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Question

Chemistry Question on Bond Parameters

The correct increasing order of bond order values for the concerned species is

A

(O2)− \left(O_{2}\right)^{-}

B

(O2)−\left(O_{2}\right)^{-}

C

(O2) \left(O_{2}\right)

D

(O2)\left(O_{2}\right)

Answer

(O2)− \left(O_{2}\right)^{-}

Explanation

Solution

O2−:KKσ2s2σ∗2s2σ2pzπ22py2O_2^-: KKσ2s^2σ^*2s^2σ2p_z\pi^22p_y^2
π∗2Px2=π∗2py1\pi^{*}2P^{2}_{x}=\pi^{*}2p^{1}_{y}
Bond order =12(8−5)=32=1.5=\frac{1}{2}(8-5)=\frac{3}{2}=1.5
O2:KKσ2S2σ∗2s2σ2pz2π2px2O_{2}: KK {\sigma}2S^{2}{\sigma}^{*}2s^{2}{\sigma}2p_{z} ^{2}\pi2p^{2}_{x}
=π2py2π∗2Px1=π∗2py1=\pi2p_{y} ^{2}\pi ^{*}2P^{1}_{x}=\pi^{*}2p^{1}_{y}
Bond order =12(8−4)=2=\frac{1}{2}(8-4)=2
N2:KKσ2S2σ∗2s2π2px2N_{2}: KK{\sigma}2S^{2}{\sigma} ^{*}2s^{2}\pi 2p^{2}_{x}
=π2py2σ2pz2=\pi 2p^{2}_{y} {\sigma}2p^{2}_{z}
Bond order =12(8−2)=3=\frac{1}{2}(8-2)=3
N2+:KKσ2s2σ∗2s2π2px2N^{+}_{2} : KK{\sigma}2s^{2}{\sigma} ^{*}2 s^{2}\pi2p^{2}_{x}
=π2py2σ2pz1=\pi 2p^{2}_{y} {\sigma}2p^{1}_{z}
Bond order =12(7−2)=\frac{1}{2}(7-2)
=52=2.5=\frac{5}{2}=2.5
Hence, the increasing order of bond order values is :
O2−O^{-}_{2}