Two pi and half sigma bonds are present in:

Two pi and half sigma bonds are present in:
  1.  O2-
  2. N2+
  3. N2
  4.  O2

Solution

O 2 − =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p x 2 =π2 p y 2 =π2 p z 2 < π * 2 p y 2 = π * 2 p z 1

Bond Order = 6−3 2 =1.5

O 2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p x 2 <π2 p y 2 =π2 p z 2 < π * 2 p y 1 = π * 2 p z 1

Bond order = 6−2 2 =2

N 2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <π2 p z 2 =π2 p z 2 <σ2 p x 2

Bond order = 6−0 2 =3

According to MOT ; electronic configuration of N2+ is:

N 2 + =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <π2 p x 2 =π2 p y 2 <σ2 p z 1

Bond order =129-4=2.5

i.e., 2π bonds and 0.5 σ bonds.

Asked in: JEE Main 2019 (10 Jan Shift 1)

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