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Assertion : $\mathrm{B}_2$ molecule is diamagnetic.
Reason : The highest occupied molecular orbital is of $\sigma$ type.
Options:
Reason : The highest occupied molecular orbital is of $\sigma$ type.
Solution:
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Verified Answer
The correct answer is:
If both assertion and reason are false.
In $\mathrm{B}_2$, total number of electrons $=10$
$$
\mathrm{B}_2 \rightarrow \sigma(1 s)^2 \sigma^*\left(1 s^2\right) \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_x\right)^1 \pi\left(2 p_y\right)^1
$$
Presence of unpaired electron shows the paramagnetic nature.
The highest occupied molecular orbital is of $\pi$-type.
$$
\mathrm{B}_2 \rightarrow \sigma(1 s)^2 \sigma^*\left(1 s^2\right) \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_x\right)^1 \pi\left(2 p_y\right)^1
$$
Presence of unpaired electron shows the paramagnetic nature.
The highest occupied molecular orbital is of $\pi$-type.
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