Question

According to MO theory the bond orders for \(O_2^{2−},CO\,\, \text {and} \,\,NO^+\) respectively, are

• A. 1, 2 and 3
• B. 1,3 and 2
• C. 2,3 and 3
• D. 1, 3 and 3

Answer 1

The Correct Option is D

To determine the bond order for each of the given molecules \(O_2^{2-}, CO, \text{and} \, NO^+\), we will apply Molecular Orbital (MO) Theory. Bond order is calculated as follows:

\text{Bond Order} = \frac{1}{2} (\text{Number of electrons in bonding orbitals} - \text{Number of electrons in anti-bonding orbitals})

1. For \((O_2^{2-})\):
   • The electron configuration for \(O_2\) is \( (1\sigma_g)^2 (1\sigma_u^*)^2 (2\sigma_g)^2 (2\sigma_u^*)^2 (3\sigma_g)^2 (1\pi_u)^4 (1\pi_g^*)^2 \).
   • \(O_2^{2-}\) implies adding 2 more electrons: added to \(1\pi_g^*\) making it \((1\pi_g^*)^4\).
   • Calculate the bond order: \text{Bond Order} = \frac{1}{2} (10 - 8) = 1 \,.
2. For \(CO\):
   • Total electrons = 14. Configuration resembles \(N_2\): \( (1\sigma)^2 (2\sigma^*)^2 (2\sigma)^2 (\pi)^4 (3\sigma)^2 \).
   • Calculate the bond order: \text{Bond Order} = \frac{1}{2} (10 - 4) = 3 \,.
3. For \(NO^+\):
   • The electron configuration for \(NO\) is \( (1\sigma)^2 (2\sigma)^2 (2\pi)^4 (3\sigma)^2 \).
   • \(NO^+\) implies removing 1 electron from the last occupied \(3\sigma\) orbital making it \( (3\sigma)^1 \). Total = 14 electrons similar to \(N_2\).
   • Calculate the bond order: \text{Bond Order} = \frac{1}{2} (10 - 4) = 3 \,.

Therefore, the bond orders for \(O_2^{2-}, CO\), and \(NO^+\) are 1, 3, and 3, respectively. The correct answer is 1, 3 and 3.