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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.

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  • $\begingroup$ $\ce{NO}$ is a radical that does not dimerize, because the free electron on the N atom is on an anti bonding molecular orbital ${\pi}$* . Try to draw the MO diagram of $\ce{NO}$ to be convinced about this situation. $\endgroup$
    – Maurice
    Nov 29, 2021 at 22:10
  • $\begingroup$ @Maurice, it does dimerize, but as far as I remember mainly in condensed phase. $\endgroup$ Nov 30, 2021 at 10:10

3 Answers 3

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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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The octet rule requires eight electrons around each atom. However, $\ce{ClO2​,NO2​,NO,}$ etc. are some stable molecules that disobeys this rule and are known as odd electron molecules.

If one of the atoms in a molecule has an odd number of valence electrons, the molecule will have an odd electron bond. Or we can say that the molecule has unpaired electrons. These molecules are quite stable and paramagnetic in nature. On dimerisation, the odd electron molecules are usually converted to more stable molecules with even number of electrons.

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The nitrogen oxides, such as nitric oxide ($\ce{NO}$) and nitrogen dioxide ($\ce{NO2}$) are the most common examples. In spite of being an odd-electron molecule $\ce{ClO2}$ does not dimerize because the odd electron of $\ce{Cl2}$ is delocalised.

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Usually, I would recommend to view molecules like $\ce{NO}$ as examples which demonstrate the limits of the valence structural formula we use today. Although the representations shown in the other answers are formally correct, they do not really help to understand the bond situation in the molecules.

$\ce{NO}$ for example has a bond order of 2.5 which you cannot deduce from the structural formula. Therefore, I would say it makes sense to avoid drawing such structural formula altogether, if possible, and rather stick to $\ce{NO}$ as a graphical representation.

In order to get an idea of the bond situation of $\ce{NO}$, it is worthwhile to study the molecular orbital diagram of the molecule.

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