Question

Bond order of $NO^+$ is

• A. 1
• B. 3
• C. 2
• D. 0

Hint:

ISOELECTRONIC species usually have the same bond order. For example, $N_2$, $CO$, $CN^-$, and $NO^+$ all have 14 electrons and a bond order of 3.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the bond order of the nitrosonium ion ($NO^+$). This can be determined using Molecular Orbital (MO) Theory.

Step 2: Key Formula or Approach:
Bond Order = $\frac{1}{2}$ (Bonding electrons - Antibonding electrons).
First, calculate the total number of valence electrons:
Electrons from $N$ = 7
Electrons from $O$ = 8
Subtract 1 electron due to the positive charge ($+$).
Total electrons = $7 + 8 - 1 = 14$ electrons.

Step 3: Detailed Explanation:
Since $NO^+$ has 14 electrons, it is isoelectronic with the Nitrogen molecule ($N_2$) and Carbon Monoxide ($CO$).
The Molecular Orbital (MO) electronic configuration for a 14-electron species is:
$[\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2]$
Counting the electrons:
Number of bonding electrons ($N_b$) = 2 (from $\sigma 1s$) + 2 (from $\sigma 2s$) + 4 (from $\pi 2p$) + 2 (from $\sigma 2p$) = 10
Number of antibonding electrons ($N_a$) = 2 (from $\sigma^* 1s$) + 2 (from $\sigma^* 2s$) = 4
Using the formula:
Bond Order = $\frac{10 - 4}{2} = \frac{6}{2} = 3$.
A bond order of 3 indicates a triple bond between Nitrogen and Oxygen.

Step 4: Final Answer:
The bond order is 3.