
C
O2$-$  >  O2  >  O2+
D
O2$-$  <  O2  <  O2+

## Explanation

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

Nb = No of electrons in bonding molecular orbital

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

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

$\therefore\,\,\,\,$ 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 $O_2^ -$ (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

$O_2^-$ $O_2$ $O_2^+$
B.O. : 1.5 2.0 2.5
2

### AIPMT 2015

In which of the following pairs, both the species are not isostructural ?
A
Diamond, Silicon carbide
B
NH3, PH3
C
XeF4, XeO4
D
SiCl4, PCl4+

## Explanation

3

### AIPMT 2014

Which one of the following species has plane triangular shape ?
A
N3
B
NO3$-$
C
NO2$-$
D
CO2

## Explanation

Hybridisation of N3 is sp, so it is linear in shape.

Hybridisation of NO3 is sp2, so it is plane triangular in shape.

Hybridisation of NO2 is sp, so it is linear in shape.

Hybridisation of CO2 is sp, so it is linear in shape.
4

### AIPMT 2014

Which of the following molecules has the maximum dipole moment ?
A
CO2
B
CH4
C
NH3
D
NF3

## Explanation

Even though for CO2 and CH4 the C - O and C - H bonds are polar but due to their symmetrical structure they have zero dipole moment.
In NH3, H is less electronegative than N and hence the dipole moment of each N-H bond is towards N which is in the similar direction of lone pair but for NF3 the dipole moment of N-F bond is in opposite direction of lone pair which reduce the value of net dipole moment.