Select the compound having all equivalent bond lengths:

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${IF}_{7}$

$\begin{aligned} {PCl}_{5} \end{aligned}$

${BF}_{3}$

${PF}_{5}$

answer is B.

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Detailed Solution

In ${BF}_{3}$ all bond lengths are equal.

The total of the covalent radii of boron and fluorine is shorter than each of the three B-F bond lengths, which are all equal.

The B-F bond has an extremely high bond energy compared to all other single bonds. 'p pi- p pi' might provide an explanation for the B-F bond's unique length and strength.

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