$\mathrm{CuSO}_{4}$ reacts with excees $\mathrm{KCN}$ solution to form:
- $\mathrm{Cu}(\mathrm{CN})$
- $\mathrm{Cu}(\mathrm{CN})_{2}$
- $\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}ight]$
- $\mathrm{K}_{4}\left[\mathrm{Cu}(\mathrm{CN})_{6}ight]$
Solution
$2 \mathrm{~K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}ight]+2 \mathrm{~K}_{2} \mathrm{SO}_{4}+(\mathrm{CN})_{2}$
Asked in: JEE-TOPICTESTS-CHEMISTRY
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