The Correct Answer and Explanation is:
The correct Lewis structure for hypobromous acid, HBrO, is the second option, which shows the atoms connected in the order H-O-Br.
Here is a detailed explanation of why this structure is correct:
To determine the correct Lewis structure, we follow a systematic process.
First, we calculate the total number of valence electrons for the molecule. Hydrogen (H) is in Group 1 and has 1 valence electron. Oxygen (O) is in Group 16 and has 6 valence electrons. Bromine (Br) is in Group 17 and has 7 valence electrons. The total count is 1 + 6 + 7 = 14 valence electrons. Any valid Lewis structure must account for exactly 14 electrons.
Second, we determine the arrangement of the atoms. For oxyacids, which are acids containing hydrogen, oxygen, and another element, a very common and stable arrangement is for the acidic hydrogen atom to be bonded to an oxygen atom. This means the skeletal structure is H-O-Br, with oxygen as the central atom. This is a more stable configuration than having bromine as the central atom (H-Br-O).
Third, we place the electrons. We start by drawing single bonds to connect the atoms in the H-O-Br skeleton. Each single bond uses 2 electrons, so the two bonds (H-O and O-Br) use a total of 4 electrons. We have 14 – 4 = 10 electrons remaining.
Fourth, we distribute the remaining electrons as lone pairs to satisfy the octet rule for each atom (or duet rule for hydrogen). We start with the outer atoms. Bromine needs 6 more electrons to complete its octet, so we add three lone pairs to it. This uses 6 of the remaining 10 electrons. We now have 4 electrons left. These final 4 electrons are placed on the central oxygen atom as two lone pairs.
This final structure, H-O-Br with two lone pairs on oxygen and three on bromine, satisfies all the rules. Hydrogen has a duet (2 electrons), oxygen has an octet (2 bonding pairs + 2 lone pairs = 8 electrons), and bromine has an octet (1 bonding pair + 3 lone pairs = 8 electrons). Furthermore, the formal charges for all atoms are zero, indicating a highly stable structure. The first option is incorrect because the connectivity is less likely, and the third is incorrect because hydrogen cannot have a lone pair as it violates the duet rule.
