The correct bond order in the following species is

Options

(a) O₂²⁺ < O₂⁻ < O₂⁺
(b) O₂⁺ < O₂⁻ < O₂²⁺
(c) O₂⁻ < O₂⁺ < O₂²⁺
(d) O₂²⁺ < O₂⁺ < O₂⁻

Correct Answer:

O₂⁻ < O₂⁺ < O₂²⁺

Explanation:

O₂⁺ ion -Total number of electrons (16-1)=15.

Electronic configuration σ1s² < σ*1s² < σ2s² < σ*2s² < σ2p²(x) < π2p²(y)=π2p²(z)< π*2p¹(y)

Bond order =N(b)-N(a)/2 = 10-5/2 = 5/2 =2 1/2 O⁻₂ (super oxide ion): Total number of electrons (16+1)=17

Electronic configuration σ1s² < σ*1s² < σ2s² < σ*2s² < σ2p²(x) < π2p²(y)=π2p²(z)< π*2p² (y)=π*2p¹(z)

Bond order =N(b)-N(a)/2 = 10-7/2 = 3/2 =1 1/2

O₂⁺² ion :Total number of electrons (16-2)=14.

Electronic configuration σ1s² < σ*1s² < σ2s² < σ*2s² < σ2p²(x) < π2p²(y)=π2p²(z)

Bond order =N(b)-N(a)/2 = 10-4/2 = 6/2 =3

So bond order :O⁻₂ <O₂⁺<O₂²⁺