The cations formed upon dissolving SnF₄ and AuF₃ in liquid BrF₃ separately, respectively are

Q3 · CSIR-NET Chemistry, June 2019

Paper: CSIR-NET June 2019 · Subject: Inorganic Chemistry · Chapter: Acids and Bases · Topic: Acid and Bases – General · Marks: 2 · Difficulty: Medium

The cations formed upon dissolving $\mathrm{SnF_4}$ and $\mathrm{AuF_3}$ in liquid $\mathrm{BrF_3}$ separately, respectively are

(a)$\mathrm{SnF_3^+}$ and $\mathrm{BrF_2^+}$

(b)$\mathrm{BrF_2^+}$ and $\mathrm{AuF_2^+}$

(c)$\mathrm{BrF_2^+}$ only

(d)$\mathrm{SnF_3^+}$ and $\mathrm{AuF_2^+}$

Answer

Answer: C ✓ checked by 4AB · confidence high

Explanation

Both SnF₄ and AuF₃ act as fluoride acceptors in BrF₃ (giving SnF₆²⁻ and AuF₄⁻), so the only cation formed is the solvent cation BrF₂⁺ (c).

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