# Draw the Lewis structure for sulfur dioxide, SO2, which does not require expanded octets. Which of the statements below is true for the Lewis structure of the SO2 molecule that obeys the octet rule?

Jun 23, 2017

Well, there is more than one answer... should be both $C$ and $D$. But since you FORCE sulfur to have only $8$ electrons around it, I guess it's only $C$.

${\text{SO}}_{2}$ contains ${\overbrace{6}}^{S} + 2 \times {\overbrace{6}}^{O} = 18$ valence electrons. With one sulfur atom, we place it at the center, since it is less electronegative than oxygen.

Two single bonds around sulfur use up

$\text{1 bond" xx "2 electron groups" xx "2 electrons} = 4$ electrons,

and each oxygen holds $3 \times \text{lone pairs}$ to use up

$\text{2 atoms" xx 2 xx "3 lone pairs} = 12$ more electrons.

The remaining $2$ electrons go on the central atom as a lone pair. However, to minimize formal charge, we use one lone pair from each oxygen to make $\pi$ bonds, constructing double bonds for each $\text{S"-"O}$.

This gives: Therefore, $C$ is true, but $D$ is also true without the pointless restriction of following the octet rule.

Arguably, one could construct a resonance structure where only the third lone pair on one oxygen forms a $\pi$ bond, but that is less stable of a resonance structure. The most stable resonance structure has two double bonds.

And in fact, sulfur can do this because it has $3 d$ orbitals to utilize.