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

Answer: (B)

(A)-III, (B) – IV, (C) – I, (D)-II

Answer 2

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.