MHTCET 15 May 2024 Shift 2 Paper

Concept:The oxidation number of an element in an ion is found using known oxidation numbers of other atoms and the overall charge on the ion.Explanation:Step 1: Find the oxidation number of nitrogen in $\mathrm{NO}_{3}^{-}$.Let the oxidation number of nitrogen be $x$.Oxygen has an oxidation number of $-2$.The total charge on the ion is $-1$.Therefore, $x + 3(-2) = -1$.Solving, $x - 6 = -1$, so $x = +5$.Hence, nitrogen has an oxidation number of $+5$ in $\mathrm{NO}_{3}^{-}$.Step 2: Find the oxidation number of nitrogen in $\mathrm{NH}_{4}^{+}$.Let the oxidation number of nitrogen be $y$.Hydrogen has an oxidation number of $+1$.The total charge on the ion is $+1$.Therefore, $y + 4(+1) = +1$.Solving, $y + 4 = +1$, so $y = -3$.Hence, nitrogen has an oxidation number of $-3$ in $\mathrm{NH}_{4}^{+}$.Step 3: Identify the change.The oxidation number changes from $+5$ to $-3$.Since the oxidation number decreases, this conversion is a reduction.Answer:The change in oxidation number of nitrogen is from $+5$ to $-3$.Correct option: A. $+5$ to $-3$.