# Electron Configuration for Vanadium (V and V2+, V3+ ion)

Vanadium is the 23rd element in the periodic table and its symbol is ‘V’. In this article, I have discussed in detail how to easily write the complete electron configuration of vanadium.

## What is the electron configuration of vanadium?

The total number of electrons in vanadium is twenty-three. These electrons are arranged according to specific rules in different orbitals.

The arrangement of electrons in vanadium in specific rules in different orbits and orbitals is called the electron configuration of vanadium.

The electron configuration of vanadium is [Ar] 3d^{3} 4s^{2}, if the electron arrangement is through orbitals. Electron configuration can be done in two ways.

- Electron configuration through orbit (Bohr principle)
- Electron configuration through orbital (Aufbau principle)

Electron configuration through orbitals follows different principles. For example Aufbau principle, Hund’s principle, and Pauli’s exclusion principle.

## Vanadium atom 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, and N is the name of the fourth orbit. The electron holding capacity of each orbit is 2n^{2}.

Shell Number (n) | Shell Name | Electrons Holding Capacity (2n^{2}) |

1 | K | 2 |

2 | L | 8 |

3 | M | 18 |

4 | N | 32 |

For example,

- n = 1 for K orbit.

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

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

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

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

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 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 and 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 through orbital

Atomic energy shells are subdivided into sub-energy levels. These sub-energy levels are also called orbital. The most probable region of electron rotation around the nucleus is called the orbital.

The sub-energy levels depend on the azimuthal quantum number. It is expressed by ‘l’. The value of ‘l’ is from 0 to (n – 1). The sub-energy levels are known as s, p, d, and f.

Orbit Number | Value of ‘l’ | Number of subshells | Number of orbital | Subshell name | Electrons holding capacity | Electron configuration |

1 | 0 | 1 | 1 | 1s | 2 | 1s^{2} |

2 | 0 1 | 2 | 1 3 | 2s 2p | 2 6 | 2s^{2} 2p^{6} |

3 | 0 1 2 | 3 | 1 3 5 | 3s 3p 3d | 2 6 10 | 3s^{2} 3p^{6} 3d^{10} |

4 | 0 1 2 3 | 4 | 1 3 5 7 | 4s 4p 4d 4f | 2 6 10 14 | 4s^{2} 4p^{6} 4d^{10} 4f^{14} |

For example,

- If n = 1,

(n – 1) = (1–1) = 0

Therefore, the value of ‘l’ is 0. So, the sub-energy level is 1s. - If n = 2,

(n – 1) = (2–1) = 1.

Therefore, the value of ‘l’ is 0, 1. So, the sub-energy levels are 2s, and 2p. - If n = 3,

(n – 1) = (3–1) = 2.

Therefore, the value of ‘l’ is 0, 1, 2. So, the sub-energy levels are 3s, 3p, and 3d. - If n = 4,

(n – 1) = (4–1) = 3

Therefore, the value of ‘l’ is 0, 1, 2, 3. So, the sub-energy levels are 4s, 4p, 4d, and 4f. - If n = 5,

(n – 1) = (n – 5) = 4.

Therefore, l = 0,1,2,3,4. The number of sub-shells will be 5 but 4s, 4p, 4d, and 4f in these four subshells it is possible to arrange the electrons of all the elements of the periodic table.

Subshell name | Name source | Value of ‘l’ | Value of ‘m’(0 to ± l) | Number of orbital (2l+1) | Electrons holding capacity2(2l+1) |

s | Sharp | 0 | 0 | 1 | 2 |

p | Principal | 1 | −1, 0, +1 | 3 | 6 |

d | Diffuse | 2 | −2, −1, 0, +1, +2 | 5 | 10 |

f | Fundamental | 3 | −3, −2, −1, 0, +1, +2, +3 | 7 | 14 |

The orbital number of the s-subshell is one, three in the p-subshell, five in the d-subshell and seven in the f-subshell. Each orbital can have a maximum of two electrons.

The sub-energy level ‘s’ can hold a maximum of two electrons, ‘p’ can hold a maximum of six electrons, ‘d’ can hold a maximum of ten electrons, and ‘f’ can hold a maximum of fourteen electrons.

Aufbau is a German word, which means building up. The main proponents of this principle are scientists Niels Bohr and Pauli. The Aufbau method is to do electron configuration through the sub-energy level.

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.

The energy of an orbital is calculated from the value of the principal quantum number ‘n’ and the azimuthal quantum number ‘l’. The orbital for which the value of (n + l) is lower is the low energy orbital and the electron will enter that orbital first.

Orbital | Orbit (n) | Azimuthal quantum number (l) | Orbital energy (n + l) |

3d | 3 | 2 | 5 |

4s | 4 | 0 | 4 |

Here, the energy of 4s orbital is less than that of 3d. So, the electron will enter the 4s orbital first and enter the 3d orbital when the 4s orbital is full.

The method of entering electrons into orbitals through the Aufbau principle 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 will enter the 3d orbital. Therefore, the vanadium full electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}.

Note:The abbreviated electron configuration of vanadium is [Ar] 3d^{3}4s^{2}. When writing an electron configuration, you have to write serially.

## Vanadium excited state electron configuration

Atoms can jump from one orbital to another orbital in the excited state. This is called quantum jump.

The ground state electron configuration of vanadium is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}. In the vanadium ground-state electron configuration, the three electrons of the 3d orbital are located in the d_{xy}, d_{yz}, and d_{zx} orbitals.

We already know that the d-subshell has five orbitals. The orbitals are d_{xy}, d_{yz}, d_{zx}, d_{x2-y2} and d_{z2} and each 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 3d orbital of a vanadium atom has three unpaired electrons. 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} orbital.

We already know that the p-subshell has three orbitals. The orbitals are p_{x}, p_{y}, and p_{z} and each orbital can have a maximum of two electrons.

Therefore, the electron configuration of vanadium(V*) in an 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}. The valency of the element is determined by electron configuration in the excited state.

Here, vanadium has five unpaired electrons. 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, and +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}.

This electron configuration shows that the vanadium ion(V^{3+}) has three shells and the last shell has ten electrons. For this, the valence electrons of the vanadium ion(V^{3+}) are ten.

## Compound formation of vanadium

Vanadium participates in the formation of bonds through its valence electrons. 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 oxide(V_{2}O_{5}) is formed by the exchange of electrons between two atoms of vanadium and five atoms of oxygen. Vanadium 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 have?

**Ans:**23 electrons.### How do you write the full electron configuration for vanadium?

**Ans:**1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^{2}.### How many valence electrons does vanadium have?

**Ans:**Five valence electrons.### What is the valency of vanadium?

**Ans:**The valency of vanadium is 2, 3, 4, and 5.