Consider the following molecules :
(I) $H_{2} O$
(II) $H_{2} S$
(III) $H_{2} S e$
(IV) $H_{2} T e$
Arrange these molecules in increasing order of bond angle.

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 V < I I I

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I I < I V < I I I < I

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Solution

The correct option is B $I V < I I I < I I < I$
The increasing order of bond angle is $I V < I I I < I I < I$ or $H_{2} T e < H_{2} S e < H_{2} S < H_{2} O$ . This is the same order in which, the electronegativity and the size of the central metal atom increases.

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Consider the following molecules :(I) H2O(II) H2S(III) H2Se(IV) H2TeArrange these molecules in increasing order of bond angle.

The correct option is B IV<III<II<IThe increasing order of bond angle is IV<III<II<I or H2Te<H2Se<H2S<H2O. This is the same order in which, the electronegativity and the size of the central metal atom increases.

Consider the following molecules :
(I) $H_{2} O$
(II) $H_{2} S$
(III) $H_{2} S e$
(IV) $H_{2} T e$
Arrange these molecules in increasing order of bond angle.

The correct option is B $I V < I I I < I I < I$
The increasing order of bond angle is $I V < I I I < I I < I$ or $H_{2} T e < H_{2} S e < H_{2} S < H_{2} O$ . This is the same order in which, the electronegativity and the size of the central metal atom increases.