Question

use formal charges to identify the better lewis structure between the following

match the items in the left column to the appropriate blanks in the sentences on the right.

Explanation:

Step1: Calculate formal charges for Structure I

Formal charge formula: \( FC = V - \frac{B}{2}-L \) (where \( V \) is valence electrons, \( B \) is bonding electrons, \( L \) is non - bonding electrons).
For \( S \) in Structure I: \( V = 6 \), \( B = 4 \), \( L = 0 \), \( FC=6-\frac{4}{2}-0 = 4 \).
For \( C \) in Structure I: \( V = 4 \), \( B = 2 \), \( L = 6 \), \( FC = 4-\frac{2}{2}-6=-3 \).

Step2: Calculate formal charges for Structure II

For \( C \) in Structure II: \( V = 4 \), \( B = 4 \), \( L = 0 \), \( FC=4-\frac{4}{2}-0 = 2 \).
For \( S \) in Structure II: \( V = 6 \), \( B = 2 \), \( L = 6 \), \( FC = 6-\frac{2}{2}-6=-1 \).

Step3: Compare the number of non - zero formal charges

Structure I has two atoms (\( S \) and \( C \)) with non - zero formal charges. Structure II has two atoms (\( C \) and \( S \)) with non - zero formal charges. But in Structure I, the magnitudes of formal charges (\(+4\) and \(-3\)) are larger. In Structure II, the magnitudes (\(+2\) and \(-1\)) are smaller. Also, in Structure I, \( S \) (less electronegative than \( C \)) has a positive formal charge. In Structure II, \( C \) (more electronegative than \( S \)) has a positive formal charge. A better Lewis structure has fewer atoms with non - zero formal charges and formal charges on atoms according to electronegativity (more electronegative atoms should have negative formal charges if possible). Structure II has fewer atoms with non - zero formal charges when considering the principles (Structure I has more extreme formal charges).

Answer:

The better Lewis structure is structure II because it has fewer atoms with nonzero formal charges.