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Chemistry Question on Chemical bonding and molecular structure

When O2O_2 is converted into O2+O_2^+

A

both paramagnetic character and bond order increase

B

bond order decreases

C

paramagnetic character increases

D

paramagnetic character decreases and the bond order increases

Answer

paramagnetic character decreases and the bond order increases

Explanation

Solution

(i) O2−(σ1s)2(σ∗1s)2(σ2s)2(σ⋆2s)2O _{2}-(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{\star} 2 s\right)^{2}
(σ2pz)2(π2px)2(π2py)2(π∗2px)1\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi * 2 p_{x}\right)^{1}
(π⋆2py)1\left(\pi^{\star} 2 p_{y}\right)^{1}

Bond order =Nb−Na2=\frac{N_{b}-N_{a}}{2}
=8−42=\frac{8-4}{2}
=2=2
O2O _{2} molecule having 2 unpaired electron.

(ii) O2+−(σ1s)2(σ⋆1s)2(σ2s)2(σ⋆2s)2O _{2}^{+}-(\sigma 1 s)^{2}\left(\sigma^{\star} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{\star} 2 s\right)^{2}
(σ2pz)2(π2px)2(π2py)2(π⋆2px)1\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi^{\star} 2 p_{x}\right)^{1}
(π∗2py)0\left(\pi^{*} 2 p_{y}\right)^{0}

Bond order =8−32=\frac{8-3}{2}
=2.5=2.5
O2+O _{2}^{+} ion having only 1 unpaired electron.

Hence, when O2O _{2} is converted into O2+O _{2}^{+} paramagnetic character decrease and the bond order increases.