Which of the following statements correctly describes the Lewis dot structure of the neutral compound BrF_3?

• A. BrF_3 has two single bonds and one double bond between Br and F atoms, and two lone pairs on the Br atom.
• B. BrF_3 has three double bonds between Br and F atoms, and no lone pairs on the Br atom.
• C. BrF_3 has three single bonds between Br and F atoms, and two lone pairs on the Br atom.
• D. BrF_3 has three single bonds between Br and F atoms, and one lone pair on the Br atom.

Answer: (C)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons in BrF_3. Bromine (Br) has 7 valence electrons, and each fluorine (F) atom has 7 valence electrons. Since there are three fluorine atoms, calculate the total as: \$7 + 3 \times 7$.
2. Step 2: Draw the skeletal structure with bromine as the central atom bonded to three fluorine atoms by single bonds. Each single bond represents 2 electrons shared between atoms.
3. Step 3: Distribute the remaining valence electrons to complete the octets of the fluorine atoms first, since fluorine is more electronegative and typically completes its octet with lone pairs.
4. Step 4: Assign any leftover electrons to the central bromine atom as lone pairs. Remember that bromine can have an expanded octet because it is in period 4 of the periodic table.
5. Step 5: Count the number of bonds and lone pairs on bromine. Verify that bromine has three single bonds to fluorine atoms and two lone pairs, which matches the correct Lewis structure for BrF_3.