Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule $B_{2}$ is:

1

and diamagnetic

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0

and diamagnetic

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1

and paramagnetic

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0

and paramagnetic

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Solution

The correct option is A $1$ and diamagnetic
In Hund's rule of maximum multiplicity, if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs.

$B_{2} (10) = σ_{1 s^{2}} σ_{1 s^{2}}^{} σ_{2 s^{2}} σ_{2 s^{2}}^{} π_{2 p_{x}^{2}}$

Number of bonding electrons $=$ 6

Number of Antibonding electrons $=$ 4

Bond order $=$ $\frac{(N u m b e r o f b o n d i n g e l e c t r o n s - N u m b e r o f a n t i b o n d i n g e l e c t r o n s)}{2}$

Bond order $= \frac{6 - 4}{2} = 1$

Since there are no unpaired electrons, it is diamagnetic in nature.

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