Question

Find the number of delocalized electrons in naphthalene molecules.

Answer 1

Hint: First of all, we need to draw the structure of naphthalene. The structure of naphthalene consists of two fused benzene rings. Delocalized electrons are those electrons which take part in resonance, and are distributed throughout the entire molecule.

Complete step by step answer:
Structure of naphthalene is given below,

The resonance in naphthalene molecules involves the delocalization of all the $\pi - $electrons present. As only those bonds take part in resonance which are present in conjugation with each other and in this molecule all the pi-bonds are in conjugation with each other.
The resonating structures or canonical forms of naphthalene are shown below.

The number of pi-bonds taking part in resonance are $5$.
As we know the number of $\pi - $electrons participating in resonance are double the number of pi-bonds. So,
Number of electrons participating in resonance, that is, number of electrons that are delocalized in the naphthalene molecule are
$ = 5 \times 2$
$ \Rightarrow 10$

Note:
The resonating structures show the delocalization of electrons within the molecule. In these structures the number of atoms within the molecule and the total number of electrons remain the same. The overall charge on the molecule also remains the same, while the charge separation and distribution may change.