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⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\), V³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\)

Step-by-step solution

a. Electron configurations for Cr, Cr²⁺, Cr³⁺

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 \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\).

b. Electron configurations for Cu, Cu⁺, Cu²⁺

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 \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\).

c. Electron configurations for V, V²⁺, V³⁺

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 \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\).

Key concepts

Chromium Electron Configuration

Understanding the electron configuration of chromium (Cr) is essential as it represents a notable exception to typical electron distribution rules. Chromium has an atomic number of 24, indicating 24 electrons in its neutral state.

The ground-state electron configuration for Cr is expressed as:

• Chromium (Cr): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\)

This configuration deviates from the expected \(4s^2 3d^4\). Instead, it occurs because a half-filled \(3d\) subshell (\(3d^5\)) offers additional stability compared to a completely filled \(4s\) subshell. This stability results from the symmetrical distribution of the electrons in the \(3d\) subshell, which provides a 'balance'.

When chromium ions are formed, electrons are removed first from the \(4s\) subshell,:

• Chromium (II) ion (Cr^{2+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^4\)
• Chromium (III) ion (Cr^{3+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)

This demonstrates that stability can also be achieved through specific electron arrangements within the d-orbital. After reaching the trivalent state, more electrons are removed from the \(3d\), revealing how electron configurations adapt to maintain stability.

Copper Electron Configuration

Copper (Cu), with an atomic number of 29, presents another fascinating case of electron configuration that prioritizes stability. A neutral copper atom's electron configuration is:

• Copper (Cu): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\)

This arrangement features a fully filled \(3d\) subshell, providing increased stability, unlike the expected \(4s^2 3d^9\). The electron moves from the \(4s\) to the \(3d\) orbital to achieve a complete \(3d\) block, which is considered lower in energy and more stable.

Copper ions continue to exhibit fascinating electron removal pathways:

• Copper (I) ion (Cu^{+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\)
• Copper (II) ion (Cu^{2+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\)

The formation of Cu⁺ involves removing an electron from the \(4s\) subshell, maintaining the complete \(3d\) subshell. Continuing with Cu²⁺, an additional electron is removed from the \(3d\) subshell, although this disrupts its overall symmetry slightly, making it slightly less stable compared to Cu⁺.

Vanadium Electron Configuration

Vanadium (V) is another transition metal where the awareness of electron removal order becomes crucial. With an atomic number of 23, a neutral vanadium atom's electron configuration is:

• Vanadium (V): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\)

Vanadium follows the expected filling order. However, for ions, electron removal starts from the \(4s\) before the \(3d\), aligning with the lower energy level principles.

For its ions:

• Vanadium (II) ion (V^{2+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)
• Vanadium (III) ion (V^{3+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\)

Formation of V²⁺ involves removing electrons first from the \(4s\) subshell, resulting in leaves the three \(3d\) electrons untouched, which are important for the metal's chemical characteristics. Progressing to V³⁺, a single \(3d\) electron is removed next to balance energy further, demonstrating once more how electrons arrange for optimal stability across transition metals.