Calculate the oxidation number of Cr in $\mathrm{CrO}_4^{2-}$ ion and $\mathrm{K}_2 \mathrm{Cr}_2…
- +4 and +6
- +3 and +2
- +6 and +6
- +8 and +2
Solution
Let $y$ be the oxidation number of Cr in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ $\begin{array}{ll} \therefore \quad & 2 \times(+1)+2(y)+7 \times(-2)=0 \\ & +2+2 y-14=0 \\ & 2 y=12 \\ \therefore \quad & y=+6 \end{array}$
Asked in: MHT CET 2024 (15 May Shift 1)