Which of the following is the correct decreasing order of bond angles?

N H_{3} < C H_{4} < C_{2} H_{2} < H_{2} O

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C_{2} H_{2} > C H_{4} > N H_{3} > H_{2} O

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N H_{3} > H_{2} O > C H_{4} > C_{2} H_{2}

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C H_{4} > C_{2} H_{2} > N H_{3} > H_{2} O

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Solution

The correct option is B $C_{2} H_{2} > C H_{4} > N H_{3} > H_{2} O$
$N H_{3}$ has a trigonal pyramidal shape and a lone pair of electrons. It has a bond angle of $107^{o}$ .

$C H_{4}$ has a tetrahedral shape and has a bond angle of ${109.28}^{o}$ .

$C_{2} H_{2}$ has a linear shape having a hybridisation of $s p$ and has a bond angle of $180^{o}$ .

$H_{2} O$ has a hybridisation of $s p^{3}$ and tetrahedral geometry. Since, it has two lone pairs, it has bent shape and a bond angle of ${104.5}^{o}$

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