Question

Which of the following molecular geometry is distorted geometry from their ideal geometry.

• A. $\text{PCl}_3\text{F}_2$
• B. $\text{SOF}_4$
• C. $\text{XeF}_5^{\ominus}$
• D. $\text{XeO}_3\text{F}_2$

Answer 1

Answer: (C)

C

Answer 2

To determine if a molecular geometry is distorted from its ideal electron geometry, we apply VSEPR theory. A "distorted geometry" typically refers to a molecular shape that differs in name from its parent electron domain geometry due to the presence of lone pairs on the central atom. While all real molecules experience some degree of distortion from ideal bond angles due to differing bond lengths, electronegativities, or multiple bonds, the term "distorted geometry" in VSEPR context usually implies a change in the fundamental shape caused by lone pairs.

Let's analyze each option:

(A) $\text{PCl}_3\text{F}_2$

• Central atom: P
• Valence electrons on P: 5
• Bonds: P forms 3 single bonds with Cl and 2 single bonds with F.
• Number of bonding pairs (BP): 5
• Number of lone pairs (LP): (5 - 5)/2 = 0
• Total electron domains: 5 (5 BP + 0 LP)
• Electron geometry: Trigonal Bipyramidal (TBP)
• Molecular geometry: Trigonal Bipyramidal. According to Bent's rule and VSEPR, more electronegative atoms (F) prefer axial positions, and less electronegative atoms (Cl) prefer equatorial positions. So, the F atoms will be axial, and Cl atoms will be equatorial. The overall shape remains TBP. While bond angles might deviate slightly from ideal 90° and 120° due to different substituents, the molecular geometry name is not changed.

(B) $\text{SOF}_4$

• Central atom: S
• Valence electrons on S: 6
• Bonds: S forms 1 double bond with O and 4 single bonds with F.
• Number of bonding domains: 1 (S=O) + 4 (S-F) = 5
• Number of lone pairs (LP): (6 - 2(for S=O) - 4(for S-F))/2 = 0
• Total electron domains: 5 (5 BP + 0 LP)
• Electron geometry: Trigonal Bipyramidal (TBP)
• Molecular geometry: Trigonal Bipyramidal. The double bond (S=O) will occupy an equatorial position, and the F atoms will occupy the remaining positions (axial and equatorial). The presence of the double bond will cause distortions in bond angles, but the fundamental TBP shape is retained.

(C) $\text{XeF}_5^{\ominus}$

• Central atom: Xe
• Valence electrons on Xe: 8
• Charge: -1 (add 1 electron)
• Total electrons for VSEPR: 8 + 1 = 9
• Bonds: Xe forms 5 single bonds with F.
• Number of bonding pairs (BP): 5
• Number of lone pairs (LP): (9 - 5)/2 = 2
• Total electron domains: 7 (5 BP + 2 LP)
• Electron geometry: Pentagonal Bipyramidal
• Molecular geometry: For 7 electron domains with 2 lone pairs, the lone pairs occupy the axial positions (to maximize their separation, 180° apart). This forces the 5 bonding pairs (F atoms) into the equatorial plane, forming a Pentagonal Planar molecular geometry. This is a clear case where the presence of lone pairs changes the molecular geometry name from its ideal electron geometry. Thus, it is a distorted geometry.

(D) $\text{XeO}_3\text{F}_2$

• Central atom: Xe
• Valence electrons on Xe: 8
• Bonds: Xe forms 3 double bonds with O and 2 single bonds with F.
• Number of bonding domains: 3 (Xe=O) + 2 (Xe-F) = 5
• Number of lone pairs (LP): (8 - 32(for Xe=O) - 21(for Xe-F))/2 = (8 - 6 - 2)/2 = 0
• Total electron domains: 5 (5 BP + 0 LP)
• Electron geometry: Trigonal Bipyramidal (TBP)
• Molecular geometry: Trigonal Bipyramidal. The double bonds (Xe=O) will prefer equatorial positions, and the more electronegative F atoms will prefer axial positions. The overall shape remains TBP. Similar to (B), there will be angular distortions due to double bonds, but the molecular geometry name is not changed.

Conclusion: Among the given options, $\text{XeF}_5^{\ominus}$ is the only one where the molecular geometry (pentagonal planar) is fundamentally different from its ideal electron geometry (pentagonal bipyramidal) due to the presence of lone pairs. This is the common interpretation of "distorted geometry" in VSEPR theory.