Question

In which of the following pairs, the outer most electronic configuration will be the same ?

• A. V$^{2+}$ and Cr$^+$
• B. Cr$^+$ and Mn$^{2+}$
• C. Ni$^{2+}$ and Cu$^+$
• D. Fe$^{2+}$ and Co$^+$

Hint:

When forming transition metal cations, {always remove electrons from the ns orbital first} before removing them from the (n-1)d orbital.
$Cr^+ = [Ar]3d^5$ (half-filled stability)
$Mn^{2+} = [Ar]3d^5$

Answer 1

The Correct Option is B

Step 1: Cr ($Z=24$): $[Ar] 3d^5 4s^1$. For Cr$^+$, remove one $s$ electron: $[Ar] 3d^5$.
Step 2: Mn ($Z=25$): $[Ar] 3d^5 4s^2$. For Mn$^{2+}$, remove two $s$ electrons: $[Ar] 3d^5$.
Step 3: Both have the same $3d^5$ configuration.
Step 4: For other pairs: V$^{2+}$ is $3d^3$, Cr$^+$ is $3d^5$. Ni$^{2+}$ is $3d^8$, Cu$^+$ is $3d^{10}$. Fe$^{2+}$ is $3d^6$, Co$^+$ is $3d^8$.