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:
Answer & explanation
Correct answer: option 3
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.