JEE Chemistry 2024 Question with Solution

Given: List-I contains $\text{Cr}^{2+}$, $\text{Mn}^{+}$, $\text{Ni}^{2+}$ and $\text{V}^{+}$. List-II contains the electronic distributions $3d^8$, $3d^5 4s^1$, $3d^4$ and $3d^3 4s^1$.

Find: The correct matching between the species and the electronic distributions.

For transition-metal cations, electrons are removed from the $4s$ orbital before the $3d$ orbital.

1. For $\text{Cr}^{2+}$: Neutral chromium is

Removing two electrons gives

So, $\text{Cr}^{2+} \rightarrow \text{III}$.

2. For $\text{Mn}^{+}$: Neutral manganese is

Removing one electron gives

So, $\text{Mn}^{+} \rightarrow \text{IV}$.

3. For $\text{Ni}^{2+}$: Neutral nickel is

Removing two electrons gives

So, $\text{Ni}^{2+} \rightarrow \text{I}$.

4. For $\text{V}^{+}$: Neutral vanadium is

Removing one electron gives

So, $\text{V}^{+} \rightarrow \text{II}$.

Thus the correct matching is (A)-III, (B)-IV, (C)-I, (D)-II. Therefore, the correct option is B.

The answer key lists option (1), but the solution working clearly establishes option B as correct.

Given: The ions are $\text{Cr}^{2+}$, $\text{Mn}^{+}$, $\text{Ni}^{2+}$ and $\text{V}^{+}$.

Find: Which option matches all four configurations correctly.

Use the shortcut rule: write the neutral configuration of the metal and remove electrons from $4s$ before $3d$.

• $\text{Cr}^{2+} : 3d^4$
• $\text{Mn}^{+} : 3d^5 4s^1$
• $\text{Ni}^{2+} : 3d^8$
• $\text{V}^{+} : 3d^3 4s^1$

This gives the sequence III, IV, I, II, which matches option B.

Therefore, the correct option is B.