The correct order of boiling points of hydrides of 15^th group elments is

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${PH}_{3} < {AsH}_{3} < {SbH}_{3} < {BiH}_{3} < {NH}_{3}$

${BiH}_{3} < {SbH}_{3} < {AsH}_{3} < {PH}_{3} < {NH}_{3}$

${PH}_{3} < {AsH}_{3} < {SbH}_{3} < {NH}_{3} < {BiH}_{3}$

${PH}_{3} < {AsH}_{3} < {NH}_{3} < {SbH}_{3} < {BiH}_{3}$

answer is A.

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Detailed Solution

NH₃ molecules are associated by strong intermolecular H-bonds.
As a result, its boiling point in exceptionally high.
The intermolecular forces in PH₃ are vander Waal's forces, due to which its boiling point in lower than NH3.
In moving from PH3, to BiH3, boiling points increase.
This is due to the increase in the magnitude of vander Waal's forces, owing to the increase in the molecular size (or increase in molar mass).
The vander Waal's forces in SbH₃, become stronger than intermolecular H-bonds in NH_3.
As a result, boiling point of SbH3, becomes more than that of NH₃. Hence, the correct sequence of boiling points is PH₃<AsH₃<NH₃<SbH₃<BiH₃