The reaction between SbF5 and two equivalents of HF leads to the formation of

ABC26IN1273 · Main Group Elements

Subject: Inorganic Chemistry · Chapter: Main Group Elements · Topic: Main Group Elements – General · Exam: NET DEC 2014 · Marks: 2 · Difficulty: Easy

The reaction between $\mathrm{SbF}_{5}$ and two equivalents of HF leads to the formation of

(a)$\mathrm{H}_{2} \mathrm{SbF}_{3}+2 \mathrm{F}_{2}$

(b)$\mathrm{HSbF}_{2}+3 \mathrm{F}_{2}$

(c)$\mathrm{SbF}_{3}+\mathrm{H}_{2}+2 \mathrm{F}_{2}$

(d)$[\mathrm{SbF_6}]^{-}[\mathrm{H_2F}]^{+}$

Answer

Answer (as printed): D

Explanation

No explanation was printed for this question.