1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
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.
4
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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