Question

Dissolution of SbF\(_5\) in BrF\(_3\) produces

• A. BrF\(_5\) and SbF\(_3\)
• B. BrF and SbF\(_7\)
• C. [BrF\(_2\)]\(^{+}\) [SbF\(_6\)]\(^{-}\)
• D. [SbF\(_4\)]\(^{+}\) [BrF\(_4\)]\(^{-}\)

Hint:

The dissolution of SbF\(_5\) in BrF\(_3\) results in the formation of an ionic pair: [BrF\(_2\)]\(^{+}\) and [SbF\(_6\)]\(^{-}\).

Answer 1

The Correct Option is C

Step 1: Understand the dissolution process.
When SbF\(_5\) dissolves in BrF\(_3\), it dissociates to form ionic pairs. The SbF\(_5\) acts as a Lewis acid, while BrF\(_3\) can donate a fluoride ion. The result is the formation of [BrF\(_2\)]\(^{+}\) and [SbF\(_6\)]\(^{-}\).
Step 2: Explanation of options.
- Option (A): This is incorrect because SbF\(_5\) and SbF\(_3\) do not form as products.
- Option (B): This option is incorrect as BrF and SbF\(_7\) are not formed.
- Option (C): This is the correct option. The dissolution produces [BrF\(_2\)]\(^{+}\) and [SbF\(_6\)]\(^{-}\).
- Option (D): This is incorrect because [SbF\(_4\)]\(^{+}\) and [BrF\(_4\)]\(^{-}\) are not the products.
Step 3: Conclusion.
The correct products are [BrF\(_2\)]\(^{+}\) and [SbF\(_6\)]\(^{-}\).
Final Answer: \[ \boxed{[BrF_2]^+ [SbF_6]^-} \]