Question

Assign formal charges to all atoms in the following Lewis

diagrams.

a).${{\mathbf{ClO}}^{\mathbf{-}}}_{\mathbf{4}}$

b). ${\mathbf{SO}}_{\mathbf{2}}$

c).${{\mathbf{BrO}}^{\mathbf{-}}}_{\mathbf{2}}$

d). ${{\mathbf{NO}}^{\mathbf{-}}}_{\mathbf{3}}$

Short answer

1. The formal charge of chlorine is +3, and each oxygen atom is -1.

Step-by-step solution

Formal charge

Formal charge of an atom gives an idea how the valence electrons of it change when the atom participates in bond formation.

The formula to calculate formal charge of an atom

The formula to calculate the formal charge of an atom is shown below: $\mathrm{FC}=\mathrm{VE}-\mathrm{LP}-\frac{1}{2}\left(\mathrm{BP}\right)$

Where FC is the formal charge, VE is the number of valence electrons, LP is the number of non-bonding electrons/lone pair of electrons and BP is the bonding electrons.

The formal charge of each atom in ClO-4

${\mathrm{ClO}}_4^{-}$has one chlorine and four oxygen atoms. Each oxygen is single-bonded to the chlorine atom.

The structure of is given below:

The following equation is used to calculate the formal charge:

Formal charge, $\mathrm{FC}=\mathrm{VE}-\mathrm{LP}-\frac{1}{2}\left(\mathrm{BP}\right)$, where VE is the number of valence electrons, LP is the lone pair of electrons, and BP is the electrons in the bonding pair.

The formal charge of oxygen, $\mathrm{FC}=\mathrm{VE}-\mathrm{LP}-\frac{1}{2}\left(\mathrm{BP}\right)$

VE of oxygen is 6, LP of oxygen is 6 and BP of oxygen is 2. Substituting these values,

$\begin{array}{c}\mathrm{FC}=6-6-\frac{1}{2}\left(2\right)\\ =-1\end{array}$

Since each chlorine atom is similar, theformal charge of each oxygen is -1.

VE of chlorine is 7, LP of chlorine is 3 and BP of chlorine is 2. Substituting these values,

$\begin{array}{c}\mathrm{FC}=7-0-\frac{1}{2}\left(8\right)\\ =+3\end{array}$

The formal charge of chlorine is +3.