Question

The correct order of bond orders of \(C_2^{2-}, N_2^{2-}, O_2^{2-}\) is, respectively

• A. \(C_2^{2-}< N_2^{2-}< O_2^{2-}\)
• B. \(O_2^{2-}< N_2^{2-}< C_2^{2-}\)
• C. \(C_2^{2-}< O_2^{2-}< N_2^{2-}\)
• D. \(N_2^{2-}< C_2^{2-}< O_2^{2-}\)

Answer 1

The Correct Option is B

 To determine the bond order of the given diatomic molecular ions \(C_2^{2-}\), \(N_2^{2-}\), and \(O_2^{2-}\), we need to use Molecular Orbital Theory (MOT). The bond order is given by the formula:

\(\text{Bond Order} = \frac{\text{Number of electrons in bonding molecular orbitals} - \text{Number of electrons in antibonding molecular orbitals}}{2}\)

Electron Configuration and Bond Order Calculation:

1. For \(C_2^{2-}\): \(C_2^{2-}\)\)has 8 valence electrons since each carbon atom contributes 4 electrons and the \(2-\) charge adds 2 more electrons.

\(2^{2-}\) electron configuration in molecular orbitals is \(\sigma_{1s}^2\sigma_{1s}^2\sigma_{2s}^2\sigma_{2s}^2\pi_{2p_x}^2\pi_{2p_y}\)

Bond order = \(\(\frac{8-2}{2} = 3 \)\)

2. For \(N_2^{2-}\): \(N_2^{2-}\)\)has 16 valence electrons (each nitrogen has 5 electrons and the \(2-\) charge gives extra 2 electrons).

\(2^{2-}\) electron configuration in molecular orbitals is \(\sigma_{1s}^2\sigma_{1s}^2\sigma_{2s}^2\sigma_{2s}^2\pi_{2p_x}^2\pi_{2p_y}^2\sigma_{2p_z}^2\right)\)

Bond order = \(\frac{10-4}{2} = 3 \)

3. For \(O_2^{2-}\): \(O_2^{2-}\)\)has 18 valence electrons (each oxygen has 6 electrons and the \(2-\) charge gives an extra 2 electrons).

\(2^{2-}\) electron configuration in molecular orbitals is \(\sigma_{1s}^2\sigma_{1s}^2\sigma_{2s}^2\sigma_{2s}^2\pi_{2p_x}^2\pi_{2p_y}^2\pi_{2p_z}^4\sigma_{2p_z}^2\right)\)

Bond order = \(\frac{10-8}{2} = 1\)

Hence, the bond orders are \(O_2^{2-} = 1\), \(N_2^{2-} = 2\), \(C_2^{2-} = 3\).

The correct order of bond orders is \(O_2^{2-} < N_2^{2-} < C_2^{2-}\).