Question

How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\)

Short answer

The number of unpaired electrons in the given complex ions are: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 unpaired electrons b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 unpaired electrons c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): 1 unpaired electron

Step-by-step solution

Identify the oxidation state for each metal

In each complex, we first need to determine the oxidation state of the central metal atom: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) : The total charge of the complex is 2+, and ammonia (NH3) is neutral, so the oxidation state of Ru is +2. b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) : The total charge of the complex is 2+, and water (H2O) is neutral, so the oxidation state of Ni is +2. c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\) : The total charge of the complex is 3+, and ethylenediamine (en) is neutral, so the oxidation state of V is +3.

Write the electron configurations

Next, we will write the electron configurations of the central metal atoms in their corresponding oxidation states: a. Ru+2: Ru has an atomic number of 44, and its electron configuration is [Kr] 4d7 5s1. After losing 2 electrons, we have [Kr] 4d6. b. Ni+2: Ni has an atomic number of 28, and its electron configuration is [Ar] 3d8 4s2. After losing 2 electrons, we have [Ar] 3d8. c. V+3: V has an atomic number of 23, and its electron configuration is [Ar] 3d3 4s2. After losing 3 electrons, we have [Ar] 3d2.

Determine the number of unpaired electrons

For each electron configuration, count the number of unpaired electrons in the d orbitals: a. [Kr] 4d6 (low-spin case): In the low-spin case, electrons preferentially occupy the lower energy orbitals first. Here, the d orbitals are completely filled before any electron jumps into a higher energy orbital. Thus, there are no unpaired electrons, and the answer is 0. b. [Ar] 3d8: The d orbitals fill in the following order: ↑↓↑↓↑↓↑↑, so there are 2 unpaired electrons. c. [Ar] 3d2: The d orbitals fill in the following order: ↑↓↑, so there is 1 unpaired electron.

Final Answers

The number of unpaired electrons in the complex ions are: a. Ru(NH3)6^2+ (low-spin case): 0 unpaired electrons b. Ni(H2O)6^2+: 2 unpaired electrons c. V(en)3^3+: 1 unpaired electron