Question

Match List-I with List-II
\[ \begin{array}{|c|c|} \hline \text{List-I (Species)} & \text{List-II (Electronic distribution)} \\ \hline \text{(A) Cr}^{+2} & \text{(I) } 3d^8 \\ \text{(B) Mn}^{+} & \text{(II) } 3d^3 4s^1 \\ \text{(C) Ni}^{+2} & \text{(III) } 3d^4 \\ \text{(D) V}^{+} & \text{(IV) } 3d^5 4s^1 \\ \hline \end{array} \]

• A. (A)-I, (B)-II, (C)-III, (D)-IV
• B. (A)-III, (B) – IV, (C) – I, (D)-II
• C. (A)-IV, (B)-III, (C)-I, (D)-II
• D. (A)-II, (B)-I, (C)-IV, (D)-III

Answer 1

The Correct Option is B

To answer this question, we will determine the electronic configuration for each ion in List-I and match it with its corresponding electronic distribution in List-II. We will analyze each species individually:

1. Cr⁺²: Neutral Chromium (Cr) has the electronic configuration \([Ar] 3d^5 4s^1\). Upon losing two electrons to form Cr⁺², the 4s electron is removed first, followed by one 3d electron. This results in the configuration \([Ar] 3d^4\), which matches option III.
2. Mn⁺: Neutral Manganese (Mn) has the electronic configuration \([Ar] 3d^5 4s^2\). When Mn loses one electron to form Mn⁺, the electron is removed from the 4s orbital, yielding the configuration \([Ar] 3d^5 4s^1\), corresponding to option IV.
3. Ni⁺²: Neutral Nickel (Ni) has the electronic configuration \([Ar] 3d^8 4s^2\). To form Ni⁺², Nickel loses two electrons, both of which are removed from the 4s orbital, resulting in the configuration \([Ar] 3d^8\). This matches option I.
4. V⁺: Neutral Vanadium (V) has the electronic configuration \([Ar] 3d^3 4s^2\). When Vanadium loses one electron to form V⁺, the electron is removed from the 4s orbital, resulting in the configuration \([Ar] 3d^3 4s^1\), which corresponds to option II.

The derived matching is as follows:

List-I (Species)	List-II (Electronic distribution)
(A) Cr+2	III – 3d4
(B) Mn+	IV – 3d5 4s1
(C) Ni+2	I – 3d8
(D) V+	II – 3d3 4s1

Therefore, the correct answer is: (A)-III, (B) – IV, (C) – I, (D)-II.