Nitrogen is restricted to a maximum covalency of $4$ since only four (one s and three p) orbitals are available for bonding. The heavier elements have vacant d orbitals in the outermost shell which can be used for bonding (covalency) and hence, expand their covalence as in $P F_{6}^{-}$ .

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Solution

Nitrogen does not have vacant d-orbital to exceed its covalency beyond $4$ and hence have only $4$ orbitals available for bonding.
Whereas heavier elements have vacant d-orbitals which can be used for bonding and hence expand their covalence as in ${P F_{6}}^{-}$
$P = 3 s^{2} 3 p^{3}$
(Refer to Image)

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Nitrogen is restricted to a maximum covalency of 4 since only four (one s and three p) orbitals are available for bonding. The heavier elements have vacant d orbitals in the outermost shell which can be used for bonding (covalency) and hence, expand their covalence as in PF−6.

Nitrogen does not have vacant d-orbital to exceed its covalency beyond 4 and hence have only 4 orbitals available for bonding.Whereas heavier elements have vacant d-orbitals which can be used for bonding and hence expand their covalence as in PF6−P=3s23p3(Refer to Image)

Nitrogen is restricted to a maximum covalency of $4$ since only four (one s and three p) orbitals are available for bonding. The heavier elements have vacant d orbitals in the outermost shell which can be used for bonding (covalency) and hence, expand their covalence as in $P F_{6}^{-}$ .