MCQ
Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule $B_2$ is
- ✓$1$ and diamagnetic
- B$0$ and diamagnetic
- C$1$ and paramagnetic
- D$0$ and paramagnetic
✓
Answer
Correct option: A.
$1$ and diamagnetic
a
$\mathrm{B}_2(10)=\sigma_{1 \mathrm{~s}^2} \sigma_{1 \mathrm{~s}^2}^* \sigma_{2 \mathrm{~s}^2} \sigma_{2 \mathrm{~s}^2}^* \pi_{2 \mathrm{p}_2^2}$
$\mathrm{B}_2(10)=\sigma_{1 \mathrm{~s}^2} \sigma_{1 \mathrm{~s}^2}^* \sigma_{2 \mathrm{~s}^2} \sigma_{2 \mathrm{~s}^2}^* \pi_{2 \mathrm{p}_2^2}$
Bond order $=\frac{6-4}{2}=1$
(nature diamagnetic as no unpaired electron)
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