Problem 22 - Chemistry An Atoms First Approach

Question

Write electron configurations for the following ions. a. \(\mathrm{Ni}^{2+}\) b. \(\mathrm{Cd}^{2+}\) c. \(\mathrm{Zr}^{3+}\) and \(\mathrm{Zr}^{4+}\) d. \(\mathrm{Os}^{2+}\) and \(\mathrm{Os}^{3+}\)

Step-by-Step Solution

Answer

The electron configurations of the ions are: a. \(\mathrm{Ni}^{2+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^8\) b. \(\mathrm{Cd}^{2+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\) c. \(\mathrm{Zr}^{3+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{7}\) and \(\mathrm{Zr}^{4+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{6}\) d. \(\mathrm{Os}^{2+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{6}\) and \(\mathrm{Os}^{3+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{5}\)

Step 1: a. Electron Configuration of \(\mathrm{Ni}^{2+}\)

The atomic number of nickel (Ni) is 28. In its neutral state, it has 28 electrons. As \(\mathrm{Ni}^{2+}\), it has lost 2 electrons, leaving it with 26 electrons. The electron configuration for 26 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^8\]

Step 2: b. Electron Configuration of \(\mathrm{Cd}^{2+}\)

The atomic number of cadmium (Cd) is 48. In its neutral state, it has 48 electrons. As \(\mathrm{Cd}^{2+}\), it has lost 2 electrons, leaving it with 46 electrons. The electron configuration for 46 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\]

Step 3: c. Electron Configuration of \(\mathrm{Zr}^{3+}\) and \(\mathrm{Zr}^{4+}\)

The atomic number of zirconium (Zr) is 40. In its neutral state, it has 40 electrons. For \(\mathrm{Zr}^{3+}\), it has lost 3 electrons, leaving it with 37 electrons. The electron configuration for 37 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{7}\] For \(\mathrm{Zr}^{4+}\), it has lost 4 electrons, leaving it with 36 electrons. The electron configuration for 36 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{6}\]

Step 4: d. Electron Configuration of \(\mathrm{Os}^{2+}\) and \(\mathrm{Os}^{3+}\)

The atomic number of osmium (Os) is 76. In its neutral state, it has 76 electrons. For \(\mathrm{Os}^{2+}\), it has lost 2 electrons, leaving it with 74 electrons. The electron configuration for 74 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{6}\] For \(\mathrm{Os}^{3+}\), it has lost 3 electrons, leaving it with 73 electrons. The electron configuration for 73 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{5}\]