Question

Is $BrF_5$ and $XeOF_4$ isostructural or not?

• A. Yes, they are isostructural.
• B. No, they are not isostructural.

Answer 1

Answer: (A)

Yes, they are isostructural.

Answer 2

To determine if $BrF_5$ and $XeOF_4$ are isostructural, we analyze their molecular structures using VSEPR theory. Isostructural species have the same molecular geometry and the same number of valence electrons.

Analysis of $BrF_5$:

• Central Atom: Bromine (Br)
• Valence Electrons: Br (Group 17) has 7, F (Group 17) has 7.
• Total Valence Electrons: $7 (\text{Br}) + 5 \times 7 (\text{F}) = 42$.
• Electron Domains: 5 bonding pairs (Br-F) + 1 lone pair on Br = 6 electron domains.
• Hybridization: $sp^3d^2$.
• Electron Geometry: Octahedral.
• Molecular Geometry: Square Pyramidal ($AX_5E_1$).

Analysis of $XeOF_4$:

• Central Atom: Xenon (Xe)
• Valence Electrons: Xe (Group 18) has 8, O (Group 16) has 6, F (Group 17) has 7.
• Total Valence Electrons: $8 (\text{Xe}) + 6 (\text{O}) + 4 \times 7 (\text{F}) = 8 + 6 + 28 = 42$.
• Electron Domains: 4 bonding pairs (Xe-F) + 1 double bond (Xe=O) + 1 lone pair on Xe = 6 electron domains.
• Hybridization: $sp^3d^2$.
• Electron Geometry: Octahedral.
• Molecular Geometry: Square Pyramidal ($AX_5E_1$, considering the double bond as one group).

Comparison: Both $BrF_5$ and $XeOF_4$ have 42 valence electrons, $sp^3d^2$ hybridization, octahedral electron geometry, and crucially, the same Square Pyramidal molecular geometry. Therefore, they are isostructural.