Final answer:
The correct Lewis structure for CHBr3 shows that the central carbon atom is bonded to one hydrogen and three bromine atoms. There are no lone pairs on the hydrogen atom and three lone pairs on each of the three bromine atoms, for a total of 9 lone pairs in the structure.
Step-by-step explanation:
The molecule CHBr3, also known as bromoform, has a central carbon atom bonded to one hydrogen and three bromine atoms. The Lewis structure for CHBr3 requires us to know the number of valence electrons for each atom involved. Carbon has 4 valence electrons, hydrogen has 1, and each bromine atom has 7. Carbon and hydrogen will share one pair of electrons to form a single bond, while each bromine atom will also form a single bond with carbon, sharing a pair of electrons for each bond.
Bromine, being a member of the halogen group, typically has one single bond when it forms compounds, which leaves it with three lone pairs of electrons per bromine atom. There are therefore 3 bromine atoms each with 3 lone pairs, totaling 9 lone pairs on the bromines. The hydrogen atom does not have lone pairs. Summing up, CHBr3 has a total of 9 lone pairs of electrons in the correct Lewis structure.
The best way to visualize this is by creating the Lewis structure: Carbon in the center with four electrons, hydrogen with one, and each bromine with seven, then draw single bonds between C-H and C-Br, placing three lone pairs around each Br atom.