Vanadium (V) is a transition metal with an atomic number of 23, meaning a neutral atom contains 23 protons and 23 electrons. Understanding how these electrons are arranged around the nucleus is the foundation for predicting the element’s chemical behavior and magnetic properties. To determine the number of unpaired electrons Vanadium possesses, we must systematically apply the fundamental rules of electron configuration.
What Defines an Unpaired Electron?
Electrons exist in specific regions of space around the nucleus called orbitals, and each orbital can hold a maximum of two electrons. When an orbital contains two electrons, they must have opposite spins, a condition known as being “paired.”
An unpaired electron is simply a single electron occupying an orbital by itself. The presence of these solitary electrons determines an atom’s physical and chemical identity. Atoms with unpaired electrons are attracted to external magnetic fields and are classified as paramagnetic. In contrast, atoms where all electrons are paired are slightly repelled by magnetic fields, a property called diamagnetism. Unpaired electrons are also often involved when an atom forms a chemical bond.
Mapping the Neutral Vanadium Atom
Determining the number of unpaired electrons begins with mapping the location of all 23 electrons in the neutral Vanadium atom. The process of filling orbitals follows the Aufbau principle, which dictates that electrons occupy the lowest energy levels first.
The first 18 electrons fill the 1s, 2s, 2p, 3s, and 3p subshells completely, matching the configuration of the noble gas Argon (\([Ar]\)). This allows us to use the condensed notation \([Ar]\) to represent the core electrons. We must account for the remaining five electrons.
Following the energy progression, the next orbital to fill is the \(4s\) orbital, which holds two electrons, making the configuration \([Ar] 4s^2\). The final three electrons are placed into the \(3d\) subshell. The complete electron configuration for neutral Vanadium is \([Ar] 4s^2 3d^3\).
The Role of Hund’s Rule in Orbital Filling
To determine the final count of unpaired electrons, we must apply Hund’s Rule to the outermost partially filled subshell, the \(3d\) subshell containing three electrons. The \(d\) subshell is composed of five separate orbitals, all having the same energy level, known as degenerate orbitals.
Hund’s Rule states that electrons will occupy these degenerate orbitals singly before they begin to pair up. Because the Vanadium atom has three electrons in the \(3d\) subshell, each of these three electrons will occupy its own \(3d\) orbital, maintaining parallel spins.
Since the two electrons in the \(4s\) orbital are paired, and all the core electrons are paired, the total number of unpaired electrons comes solely from the \(3d\) subshell. Consequently, a neutral Vanadium atom has three unpaired electrons.
How Vanadium’s Ions Affect Electron Count
As a transition metal, Vanadium is capable of losing electrons to form positively charged ions, which changes the number of unpaired electrons. The most common ions are \(V^{2+}\) and \(V^{3+}\). Transition metals lose electrons from the highest principal quantum number shell first, meaning electrons are removed from the \(4s\) orbital before the \(3d\) orbital.
Vanadium (II) Ion (\(V^{2+}\))
The \(V^{2+}\) ion is formed by the loss of two electrons from the neutral atom. These two electrons are removed from the outermost \(4s\) orbital, resulting in the electron configuration \([Ar] 3d^3\). With three electrons remaining in the \(3d\) subshell, applying Hund’s Rule shows the \(V^{2+}\) ion also has three unpaired electrons.
Vanadium (III) Ion (\(V^{3+}\))
The \(V^{3+}\) ion is formed by the loss of a total of three electrons. The first two are lost from the \(4s\) orbital, and the third is lost from the \(3d\) subshell. This leaves the \(V^{3+}\) ion with the electron configuration \([Ar] 3d^2\). Applying Hund’s Rule to the two remaining electrons in the \(3d\) subshell, these electrons will occupy two separate \(3d\) orbitals. Therefore, the Vanadium \(V^{3+}\) ion has two unpaired electrons.