MCQ
The oxidation state of $Fe$ in ${K_4}[Fe{(CN)_6}]$ is
- ✓$2$
- B$-2$
- C$3$
- D$4$
or $x = + 6 - 4 = + 2$
Oxidation state of $Fe = + 2$
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Sucrose $\xrightarrow[{Cleavage\,\,(Hydrolysis)}]{{Gly\cos idic\,bond}}A + B\xrightarrow[{{\text{reagent}}}]{{{\text{Seliwanoff 's}}}}?$