Arrange the following ions in increasing order of number of 3d-electrons

(A) $Cr^{2+}$
(B) $Cu^+$
(C) $Ti^{3+}$
(D) $Mn^+$

Choose the correct answer from the options given below:

(C), (A), (D), (B)

The correct answer is Option (2) → (C), (A), (D), (B)

1. Determine the number of $3d$-electrons for each ion

The neutral atoms have the following atomic numbers ($Z$) and ground-state electron configurations:

• $\text{Ti} (Z=22): [\text{Ar}] 4s^2 3d^2$
• $\text{Cr} (Z=24): [\text{Ar}] 4s^1 3d^5$ (Exception)
• $\text{Mn} (Z=25): [\text{Ar}] 4s^2 3d^5$
• $\text{Cu} (Z=29): [\text{Ar}] 4s^1 3d^{10}$ (Exception)

When forming ions, electrons are always removed from the outermost shell first, which is the $4s$ orbital, before the $3d$ orbital.

2. Arrange in Increasing Order

Now, arrange the ions based on the number of $3d$-electrons in ascending order:

The increasing order is (C), (A), (D), (B).