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Chemistry Question on Chemical bonding and molecular structure

Which of the following represents the arrangement in increasing order of bond order and bond dissociation energy?

A

O2+<O22−<O2−<O2O _{2}^{+}< O _{2}^{2-}< O _{2}^{-}< O _{2}

B

O22−<O2−<O2<O2+O _{2}^{2-}< O _{2}^{-}< O _{2}< O _{2}^{+}

C

O2<O2+<O22−<O2−O _{2}< O _{2}^{+}< O _{2}^{2-}< O _{2}^{-}

D

O22−<O2−<O2+<O2O _{2}^{2-}< O _{2}^{-}< O _{2}^{+}< O _{2}

Answer

O22−<O2−<O2<O2+O _{2}^{2-}< O _{2}^{-}< O _{2}< O _{2}^{+}

Explanation

Solution

The MOMO configuration of O2O _{2} can be written as
O2(8+8=16)=σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2pz2,π2px2≈π∗2py2,π∗2px1≈π2py1O _{2}(8+8=16)=\sigma 1 s^{2}, \overset{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \overset{*}{\pi} 2 p_{y}^{2}, \overset{*}{\pi} 2 p_{x}^{1} \approx \pi 2 p_{y}^{1}
∴\therefore Bond order
=Nb−Na2=10−62=2=\frac{N_{b}-N_{a}}{2}=\frac{10-6}{2}=2
Similarly,
O2+(8+8−1=15),O _{2}^{+}(8+8-1=15),
BO=10−52=2.5BO =\frac{10-5}{2}=2.5
O2−(8+8+1=17)O _{2}^{-}(8+8+1=17),
BO=10−72=1.5BO =\frac{10-7}{2}=1.5
O22−(8+8+2=18)O _{2}^{2-}(8+8+2=18)
BO=10−82=1BO=\frac{10-8}{2}=1
Thus, the order of bond order (BO)(BO) and bond dissociation energy is
O22−<O2−<O2<O2+O _{2}^{2-}< O _{2}^{-}< O _{2}< O _{2}^{+}