Question

What is the Lewis structure of `Cl_3`?

Answer 1

Explanation:

(Assuming that `Cl_3^-` was meant by `Cl_3`.)

1. The total valence electrons for `Cl_3^-` is `7xx3+1=22`, since `Cl` has `7` valence electrons and there's a negative charge (which adds `1` electron).
2. Let's draw a single bond between the `3` `Cl` 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 `Cl` atoms:
   
4. We still have `2` electrons left after placing all those valence electrons. `Cl` 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 `Cl` atom, then the formal charge on that `Cl` atom would be `7-9=-2`.
7. If it were placed on the right `Cl` atom, then that `Cl` atom would have a formal charge of `7-9=-2`.
8. If it were placed on the central `Cl` atom, then that `Cl` atom would have a formal charge of `7-8=-1`.
9. So, the best placement for that extra electron pair would be the central `Cl`, 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. :)