To draw the Lewis structure of tin(II) chloride (SnCl2), we start by determining the total number of valence electrons available in the molecule. Tin (Sn) is in group 14 of the periodic table, so it has 4 valence electrons. Chlorine (Cl), being in group 17, has 7 valence electrons, and since there are two chlorine atoms, we have a total of 14 (2 × 7) valence electrons from chlorine.
Adding these together gives us:
4 (from Sn) + 14 (from 2 Cl) = 18 valence electrons.
Next, we start by placing the tin atom in the center, as it is the less electronegative atom compared to chlorine. We then surround it with two chlorine atoms. With the Sn atom at the center, we initially place two single bonds – one between Sn and each Cl:
Cl
|
Sn — Cl
Now, each bond accounts for 2 electrons (one from Sn and one from Cl), using up 4 of our 18 total valence electrons. This leaves us with:
18 – 4 = 14 valence electrons remaining.
Next, we need to fulfill the octet rule for the chlorine atoms. Each chlorine atom will require 6 more electrons to complete its octet. We do this by placing 3 lone pairs on each chlorine atom:
Cl: ..
|
Sn — Cl:
..
This uses up an additional 12 valence electrons (6 for each Cl), leaving:
14 – 12 = 2 valence electrons.
Finally, the remaining 2 electrons can be added as a lone pair on the tin atom:
Cl: ..
|
Sn : — Cl:
..
Now, the Sn atom has a total of 4 electrons (from the two bonds with Cl) + 2 lone electrons = 6 electrons around it, which is acceptable since Sn can exceed the octet rule. In summary, the Lewis structure of SnCl2 shows each chlorine atom fulfilling its octet, while tin has more than 4 electrons, consistent with its position in the periodic table.