Bohr's Model of Atom

Rutherford's atom had a fatal flaw: an electron circling a nucleus should continuously radiate energy and spiral inward within a fraction of a second, yet atoms are stable for billions of years. In 1913 Niels Bohr fixed this by proposing that electrons are not free to orbit anywhere — they can only travel in certain fixed, discrete paths called orbits, shells, or energy levels, and while confined to one of these paths an electron neither gains nor loses energy. This single idea rescued atomic stability and, through the accompanying Bohr–Bury rules for filling shells, gave chemistry its everyday shell-based picture of the atom: K, L, M, N shells holding a fixed maximum of electrons each, with the outermost shell capped at eight. This article builds that picture from Bohr's postulates through the $2n^2$ rule to full worked electron distributions, the same skill you will use constantly when studying valency and chemical bonding.

The core insight

Electrons occupy only fixed, numbered shells ( $n = 1, 2, 3, \dots$ , labelled K, L, M, N) in which they do not radiate energy, each shell holding at most $2n^2$ electrons subject to an outermost-shell cap of 8 and a next-to-outermost cap of 18, filled strictly from the inside outward.

The core idea

The stability problem Bohr solved. In Rutherford's nuclear model, electrons revolving around a nucleus should behave like any accelerating charge and lose energy as electromagnetic radiation, spiralling into the nucleus almost instantly. Since atoms plainly do not collapse, something in this picture had to be wrong — not the nucleus, but the assumption that electrons can occupy any orbit at any distance.

Bohr's postulates. Niels Bohr proposed a simple but far-reaching fix, built on three ideas:

Electrons revolve around the nucleus only in certain fixed, discrete circular paths, called orbits, shells, or energy levels — not just anywhere.
As long as an electron stays within one of these permitted orbits, it does not lose or radiate energy, so the atom remains stable indefinitely.
The shells are numbered outward from the nucleus as $n = 1, 2, 3, 4, \dots$ and carry the letter labels K ( $n=1$ ), L ( $n=2$ ), M ( $n=3$ ), N ( $n=4$ ), and so on.

The Bohr–Bury scheme for filling shells. Once shells exist as fixed slots, chemists needed a rule for how many electrons each slot can hold and in what order they fill. The maximum capacity of a shell is:

2n^2

where: $n$ is the shell number counting from the nucleus (K $=1$ , L $=2$ , M $=3$ , N $=4$ ). This gives maximum capacities of K $= 2$ , L $= 8$ , M $= 18$ , N $= 32$ . Two further restrictions sit on top of this formula and matter a great deal in practice:

The outermost shell of an atom can never hold more than 8 electrons, even where $2n^2$ would allow more.
The shell next to the outermost cannot exceed 18 electrons under this simplified school scheme, whether or not it has become the outermost shell yet.
Shells fill step by step from the inside out: an inner shell must reach its permitted capacity before any electron is placed in the next shell outward.

This shell-and-capacity picture is not just bookkeeping — the number of electrons left in the outermost shell (the valence electrons) is exactly what determines how an atom bonds with others, the subject of the valency article that follows this one.

Figure

Bohr shell diagram of a chlorine atom showing the nucleus and three shells K L M holding 2 8 and 7 electrons

Derivation AWorking out an atom's electron shell distribution

Turning a bare atomic number into a Bohr shell diagram means repeatedly filling one shell to its allowed capacity and carrying the leftover electrons outward, checking at each stage whether the shell you just filled has become the outermost one and so must obey the 8-electron cap rather than the full $2n^2$ value. Consider an atom of aluminium, atomic number 13, so it has 13 electrons to place.

Step 1. Fill the K shell first, since filling always starts from the shell nearest the nucleus. Its maximum capacity is $2n^2 = 2(1)^2 = 2$ . Place 2 electrons in K. Electrons remaining: $13 - 2 = 11$ .

Step 2. Move to the L shell ( $n = 2$ ), whose maximum capacity is $2(2)^2 = 8$ . Since 11 electrons remain and L can take up to 8, fill L completely with 8 electrons. Electrons remaining: $11 - 8 = 3$ .

