Question

10. consider the following molecule, so₂**
a. draw the lewis structure
b. write the number of valence electrons in the molecule
c. calculate the formal charge of each atom

Explanation:

A. Lewis Structure

Sulfur (\(S\)) is the central atom. Sulfur has \(6\) valence electrons and each oxygen (\(O\)) has \(6\) valence electrons. The total number of valence electrons is \(6 + 2\times6=18\). Sulfur forms a double - bond with one oxygen and a single - bond with the other oxygen, and there is a lone pair on sulfur. The Lewis structure is \(O = S - O\) with lone pairs on \(O\) and \(S\) (in a bent geometry).

B. Number of Valence Electrons

Sulfur (\(S\)) is in group \(16\) (\(n s^{2}n p^{4}\)), so it has \(6\) valence electrons. Oxygen (\(O\)) is also in group \(16\), so each \(O\) atom has \(6\) valence electrons. For \(SO_{2}\), the total number of valence electrons \(=6+(2\times6)=18\)

C. Formal Charge Calculation

The formula for formal charge is \(FC = V-\frac{1}{2}B - L\), where \(V\) is the number of valence electrons in the free atom, \(B\) is the number of bonding electrons, and \(L\) is the number of non - bonding electrons.

• For the double - bonded \(O\): \(V = 6\), \(B = 4\), \(L = 4\). \(FC=6-\frac{4}{2}-4 = 0\)
• For the single - bonded \(O\): \(V = 6\), \(B = 2\), \(L = 6\). \(FC=6 - 1-6=- 1\)
• For \(S\): \(V = 6\), \(B = 6\), \(L = 2\). \(FC=6-\frac{6}{2}-2=1\)

Answer:

A. Lewis structure: \(O = S - O\) (with lone pairs, bent geometry)
B. Number of valence electrons: \(18\)
C. Formal charges: Double - bonded \(O\): \(0\); Single - bonded \(O\): \(-1\); \(S\): \(+1\)