Electron Configuration Calculator

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quantum numbers of the last electron

Quantum numbers of the last electron: n, ℓ, mℓ and ms step by step

The last electron is the one in the highest-energy subshell, and the four quantum numbers give the shell, the subshell, the orbital orientation and the spin. The tool calculates all four and lets you choose your book's spin convention.

How to read each number from the configuration

First find the highest-energy subshell, the last in the diagram order. Two numbers come from it: n is the shell number and ℓ is 0, 1, 2 or 3 for s, p, d or f. The magnetic number mℓ runs from -ℓ to +ℓ and identifies the orbital within the subshell: an s subshell has 1 orbital (mℓ = 0), p has 3 (-1, 0, +1), d has 5 (-2 to +2) and f has 7 (-3 to +3). The spin ms is +1/2 or -1/2. These four numbers identify the electron, and the Pauli exclusion principle guarantees that two electrons in an atom never share all four.
To find mℓ and ms, draw the subshell's boxes and fill them by Hund's rule: one electron in each box, from left (mℓ = -ℓ) to right (+ℓ), all with the arrow up, and only then pair again from left to right. The last electron is the last one drawn. In nitrogen, 2p3, the three boxes get one arrow each and the last is at mℓ = +1 with the arrow up. In oxygen, 2p4, the fourth electron closes the first box (mℓ = -1) with the arrow down. In iron, 3d6, it is the sixth: the first box (mℓ = -2) gets the down arrow.
The sign of the spin depends on the material's convention. The most widely used internationally gives ms = +1/2 to the up arrow; several Brazilian books do the opposite and give -1/2 to the first arrow. Both are right as long as the same one is used throughout, and the tool has a switch that flips only the sign of ms, leaving n, ℓ and mℓ intact. The most common mistake is taking the subshell with the highest n as the last: in chromium, 3d5 4s1, the highest-energy subshell is 3d (n + ℓ = 5), not 4s; in palladium, 4d10, the last electron is the 10th in 4d, mℓ = +2, and there is no electron in 5s. For ions, the last electron is the one in the highest-energy subshell that remains.

Quantum numbers (convention ↑ = +1/2)

Sodium, 3s1

Input: Na
Expected output: n = 3, l = 0, ml = 0, ms = +1/2

A single s orbital, with the arrow up.

Nitrogen, 2p3

Input: N
Expected output: n = 2, l = 1, ml = +1, ms = +1/2

Third electron: third box, unpaired.

Oxygen, 2p4

Input: O
Expected output: n = 2, l = 1, ml = -1, ms = -1/2

Fourth electron: back to the first box, pairing.

Chlorine, 3p5

Input: Cl
Expected output: n = 3, l = 1, ml = 0, ms = -1/2

Fifth electron: second box paired.

Iron, 3d6

Input: Fe
Expected output: n = 3, l = 2, ml = -2, ms = -1/2

Sixth electron: the first of five boxes gets its second.

Copper, 3d10

Input: Cu
Expected output: n = 3, l = 2, ml = +2, ms = -1/2

The highest-energy subshell is 3d, with 10 electrons; 4s1 does not count.

Full tool FAQ

How do you write the electron configuration of an element?

Take the atomic number (the number of electrons in the neutral atom) and distribute them over the subshells in Madelung order, filling each to its maximum (s 2, p 6, d 10, f 14). For chlorine (Z = 17): 1s2 2s2 2p6 3s2 3p5, or [Ne] 3s2 3p5, with 2, 8 and 7 electrons in shells K, L and M.

What is the filling order of the Madelung diagram?

What are the exceptions to the filling order?

How do you write the configuration of an ion?

What are the valence shell and the highest-energy subshell?

How do you find the quantum numbers of the last electron?

What is Hund's rule and what is paramagnetism?

What does [Ar] 4s2 3d6 mean?

How do you know how many electrons each shell has?

Where does the data come from, and does it cover 109 to 118?

Frequently asked questions

Is the last electron the one in the highest n or the highest energy?

The highest-energy one, the last in the diagram order. In most cases it matches the highest n, but not in iron, where the highest-energy subshell is 3d (n = 3) even though 4s has n = 4.

