# Vanadium(V) electron configuration and orbital diagram

Vanadium(V) is the 23rd element in the periodic table and its symbol is ‘V’. The electron configuration of vanadium and the orbital diagram is the main topic in this article. Also, valency and **valence electrons of vanadium**, and compound formation, bond formation have been discussed. Hopefully, after reading this article you will know in detail about this.

The total number of electrons in vanadium is twenty-three. These electrons are arranged according to specific rules of different orbits. The arrangement of electrons in different orbits and orbitals of an atom in a certain order is called electron configuration. The electron configuration of vanadium(V) atoms can be done in two ways.

- Electron configuration through orbit
- Electron configuration through orbital

Electron configuration through orbitals follows different principles. For example Aufbau principle, Hund’s principle, Pauli’s exclusion principle. There is an article published on this site detailing the **electron configuration**, you can read it if you want.

Table of Contents

## Vanadium(V) electron configuration through orbit

Scientist Niels Bohr was the first to give an idea of the atom’s orbit. He provided a model of the atom in 1913. The complete idea of the orbit is given there. The electrons of the atom revolve around the nucleus in a certain circular path. These circular paths are called orbit(shell). These orbits are expressed by n. [n = 1,2,3,4 . . . The serial number of the orbit]

K is the name of the first orbit, L is the second, M is the third, N is the name of the fourth orbit. The electron holding capacity of each orbit is 2n^{2}.

For example,

- n = 1 for K orbit.

The electron holding capacity of K orbit is 2n^{2}= 2 × 1^{2}= 2 electrons. - For L orbit, n = 2.

The electron holding capacity of the L orbit is 2n^{2}= 2 × 2^{2}= 8 electrons. - n=3 for M orbit.

The maximum electron holding capacity in M orbit is 2n^{2}= 2 × 3^{2 }= 18 electrons. - n=4 for N orbit.

The maximum electron holding capacity in N orbit is 2n^{2}= 2 × 4^{2}= 32 electrons.

Therefore, the maximum electron holding capacity in the first shell is two, the second shell is eight and the 3rd shell can have a maximum of eighteen electrons. The atomic number is the number of electrons in that element. The atomic number of vanadium(V) is 23. That is, the number of electrons in vanadium is twenty-three. Therefore, the vanadium atom will have two electrons in the first shell, eight in the 2nd orbit. According to Bohr’s formula, the third orbit will have thirteen electrons but the third orbit of vanadium will have eleven electrons and the remaining two electrons will be in the fourth orbit. Therefore, the order of the number of electrons in each shell of the vanadium(V) atom is 2, 8, 11, 2.

Electrons can be arranged correctly through orbits from elements 1 to 18. The electron configuration of an element with an atomic number greater than 18 cannot be properly determined according to the Bohr atomic model. The **electron configuration of all the elements** can be done through the orbital diagram.

## Electron configuration of vanadium(V) through orbital

The German physicist Aufbau first proposed an idea of electron configuration through sub-orbits. The Aufbau method is to do electron configuration through the sub-energy level. These sub-orbitals are expressed by ‘l’. The Aufbau principle is that the electrons present in the atom will first complete the lowest energy orbital and then gradually continue to complete the higher energy orbital. These orbitals are named s, p, d, f. The electron holding capacity of these orbitals is s = 2, p = 6, d = 10 and f = 14. The Aufbau electron configuration method is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d.

The first two electrons of vanadium enter the 1s orbital. The s-orbital can have a maximum of two electrons. Therefore, the next two electrons enter the 2s orbital. The p-orbital can have a maximum of six electrons. So, the next six electrons enter the 2p orbital. The second orbit is now full. So, the remaining electrons will enter the third orbit. Then two electrons will enter the 3s orbital of the third orbit and the next six electrons will be in the 3p orbital. The 3p orbital is now full. So, the next two electrons will enter the 4s orbital and the remaining three electrons will enter the 3d orbital. **Therefore, the vanadium(V) electron configuration will be 1s ^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}.**

## How to write the orbital diagram for vanadium(V)?

To create an orbital diagram of an atom, you first need to know Hund’s principle and Pauli’s exclusion principle. Hund’s principle is that electrons in different orbitals with the same energy would be positioned in such a way that they could be in the unpaired state of maximum number and the spin of the unpaired electrons will be one-way. And Pauli’s exclusion principle is that the value of four quantum numbers of two electrons in an atom cannot be the same. To write the orbital diagram of vanadium(V), you have to do the electron configuration of vanadium. Which has been discussed in detail above. 1s is the closest and lowest energy orbital to the nucleus. Therefore, the electron will first enter the 1s orbital. According to Hund’s principle, the first electron will enter in the clockwise direction and the next electron will enter the 1s orbital in the anti-clockwise direction. The 1s orbital is now filled with two electrons. Then the next two electrons will enter the 2s orbital just like the 1s orbital.

The next three electrons will enter the 2p orbital in the clockwise direction and the next three electrons will enter the 2p orbital in the anti-clockwise direction. The next two electrons will enter the 3s orbital. Then the next three electrons will enter the 3p orbital in the clockwise direction and the next three electrons will enter the 3p orbital in the anti-clockwise direction. The 3p orbital is now full. So, the next two electrons will enter the 4s orbital just like the 1s orbital and the remaining three electrons will enter the 3d orbital in the clockwise direction. This is clearly shown in the figure of the orbital diagram of vanadium(V).

## Vanadium(V) excited state electron configuration

Atoms can jump from one orbital to another orbital by excited state. This is called quantum jump. Ground state electron configuration of vanadium is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}. The valency of the element is determined by electron configuration in the excited state. The p-orbital has three sub-orbitals. The sub-orbitals are p_{x}, p_{y}, and p_{z}. Each sub-orbital can have a maximum of two electrons.

In the vanadium(V) ground-state electron configuration, the three electrons of the 3d orbital are located in the d_{xy}, d_{yz}, and d_{zx} sub-orbitals. The d-orbital has five sub-orbitals. The sub-orbitals are d_{xy}, d_{yz}, d_{zx}, d_{x2-y2}, and d_{z2}. Each sub-orbital can have a maximum of two electrons. Then the correct electron configuration of vanadium in the ground state will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d_{xy}^{1} 3d_{yz}^{1} 3d_{zx}^{1} 4s^{2}. This electron configuration shows that the last shell of the vanadium atom has three unpaired electrons(3d_{xy}^{1} 3d_{yz}^{1} 3d_{zx}^{1}). So the valency of vanadium is 3.

When a vanadium atom is excited, then the vanadium atom absorbs energy. As a result, an electron in the 4s orbital jumps to the 4p_{x} sub-orbital. Therefore, the electron configuration of vanadium(V*) in excited state will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2} 3d_{xy}^{1} 3d_{yz}^{1} 3d_{zx}^{1} 4s^{1} 4p_{x}^{1}. Here, vanadium has five unpaired electrons(3d_{xy}^{1} 3d_{yz}^{1} 3d_{zx}^{1} 4s^{1} 4p_{x}^{1}). Therefore, the valency of vanadium is 5.

From the above information, we can say that vanadium exhibits variable valency. Therefore, the valency of vanadium is 3, 5. Due to this, the oxidation states of vanadium are +3, +5. Vanadium also exhibits +2 and +4 oxidation states.

## Vanadium ion(V^{2+},V^{3+}) electron configuration

The electron configuration of vanadium shows that the last shell of vanadium has two electrons and the d-orbital has a total of three electrons. In this case, the **valence electrons** of vanadium are five. There are two types of vanadium ions. The ionic state of the element changes depending on the bond formation. The vanadium atom exhibits 2+ and 3+ ions. The vanadium atom donates two electrons from the last shell to form the vanadium ion(V^{2+}).

V – 2e^{–} → V^{2+}

Here, The electron configuration of vanadium ion(V^{2+}) is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3}. The vanadium atom donates two electrons in 4s orbital and an electron in 3d orbital to convert to vanadium ion(V^{3+}).

V – 3e^{–} → V^{3+}

The electron configuration of vanadium ion(V^{3+}) is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}. The electron configuration of vanadium ion(V^{3+}) shows that a vanadium ion has three shells and that shell has ten electrons. For this, the valence electrons of the vanadium ion(V^{3+}) are ten.

## Compound formation of vanadium(V)

Vanadium participates in the formation of bonds through its valence electrons. We know that the valence electrons in vanadium are five. These valence electrons participate in the formation of bonds with atoms of other elements. The **electron configuration of oxygen** shows that the **valence electrons of oxygen** are six. The vanadium atom donates its valence electrons to the oxygen atom and the oxygen atom receives those electrons. As a result, oxygen acquires the **electron configuration of neon**, and vanadium atoms acquire the **electron configuration of argon**. Vanadium(V) oxide(V_{2}O_{5}) is formed by the exchange of electrons between two atoms of vanadium and five atoms of oxygen. Vanadium(V) oxide(V_{2}O_{5}) is ionic bonding.

## FAQs

What is the symbol for vanadium?**Ans:** The symbol for vanadium is ‘V’.

How many electrons does vanadium(V) have?**Ans:** 23 electrons.

How do you write the electron configuration for vanadium?**Ans:** Vanadium(V) electron configuration is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}.

How many valence electrons does vanadium(V) have?**Ans:** Five valence electrons.

What is the valency of vanadium(V)?**Ans:** The valency of vanadium are 2, 3, 4, 5.

**Reference**