Question

Two possible Lewis diagrams for sulfine $\left({\mathrm{H}}_2\mathrm{CSO}\right)$are

(a)Compute the formal charges on all atoms.

(b) Draw a Lewis diagram for which all the atoms in sulfine have formal charges of zero.

Short answer

• a)In the left structure, the formal charge of two hydrogen atoms is 0, sulfur is +1, oxygen is -1. In the right structure, the formal charge of hydrogen is 0, oxygen is 0, carbon is -1, and sulfur is +1.

Lewis diagram of sulfine in which all the atoms have 0 formal charges

Step-by-step solution

Formal charges of each atom in the first structure

Lewis structure of sulfine is given below:

The following formula calculates the formal charge:

Formal charge, where VE is the valence electrons, LP is the lone pair, and BP is the bonding pair of electrons.

Let us consider the formal charge of hydrogen.

VE is 1; LP is 0, and BP is 2.

The formal charge of hydrogen,

Both hydrogens have the same formal charge 0.

Consider the formal charge of carbon.

VE is 4; LP is 0, and BP is 8.

The formal charge of carbon,

The formal charge of carbon is 0.

Consider the formal charge of sulfur.

VE is 6; LP is 2, and BP is 6.

The formal charge of carbon,

The formal charge of sulfur is +1.

Consider the formal charge of oxygen.

VE is 6; LP is 6, and BP is 2.

The formal charge of carbon,

The formal charge of oxygen is -1.

Therefore, the formal charges of hydrogen, carbon, sulfur and oxygen are 0, 0, +1, and -1, respectively.

Formal charge of each atom in the second structure

Lewis structure of sulfine is given below:

The formal charge$\mathrm{FC}=\mathrm{VE}-\mathrm{LP}-\frac{\mathrm{BP}}{2}$

Let us consider the formal charge of hydrogen.

VE is 1; LP is 0, and BP is 2.

The formal charge of hydrogen,

Both hydrogens have the same formal charge 0.

Consider the formal charge of carbon.

VE is 4; LP is 2, and BP is 6.

The formal charge of carbon $\begin{array}{c}{\mathrm{FC}}_{\mathrm{C}}=4-2-\frac{6}{2}\\ =-1\end{array}$,

The formal charge of carbon is -1.

Consider the formal charge of sulfur.

VE is 6; LP is 2, and BP is 6.

The formal charge of sulfur,$\begin{array}{c}{\mathrm{FC}}_{\mathrm{S}}=6-2-\frac{6}{2}\\ =+1\end{array}$

The formal charge of sulfur is +1.

Consider the formal charge of oxygen.

VE is 6; LP is 4, and BP is 4.

The formal charge of oxygen ,

The formal charge of oxygen is 0.

Therefore, the formal charges of hydrogen, carbon, sulfur, and oxygen are 0, -1, +1, and 0, respectively.

Lewis diagram of sulfine

We can draw the Lewis diagram of sulfine in which all the atoms have zero formal charges by violating the octet rule of sulfur. The Lewis structure of sulfine is given below: