Isostructural Species with Sulfur Tetrafluoride

To determine which of the given species are isostructural with $\text{SF}_4$, we need to analyze their hybridization, number of bond pairs, number of lone pairs, and resulting molecular geometries using VSEPR (Valence Shell Electron Pair Repulsion) theory.

Two species are considered isostructural if they have the same molecular shape and geometry.

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1. Structure of $\text{SF}_4$:

• Central atom: Sulfur ($\text{S}$) has $6$ valence electrons.
• Number of $\sigma$-bonds: $4$ (with four $\text{F}$ atoms).
• Number of lone pairs: $\frac{6 - 4}{2} = 1$ lone pair.
• Steric Number: $\text{Bond pairs} + \text{Lone pairs} = 4 + 1 = 5$.
• Hybridization: $sp^3d$
• Electron Geometry: Trigonal bipyramidal
• Molecular Shape: See-saw

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2. Analysis of the Given Options:

• A. $\text{BrF}_4^-$:

  • Valence electrons on $\text{Br}^-$: $7 + 1 = 8$.
  • Number of $\sigma$-bonds: $4$.
  • Lone pairs: $\frac{8 - 4}{2} = 2$.
  • Steric number: $4 + 2 = 6$ ($sp^3d^2$ hybridization).
  • Molecular Shape: Square planar

• B. $\text{CH}_4$:

  • Valence electrons on $\text{C}$: $4$.
  • Number of $\sigma$-bonds: $4$.
  • Lone pairs: $0$.
  • Steric number: $4 + 0 = 4$ ($sp^3$ hybridization).
  • Molecular Shape: Tetrahedral

• C. $\text{IF}_4^+$:

  • Valence electrons on $\text{I}^+$: $7 - 1 = 6$.
  • Number of $\sigma$-bonds: $4$.
  • Lone pairs: $\frac{6 - 4}{2} = 1$.
  • Steric number: $4 + 1 = 5$ ($sp^3d$ hybridization).
  • Molecular Shape: See-saw

• D. $\text{XeF}_4$:

  • Valence electrons on $\text{Xe}$: $8$.
  • Number of $\sigma$-bonds: $4$.
  • Lone pairs: $\frac{8 - 4}{2} = 2$.
  • Steric number: $4 + 2 = 6$ ($sp^3d^2$ hybridization).
  • Molecular Shape: Square planar

• E. $\text{XeO}_2\text{F}_2$:

  • Valence electrons on $\text{Xe}$: $8$.
  • Electrons used in bonding: $2 \times 2\ (\text{for two } \text{O=Xe} \text{ double bonds}) + 2 \times 1\ (\text{for two } \text{F-Xe} \text{ single bonds}) = 6$ electrons.
  • Number of $\sigma$-bonds: $2\ (\text{with } \text{O}) + 2\ (\text{with } \text{F}) = 4$.
  • Lone pairs: $\frac{8 - 6}{2} = 1$.
  • Steric number: $4 + 1 = 5$ ($sp^3d$ hybridization).
  • Molecular Shape: See-saw

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Conclusion:

Both $\text{IF}_4^+$ (C) and $\text{XeO}_2\text{F}_2$ (E) have a see-saw geometry with $sp^3d$ hybridization, making them isostructural with $\text{SF}_4$.

Thus, the correct answer is C and E Only.