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

180^{\circ}

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120^{\circ}

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109^{\circ}

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90^{\circ}

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Solution

The correct option is C $109^{\circ}$
$a p^{3} - h y b r i d i s e d$
$∠ H - B - H = {109.28}^{\circ}$

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