MHT CET · Chemistry · Chemical Bonding and Molecular Structure
What is bond order of \(\mathrm{F}_2\) molecule?
- A \(\frac{1}{2}\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
The electronic configuration of \(\mathrm{F}_2\)
\((\sigma 1 \mathrm{~s})^2\left(\sigma^* 1 \mathrm{~s}\right)^2(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}\right)^2 \)
\( \left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)^2\left(\pi^* 2 \mathrm{p}_{\mathrm{x}}\right)^2\left(\pi^* 2 \mathrm{p}_{\mathrm{y}}\right)^2 \)
\( \text {Bond order of } \mathrm{F}_2 \text { molecule }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)\(=\frac{10-8}{2}=1\)
\((\sigma 1 \mathrm{~s})^2\left(\sigma^* 1 \mathrm{~s}\right)^2(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}\right)^2 \)
\( \left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)^2\left(\pi^* 2 \mathrm{p}_{\mathrm{x}}\right)^2\left(\pi^* 2 \mathrm{p}_{\mathrm{y}}\right)^2 \)
\( \text {Bond order of } \mathrm{F}_2 \text { molecule }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)\(=\frac{10-8}{2}=1\)
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