1

AIPMT 2001

Which of the following two are isostructural?
A
XeF2, IF2$-$
B
NH3, BF3
C
CO32$-$, SO32$-$
D
PCl5, ICl5

Explanation

Compounds with same shape and same hybridisation are known as isostructural.

XeF2, IF2- $\to$ both are sp3 hybridised linear molecules.
2

AIPMT 2000

Right order of dissociation energy N2 and N2+ is
A
N2 > N2+
B
N2 = N2+
C
N2+ > N2
D
none

Explanation

$N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

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

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

Na = 4

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

N2+ has 13 electrons.

Moleculer orbital configuration of N2+

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

$\therefore\,\,\,\,$ Nb = 9

Na = 4

$\therefore$ BO = ${1 \over 2}\left[ {9 - 4} \right]$ = 2.5

As the bond order in N2 is more than N2+ so the dissociation energy of N2 is higher than N2+.
3

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.
4

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.