Which of the following is a false statement?

B H_{3}

is a electron deficient compound

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Boron hydrides are formed when

H_{3} P O_{4}

reacts with

M g_{3} B_{2}

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All the

B - H

bond distances in

B_{2} H_{6}

are equal

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The boron hydrides are readily hydrolysed.

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Solution

The correct option is C All the $B - H$ bond distances in $B_{2} H_{6}$ are equal
In $B H_{3}$ boron has only 6 valence electrons, hence $B H_{3}$ is a electron deficient compound.
Boron hydrides are formed when $H_{3} P O_{4}$ reacts with $M g_{3} B_{2}$
$M g_{3} B_{2} + H_{3} P O_{4} \to Mixture of boranes mainly B_{4} H_{10} Heat - - \to B_{2} H_{6}$
In diborane $B$ is $s p^{3}$ hybridised and $B - H - B$ bonds are ' $3$ centre $- 2$ electrons' bond, whereas $B - H$ bonds are ' $2$ centre $- 2$ electrons' bond. Therefore, all bonds are not of same length.

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