Write electron configurations for each of the following. a. \(\mathrm{Cr}, \mathrm{Cr}^{2+}, \mathrm{Cr}^{3+}\) b. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}\)

First, we need to determine the atomic number of Chromium (Cr). Cr has an atomic number of 24, which means it has 24 electrons. Now, let's find the electron configurations for each species: 1. Cr: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\). This is an exception to the rule when writing electron configurations as the half-filled 3d subshell is more stable than the fully-filled 4s subshell. 2. Cr²⁺: To form Cr²⁺, two electrons are removed from Cr. Electrons are removed starting from the 4s subshell. The resulting configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^4\). 3. Cr³⁺: To form Cr³⁺, three electrons are removed from Cr. Again, after removing two electrons from the 4s subshell, the next electron will be removed from the 3d subshell. The final configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\).

Copper (Cu) has an atomic number of 29, which means it has 29 electrons. Now, let's find the electron configurations for each species: 1. Cu: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\). Just like Cr, Cu has a more stable configuration with a completely-filled 3d subshell. 2. Cu⁺: To form Cu⁺, one electron is removed from Cu. The electron is removed from the 4s subshell, resulting in the configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\). 3. Cu²⁺: To form Cu²⁺, two electrons are removed from Cu. After removing one electron from the 4s subshell, the other electron is removed from the 3d subshell. The final configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\).

Vanadium (V) has an atomic number of 23, which means it has 23 electrons. Let's find the electron configurations for each species: 1. V: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\). 2. V²⁺: To form V²⁺, two electrons are removed from V. The electrons are removed from the 4s subshell first. The resulting configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\). 3. V³⁺: To form V³⁺, three electrons are removed from V. After removing two electrons from the 4s subshell, the next electron is removed from the 3d subshell. The final configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\).