Write a note on oxidation states of group $-15$ elements. Write a note on the nature of bonding of group $-15$ elements.

(N/A) The common oxidation states of group $-15$ elements are $-3, +3$ and $+5$. The tendency to exhibit $-3$ oxidation state decreases down the group due to an increase in atomic size and metallic character. In fact,$Bi$ hardly exists in the $(-3)$ oxidation state.
The stability of the $(+5)$ oxidation state decreases down the group,and that of the $(+3)$ oxidation state increases due to the inert pair effect. The only known compound of bismuth in the $(+5)$ oxidation state is $BiF_{5}$.
Nitrogen exhibits $(+1), (+2), (+4)$ oxidation states with oxygen in addition to the $(+5)$ oxidation state. However,it does not form compounds in the $(+5)$ oxidation state with halogens because nitrogen lacks $d$-orbitals to accommodate electrons from other elements to form bonds.
Nitrogen in all oxidation states from $(+1)$ to $(+4)$ disproportionates in acidic medium. For example: $3HNO_{2} \rightarrow HNO_{3} + H_{2}O + 2NO$.
Phosphorus shows $(+1)$ and $(+4)$ oxidation states in some oxo-acids. Phosphorus disproportionates in all intermediate states from $(-3)$ to $(+5)$ in acidic or alkaline medium. For example: $2H_{3}PO_{2} \stackrel{\Delta}{\longrightarrow} H_{3}PO_{4} + PH_{3}$.
Arsenic,Antimony,and Bismuth in $(+3)$ oxidation states are stable with respect to disproportionation.

Element	Oxidation states
$1.$ Nitrogen	$(-3)$ to $(+5)$
$2.$ Phosphorus	$(-3), (+3), (+5)$
$3.$ Arsenic	$(-3), (+3), (+5)$
$4.$ Antimony	$(+3), (+5)$
$5.$ Bismuth	$(+3)$

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 can expand their covalence. Example: $PCl_{5}, [PCl_{6}]^{-}, [PF_{6}]^{-}$.
Nitrogen forms $p\pi-p\pi$ bonds due to its small size,while other elements do not form $p\pi-p\pi$ bonds. Hence,in their elemental state,Phosphorus,Arsenic,and Antimony exist as $P_{4}, As_{4},$ and $Sb_{4}$ respectively as tetrahedral molecules.
Due to the non-availability of $d$-orbitals in nitrogen,it cannot form $d\pi-p\pi$ bonds,which are possible with other elements. Example: $R_{3}P=O$ or $R_{3}P=CH_{2}$ $(R = \text{alkyl group})$. Phosphorus and Arsenic form $d\pi-d\pi$ bonds with transition metals when their compounds act as ligands. For example: $P(C_{2}H_{5})_{3}$.