MHT CET · Chemistry · Chemical Bonding and Molecular Structure
What is the bond order in \(\mathrm{N}_2^*\) ?
- A 0
- B 1
- C 2
- D 2.5
Answer & Solution
Correct Answer
(D) 2.5
Step-by-step Solution
Detailed explanation
Electronic configuration of \(\mathrm{N}_2^{+}:(\sigma 1 \mathrm{~s})^2\left(\sigma^* 1 \mathrm{~s}\right)^2\)
\((\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)^1\)
Bond order \(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{1}{2}(9-4)=2.5\)
\((\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)^1\)
Bond order \(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{1}{2}(9-4)=2.5\)
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