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

(A) Cr²⁺

(B) Cu⁺

(C) Ti³⁺

(D) Mn⁺

Choose the correct answer from the options given below:

The correct answer is (C). (A). (D). (B).

To determine the correct increasing order of the number of 3d electrons, we write down the electronic configurations of the neutral transition metal atoms and then derive the configurations of their respective ions by removing the outer electrons first.

1. Titanium ion (Ti³⁺):
The atomic number of Titanium (Ti) is 22. Its ground-state electronic configuration is:
Ti: [Ar] 3d² 4s²
To form Ti³⁺, we remove three electrons (two from the 4s orbital and one from the 3d orbital):
Ti³⁺: [Ar] 3d¹
Number of 3d electrons = 1

2. Chromium ion (Cr²⁺):
The atomic number of Chromium (Cr) is 24. Its ground-state electronic configuration is:
Cr: [Ar] 3d⁵ 4s¹
To form Cr²⁺, we remove two electrons (one from the 4s orbital and one from the 3d orbital):
Cr²⁺: [Ar] 3d⁴
Number of 3d electrons = 4

3. Manganese ion (Mn⁺):
The atomic number of Manganese (Mn) is 25. Its ground-state electronic configuration is:
Mn: [Ar] 3d⁵ 4s²
To form Mn⁺, we remove one electron from the outer 4s orbital:
Mn⁺: [Ar] 3d⁵ 4s¹
Number of 3d electrons = 5

4. Copper ion (Cu⁺):
The atomic number of Copper (Cu) is 29. Its ground-state electronic configuration is:
Cu: [Ar] 3d¹⁰ 4s¹
To form Cu⁺, we remove one electron from the 4s orbital:
Cu⁺: [Ar] 3d¹⁰
Number of 3d electrons = 10

Now, let us arrange the ions in the increasing order of their 3d electrons:

$${\mathrm{Ti}}^{3+}(1)<{\mathrm{Cr}}^{2+}(4)<{\mathrm{Mn}}^{+}(5)<{\mathrm{Cu}}^{+}(10)$$

This corresponds to the sequence: (C) < (A) < (D) < (B), which is represented by the option (C). (A). (D). (B).