Consider the following hydrides and arrange them in increasing order of bond angles :
(I) $N H_{3}$
(II) $P H_{3}$
(III) $A s H_{3}$
(IV) $S b H_{3}$

I < I I < I I I < I V

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I V < I I I < I I < I

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I < I I I < I I < I V

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I V < I I < I I I < I

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Solution

The correct option is A $I V < I I I < I I < I$
The increasing order of bond angles is $I V < I I I < I I < I$ or $S b H_{3} < A s H_{3} < P H_{3} < N H_{3}$ . This is the same order in which the electronegativity of the central metal atom increases.

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Consider the following hydrides and arrange them in increasing order of bond angles :(I) NH3(II) PH3(III) AsH3(IV) SbH3

The correct option is A IV<III<II<IThe increasing order of bond angles is IV<III<II<I or SbH3<AsH3<PH3<NH3. This is the same order in which the electronegativity of the central metal atom increases.

Consider the following hydrides and arrange them in increasing order of bond angles :
(I) $N H_{3}$
(II) $P H_{3}$
(III) $A s H_{3}$
(IV) $S b H_{3}$

The correct option is A $I V < I I I < I I < I$
The increasing order of bond angles is $I V < I I I < I I < I$ or $S b H_{3} < A s H_{3} < P H_{3} < N H_{3}$ . This is the same order in which the electronegativity of the central metal atom increases.