Which ml values are possible for each subshell?

From -ℓ to +ℓ in steps of 1: s has 0; p has -1, 0, +1; d has -2, -1, 0, +1, +2; f runs from -3 to +3. The number of values, 2ℓ + 1, is the number of orbitals, and each orbital holds 2 electrons.

Why does my book give -1/2 to the first electron?

Because it adopts the opposite convention, in which the up arrow is -1/2. Choose the "arrow ↑ = -1/2" option in the tool and only the sign of ms changes. Always follow the convention of your class material.

Next steps

How to read each number from the configuration

First find the highest-energy subshell, the last in the diagram order. Two numbers come from it: n is the shell number and ℓ is 0, 1, 2 or 3 for s, p, d or f. The magnetic number mℓ runs from -ℓ to +ℓ and identifies the orbital within the subshell: an s subshell has 1 orbital (mℓ = 0), p has 3 (-1, 0, +1), d has 5 (-2 to +2) and f has 7 (-3 to +3). The spin ms is +1/2 or -1/2. These four numbers identify the electron, and the Pauli exclusion principle guarantees that two electrons in an atom never share all four.

To find mℓ and ms, draw the subshell's boxes and fill them by Hund's rule: one electron in each box, from left (mℓ = -ℓ) to right (+ℓ), all with the arrow up, and only then pair again from left to right. The last electron is the last one drawn. In nitrogen, 2p3, the three boxes get one arrow each and the last is at mℓ = +1 with the arrow up. In oxygen, 2p4, the fourth electron closes the first box (mℓ = -1) with the arrow down. In iron, 3d6, it is the sixth: the first box (mℓ = -2) gets the down arrow.

The sign of the spin depends on the material's convention. The most widely used internationally gives ms = +1/2 to the up arrow; several Brazilian books do the opposite and give -1/2 to the first arrow. Both are right as long as the same one is used throughout, and the tool has a switch that flips only the sign of ms, leaving n, ℓ and mℓ intact. The most common mistake is taking the subshell with the highest n as the last: in chromium, 3d5 4s1, the highest-energy subshell is 3d (n + ℓ = 5), not 4s; in palladium, 4d10, the last electron is the 10th in 4d, mℓ = +2, and there is no electron in 5s. For ions, the last electron is the one in the highest-energy subshell that remains.

Quantum numbers (convention ↑ = +1/2)

Sodium, 3s1

Input: Na
Expected output: n = 3, l = 0, ml = 0, ms = +1/2

A single s orbital, with the arrow up.

Nitrogen, 2p3

Input: N
Expected output: n = 2, l = 1, ml = +1, ms = +1/2

Third electron: third box, unpaired.

Oxygen, 2p4

Input: O
Expected output: n = 2, l = 1, ml = -1, ms = -1/2

Fourth electron: back to the first box, pairing.

Chlorine, 3p5

Input: Cl
Expected output: n = 3, l = 1, ml = 0, ms = -1/2

Fifth electron: second box paired.

Iron, 3d6

Input: Fe
Expected output: n = 3, l = 2, ml = -2, ms = -1/2

Sixth electron: the first of five boxes gets its second.

Copper, 3d10

Input: Cu
Expected output: n = 3, l = 2, ml = +2, ms = -1/2

The highest-energy subshell is 3d, with 10 electrons; 4s1 does not count.

Full tool FAQ

How do you write the electron configuration of an element?

What is the filling order of the Madelung diagram?

What are the exceptions to the filling order?

How do you write the configuration of an ion?

What are the valence shell and the highest-energy subshell?

How do you find the quantum numbers of the last electron?

What is Hund's rule and what is paramagnetism?

What does [Ar] 4s2 3d6 mean?

How do you know how many electrons each shell has?

Where does the data come from, and does it cover 109 to 118?

Frequently asked questions

Is the last electron the one in the highest n or the highest energy?

The highest-energy one, the last in the diagram order. In most cases it matches the highest n, but not in iron, where the highest-energy subshell is 3d (n = 3) even though 4s has n = 4.