To draw the Lewis structure for BBr3 (Boron Tribromide), we start by determining the total number of valence electrons. Boron (B) has 3 valence electrons, and each Bromine (Br) has 7 valence electrons. Since there are three Bromine atoms, we calculate:
Valence electrons = 3 (from B) + 3 × 7 (from Br) = 3 + 21 = 24 valence electrons.
Next, we place the Boron atom in the center because it is the least electronegative element. We then surround Boron with the three Bromine atoms. Each Bromine will form a single bond with Boron, using 2 electrons per bond:
B—Br
B—Br
B—Br
Now, we have used 6 electrons for the three B—Br bonds. Next, we need to place the remaining electrons. After the bonds, we have:
24 total – 6 used = 18 valence electrons left.
Each Bromine needs to complete its octet. We will place 6 electrons around each Bromine, forming three lone pairs on each Bromine atom.
This is how the Lewis structure will look:
In this structure, Boron has 3 bonds but only 6 electrons in its valence shell, which is acceptable for Boron as it often has an incomplete octet. Each Bromine atom has 8 electrons, thereby satisfying the octet rule.
In conclusion, the Lewis structure for BBr3 is adequately represented by a central Boron atom bonded to three Bromine atoms, with each Bromine carrying three lone pairs of electrons.