1
MCQ (Single Correct Answer)

AIPMT 2001

In which of the following bond angle is maximum ?
A
NH3
B
NH4+
C
PCl3
D
SCl2

Explanation

Bond angle is maximum in NH4+ tetrahedral molecule with bond angle 109°.
2
MCQ (Single Correct Answer)

AIPMT 2001

In X - H --- Y, X and Y both are electronegative elements. Then
A
electron density on X will increase and on H will decrease
B
in both electron density will increase
C
in both electron density will decrease
D
on X electron density will decrease and on H increases.

Explanation

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.
3
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.
4
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+.

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