Which of the following best describes the electron domain geometry and molecular geometry of the ion BrF4−?

• A. Electron domain geometry: trigonal bipyramidal; Molecular geometry: seesaw
• B. Electron domain geometry: tetrahedral; Molecular geometry: trigonal pyramidal
• C. Electron domain geometry: square pyramidal; Molecular geometry: square pyramidal
• D. Electron domain geometry: octahedral; Molecular geometry: square planar

Answer: (D)

Verified step by step guidance

1. Determine the total number of valence electrons for the BrF4⁻ ion. Bromine (Br) has 7 valence electrons, each fluorine (F) has 7 valence electrons, and the negative charge adds 1 extra electron. Calculate the sum: \$7 + 4 \times 7 + 1$.
2. Draw the Lewis structure for BrF4⁻ using the total valence electrons. Place bromine as the central atom bonded to four fluorine atoms. Distribute the remaining electrons to satisfy the octet rule, and identify any lone pairs on the central atom.
3. Count the number of electron domains around the central bromine atom. Electron domains include bonding pairs and lone pairs. This total number determines the electron domain geometry according to VSEPR theory.
4. Use the number of electron domains to identify the electron domain geometry. For example, 6 electron domains correspond to an octahedral electron domain geometry.
5. Determine the molecular geometry by considering the positions of the atoms only (ignoring lone pairs). For BrF4⁻, with 6 electron domains and 2 lone pairs, the molecular geometry is square planar.