Vanadium Electron Configuration and Atomic Orbital Diagram
Vanadium is the 23th element in the periodic table and the symbol is ‘V’. Vanadium has an atomic number of 23, which means that its atom has 23 electrons around its nucleus.
The electron configuration of vanadium is 1s2 2s2 2p6 3s2 3p6 4s2 3d3 which means that the first two electrons enter the 1s orbital. Since the 1s orbital can hold only two electrons, the next two will enter the 2s orbital. The next six electrons enter the 2p subshell. The p subshell can hold a maximum of six electrons. So first we put six electrons in the 2p subshell and then the next two electrons in the 3s orbital.
Since the 3s is now full, the electrons will move to the 3p subshell, where the next six electrons will enter. The 3p subshell is now full. Consequently, the following two electrons will enter the 4s orbital. Since the 4s orbital is now full, the remaining three electrons will move into the 3d subshell. Hence, the electron configuration of vanadium will be 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
The electron configuration of vanadium refers to the arrangement of electrons in the vanadium atom’s orbitals. It describes how electrons are distributed among the various atomic orbitals and energy levels, and provides a detailed map of where each electron is likely to be found.
To understand the mechanism of vanadium electron configuration, you need to understand two basic things. These are orbits and orbitals. Also, you can arrange electrons in those two ways. In this article, I have discussed all the necessary points to understand the mechanism of vanadium and V²⁺, V³⁺, and V⁵⁺ ions electron configuration, ground and excited state electron configuration, valency and valence electrons. I hope this will be helpful in your study.
Electron arrangement of Vanadium through Bohr Model
Scientist Niels Bohr was the first to give an idea of the atom’s orbit. He provided a model of the atom in 1913 and provided a complete idea of orbit in that model.
The electrons of the atom revolve around the nucleus in a certain circular path. These circular paths are called orbits (shells or energy levels). These orbits are expressed by n. [n = 1,2,3,4 . . . The serial number of the orbit]
The name of the first orbit is K, 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 2n2.
| Shell Number (n) | Shell Name | Electrons Holding Capacity (2n2) |
| 1 | K | 2 |
| 2 | L | 8 |
| 3 | M | 18 |
| 4 | N | 32 |
Explanation:
- Let, n = 1 for K orbit. So, the maximum electron holding capacity in the K orbit is 2n2 = 2 × 12 = 2 electrons.
- n = 2, for L orbit. The maximum electron holding capacity in the L orbit is 2n2 = 2 × 22 = 8 electrons.
- n=3 for M orbit. The maximum electron holding capacity in the M orbit is 2n2 = 2 × 32 = 18 electrons.
- n=4 for N orbit. The maximum electron holding capacity in N orbit is 2n2 = 2 × 42 = 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 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 atom is 2, 8, 11, 2.
The Bohr atomic model has many limitations. In the Bohr atomic model, the electrons can only be arranged in different shells but the exact position, orbital shape, and spin of the electron cannot be determined.
Also, electrons can be arranged correctly from 1 to 18 elements. The electron arrangement of any element with atomic number greater than 18 cannot be accurately determined by the Bohr atomic model following the 2n2 formula. We can overcome all limitations of the Bohr model following the electron configuration through orbital.
Electron configuration of Vanadium through Aufbau Model
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 orbitals | Subshell name | Electrons holding capacity | Electron configuration |
| 1 | 0 | 1 | 1 | 1s | 2 | 1s2 |
| 2 | 0 1 | 2 | 1 3 | 2s 2p | 2 6 | 2s2 2p6 |
| 3 | 0 1 2 | 3 | 1 3 5 | 3s 3p 3d | 2 6 10 | 3s2 3p6 3d10 |
| 4 | 0 1 2 3 | 4 | 1 3 5 7 | 4s 4p 4d 4f | 2 6 10 14 | 4s2 4p6 4d10 4f14 |
Explanation:
- 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.
| Sub-shell name | Name source | Value of ‘l’ | Value of ‘m’ (0 to ± l) | Number of orbital (2l+1) | Electrons holding capacity 2(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. Following the Aufbau principle, the sequence of entry of electrons into orbitals is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.
Therefore, the complete electron configuration for vanadium should be written as 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
Note: The abbreviated electron configuration of vanadium is [Ar] 3d3 4s2. When writing an electron configuration, you have to write serially.
How to write the orbital diagram for vanadium?
Orbital diagrams are usually represented by boxes. Each box represents an orbital and the arrows within the box represent the position of the electron. The boxes are arranged in order of energy of the orbitals.
The lowest energy orbitals are closest to the nucleus and the higher energy orbitals are progressively further away from the nucleus in order of their energy levels. To write the orbital diagram of vanadium, you have to write the orbital notation of vanadium. Which has been discussed in detail above.
1s is the closest and lowest energy orbital to the nucleus. Therefore, the electrons will first enter the 1s orbital. According to Hund’s principle, the first electron will enter 1s orbital 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. 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 2p orbital is now full. Then the next two electrons will enter the 3s orbital just like the 1s 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.
Try the Orbital Diagram Calculator and get instant results for any element.
The excited state electron configuration of Vanadium
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 1s2 2s2 2p6 3s2 3p6 3d3 4s2. In the vanadium ground-state electron configuration, the three electrons of the 3d orbital are located in the dxy, dyz, and dzx orbitals.
We already know that the d-subshell has five orbitals. The orbitals are dxy, dyz, dzx, dx2-y2 and dz2 and each orbital can have a maximum of two electrons. Then the correct electron configuration of vanadium in the ground state will be 1s2 2s2 2p6 3s2 3p6 3dxy1 3dyz1 3dzx1 4s2. 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 4px orbital. We already know that the p-subshell has three orbitals. The orbitals are px, py, and pz and each orbital can have a maximum of two electrons. Therefore, the electron configuration of vanadium(V*) in an excited state will be 1s2 2s2 2p6 3s2 3p6 3d3 4s2 3dxy1 3dyz1 3dzx1 4s1 4px1. 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(V2+, V3+) 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(V2+).
V – 2e– → V2+
Here, the electron configuration of vanadium ion(V2+) is 1s2 2s2 2p6 3s2 3p6 3d3. The vanadium atom donates two electrons in 4s orbital and an electron in 3d orbital to convert to vanadium ion(V3+).
V – 3e– → V3+
The electron configuration of vanadium ion(V3+) is 1s2 2s2 2p6 3s2 3p6 3d2. This electron configuration shows that the vanadium ion(V3+) has three shells and the last shell has ten electrons. For this, the valence electrons of the vanadium ion(V3+) 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(V2O5) is formed by the exchange of electrons between two atoms of vanadium and five atoms of oxygen. Vanadium oxide(V2O5) is ionic bonding.
