# What is the Lewis structure of Cl_3?

May 13, 2018 #### Explanation:

(Assuming that $C {l}_{3}^{-}$ was meant by $C {l}_{3}$.)

1. The total valence electrons for $C {l}_{3}^{-}$ is $7 \times 3 + 1 = 22$, since $C l$ has $7$ valence electrons and there's a negative charge (which adds $1$ electron).
2. Let's draw a single bond between the $3$ $C l$ atoms to begin: Now, we have $22 - 4 = 18$ electrons left to put on the diagram, since each single bond counts as $2$ electrons.
3. Then, we should complete the octets for all of the $C l$ atoms: 4. We still have $2$ electrons left after placing all those valence electrons. $C l$ can have an expanded octet, though, so that's not too much of a problem.
5. To determine where to put this extra electron pair, let's calculate the formal charges of all the situations:
6. If the extra electron pair is placed on the left $C l$ atom, then the formal charge on that $C l$ atom would be $7 - 9 = - 2$.
7. If it were placed on the right $C l$ atom, then that $C l$ atom would have a formal charge of $7 - 9 = - 2$.
8. If it were placed on the central $C l$ atom, then that $C l$ atom would have a formal charge of $7 - 8 = - 1$.
9. So, the best placement for that extra electron pair would be the central $C l$, since that arrangement minimises formal charge.
10. Let's draw that: 11. And add a bracket around everything, with a negative sign, to indicate that this is a negatively charged ion: 12. Finally, let's do a last double-check for the number of electrons. We should end up with $22$ electrons. :)