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For example, I'm trying to write the formula for $\ce{NO}$, and $\ce{ClO2}$. 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.

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  • $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I edited your question to improve formatting, most notably by adding MathJax to display formulae. You can learn more about that in the help center and this meta post. I also added the homework tag because this is a homework-type question. $\endgroup$ – Jan Oct 13 '15 at 20:01
  • $\begingroup$ What are you asking really? There are lots of compounds with odd number of electrons. That's it. $\endgroup$ – Mithoron Oct 13 '15 at 20:22
  • $\begingroup$ Well, 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. $\endgroup$ – orthocresol Oct 13 '15 at 20:42
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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.

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