Question

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. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\)

Short answer

The electron configurations for the transition metal ions in each of the following compounds are: a. K\(_{3}\)[Fe(CN)\(_{6}\)] - The electron configuration for Fe+3 is [Ar] \(3d^5\). b. [Ag(NH\(_{3}\))\(_{2}\)]Cl - The electron configuration for Ag+1 is [Kr] \(4d^{10}\). c. [Ni(H\(_{2}\)O)\(_{6}\)]Br\(_{2}\) - The electron configuration for Ni+2 is [Ar] \(3d^8\). d. [Cr(H\(_{2}\)O)\(_{4}\)(NO\(_{2}\))\(_{2}\)]I - The electron configuration for Cr+3 is [Ar] \(3d^3\).

Step-by-step solution

Identify the transition metals and oxidation states

For each compound, find the transition metal and determine the charge or oxidation state of the central metal ion. a. \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) - The transition metal is Fe (Iron). Knowing that \(\mathrm{CN}^-\) ion has a charge of -1 and Fe is (x) from its oxidation state, the equation: 3 K (\(+1\)) + x + 6 CN (\(-1\)) = 0, where x = Fe oxidation state, can be used to find x. x = +3, so Fe has an oxidation state of +3. b. \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}\) - The transition metal is Ag (Silver). Since \(\mathrm{Cl}^-\) has a charge of -1, Ag will have to have a charge of +1 to balance out the compound. Thus, the oxidation state of Ag is +1. c. \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}\) - The transition metal is Ni (Nickel). \(\mathrm{Br}^-\) has a charge of -1, and since Br shows up twice in the compound, Ni will have to balance it out with a charge of +2. Thus, the oxidation state of Ni is +2. d. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\) - The transition metal is Cr (Chromium). The \(\mathrm{NO}_{2}^-\) ion has a charge of -1 and \(\mathrm{I}^-\) has a charge of -1, so Cr will need to balance out the charges with an oxidation state of +3.

Write electron configurations

Use the transition metals and their oxidation states to write electron configurations for each compound: a. Fe+3: Starting with the electron configuration for neutral Fe - [\(\ce{Ar}\)] \(3d^6\) \(4s^2\), remove 3 electrons. Electrons are removed first from the outermost shell (4s) and then from the inner d orbital. Thus, the electron configuration for Fe+3 is [\(\ce{Ar}\)] \(3d^5\). b. Ag+1: Starting with the electron configuration for neutral Ag - [\(\ce{Kr}\)] \(4d^{10}\) \(5s^1\), remove 1 electron from the outermost shell (5s). The electron configuration for Ag+1 is [\(\ce{Kr}\)] \(4d^{10}\). c. Ni+2: Starting with the electron configuration for neutral Ni - [\(\ce{Ar}\)] \(3d^8\) \(4s^2\), remove 2 electrons. Electrons are removed first from the outermost shell (4s) and then from the inner d orbital. The electron configuration for Ni+2 is [\(\ce{Ar}\)] \(3d^8\). d. Cr+3: Starting with the electron configuration for neutral Cr - [\(\ce{Ar}\)] \(3d^5\) \(4s^1\), remove 3 electrons. Electrons are removed first from the outermost shell (4s) and then from the inner d orbital. The electron configuration for Cr+3 is [\(\ce{Ar}\)] \(3d^3\).