General physical and chemical properties of the first row of transition elements, titanium to copper — practice questions

(a) Let the oxidation state of V be $x$. Oxygen is -2. The overall charge is -1. Equation: $x + 3(-2) = -1$ $x - 6 = -1$ $x = +5$ The oxidation state of vanadium is +5.

(b) A neutral vanadium atom has 23 electrons. The electronic configuration is filled in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d. Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³ or in shorthand: [Ar] 4s² 3d³. (Note: It is common practice to write 3d before 4s once filled: [Ar] 3d³ 4s²)

(c) The vanadium ion is V⁵⁺. To form the ion, we remove electrons from the neutral atom, starting with the outermost shell (4s) first. Remove two 4s electrons: [Ar] 3d³ Remove three 3d electrons: [Ar] 3d⁰ The electronic configuration of V⁵⁺ is [Ar] or 1s² 2s² 2p⁶ 3s² 3p⁶.

1. Electronic Configurations: First, state the electronic configurations of the ions.
   • Cu atom: [Ar] 3d¹⁰ 4s¹ → Cu²⁺ ion: [Ar] 3d⁹
   • Zn atom: [Ar] 3d¹⁰ 4s² → Zn²⁺ ion: [Ar] 3d¹⁰
2. Explanation for Cu²⁺ (Colour): The Cu²⁺ ion has a partially filled d-subshell (3d⁹). In an aqueous solution, it forms the complex ion [Cu(H₂O)₆]²⁺. The water ligands cause the d-orbitals to split into different energy levels. An electron can absorb energy from the visible light spectrum to be promoted from a lower to a higher energy d-orbital (a d-d transition). The complementary colour (blue) is observed.
3. Explanation for Zn²⁺ (Colourless): The Zn²⁺ ion has a full d-subshell (3d¹⁰). Since there are no vacancies in the d-orbitals, electrons cannot be promoted. Therefore, d-d transitions are not possible, no visible light is absorbed, and the solution appears colourless.