MCQ
The $H - C - H$ bond angle in $C{H_4}$ is
- ✓${109^o}28'$
- B${107^o}28'$
- C${90^o}$
- D${180^o}$
✓
Answer
Correct option: A.
${109^o}28'$
a
The shape of $CH _4$ is perfect tetrahedron and hence bond angle is $109^{\circ} 28^{\prime}$.
The shape of $CH _4$ is perfect tetrahedron and hence bond angle is $109^{\circ} 28^{\prime}$.
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