# How is the Lewis structure of an ion written?

Jul 31, 2015

The same way you draw the structure of the original molecule or atom, except with $X$ more or $X$ less electrons, depending on the charge.

i.e. if charge is $- 2$, then add $2$ electrons (because the charge of an electron is $- 1.602 \times {10}^{- 19} C$ and is thus the same sign as the charge), and vice versa for charges of $\pm X$.

So if you use the "counting valence electrons" method, you can draw $S {O}_{4}^{2 -}$ by saying that:

$S : 6$
${O}_{4} : 6 \cdot 4 = 24$
$\text{From 2- charge: } 2$

Total: $6 + 24 + 2 = \textcolor{g r e e n}{32}$

Count the electrons in here:

where the above is all of the resonance structures of $S {O}_{4}^{2 -}$.

With formal charge defined here as $\text{Actual " e^(-) - "owned } {e}^{-}$:

"Owned" electrons:

• $6$ per thionyl oxygen ($S = O$), so $12$ total.
• $7$ per oxygen with a formal charge of $- 1$ (from $6 - 7$), so $14$ total.
• $6$ per sulfur, so $6$ total.

Sure enough, it matches:
$12 + 14 + 6 = \textcolor{g r e e n}{32}$