The correct order of decreasing H-C-H angle in the following molecule is

Correct answer: (B) II>I>III. H-C-H bond angle is about 120^° , when C is bonded through a double bond but in case of CH_4 the H-C-H bond angle is only 109^° 2B^′ Cyclic compounds have a H-C-H bond angle Feater than 109^° 28^′ due to hindrance. Thus, the correct order of 1 H-C-H bond angle is

Q8 / 200 · WBJEE 2014 · Chemistry

The correct order of decreasing $\mathrm{H}-\mathrm{C}-\mathrm{H}$ angle in the following molecule is

A
I>II>III
Correct answer
B
II>I>III
C
III>II>I
D
I>III>II

Solution

$\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is about $120^{\circ}$, when $\mathrm{C}$ is bonded through a double bond but in case of $\mathrm{CH}_{4}$ the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is only $109^{\circ} \mathrm{2B}^{\prime}$ Cyclic compounds have a $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle Feater than $109^{\circ} 28^{\prime}$ due to hindrance. Thus, the correct order of $1 \mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is

Hence, the correct answer is option: 'B'.

The correct order of decreasing $\mathrm{H}-\mathrm{C}-\mathrm{H}$ angle in the following molecule is

Solution

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