H - B - H bond angle in $B H_{4}^{-}$ is:

180^{o}

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120^{o}

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109^{o}

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90^{o}

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Solution

The correct option is C $109^{o}$
$B H_{4}^{-}$ is tetrahedral

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Similar questions

HB H bond angle in $B H_{4}^{-}$ is

B H_{4}^{-}

Consider the molecules $C H_{4}$ , $N H_{3}$ and $H_{2} O$ . Which of the given statements is false?

H - B - H

{B H}_{4}^{-}

(B_{2} H_{6})

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