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

I > II > III

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II > I > III

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III > II > I

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I > III > II

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Solution

The correct option is A II > I > III
$H - C - H$ bond angle is about $120$ , when $C$ is bonded through a double bond but in case of $C H_{4}$ , the $H - C - H$ bond angle is only $109^{\circ} 28^{'}$ due to hindrance.
Thus, the correct order of $H - C - H$ bond angle is

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C_{6} H_{5} C H_{3} I, C_{6} H_{5} C O O H I I, C_{6} H_{6} I I I, C_{6} H_{5} N O_{2} I V

π

H_{2} S_{2} O_{6}

H_{2} S O_{3}

H_{2} S_{2} O_{5}

(i) C H_{3} - + C H - C H_{2} - O C H_{3}

(i i) C H_{3} - + C H - C H_{2} - C H_{3}

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The decreasing nucleophilic order of the following compound is:

$(i) Θ C H_{3}$

$(i i) Θ N H_{2}$

$(i i i) Θ O H$

$(i v) Θ F$

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