MCQ
Silicon dissolves in excess of $\ce{HF}$ due to formation of
- A$SiF_4$
- B$SiH_4$
- C$H_2SiF_6$
- D$H_2SiF_4$
✓
Answer
$\ce{HF}$ is a very strong acid and can also react with silicon and corrode its lattice
$\mathrm{Si}+4 \mathrm{HF} \rightarrow \mathrm{SiF}_{4}+4 \mathrm{H}^{+}$
$\downarrow \mathrm{HF}$
$\mathrm{H}_{2} \mathrm{SiF}_{6}$
$\mathrm{Si}+4 \mathrm{HF} \rightarrow \mathrm{SiF}_{4}+4 \mathrm{H}^{+}$
$\downarrow \mathrm{HF}$
$\mathrm{H}_{2} \mathrm{SiF}_{6}$
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