MHTCET 15 May 2024 Shift 1 Paper

Concept:The oxidation number of an element in an ion or compound is found by equating the sum of all oxidation numbers to the total charge of the species.Explanation:For the $\mathrm{CrO}_{4}^{2-}$ ion, let the oxidation number of Cr be $x$.Oxygen has an oxidation number of $-2$, and the total charge on the ion is $-2$.Therefore, $x + 4(-2) = -2$.Simplifying: $x - 8 = -2$, which gives $x = +6$.Thus, the oxidation number of Cr in $\mathrm{CrO}_{4}^{2-}$ is $+6$.For $\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}$, potassium has an oxidation number of $+1$ and oxygen has $-2$.Let the oxidation number of each Cr atom be $x$. The compound is neutral, so:$2(+1) + 2x + 7(-2) = 0$.Simplifying: $2 + 2x - 14 = 0$, which gives $2x = 12$.Therefore, $x = +6$.Thus, the oxidation number of Cr in $\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}$ is also $+6$.Answer:The oxidation numbers are $+6$ and $+6$ respectively.Therefore, the correct option is C.