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Chemistry Question on Chemical bonding and molecular structure

What is the correct order of bond strength for NO,NO+NO, NO^+ and NO−NO^- ?

A

NO>NO+>NO−NO > NO^+ > NO^-

B

NO+>NO>NO−NO^+ > NO > NO^-

C

NO−>NO>NO+NO^- > NO > NO^+

D

NO+>NO−>NONO^+ > NO^- >NO

Answer

NO+>NO>NO−NO^+ > NO > NO^-

Explanation

Solution

Electronic configuration of NO,NO+NO, NO^{+} and NO−:NO^{-}:
NO : KKσ2s2σ∗2s2σ2pz2π2px2 π2py2π∗2px1KK{\sigma}2s^{2}{\sigma}^{*}2s^{2}{\sigma}2p^{2}_{z} \pi 2p^{2}_{x}\, \pi 2p^{2}_{y}\pi^{*}2p^{1}_{x}
NO+:KKσ2s2σ∗2s2σ2pz2π2px2π2py2NO^{+}: KK{\sigma}2s^{2}{\sigma}^{*} 2s^{2}{\sigma} 2p^{2}_{z} \pi 2p^{2}_{x} \pi 2 p^{2}_{y}
NO−:KKσ 2s2σ∗2s2σ2pz2π2px2π2py2π∗2px1NO^{-}: KK{\sigma}\, 2s^{2} {\sigma}^{*} 2s^{2}{\sigma}2p^{2}_{z}\pi 2p^{2}_{x} \pi 2p^{2}_{y} \pi^{*} 2 p^{1}_{x}
=π∗2py1= \pi^{*} 2p^{1}_{y}
Bond order =12(Nb−Na)=\frac{1}{2} (N_{b}-N_{a})
B.O. of NO =12(8−3)=2.5=\frac{1}{2} (8-3)=2.5
B.O. of NO+=12(8−2)=3NO^{+}=\frac{1}{2}(8-2)=3
B.O. of NO−=12(8−4)=2NO^{-}=\frac{1}{2}(8-4)=2
Bond strength ∝\propto Bond order Thus order of bond strength is :
NO+> NO> NO−NO^{+} > \, NO >\, NO^{-}