Which of the following ions will be coloured in the aqueous solution?

(A) $Ti^{3+}$
(B) $Nb^{3+}$
(C) $Cu^{+}$
(D) $Y^{3+}$

Choose the correct answer from the options given below:

(A) and (B) only

The correct answer is Option (3) → (A) and (B) only.

To determine which ions will be colored in aqueous solution, we need to consider the presence of unpaired d-electrons in the transition metals. Transition metal ions can exhibit color due to d-d transitions when unpaired electrons absorb visible light, leading to electronic excitations between different d-orbitals.

Let us analyze each ion:

(A) Ti³⁺:

Electronic configuration of Ti³⁺ (Titanium): \( [Ar] \, 3d^1 \)

Ti³⁺ has one unpaired electron in the 3d orbital.

Since it has unpaired d-electrons, it can undergo d-d transitions and will be colored in aqueous solution.

(B) Nb³⁺:

Electronic configuration of Nb³⁺ (Niobium): \( [Kr] \, 4d^2 \)

Nb³⁺ has two unpaired electrons in the 4d orbital.

This ion also has unpaired d-electrons, so it can exhibit d-d transitions and will be colored in aqueous solution.

(C) Cu⁺:

Electronic configuration of Cu⁺ (Copper): \( [Ar] \, 3d^{10} \)

Cu⁺ has a completely filled 3d orbital (10 electrons), so no unpaired electrons are present.

Without unpaired d-electrons, Cu⁺ cannot undergo d-d transitions and will be colorless in aqueous solution.

(D) Y³⁺:

Electronic configuration of Y³⁺ (Yttrium): \( [Kr] \)
- Y³⁺ has no d-electrons because it loses all of its d-electrons upon ionization.

Since there are no d-electrons to participate in d-d transitions, Y³⁺ will be colorless in aqueous solution.

Conclusion:

Ti³⁺ and Nb³⁺ are colored in aqueous solution because they have unpaired d-electrons.

Cu⁺ and Y³⁺ are colorless due to the absence of unpaired d-electrons.