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. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\)

a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): Since NH3 is a neutral ligand, the oxidation state of Ru is +2. b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): H2O is a neutral ligand, so the oxidation state of Ni is +2. c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): 'en' refers to ethylenediamine, which is a bidentate neutral ligand. Therefore, the oxidation state of V is +3.

We will consider the electron configurations of the metal ions in their ground state and then remove the required number of electrons from the outermost orbitals based on their oxidation state: a. Ru has an atomic number of 44. Its electron configuration is [Kr] 4d7 5s1. In an oxidation state of +2, we have Ru(II), with the electron configuration [Kr] 4d6. b. Ni has an atomic number of 28. Its electron configuration is [Ar] 3d8 4s2. In an oxidation state of +2, we have Ni(II), with the electron configuration [Ar] 3d8. c. V has an atomic number of 23. Its electron configuration is [Ar] 3d3 4s2. In an oxidation state of +3, we have V(III), with the electron configuration [Ar] 3d2.

Now we calculate the unpaired electrons considering the electron distribution in d orbitals for each complex ion: a. For \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): It is a low spin complex due to the strong field ligand (NH3) present in the complex ion. Therefore, the d orbitals will pair up before filling, resulting in the electron configuration t2g6 eg0. The number of unpaired electrons is 0. b. For \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): It is a high spin complex due to the weak field ligand (H2O) present in the complex ion. The electron configuration is t2g6 eg2. The number of unpaired electrons is 2. c. For \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): Ethylenediamine is a strong field ligand, making it a low spin complex. The electron configuration is t2g2 eg0. The number of unpaired electrons is 0. Thus, the number of unpaired electrons in the complex ions are as follows: a. 0 unpaired electrons b. 2 unpaired electrons c. 0 unpaired electrons