IIT JEE Advanced 2009 Paper 1

(A) B2 → σ1s2σ*1s2σ2s2σ*2s22px1 = π2py1
It is paramagnetic due to two unpaired electrons. The bond order is

Nb−Na

2

=

6−4

2

= 1
The gain of electron increases bond order, so reduction is possible.
(B) N2 → σ1s2σ*1s2σ2s2σ*2s2π2px2 = π2py2σ2pz2
There are no unpaired electrons. Bond order is

b−Na

2

=

10−4

2

=

6

2

= 3
Mixing of 2s and 2 p orbitals is possible because of similar energies.
(C) O2− → σ1s2σ1s2σ2s2σ*2s2σ2pz2π2px2
= π2py2π*2px2π*2py1
The molecule is paramagnetic due to presence of unpaired electron.
Bond order is less than 2.

Nb−Na

2

=

10−7

2

=

3

2

Loss of electron increases the bond order so oxidation is possible.
(D) O2 → σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2
The molecule is paramagnetic due to presence of unpaired electrons. Bond order is

Nb−Na

2

=

10−6

2

=

4

2

= 2
Loss of electron causes increase in bond order, so it undergoes oxidation.