What is the electron configuration for the transition metal ion in each of the following compounds? a. \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) b. \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}\) c. \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}\) d. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\)

: For each compound, we will determine the charge of the transition metal ion based on the charges of the coordination complex and counterions. a. In \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), the potassium ions have a charge of +1, and the coordination complex has a charge of -3. Thus, the charge on the Fe ion must be +3. b. In \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}\), the chloride ion has a charge of -1, and the coordination complex has a charge of +1. Thus, the charge on the Ag ion must be +1. c. In \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}\), the bromide ions have a charge of -1 each, and the coordination complex has a charge of +2. Thus, the charge on the Ni ion must be +2. d. In \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\), the iodide ion has a charge of -1, and the coordination complex has a charge of +1. Thus, the charge on the Cr ion must be +1.

: Now, we will determine the electron configuration of the neutral atoms of each transition metal. a. Fe: \([Ar] 3d^6 4s^2 \) b. Ag: \([Kr] 4d^{10} 5s^1 \) c. Ni: \([Ar] 3d^8 4s^2 \) d. Cr: \([Ar] 3d^5 4s^1 \)

: Finally, we'll determine the electron configuration of each transition metal ion based on the oxidations states we found in step 1. Remember to remove electrons from the outermost shell (higher n value) first. a. Fe³⁺: \( [Ar] 3d^6 4s^2 \) --> remove three electrons: \( [Ar] 3d^5 \) b. Ag¹⁺: \( [Kr] 4d^{10} 5s^1 \) --> remove one electron: \( [Kr] 4d^{10} \) c. Ni²⁺: \( [Ar] 3d^8 4s^2 \) --> remove two electrons: \( [Ar] 3d^8 \) d. Cr¹⁺: \( [Ar] 3d^5 4s^1 \) --> remove one electron: \( [Ar] 3d^5 \) The resulting electron configurations are: a. Fe³⁺: \( [Ar] 3d^5 \) b. Ag¹⁺: \( [Kr] 4d^{10} \) c. Ni²⁺: \( [Ar] 3d^8 \) d. Cr¹⁺: \( [Ar] 3d^5 \)