The correct order for increasing the boiling point of hydrides (given below) of group 15 is:

(A) \(NH_3\)

(B) \(PH_3\)

(C) \(AsH_3\)

(D) \(SbH_3\)

(E) \(BiH_3\)

Choose the correct answer from the options given below :

B < C < D < A < E

The correct answer is option 4. B < C < D < A < E.

\(NH_3\) molecules are associated by strong intermolecular H-bonds. As a result, its boiling point in exceptionally high. The intermolecular forces in \(PH_3\) are van der Waal's forces, due to which its boiling point in lower than \(NH_3\). In moving from \(PH_3\), to \(BiH_3\), boiling points increase. This is due to the increase in the magnitude of van der Waal's forces, owing to the increase in the molecular size (or increase in molar mass). The van der Waal's forces in \(SbH_3\), become stronger than intermolecular H-bonds in \(NH_3\). As a result, boiling point of \(SbH_3\), becomes more than that of \(NH_3\). Hence, the correct sequence of boiling points \(PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3\).