Question

Maximum covalency of an element of atomic number \[7\] is
(A) $4$
(B) $2$
(C) $5$
(D) $3$

Answer 1

Hint: From left to right of the periodic table, the elements are arranged in an increasing order of atomic number. Covalency is defined as the number of covalent bonds formed by an atom. A covalent bond is formed when an atom shares electrons with another atom. It is not a complete transfer of electrons but mutual sharing of electrons, if this sharing is equal between both the atoms meaning that each atom shares one electron each, then it is covalent type of bonding. But if a single atom shares a pair of electrons with another atom, then it is called as a coordinate bond or dative bond. It is also a type of covalent bond.

Complete Step by Step Solution:
Atomic number \[7\] means the electronic configuration of the element is $1{s^2}2{s^2}2{p_x}^12{p_y}^12{p_z}^1$. It is clear from the electronic configuration that the element is nitrogen.

It is only the valence shell that is involved in bond formation so the concern is only for valence shell electronic configuration given as $2{s^2}2{p_x}^12{p_y}^12{p_z}^1$.
It is clear from the electronic configuration that there are three unpaired electrons in each \[2p\] orbital. Thus, this unpaired electron is used to form a covalent bond with another unpaired electron of another atom.

A fourth covalent bond is formed by the donation of a pair of electrons present in a $2s$ orbital to any vacant orbital of another atom. This is the coordinate bond which is a type of covalent bond.
Hence covalency is \[3 + 1 = 4\].
Thus, the correct option is A.

Note: we would have expected a covalency of \[5\] if there was a low-lying vacant orbital to unpair the pair of electrons of s orbital. Then, the number of unpaired electrons would be \[5\] hence formation of \[5\] covalent bonds would have been possible. Such a case is observed in phosphorus because of the availability of low-lying vacant orbital \[3d\]. phosphorus has maximum covalency \[5\].