Term Symbols and Spectroscopic Notation for Beginners

Open a table of atomic data and next to each element you’ll see something like ²S₁/₂, ³P₀ or ⁵D₄. These term symbols pack a surprising amount of information into a few characters: how the electrons’ spins add up, how their orbital motions combine, and how the two couple together. Once you know the method, you can write the ground-state term symbol for most atoms in under a minute.

This guide assumes you’re comfortable with quantum numbers and electron configurations.

What a term symbol looks like

A term symbol has the form:

²ˢ⁺¹L_J

• S is the total spin quantum number of the atom. The superscript is the spin multiplicity, 2S + 1.
• L is the total orbital angular momentum quantum number, written as a capital letter.
• J is the total angular momentum quantum number, written as a subscript.

The letters for L follow the same pattern as orbital letters, just in capitals:

L	0	1	2	3	4	5
Letter	S	P	D	F	G	H

So ³P₀ means 2S + 1 = 3 (so S = 1), L = 1, and J = 0.

Don’t confuse the letter S (meaning L = 0) with the spin quantum number S. It’s an unfortunate clash of notation that everyone learns to live with.

The key simplification: ignore filled shells

A completely filled shell or subshell always has S = 0 and L = 0: every spin-up electron is paired with a spin-down one, and every positive mₗ is cancelled by a negative one. So you only need to look at partly filled subshells. For carbon, 1s² 2s² 2p², only the two 2p electrons matter.

This also tells you that every noble gas, and any atom with only closed subshells (like beryllium, magnesium or zinc), has the ground-state term ¹S₀.

The method: ground-state term symbol in five steps

The procedure uses Hund’s rules, which pick out the lowest-energy arrangement.

Step 1. Draw boxes for the open subshell, labelled with mₗ values from +l down to −l. For p that’s +1, 0, −1; for d it’s +2, +1, 0, −1, −2.

Step 2. Fill the electrons following Hund’s rule: put one electron, spin up, in each box starting from the highest mₗ, and only then start pairing (spin down), again from the highest mₗ.

Step 3. Find S. Count unpaired electrons and multiply by ½. Then write the multiplicity 2S + 1. (A quick shortcut: the multiplicity is the number of unpaired electrons plus one.)

Step 4. Find L. Add up the mₗ values of all the electrons. Because you filled from the top, this sum is the maximum possible value, which equals L. Convert it to a letter.

Step 5. Find J. J can take values from |L − S| to L + S. For the ground state:

• if the subshell is less than half full, J = |L − S| (the lowest value);
• if it is more than half full, J = L + S (the highest value);
• if it is exactly half full, L = 0, so J = S.

Then assemble ²ˢ⁺¹L_J.

Worked example 1: carbon (2p²)

• Boxes: +1, 0, −1.
• Electrons: one up in +1, one up in 0.
• Unpaired electrons = 2, so S = 1 and the multiplicity is 3.
• ML = (+1) + (0) = 1, so L = 1 → P.
• Two electrons in a subshell that holds six: less than half full, so J = |1 − 1| = 0.

Ground term: ³P₀.

Worked example 2: oxygen (2p⁴)

• Four electrons in the p boxes: up in +1, 0, −1, then the fourth pairs (down) in +1.
• Unpaired electrons = 2, so S = 1, multiplicity 3.
• ML = (+1) + 0 + (−1) + (+1) = 1, so L = 1 → P.
• More than half full, so J = L + S = 2.

Ground term: ³P₂. Notice that carbon and oxygen have the same L and S but different J. This reflects the “hole” rule: four electrons in p behave like two missing electrons, but spin–orbit coupling flips the order of the J levels.

Worked example 3: nitrogen (2p³)

• One electron up in each of +1, 0, −1.
• Unpaired = 3, so S = 3/2 and the multiplicity is 4.
• ML = 1 + 0 − 1 = 0, so L = 0 → S.
• Half full: J = S = 3/2.

Ground term: ⁴S₃/₂. Half-filled subshells always give an S term, which is part of why they’re especially stable.

Worked example 4: iron (3d⁶ 4s²)

Only the 3d⁶ part counts.

