Arrange in decreasing order of boiling point.

$S i H_{4}, G e H_{4}, S n H_{4}, C H_{4}$

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

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

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

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

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Solution

The correct option is B $S n H_{4} > G e H_{4} > S i H_{4} > C H_{4}$
Observe that H is attached to carbon group elements. None of them enough electronegativity to reflect as H-bonding

Thus, only molecular weight is the criteria.

From periodic table, size and weight in order Sn > Ge > Si > C

Thus molecular weight order is $S n H_{4} > G e H_{4} > S i H_{4} > C H_{4}$ .

Boiling point also follows the same order. So, $S n H_{4}$ has highest boiling point.

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