Question

The ${\mathit{N}}_{\mathbf{2}}\mathit{O}$molecule is linear and polar.

a. On the basis of this experimental evidence, which arrangement is correct, NNO or NON? Explain your answer.

b. On the basis of your answer in part a, write the Lewis structure of ${\mathit{N}}_{\mathbf{2}}\mathit{O}$(including resonance forms). Give the formal charge on each atom and the hybridization of the central atom.

c. How would the multiple bonding inbe described in terms of orbitals?

Short answer

Dinitrogen oxide consists of two nitrogen atoms and one oxygen atom which are together bonded with two sigma and two pie bonds and the most stable form is NNO in which the oxygen has 0, central nitrogen has +1 and terminal nitrogen has -1 formal charge, with hybridization , sp³, and sp respectively.

Step-by-step solution

(a)

NNO is the correct arrangement as nitrogen have a tendency to form four bond whereas oxygen can form only two bonds to become stable. That’s why one nitrogen is in between so that it can make two bonds with nitrogen and another two bonds with oxygen to form a stable structure

(b) Lewis and resonance structure of N2O

The formal charge on each atom:

The formal charge of an atom can be calculated by the given formula i.e.

$Formalcharge=Valenceelectron-Non-bondingelectron-\frac{Bondingelectron}{2}$

As the central nitrogen has 5 valence electrons, 0 nonbonding electrons and 8 bonding electrons, the formal charge will be

$Formalcharge=5-0-\frac{8}{2}=+1$

Similarly, in oxygen 6 valence electrons, 4nonbonding electrons and 4 bonding electrons are present, so the formal charge will be

$Formalcharge=6-4-\frac{4}{2}=0$

In the second nitrogen atom, 5 valence electrons, 4 nonbonding electrons and 4 bonding electrons are present, so the formal charge will be

$Formalcharge=5-4-\frac{4}{2}=-1$

Hybridisation on central nitrogen:

As the central nitrogen forms 2 sigma bond and has no valence electrons, so the hybridization of nitrogen will be sp.

(c) Multiple bonding.

In the given structure, two sigma and two pie bonds are formed. As oxygen can form two bonds and has 6 valence electrons, it forms one bond with nitrogen and gains one negative charge which is responsible for its stability. Whereas nitrogen has 5 valence electrons, that’s why central nitrogen will form 4 bonds and gain a positive charge as it is less electronegative.