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Question

Match the species im Column I with the bond order in Column II.

Column I Column II
A \(NO\) \((i)\) 1.5
B \(CO\) \((ii)\) 2.0
C \(O_2^-\) \((iii)\) 2.5
D \(O_2\) \((iv)\) 3.0

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Solution

\(NO~\text{total electrons}=15\)

\(CO~\text{total electrons}=14\)

\(O_2^-~\text{total electrons}=17\)

\(O_2~\text{total electrons}=16\)


Calculation of bond order

Electroni configuration of \(NO\) molecule is:

\(\sigma 1s^2\sigma^\ast 1s^2\sigma 2s^2\sigma^{\ast}2s^2\pi2p_x^2=\pi 2p_y^2\sigma 2p_z^2\)
\(\pi^\ast2p_x^1=\pi^\ast 2p_y^0\)
Bond order of \(NO=\left [ \dfrac{10-5}{2} \right ]=2.5\)

Electronic configuration of \(CO\) molecule is:

\(\sigma 1s^2\sigma^\ast 1s^2\sigma 2s^2\sigma^{\ast}2s^2\pi2p_x^2=\pi 2p_y^2\sigma 2p_z^2\)

Bond order \(CO=\left [ \dfrac{10-4}{2} \right ]=3\)

Electronic configuration of \(O_2\) molecule is:
\(\sigma 1s^2\sigma^\ast 1s^2\sigma 2s^2\sigma^{\ast}2s^2\sigma 2p_z^2\pi2p_x^2=\pi 2p_y^2\)
\(\pi^\ast 2p_x^1=\pi^{\ast}P_y^1\)

Bond order of \(O_2=\left [ \dfrac{10-6}{2} \right ]=2\)

Electronic configuration of \(O_2^-\) is:
\(\sigma 1s^2\sigma^\ast 1s^2\sigma 2s^2\sigma^{\ast}2s^2\sigma 2p_z^2\pi2p_x^2=\pi 2p_y^2\)
\(\pi^\ast 2p_x^2=\pi^{\ast}P_y^1\)

Bond Order of \(O_2^-=\left [ \dfrac{10-7}{2} \right ]=1.5\)

Column I Column IUI
A \(NO\) \((iii)\) 2.5
B \(CO\) \((iv)\) 3
C \(O_2^-\) \((i)\) 1.5
D \(O_2\) \((ii)\) 2

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