To draw the Lewis structure of tellurium tetrachloride (TeCl4), we first need to determine the total number of valence electrons available for bonding.
Tellurium (Te) is in group 16, so it has 6 valence electrons. Each chlorine (Cl) atom, being in group 17, has 7 valence electrons. Since there are four chlorine atoms, the total number of valence electrons from chlorine is 4 x 7 = 28 electrons. Therefore, the total number of valence electrons for TeCl4 is:
6 (from Te) + 28 (from Cl) = 34 valence electrons.
Next, we place the central atom, which is Te, in the center and surround it with the four chlorine atoms. We connect each Cl atom to Te with a single bond. Each bond accounts for 2 electrons, so with four bonds:
4 bonds x 2 electrons = 8 electrons used.
We then subtract these 8 electrons from the total number of valence electrons:
34 – 8 = 26 electrons remaining.
Now, we distribute the remaining electrons to the outer atoms (the Cl atoms) to satisfy their octet. Each Cl atom needs 8 electrons to be stable. Each Cl already has 2 from the bond with Te, so we need to add 6 more to each Cl atom. There are four Cl atoms:
4 Cl x 6 electrons = 24 electrons used for the outer atoms.
26 (remaining electrons) – 24 (used for Cl) = 2 electrons left, which we place on the central atom (Te).
This gives us the complete Lewis structure with each chlorine atom holding 8 electrons in total and Te having 2 lone electrons. Now we evaluate formal charges:
For Te: Valence electrons = 6; Non-bonding electrons = 2; Bonding electrons = 8 (4 bonds x 2 electrons), so:
Formal charge = 6 – (2 + 8/2) = 6 – 2 – 4 = 0.
For each Cl: Valence electrons = 7; Non-bonding electrons = 6; Bonding electrons = 2 (from 1 bond), so:
Formal charge = 7 – (6 + 2/2) = 7 – 6 – 1 = 0.
Since all formal charges are zero, the structure of TeCl4 is stable, confirming our Lewis structure is correct.