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The $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle in $\mathrm{CH}_4$ molecule is
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The correct answer is:
$109^{\circ} 28^{\prime}$
$\mathrm{CH}_4$ has $\mathrm{sp}^3$ Hybridization and bond angle
$109^{\circ} 28^{\prime}$
$109^{\circ} 28^{\prime}$
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