Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right (\mathbf{b})\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\) (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4},(\mathbf{e})[\operatorname{Sr}(\mathrm{EDTA})]^{2-}\)

Use the charge on the complex and the charges on ligands to find the oxidation state of the central metal ion in each complex. (a) K3[Fe(CN)6] (b) [Mn(H2O)6](NO3)2 (c) Na[Ag(CN)2] (d) [Cr(NH3)4Br2]ClO4 (e) [Sr(EDTA)]^2−

(a) In K3[Fe(CN)6], the overall charge on the complex is 0. Therefore, the charge on Fe is -3. The atomic number of Fe is 26, and its ground state electronic configuration is [Ar]3d^64s^2. When Fe loses 3 electrons, it forms Fe(III) with the electronic configuration: [Ar]3d^5. The number of valence d electrons is 5. (b) In [Mn(H2O)6](NO3)2, the overall charge on the complex is +2. The charge on Mn is +2 since the complex is neutral. The atomic number of Mn is 25, and its ground state electronic configuration is [Ar]3d^54s^2. When Mn loses 2 electrons, it forms Mn(II) with the electronic configuration: [Ar]3d^5. The number of valence d electrons is 5. (c) In Na[Ag(CN)2], the overall charge on the complex is 0. Therefore, the charge on Ag is +1. The atomic number of Ag is 47, and its ground state electronic configuration is [Kr]4d^105s^1. When Ag loses 1 electron, it forms Ag(I) with the electronic configuration: [Kr]4d^10. The number of valence d electrons is 10. (d) In [Cr(NH3)4Br2]ClO4, the overall charge on the complex is +1. The charge on Cr is +3 since NH3 ligands are neutral, and Br has a charge of -1. The atomic number of Cr is 24, and its ground state electronic configuration is [Ar]3d^54s^1. When Cr loses 3 electrons, it forms Cr(III) with the electronic configuration: [Ar]3d^3. The number of valence d electrons is 3. (e) In [Sr(EDTA)]^2−, Sr has a charge of +2, and the charge on the EDTA ligand is -4. Therefore, the overall charge on the complex is -2. The atomic number of Sr is 38, and its ground state electronic configuration is [Kr]5s^2. Sr has no valence d electrons with the electronic configuration: [Kr]. The number of valence d electrons is 0. So, the number of valence d electrons of central metal ions in the given complexes are: (a) 5 (b) 5 (c) 10 (d) 3 (e) 0