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⁺