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Question: Which of the following pairs will have same bond order?...

Which of the following pairs will have same bond order?

A

F2F_{2}and O22−O_{2}^{2 -}

B

N2andCO2N_{2}andCO_{2}

C

O2andO2−O_{2}andO_{2}^{-}

D

N2andN2+N_{2}andN_{2}^{+}

Answer

F2F_{2}and O22−O_{2}^{2 -}

Explanation

Solution

: F2andO22−F_{2}andO_{2}^{2 -}are isoelectronic.

F2:(σ1s2)(σ∗1s2)(σ2s2)(σ∗2s2)(σ2pz2)(π2pz2=π2py2)F_{2}:(\sigma 1s^{2})(\sigma*1s^{2})(\sigma 2s^{2})(\sigma*2s^{2})(\sigma 2p_{z}^{2})(\pi 2p_{z}^{2} = \pi 2p_{y}^{2}) π∗2px2=π∗2py2\pi*2p_{x}^{2} = \pi*2p_{y}^{2}

B.O. =12×(10−8)=1= \frac{1}{2} \times (10 - 8) = 1

O22−:(σ1s2)(σ∗1s2)(σ2s2)(σ∗2s2)(σ2p2z)O_{2}^{2 -}:(\sigma 1s^{2})(\sigma*1s^{2})(\sigma 2s^{2})(\sigma*2s^{2})(\sigma 2{p^{2}}_{z})

(π2px2=π2py2)(\pi 2p_{x}^{2} = \pi 2p_{y}^{2}) (π∗2px2=π∗2py2\pi*2p_{x}^{2} = \pi*2p_{y}^{2})

B.O. =10−82=1= \frac{10 - 8}{2} = 1