To draw the Lewis structure for sulfur dibromide (SBr₂), we start by determining the total number of valence electrons. Sulfur has 6 valence electrons, and each bromine (Br) atom has 7 valence electrons. Since there are two bromine atoms, we calculate the total as follows:
- Valence electrons from sulfur: 6
- Valence electrons from two bromine atoms: 2 × 7 = 14
- Total valence electrons: 6 + 14 = 20
Next, we place the sulfur atom in the center and surround it with the two bromine atoms. Initially, we connect sulfur to each bromine with a single bond, which uses up 4 of the total 20 valence electrons (2 electrons per bond). At this point, we have 16 valence electrons remaining.
Now, we distribute the remaining electrons to complete the octets for the bromine atoms. Each bromine atom needs 6 more electrons (3 lone pairs) to complete its octet. This allocation uses 12 of the 16 remaining electrons, leaving us with 4 electrons.
The last 4 electrons can be placed on the sulfur atom as 2 lone pairs, which allows sulfur to have an expanded octet (a common property for third-period elements). Now, the completed Lewis structure looks like this:
In summary, the Lewis structure for SBr₂ shows sulfur bonded to two bromine atoms with two lone pairs on the sulfur and three lone pairs on each bromine. This arrangement satisfies the octet rule for the bromine atoms and accommodates sulfur’s ability to hold more than eight electrons in its valence shell. The molecule is bent due to the repulsion between the lone pairs, which affects its shape and polarity.