Question

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

Short answer

The electron configurations for the given elements and ions are as follows: a. Chromium: \(\mathrm{Cr: [Ar]4s^1 3d^5}\), \(\mathrm{Cr^{2+}: [Ar]3d^4}\), \(\mathrm{Cr^{3+}: [Ar]3d^3}\). b. Copper: \(\mathrm{Cu: [Ar]4s^1 3d^{10}}\), \(\mathrm{Cu^{+}: [Ar]3d^9}\), \(\mathrm{Cu^{2+}: [Ar]3d^8}\). c. Vanadium: \(\mathrm{V: [Ar]4s^2 3d^3}\), \(\mathrm{V^{2+}: [Ar]3d^3}\), \(\mathrm{V^{3+}: [Ar]3d^2}\).

Step-by-step solution

Determine the atomic number of Chromium

Chromium (Cr) has an atomic number of 24 which means it has 24 electrons in the neutral form.

Write the electron configuration for Chromium

The electron configuration of Chromium is \([Ar]4s^1 3d^5\). This is an exception to the Aufbau's principle. Instead of the expected \([Ar]4s^2 3d^4\), one electron from 4s promotes to the 3d orbital to make a half-filled stable configuration.

Write the electron configuration for Chromium ions

For \(\mathrm{Cr^{2+}}\), we lose 2 electrons: \([Ar]3d^4\). For \(\mathrm{Cr^{3+}}\), we lose 3 electrons: \([Ar]3d^3\). b. Copper

Determine the atomic number of Copper

Copper (Cu) has an atomic number of 29 which means it has 29 electrons in the neutral form.

Write the electron configuration for Copper

The electron configuration of Copper is \([Ar]4s^1 3d^{10}\). This is another exception to the Aufbau's principle. Instead of the expected \([Ar]4s^2 3d^9\), one electron from 4s promotes to the 3d orbital to make a fully filled stable configuration.

Write the electron configuration for Copper ions

For \(\mathrm{Cu^{+}}\), we lose 1 electron: \([Ar]3d^9\). For \(\mathrm{Cu^{2+}}\), we lose 2 electrons: \([Ar]3d^8\). c. Vanadium

Determine the atomic number of Vanadium

Vanadium (V) has an atomic number of 23 which means it has 23 electrons in the neutral form.

Write the electron configuration for Vanadium

The electron configuration of Vanadium follows the Aufbau's principle: \([Ar]4s^2 3d^3\).

Write the electron configuration for Vanadium ions

For \(\mathrm{V^{2+}}\), we lose 2 electrons: \([Ar]3d^3\). For \(\mathrm{V^{3+}}\), we lose 3 electrons: \([Ar]3d^2\).

Key concepts

Chromium

Chromium is a fascinating element when it comes to electron configurations due to its unique position in the periodic table. It has 24 protons, which means a neutral chromium atom also contains 24 electrons.
The expected electron configuration following the Aufbau principle would be \( [Ar]4s^2 3d^4 \). However, chromium defies this norm.
The actual configuration is \( [Ar]4s^1 3d^5 \), achieved by promoting an electron from the 4s orbital to the 3d orbital. This creates a half-filled d sublevel, which increases stability.
This is a lovely showcase of nature's tendency to seek stability through symmetry.- **Chromium Ion Configurations**: - \( \text{Cr}^{2+} \): By removing two electrons, we get \( [Ar]3d^4 \). - \( \text{Cr}^{3+} \): By removing three electrons, it becomes \( [Ar]3d^3 \).
The electron removal occurs primarily from the outermost shell—typically the 4s orbital first, followed by the 3d orbitals for transition metals. This further underlines the stability chromium attains through its configuration adjustments.

Copper

Copper offers another intriguing look into the world of electron configurations. With its atomic number of 29, copper has 29 electrons in its neutral state. Normally, the configuration might be expected as \( [Ar]4s^2 3d^9 \), but copper takes an unconventional path.
It actually stabilizes itself by adopting \( [Ar]4s^1 3d^{10} \). This configuration is possible by promoting one electron from the 4s orbital to fill the 3d orbital completely.
A fully filled d sublevel is particularly stable, making copper's electron configuration an excellent example of achieving stability through electron rearrangement. It’s one of nature's clever tricks for element stability.- **Copper Ion Configurations**: - \( \text{Cu}^{+} \): Losing a single electron gives us \( [Ar]3d^{10} \). - \( \text{Cu}^{2+} \): Losing two electrons results in \( [Ar]3d^9 \).
For copper ions, the 4s electrons are removed first, which is typical for transition metals as they generally lose s electrons before d electrons.

Vanadium

Vanadium stands out because it adheres to the Aufbau principle more closely when compared to chromium and copper. With an atomic number of 23, a neutral vanadium atom has 23 electrons.
Its electron configuration is \( [Ar]4s^2 3d^3 \), which follows the expected order of filling orbitals.
Unlike chromium and copper, vanadium does not need to promote electrons for achieving stability. This makes it quite straightforward in terms of electronic structure.- **Vanadium Ion Configurations**: - \( \text{V}^{2+} \): After losing two electrons, it has the configuration \( [Ar]3d^3 \). - \( \text{V}^{3+} \): With the loss of three electrons, it becomes \( [Ar]3d^2 \).
This stability and straightforward adherence to the Aufbau principle make vanadium a classic transition metal. However, like all transition metals, during ion formation, s electrons are lost before the d electrons, showing the complex nature of these elements' chemistry.