The hybridisation of central atom in ${N_3}^ - $, NOCl and ${N_2}O$ respectively are:

(A) sp, $s{p^2}$, sp

(B) sp, sp, $s{p^3}$

(C) $s{p^2}$, sp, sp

(D) $s{p^2}$, $s{p^2}$, sp

To determine the hybridization of the central atom in the given molecules—${N_3}^-$ (azide ion), NOCl (nitrosyl chloride), and ${N_2}O$ (nitrous oxide)—we need to analyze the molecular structure and the bonding characteristics of each compound. Let's break this down step by step.

Hybridization of ${N_3}^-$

The azide ion, ${N_3}^-$, consists of three nitrogen atoms. The central nitrogen atom is bonded to two other nitrogen atoms. To find the hybridization, we can look at the number of regions of electron density around the central nitrogen. In this case, the central nitrogen has:

• Two single bonds (to the terminal nitrogen atoms)
• One lone pair of electrons

This gives a total of three regions of electron density. According to VSEPR theory, three regions of electron density correspond to sp² hybridization. Therefore, the hybridization of the central atom in ${N_3}^-$ is sp².

Hybridization of NOCl

Next, let's examine NOCl. In this molecule, nitrogen is the central atom bonded to one oxygen atom and one chlorine atom. The nitrogen atom has:

• One double bond with oxygen
• One single bond with chlorine
• One lone pair of electrons

This results in three regions of electron density around the nitrogen atom. Therefore, the hybridization of nitrogen in NOCl is also sp².

Hybridization of ${N_2}O$

Finally, let's consider nitrous oxide, ${N_2}O$. In this molecule, the structure can be represented as N≡N-O, where the first nitrogen is triple-bonded to the second nitrogen, which is then single-bonded to the oxygen. The central nitrogen atom (the second nitrogen) has:

• One triple bond with the first nitrogen
• One single bond with oxygen

This gives us two regions of electron density, indicating that the hybridization of the central nitrogen in ${N_2}O$ is sp.

Summary of Hybridizations

Now that we have analyzed each molecule, we can summarize the hybridizations:

• ${N_3}^-$: sp²
• NOCl: sp²
• ${N_2}O$: sp

Based on this analysis, the correct answer to the question is (D) sp², sp², sp.