• Boxes: +2, +1, 0, −1, −2.
• Five electrons up (one per box), sixth pairs in +2.
• Unpaired = 4, so S = 2, multiplicity 5.
• ML = (2 + 1 + 0 − 1 − 2) + 2 = 2, so L = 2 → D.
• Six electrons in a subshell that holds ten: more than half full, so J = L + S = 4.

Ground term: ⁵D₄.

Worked example 5: titanium (3d² 4s²)

• Two electrons up, in +2 and +1.
• S = 1, multiplicity 3.
• ML = 3, so L = 3 → F.
• Less than half full: J = |3 − 1| = 2.

Ground term: ³F₂.

Worked example 6: chromium (3d⁵ 4s¹)

Chromium is one of the famous configuration exceptions. Here both 3d and 4s are open.

• Five d electrons, all spin up: ML = 0 from d.
• One s electron, spin up: s has mₗ = 0 only.
• Six unpaired electrons, so S = 3 and the multiplicity is 7.
• L = 0 → S. With L = 0, J = S = 3.

Ground term: ⁷S₃, a very high multiplicity that goes with chromium’s many unpaired electrons.

Beyond the ground state: all the terms of p²

The ground state is only one of several terms that a configuration can produce. For p², the two electrons can be arranged in 15 different ways (microstates). Grouping them by L and S gives three terms:

Term	Degeneracy (2S + 1)(2L + 1)
³P	3 × 3 = 9
¹D	1 × 5 = 5
¹S	1 × 1 = 1

9 + 5 + 1 = 15, which is a useful check. Hund’s rules put ³P lowest, then ¹D, then ¹S. The excited terms show up as spectral lines and as different reactive forms of molecules. The famous “singlet oxygen” of photochemistry, for example, is a molecular cousin of this idea. Working out every term for a configuration takes a microstate table, which is a topic for a later guide.

Pitfalls to avoid

• Counting filled subshells. Only open subshells contribute. Including the 2s² of carbon changes nothing, but it wastes time and invites errors.
• Filling from the wrong end. Start at the highest mₗ, or your ML sum won’t equal L.
• Using the wrong J rule. Remember: less than half full → lowest J; more than half full → highest J.
• Confusing the multiplicity with S. A “triplet” has S = 1, not S = 3.
• Applying the method to heavy atoms without care. The scheme above, called Russell–Saunders or LS coupling, works well for light atoms. For very heavy elements, spin–orbit coupling becomes so strong that L and S stop being good quantum numbers, and a different scheme (jj coupling) is used. Real ground terms for some heavy atoms don’t follow the simple rules; see our note on relativistic effects.

Practice questions

Write the ground-state term symbol for each:

1. Boron (2p¹)
2. Fluorine (2p⁵)
3. Sodium (3s¹)
4. Neon
5. Vanadium (3d³ 4s²)
6. Copper (3d¹⁰ 4s¹)
7. Nickel (3d⁸ 4s²)

Answers

1. One electron in +1: S = ½, L = 1, less than half full → J = ½. ²P₁/₂
2. Five electrons: ML = 1, one unpaired, more than half full → J = 3/2. ²P₃/₂
3. One s electron: L = 0, S = ½, J = ½. ²S₁/₂
4. All subshells full. ¹S₀
5. Three d electrons up in +2, +1, 0: S = 3/2, L = 3, less than half full → J = 3/2. ⁴F₃/₂
6. Filled 3d, one 4s electron. ²S₁/₂
7. d⁸: five up, three paired in +2, +1, 0; ML = 3, S = 1, more than half full → J = 4. ³F₄

Key takeaways

• A term symbol ²ˢ⁺¹L_J summarises an atom’s total spin (S), orbital angular momentum (L) and their combination (J).
• Only partly filled subshells matter; closed shells contribute nothing.
• For the ground state: fill boxes by Hund’s rule from the highest mₗ, read off S and L, then choose J as lowest (less than half full) or highest (more than half full).
• Half-filled subshells give S terms; closed shells give ¹S₀.
• The simple method assumes LS coupling, which works best for lighter atoms.
• For the pictures behind the numbers, see atomic orbitals and their shapes.