Question

Number of unpaired electron in highest occupied molecular orbital of following species is :\(N_2, N ^+_ 2 , O_2, O^+_ 2 \) ? 

• A. 0, 1, 2, 1
• B. 2, 1, 2, 1
• C. 0, 1, 0, 1
• D. 2, 1, 0, 1

Hint:

For molecular species, determine the number of unpaired electrons using the molecular orbital theory. Focus on the HOMO.

Answer 1

The Correct Option is A

Molecular Orbital Configurations:
    \(N_2\): The molecular orbital configuration is:
   \[\sigma(1s)^2 \, \sigma^*(1s)^2 \, \sigma(2s)^2 \, \sigma^*(2s)^2 \, \pi(2p_x)^2 \, \pi(2p_y)^2\]
   Here, the highest occupied molecular orbital (HOMO) is \(\pi(2p_x)^2 \, \pi(2p_y)^2\) with no unpaired electrons.
   \[\text{Unpaired electrons in } \text{N}_2 = 0.\]
\(N_2^+\): The configuration is:
   \[\sigma(1s)^2 \, \sigma^*(1s)^2 \, \sigma(2s)^2 \, \sigma^*(2s)^2 \, \pi(2p_x)^2 \, \pi(2p_y)^1\]
   The HOMO is \(\pi(2p_y)^1\) with 1 unpaired electron.
   \[\text{Unpaired electrons in } \text{N}_2^+ = 1.\]
\(O_2\): The configuration is:
   \[\sigma(1s)^2 \, \sigma^*(1s)^2 \, \sigma(2s)^2 \, \sigma^*(2s)^2 \, \pi(2p_x)^2 \, \pi(2p_y)^2 \, \pi^*(2p_x)^1 \, \pi^*(2p_y)^1\]
   The HOMO is \(\pi^*(2p_x)^1 \, \pi^*(2p_y)^1\) with 2 unpaired electrons.
   \[\text{Unpaired electrons in } \text{O}_2 = 2.\]
\(O_2^-\): The configuration is:
   \[\sigma(1s)^2 \, \sigma^*(1s)^2 \, \sigma(2s)^2 \, \sigma^*(2s)^2 \, \pi(2p_x)^2 \, \pi(2p_y)^2 \, \pi^*(2p_x)^2 \, \pi^*(2p_y)^1\]
   The HOMO is \(\pi^*(2p_y)^1\) with 1 unpaired electron.
   \[\text{Unpaired electrons in } \text{O}_2^- = 1.\]

Thus, the number of unpaired electrons for \(\text{N}_2\), \(\text{N}_2^+\), \(\text{O}_2\), and \(\text{O}_2^-\) are {0, 1, 2, 1} respectively.