Step 3. Move to the M shell ( $n = 3$ ). By the $2n^2$ formula alone, M could hold $2(3)^2 = 18$ electrons, but only 3 electrons are left to place, so all 3 go into M — well under both the raw $2n^2$ limit and the outermost-shell cap of 8, so no restriction is triggered here.

Step 4. Confirm the distribution and check it against both scheme rules. The full distribution is K $=2$ , L $=8$ , M $=3$ , written $2, 8, 3$ , and the total checks out as $2 + 8 + 3 = 13$ . The outermost shell (M, with 3 electrons) satisfies the $\le 8$ rule, and the next-to-outermost shell (L, with 8 electrons) satisfies the $\le 18$ rule, so the distribution is valid and complete.

Worked example

Question: An atom has the electron distribution $2, 8, 8, 2$ (from Pass-1). Extend the analysis: identify the element, state its valency, and explain why the M shell here is allowed 8 electrons even though $2n^2$ for M would permit 18.

Step 1. Add up the electrons: $2 + 8 + 8 + 2 = 20$ , so this is an atom with atomic number 20 — calcium.

Step 2. Identify the shells occupied: four shells are filled to some extent, K, L, M, N, corresponding to $n = 1, 2, 3, 4$ . The outermost occupied shell is N, holding 2 electrons.

Step 3. Check the M shell's electron count against the rules. By $2n^2$ , M ( $n=3$ ) could in principle hold $2(3)^2 = 18$ electrons, but here it holds only 8. This is not a violation — the $2n^2$ figure is only an upper ceiling; a shell may legitimately hold fewer electrons than its maximum once the outer shells begin filling in the step-wise sequence, as required by this scheme.

Step 4. Read off the valency. The outermost shell (N) has 2 electrons, which is 4 or fewer, so this atom tends to lose those 2 electrons to attain a stable outer shell, giving calcium a valency of 2.

Answer: the atom is calcium (Z = 20), with electron distribution $2, 8, 8, 2$ and valency 2.

Common traps

Applying $2n^2$ blindly to every shell — the outermost shell is capped at 8 electrons regardless of what $2n^2$ would otherwise allow.
Filling shells out of sequence, such as placing electrons in M before L has reached its permitted capacity — the scheme strictly requires inside-out, step-wise filling.
Forgetting the next-to-outermost shell's $\le 18$ condition, which only becomes relevant once electrons start occupying a fourth shell.
Confusing shell letters with shell numbers — K is $n = 1$ , not $n = 0$ , so always map the letter to the correct $n$ before applying $2n^2$ .

Ask the tutor: if you want to practise drawing Bohr shell diagrams for other elements, or want the K/L/M/N step-wise filling walked through again more slowly, ask the AI tutor for extra worked examples.

Why can't the outermost shell just follow 2n squared like every other shell?

The $2n^2$ formula gives the absolute physical capacity of a shell, but experiments on real atoms show that once a shell becomes the outermost one, it never holds more than 8 electrons, no matter how large $2n^2$ says it could be. This 8-electron cap is what later becomes the basis of the octet rule in bonding — atoms are most stable with 8 electrons in their outer shell, so nature simply never fills an outer shell past that point before starting a new one.

Do electrons really move in neat circular paths like the diagram shows?

Not exactly — Bohr's circular orbits were a simplified, visualizable model that correctly predicted shell capacities and atomic stability, which is why it is still taught for shell-filling and valency. A more accurate quantum-mechanical picture, developed later, replaces sharp orbits with probability clouds. For the Class 9 syllabus, however, the fixed-shell picture is the correct and sufficient tool, and the diagrams and $2n^2$ rule you draw from it are entirely valid.

If M can hold 18 electrons, why does potassium (Z=19) put its 19th electron in a new N shell instead of the M shell?

Because the outermost-shell rule intervenes before M can reach 18. After K takes 2 and L takes 8, M is the outermost shell with 9 electrons still to place, but the outermost-shell cap of 8 stops M there — so M holds 8, and the last electron starts a new N shell, giving potassium the distribution $2, 8, 8, 1$ . This is exactly why the 8-electron outermost cap has to be checked at every step, not just $2n^2$ alone.

Bohr's Model of Atom

The core idea

Worked example

Common traps

Before this

Where this leads