# A matrix formulation for Lewis structures

Any potential Lewis structure — valid or not — can be represented in terms of a matrix, and the properties of the structure can be read off as properties of that matrix:

Suppose you have a built a Lewis structure for a molecule composed of $n$ atoms. You can form an $n\times n$ matrix with one row and one column for each atom. Fill in the entries of the matrix with 0, 1, 2, or 3 depending on whether the structure assigns no bond, a single bond, a double bond, or a triple bond between the atom in that row and the atom in that column. Now according to this scheme, the diagonal entries of the matrix would be places where an atom bonds to itself, which doesn't make sense — instead, fill in the diagonal elements of the matrix with the number of non-bonding electrons that the Lewis structure assigns to that atom. Altogether, I call the resulting matrix the electron matrix, $E$. (Question: what quantity do you obtain if you add up all the entries in this matrix?)

$$E_{i,j} = \begin{cases}\text{bond strength between $i$ and j},&&\text{if i\neq j} \\ \text{# of non-bonding electrons at atom i}&&\text{if i=j.}\end{cases}$$

Example (Electron matrix): Nitrous oxide (N2O) has several possible Lewis structures, two of which are described by the following electron matrices (where the columns/rows correspond to atoms of nitrogen, nitrogen, and oxygen in that order.):

$$E_1 = \begin{bmatrix}2 & 3 & 0 \\ 3 & 0 & 1 \\ 0 & 1 & 6\end{bmatrix}, \quad E_2 = \begin{bmatrix}4 & 2 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & 4\end{bmatrix}$$

For practice interpreting such configurations, consider the first matrix. By reading the first row/column, we see that the first nitrogen has two (2) nonbonding electrons itself, that it is triple-bonded (3) to the other nitrogen atom, and that it is not bonded (0) to the oxygen atom. The second row/column tells us that the second nitrogen atom is triple bonded (3) to the first, that it has no non-bonding electrons itself (0), and that it is single-bonded (1) to the oxygen atom. Finally, in the third row/column, we discover that the oxygen atom is not bonded to the first nitrogen atom (0), that it's single-bonded (1) to the second nitrogen atom, and that it has six (6) non-bonding electrons.

For our purposes, we will also need a second $n\times n$ matrix which encodes the properties of the molecular components only and does not depend on the Lewis structure itself: the atomic valence matrix $V$ is a matrix whose diagonal entries are the number of valence electrons of the atom in that row (or, equivalently, column); all other entries are 0.

Example (Atomic valence matrix): The atomic valence matrix of nitrous oxide (N2O) is always

$$V = \begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6\end{bmatrix},$$

corresponding to the accessible electrons in the valence shells of nitrogen, nitrogen, and oxygen. Because $V$ encodes only atomic properties and not properties of the overall molecule, this matrix will be the same regardless of which N2O Lewis structures you're considering .

Now according to the theory, the physical feasibility of a given Lewis structure depends on two criteria: first, it must account for all the valence electrons provided by the individual atoms (if it uses too many or too few, the charge of the molecule will come out wrong.) Second, it must ensure that the formal charges of the atoms in the structure are as small in magnitude as possible, which means that each atom in the molecule “has access to” about as many electrons as it had in its valence shell before bonding to anything.

Interestingly enough, the atomic valence and electron matrices can help us account for these criteria: if you take their difference, you get what I call the charge matrix $C = V - E,$ which has the following remarkable properties:

1. The sum of the entries in any row or column in $C$ is the formal charge on that atom.
2. The sum of all the entries in $C$ is the charge of the molecule.

Example (Charge matrix): From our earlier examples with N2O, we have the atomic valence matrix

$$V = \begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6\end{bmatrix},$$

and two possible electron matrices:

$$E_1 = \begin{bmatrix}2 & 3 & 0 \\ 3 & 0 & 1 \\ 0 & 1 & 6\end{bmatrix}, \quad E_2 = \begin{bmatrix}4 & 2 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & 4\end{bmatrix}$$.

Taking the first electron matrix as an example, we obtain the following charge matrix:

$$\begin{eqnarray*}C_1 &=& V - E_1\\ &=& \begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6\end{bmatrix} - \begin{bmatrix}2 & 3 & 0 \\ 3 & 0 & 1 \\ 0 & 1 & 6\end{bmatrix}\\ &=& \begin{bmatrix}3 & -3 & 0 \\ -3 & 5 & -1 \\ 0 & -1 & 0\end{bmatrix}\end{eqnarray*}.$$

And look—by taking row/column totals, we find that the formal charges of the atoms are respectively 0, +1, and -1. By adding up these formal charges (or, equivalently, by adding up all the entries in the charge matrix), we find that the overall charge on the molecule is 0.

So this is the general result — why does it matter? On one view, we are just using a kind of adjacency matrix for book-keeping purposes when we would ordinarily use electron dot diagrams. But on another view, matrices are rather nice to work with in general, and (as we have already seen in one case) they produce succinct formulas like $C = V - E$ which, through a straightforward calculation, help you to produce a number of useful quantities simultaneously.

Afterword: I plan to extend these ideas more as I work on them. In particular, it's interesting to extend electron matrices (which we have used here only for molecules with exclusively covalent bonds) to the case where some bonds have more polar character. In this case, the matrix ceases to be symmetric — matrix symmetry corresponds to covalence(!). Computing formal charges with these new matrices still works — and in fact will produce more realistic values.

Second, I notice that the charge matrix looks like a discrete Laplacian operator, which suggests that it could be involved in a kind of diffusion equation (the diffusion of charge across a molecule); I wonder what the equilibrium solutions could be?

I also suspect that these matrices represent a sort of pixelated model of the actual quantum mechanics involved with the formation of bonds — I wonder how close the correspondence is?