Question

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. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}\)

Short answer

a. Cr: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5}\), Cr²⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{4}\), Cr³⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}\) b. Cu: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}\), Cu⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\), Cu²⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}\) c. V: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}\), V²⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}\), V³⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}\)

Step-by-step solution

Determine the atomic number of the element

The atomic number of Chromium (Cr) is 24.

Write the electron configuration for Cr

Using the periodic table and applying the rules, we find the electron configuration of Chromium to be: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5}\]

Write the electron configuration for Cr²⁺

For the Cr²⁺ ion, we need to remove two electrons. The electron configuration becomes: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{4}\]

Write the electron configuration for Cr³⁺

For the Cr³⁺ ion, we need to remove three electrons. The electron configuration becomes: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}\] b. Cu, Cu⁺, Cu²⁺

Determine the atomic number of the element

The atomic number of Copper (Cu) is 29.

Write the electron configuration for Cu

Using the periodic table and applying the rules, we find the electron configuration of Copper to be: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}\]

Write the electron configuration for Cu⁺

For the Cu⁺ ion, we need to remove one electron. The electron configuration becomes: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\]

Write the electron configuration for Cu²⁺

For the Cu²⁺ ion, we need to remove two electrons. The electron configuration becomes: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}\] c. V, V²⁺, V³⁺

Determine the atomic number of the element

The atomic number of Vanadium (V) is 23.

Write the electron configuration for V

Using the periodic table and applying the rules, we find the electron configuration of Vanadium to be: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}\]

Write the electron configuration for V²⁺

For the V²⁺ ion, we need to remove two electrons. The electron configuration becomes: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}\]

Write the electron configuration for V³⁺

For the V³⁺ ion, we need to remove three electrons. The electron configuration becomes: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}\]