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Answer» `PH_3 lt AsH_3lt SbH_3 lt BiH_3lt NH_(3)`
`PH_(3) lt AsH_(3) LT SbH_(3) lt NH_3 lt BiH_(3)`
`PH_(3) lt AsH_(3) lt NH_(3) lt SbH_(3) lt BiH_(3)`
`NH_(3) lt PH_(3) lt AsH_(3) lt SbH_(3) lt BiH_(3)` Solution :`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 vander 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 vander Waal's forces, owing to the increase in the molecular size (or increase in molar mass). The vander Waal's forces in SbH, become stronger than intermolecular H-bonds in `NH_3`. As a result, boiling point of SbH, BECOMES more than that of `NH_3`. Hence, the correct sequenceof boiling points `PH_(3) lt AsH_(3) lt NH_(3) lt SbH_(3) lt BiH_(3)`.
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