The correct decreasing order of boiling points is:

H F > H I > H B r > H C l

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H_{2} O > H_{2} T e > H_{2} S e > H_{2} S

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I_{2} > B r_{2} > C l_{2} > F_{2}

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C H_{4} > G e H_{4} > S i H_{4}

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Solution

The correct options are
A $H F > H I > H B r > H C l$
B $H_{2} O > H_{2} T e > H_{2} S e > H_{2} S$
C $I_{2} > B r_{2} > C l_{2} > F_{2}$
As molecular weight of molecule is increased, it's boiling point also increases.
But, $H F$ and $H_{2} O$ have intermolecular H-bonding. So, they have high boiling point.

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