SBr2 (sulfur dibromide) has one sulfur atom and two bromine atoms. In the lewis structure of SBr2, there are two single bonds around the sulfur atom, with two bromine atoms attached to it. Each bromine atom has three lone pairs, and the sulfur atom has two lone pairs.
Here’s how you can draw the SBr2 lewis structure step by step.
Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges (if there are)
Let’s break down each step in detail.
#1 Draw Sketch
- First, determine the total number of valence electrons
Hence, sulfur has six valence electrons and bromine has seven valence electrons.
Since SBr2 has one sulfur atom and two bromine atoms, so…
Valence electrons of one sulfur atom = 6 × 1 = 6
Valence electrons of two bromine atoms = 7 × 2 = 14
And the total valence electrons = 6 + 14 = 20
- Second, find the total electron pairs
We have a total of 20 valence electrons. And when we divide this value by two, we get the value of total electron pairs.
Total electron pairs = total valence electrons ÷ 2
So the total electron pairs = 20 ÷ 2 = 10
- Third, determine the central atom
We have to place the least electronegative atom at the center.
Since sulfur is less electronegative than bromine, assume that the central atom is sulfur.
Therefore, place sulfur in the center and bromines on either side.
- And finally, draw the rough sketch
#2 Mark Lone Pairs
Here, we have a total of 10 electron pairs. And two S — Br bonds are already marked. So we have to only mark the remaining eight electron pairs as lone pairs on the sketch.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are bromines.
So for each bromine, there are three lone pairs, and for sulfur, there are two lone pairs.
Mark the lone pairs on the sketch as follows:
#3 Mark Charges
Use the following formula to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
For sulfur atom, formal charge = 6 – 4 – ½ (4) = 0
For each bromine atom, formal charge = 7 – 6 – ½ (2) = 0
Here, both sulfur and bromine atoms do not have charges, so no need to mark the charges.
And in the above structure, you can see that the central atom (sulfur) forms an octet. Hence, the octet rule is satisfied.
Therefore, this structure is the stable lewis structure of SBr2.
Next: C3H6 Lewis Structure