Question

The bond order of \(\text{N}_{2}^{+}\) ion is:

• A. \(2.0\)
• B. \(2.5\)
• C. \(3.0\)
• D. \(1.5\)

Hint:

Count the electrons in bonding and antibonding molecular orbitals and use bond order = (Nb - Na) / 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Bond order \(=\dfrac{N_{b}-N_{a}}{2}\), where \(N_{b}\) and \(N_{a}\) are electrons in bonding and antibonding MOs.

Step 2: Electrons in \(\text{N}_{2}^{+}\)
\(\text{N}_{2}\) has 14 electrons, so \(\text{N}_{2}^{+}\) has 13.
Configuration: \(\sigma 1s^{2}\,\sigma^{*}1s^{2}\,\sigma 2s^{2}\,\sigma^{*}2s^{2}\,\pi 2p_{x}^{2}\,\pi 2p_{y}^{2}\,\sigma 2p_{z}^{1}\)

Step 3: Count
\(N_{b}=2+2+2+2+1=9\)
\(N_{a}=2+2=4\)

Step 4: Bond order
\(\dfrac{9-4}{2}=2.5\)

Final Answer:
The bond order of N2 plus is 2.5. \[ \boxed{\text{(B) }2.5} \]