Question

Which of the following are iso-structural with \( \text{SF}_4 \)?

• A. \( \text{IF}_4^+ \)
• B. \( \text{BrF}_4^+ \)
• C. \( \text{XeO}_2\text{F}_2 \)
• D. \( \text{CH}_4 \)
• E. \( \text{XeF}_4 \)

Hint:

Iso-structural compounds have the same geometry and the same number of bonding and lone pairs. Pay attention to the electron pair arrangement and molecular geometry to determine if compounds are iso-structural.

Answer 1

The Correct Option is A, C

To find the compounds iso-structural with \( \text{SF}_4 \), we need to look for compounds that have the same number of electron pairs and the same geometry. - \( \text{SF}_4 \) has a trigonal bipyramidal structure with one lone pair in the equatorial plane. This gives it a "see-saw" shape.
- For \( \text{IF}_4^+ \), the structure is also trigonal bipyramidal with a lone pair in the equatorial plane, so it is iso-structural with \( \text{SF}_4 \).
- Similarly, \( \text{BrF}_4^+ \) has a similar structure to \( \text{SF}_4 \) and is also iso-structural.
- \( \text{XeO}_2\text{F}_2 \) does not have the same geometry and electron pair arrangement as \( \text{SF}_4 \), so it is not iso-structural.
- \( \text{CH}_4 \) has a tetrahedral structure, which is different from the trigonal bipyramidal structure of \( \text{SF}_4 \), so it is not iso-structural.
- \( \text{XeF}_4 \) has a square planar structure, which is different from \( \text{SF}_4 \).
Step 1: Conclusion.
The compounds that are iso-structural with \( \text{SF}_4 \) are \( \text{IF}_4^+ \) (A), and \( \text{BrF}_4^+ \) (B). Therefore, the correct option is (C) A & C only.
Final Answer: (C) A & C only