1
MCQ (Single Correct Answer)

### AIPMT 2000

Which species does not exhibit paramagnetism?
A
N2+
B
O2$-$
C
CO
D
NO

## Explanation

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A) N2+ has 13 electrons.

Moleculer orbital configuration of N2+

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

Here 1 unpaired electrons present, so it is paramagnetic.

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

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

Here 1 unpaired electrons present, so it is paramagnetic.

(A) CO has 14 electrons.

Moleculer orbital configuration of CO

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

Here 0 unpaired electrons present, so it is diamagnetic.

(D) Molecular orbital configuration of NO (15 electrons) is

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

Here 1 unpaired electrons present, so it is paramagnetic.
2
MCQ (Single Correct Answer)

### AIPMT 2000

d$\pi$ $-$ p$\pi$ bond present in
A
CO32$-$
B
PO43-
C
NO3$-$
D
NO2$-$

## Explanation

In PO43–, P atom has vacant d-orbitals, thus it can form p$\pi$ - d$\pi$ bond. ‘N’ and ‘C’ have no vacant ‘d’ orbital in their valence shell, so they cannot form such bond.

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