What to do when there is an odd number of valence electrons left on the the central atom?

Question

For example, I'm trying to write the formula for $\ce{NO}$, and $\ce{ClO2}$. If you count the total number of electrons in these molecules, it's odd. This means that no matter how you arrange the electrons, some atom is not going to have a full octet, and some atom is going to have an unpaired electron somewhere - it's inevitable. So my advice is: don't worry about the central atom not having a full octet. Just try and get as many full octets as you can. Of course, don't ever stuff more than 8 valence electrons into a Period 2 element. That's a big no-no. Once you have a structure that you think is reasonable, just check Google for the structures.

The steps my teacher gave me always assume that the central atom has a full valence, or more than full, after the outer atoms have been bonded with it. I'm running into trouble here because my central atom in these cases does not have enough electrons!

I know I could Google what they look like, but it doesn't help me understand why.

Answer 1

There are some molecules where the full octet isn't possible (like you've found with $\text{NO}$ and $\text{ClO}_2$). In these cases, the important thing to consider is trying to keep the formal charge (Wikipedia) on each of the atoms as close to zero as possible.

For example, for the $\text{NO}$, we know that $\text{N}$ contributes 5 electrons and $\text{O}$ contributes 6 electrons for a total of 11 electrons. We essentially have to decide between the following Lewis structures:

$$ (1) \qquad :\ddot{\text{N}}::\ddot{\text{O}}. \quad \text{Single electron on O} \\ \qquad \text{vs.} \\ (2) \qquad .\ddot{\text{N}}::\ddot{\text{O}}: \quad \text{Single electron on N} $$

Here is where calculating the formal charge is important. For $(1)$ above the formal charge on the N is (5 - 4 - (4$\div$2) = -1) and for O is (6 - 3 - (4$\div$2) = +1, whereas for $(2)$ the formal charge on the N is (5 - 3 - (4$\div$2) = 0) and for O is (6 - 4 - (4$\div$2) = 0.

Therefore the second Lewis structure is more favorable.