BITSAT 2011 Paper

O2+ ion - Total number of electrons
(16 – 1) = 15.
E.C. :
σ1s2 σ* 1s2 α 2s2 α* 2s2 α 2pz2
π2px2 = π2py2π*2p)‌x1 = π*2py
Bond order =

Nb−Na

2

=

10−5

2

=

5

2

= 2

1

2

O2− (Super oxide ion): Total number of electrons (16 +1) =17 .
E.C. :
σ1s2 σ* 1s2σ2s2 σ* 2s2α2pz2
π2px2 = π2py2 π* 2px2 = π*2py1
Bond order =

(Nb−Na)

2

= 10−7 =

3

2

= 1

1

2

O22+ ion: Total number of electrons
= (16 – 2) = 14
E.C. :
σ 1s2 σ* 1s2σ2s2 σ* 2s2 π 2px2 = π2py2σ2pz2
Bond order =

(Nb−Na)

2

=

10−4

2

=

6

2

= 3
So bond order: O2− < O2+ < O22+