(i) Why does$NH _{ 3}$ form hydrogen bonding but $PH _{ 3}$ does not?
(ii) Why $PH _{ 3}$ has lower boiling point than $NH _{ 3}$?
(i) The electronegativity of N is 3.0, of P is 2.1 and that of hydrogen is 2.1. This difference in electronegativity values make N—H bond polar in nature. However, in $PH _{ 3}$, P—H bonds form non-polar bonds because both P and H have the same value of electronegativity. Hence, $PH _{ 3}$ does not form hydrogen bonds.
(ii) The electronegativity of N is much higher than that of P. So, $NH _{ 3}$ undergoes extensive H-bonding and hence, it exists as an associated molecule. To break these additional bonds, the large amount of energy is required while $PH _{ 3}$ does not form H-bond and hence, exists as discrete molecule.