Predict which of the following will be colored in its aqueous solution?

$Mn^{2+}$

The correct answer is Option (3) → $Mn^{2+}$

The color of transition metal ions in aqueous solution arises due to d–d electronic transitions.

When light falls on a transition metal ion:

• Electrons in partially filled d-orbitals absorb certain wavelengths of visible light.
• These electrons move from a lower energy d-orbital to a higher energy d-orbital.
• The remaining transmitted or reflected light appears as the observed color.

Thus, ions with partially filled d-orbitals (d¹–d⁹) generally show color, while ions with d⁰ or d¹⁰ configuration are colorless.

Option-wise Explanation

Option 1: Ti⁴⁺

Titanium atomic number = 22

Electronic configuration of Ti:

[Ar] 3d² 4s²

Ti⁴⁺ loses four electrons:

Ti⁴⁺ = [Ar] 3d⁰

Since there are no d-electrons, d–d transition cannot occur. Therefore this ion is colorless.

Option 2: Sc³⁺

Scandium atomic number = 21

Sc = [Ar] 3d¹ 4s²

Sc³⁺ = [Ar] 3d⁰

Again, no d-electrons are present, so d–d transitions cannot occur. Hence this ion is colorless.

Option 3: Mn²⁺

Manganese atomic number = 25

Mn = [Ar] 3d⁵ 4s²

Mn²⁺ = [Ar] 3d⁵

Since the d-orbitals are partially filled, d–d transitions are possible. Therefore Mn²⁺ ions can absorb visible light and appear colored (pale pink) in aqueous solution.

Thus this option is correct.

Option 4: Zn²⁺

Zinc atomic number = 30

Zn = [Ar] 3d¹⁰ 4s²

Zn²⁺ = [Ar] 3d¹⁰

The d-orbitals are completely filled, so d–d transitions cannot occur. Therefore this ion is colorless.