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Chemistry Question on Classical Idea Of Redox Reactions – Oxidation And Reduction Reactions

The correct order of bond orders of C22−,N22−,O22−C_2^{2-}, N_2^{2-}, O_2^{2-} is, respectively

A

C22−<N22−<O22−C_2^{2-}< N_2^{2-}< O_2^{2-}

B

O22−<N22−<C22−O_2^{2-}< N_2^{2-}< C_2^{2-}

C

C22−<O22−<N22−C_2^{2-}< O_2^{2-}< N_2^{2-}

D

N22−<C22−<O22−N_2^{2-}< C_2^{2-}< O_2^{2-}

Answer

O22−<N22−<C22−O_2^{2-}< N_2^{2-}< C_2^{2-}

Explanation

Solution

C22−C_2^{2-} : σ1s2σ∗1s2σ2s2σ∗2s2π2px2=π2py2σ2pz2σ^2_{1s} σ*^2_{1s} σ^2_{2s} σ*^2_{2s} \pi^2_{2px} = \pi^2_{2p_y} σ^2_{2p_z}

N22−N^{2-}_2 : σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2px2=π2py2π∗2pz2=π∗2py1σ^2_{1s} σ*^2_{1s} σ^2_{2s} σ*^2_{2s}\sigma^2_{2p_z} \pi^2_{2px} = \pi^2_{2p_y} \pi*^2_{2p_z}=\pi*^1_{2p_y}

O22−:σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2px2=π2py2π∗2px2=π∗2py2O^{2-}_2 : σ^2_{1s} σ*^2_{1s} σ^2_{2s} σ*^2_{2s} σ^2_{2p_z} \pi^2_{2p_x} = \pi^2_{2p_y} \pi*^2_{2p_x} = \pi*^2_{2p_y}

B.O.(C22−)=3;B.O.(N22−)=2;B.O.(O22−)=1B.O.(C^{2-}_2) = 3; B.O. (N^{2-}_2) = 2; B.O.(O^{2-}_2) = 1