The order of increasing bond angle in the molecules
$B e C l_{2}, B C l_{3}, C C l_{4}$ and $S F_{6}$ is :

S F_{6} < C C l_{4} < B C l_{3} < B e C l_{2}

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B e C l_{2} < B C l_{3} < C C l_{4} < S F_{6}

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S F_{6} < C C l_{4} < B e C l_{2} < B C l_{3}

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B C l_{3} < B e C l_{2} < S F_{6} < C C l_{4}

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Solution

The correct option is A $S F_{6} < C C l_{4} < B C l_{3} < B e C l_{2}$
The hybridisation of the molecules respectively is:
(1) $s p$
(2) ${s p}^{2}$
(3) ${s p}^{3}$
(4) ${s p}^{3} d^{2}$
which explains the order of bond angle.

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