Electron Configuration for Aluminum (Al, Al3+ ion)
Aluminum is the 13th element in the periodic table and its symbol is ‘Al’. In this article, I have discussed in detail how to easily write the complete electron configuration of aluminum.
What is the electron configuration of aluminum?
The total number of electrons in aluminum is thirteen. These electrons are arranged according to specific rules in different orbitals.
The arrangement of electrons in aluminum in specific rules in different orbits and orbitals is called the electron configuration of aluminum.
The electron configuration of aluminum is [Ne] 3s2 3p1, 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.
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 2n2.
| Shell Number (n) | Shell Name | Electrons Holding Capacity (2n2) |
| 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 2n2 = 2 × 12 = 2. - For L orbit, n = 2.
The maximum electron holding capacity in L orbit is 2n2 = 2 × 22 = 8. - n=3 for M orbit.
The maximum electrons holding capacity in M orbit is 2n2 = 2 × 32 = 18. - n=4 for N orbit.
The maximum electrons holding capacity in N orbit is 2n2 = 2 × 42 = 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 aluminum is 13. That is, the number of electrons in aluminum is thirteen. Therefore, the aluminum atom will have two electrons in the first shell, eight in the 2nd orbit, and three electrons in the 3rd shell.
Therefore, the order of the number of electrons in each shell of an aluminum(Al) atom is 2, 8, 3. 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 orbital diagrams.
Electron configuration 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 | 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 |
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.
| 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. 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 aluminum 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 remaining one electron will be in the 3p orbital. Therefore, the aluminum full electron configuration will be 1s2 2s2 2p6 3s2 3p1.
Note: The unabbreviated electron configuration of aluminum is [Ne] 3s2 3p1. When writing an electron configuration, you have to write serially.
Aluminum ion(Al3+) electron configuration
The ground state electron configuration of aluminum is 1s2 2s2 2p6 3s2 3p1. After the electron configuration, the last shell of the aluminum atom has three electrons. Therefore, the valency and valence electrons of aluminum are 3.
The elements that have 1, 2, or 3 electrons in the last shell donate the electrons in the last shell during bond formation. Aluminum donates the electron of the last shell to form bonds and turns into an aluminum ion(Al3+).
The elements that form bonds by donating electrons are called cations. That is, aluminum is a cation element.
Al – 3e– → Al3+
The electron configuration of aluminum ion(Al3+) is 1s2 2s2 2p6. This electron configuration shows that aluminum ion(Al3+) has acquired the electron configuration of neon and it achieves an octave full stable electron configuration.
FAQs
How do you write the complete electron configuration for aluminum?
Ans: The complete electron configuration for aluminum is 1s2 2s2 2p6 3s2 3p1.
How many full electron levels does aluminum have?
Aluminum has two full electron levels. The electron configuration of aluminum 1s2 2s2 2p6 3s2 3p1, indicates that the 1s, 2s, 2p, 3s, and 3p subshells are all occupied. The first two levels (1s and 2s/2p) are completely filled, while the third level (3s/3p) has two electrons in the 3s subshell and one electron in the 3p subshell.
What is the valence electron configuration for the aluminum atom?
The valence electron configuration for the magnesium atom is [Ne] 3s2 3p1.
What is the noble gas electron configuration of aluminum (Al)?
The noble gas electron configuration of aluminum (Al) is [Ne] 3s2 3p1.
What is the complete ground state electron configuration for the aluminum atom?
The complete ground state electron configuration for the aluminum atom is 1s2 2s2 2p6 3s2 3p1.
How many electrons does an atom of aluminum have in its outer electron configuration?
An atom of aluminum has three electrons in its outer electron configuration. The outer electron configuration refers to the electrons present in the outermost energy level (valence shell). In the case of aluminum, the outer electron configuration is 3s2 3p1, which means there are two electrons in the 3s subshell and one electron in the 3p subshell.
What is the electron configuration of Al3+?
The electron configuration of aluminum ion(Al3+) is 1s2 2s2 2p6. This electron configuration shows that aluminum ion(Al3+) has acquired the electron configuration of neon and it achieves an octave full stable electron configuration.
How many electrons in the last level does aluminum have?
Aluminum has 3 electrons in the last energy level (valence level). The last energy level refers to the outermost electron shell, which, in the case of aluminum, is the third energy level (n = 3).
How many electrons will be in the n=3 shell of aluminum?
In the n=3 shell of aluminum, there will be a total of 8 electrons. The n=3 shell consists of the 3s and 3p subshells. The 3s subshell can hold a maximum of 2 electrons, while the 3p subshell can hold a maximum of 6 electrons. Adding these together, 2 + 6 = 8 electrons in the n=3 shell of aluminum.
How many unpaired electrons does aluminum have?
Aluminum has one unpaired electron. In its electron configuration, the 3p subshell of aluminum contains one electron.
What is the ground-state electron configuration of an Al3+ ion?
The ground-state electron configuration of an Al3+ ion is 1s2 2s2 2p6.
