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

Each ion has a specific electron configuration:

• Cr⁺²: After losing two electrons, Cr has an electronic configuration of 3d⁴.
• Mn⁺: Losing one electron results in the configuration 3d⁵4s¹.
• Ni⁺²: Removal of two electrons leads to a configuration of 3d⁸.
• V⁺: Removing one electron yields 3d³4s¹.