Why Chromium Breaks the Filling Order Everyone Memorises

Bohr's orbits explained hydrogen's spectrum but failed for every other atom. The deeper problem is that an electron behaves like a wave, and a wave cannot follow a sharp, fixed path.

This part covers the wave nature of electrons and the uncertainty principle, quantum numbers, the shapes and nodes of orbitals, and electronic configurations. Take $h=6.626\times 10^{-34}$ J s and $m_e=9.11\times 10^{-31}$ kg.

**Every moving particle has a wavelength $\lambda =\frac{h}{mv}$, and the uncertainty principle, $\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$, says an electron's position and momentum cannot both be known exactly — so an exact orbit is meaningless and only the probability of finding the electron can be given.

Worked example 1 — an electron in hydrogen.** At $2.2\times 10^6$ m/s:

$$\lambda =\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 2.2\times 10^6}\approx 3.3\times 10^{-10}\text{ m}$$

That is about the circumference of hydrogen's first Bohr orbit, $2\pi \times 52.9$ pm $\approx 332$ pm — the wave nature matters at atomic scale.

Worked example 2 — uncertainty. If an electron's position is known to within $10^{-10}$ m:

$$\Delta v\ge \frac{h}{4\pi m\Delta x}=\frac{6.626\times 10^{-34}}{4\pi \times 9.11\times 10^{-31}\times 10^{-10}}\approx 5.8\times 10^5\text{ m/s}$$

This uncertainty is comparable to the electron's speed itself, so its path cannot be traced.

An everyday example. Photographing a speeding scooter shows the trade-off in spirit: a sharp position means a frozen image with no sense of speed, and a streak shows motion but no single position.

The substance. The uncertainty is a law of nature, not a limit of measuring instruments.

**The principal quantum number $n$ gives the shell and energy level, the azimuthal quantum number $l$ gives the subshell and shape, the magnetic quantum number $m_l$ gives the orbital's orientation, and the spin quantum number $m_s$ gives the electron's spin.

- $n$** $=1,2,3,\dots$ — larger $n$ means a larger, higher-energy shell
- **$l$** $=0$ to $n-1$ — $0,1,2,3$ are s, p, d, f subshells
- **$m_l$** $=-l$ to $+l$ — giving $2l+1$ orbitals in a subshell
- **$m_s$** $=+\frac{1}{2}$ or $-\frac{1}{2}$

Worked example 1 — a 3p electron. $n=3$, $l=1$, $m_l=-1,0$ or $+1$, and $m_s=\pm \frac{1}{2}$.

Worked example 2 — counting. Shell $n=3$ has $n^2=9$ orbitals (one 3s, three 3p, five 3d) holding $2n^2=18$ electrons.

Worked example 3 — invalid sets.

- $n=2$, $l=2$ — not allowed, since $l$ must be less than $n$
- $n=3$, $l=1$, $m_l=+2$ — not allowed, since $|m_l|\le l$

An everyday example. A postal address narrows down from state to city to street to house — just as $n$, $l$, $m_l$ and $m_s$ narrow down to one electron.

The substance. No two electrons in an atom share all four quantum numbers — the Pauli exclusion principle.

**s orbitals are spherical, p orbitals are dumbbell-shaped along the x, y or z axis, and d orbitals are mostly four-lobed; an orbital has $n-l-1$ radial nodes and $l$ angular nodes, giving $n-1$ nodes in all.

- s** — a sphere around the nucleus; larger for higher $n$
- p — two lobes on opposite sides of the nucleus, with a nodal plane through it: $p_x$, $p_y$, $p_z$
- d — four lobes for $d_{xy}$, $d_{yz}$, $d_{xz}$ and $d_{x^2-y^2}$; $d_{z^2}$ has two lobes along $z$ with a ring around its middle

Worked example — counting nodes.

- 2s: radial $=2-0-1=1$, angular $=0$
- 3p: radial $=3-1-1=1$, angular $=1$
- 3d: radial $=3-2-1=0$, angular $=2$
- 4d: radial $=4-2-1=1$, angular $=2$, total $3$

An everyday example. A gym dumbbell is a good picture of a p orbital — two bulges on either side of a narrow middle where the electron is never found.

The substance. **Orbital diagrams show a boundary enclosing about $90\%$ of the probability**, not a hard edge.

**Electrons fill orbitals in order of increasing energy by the $(n+l)$ rule (Aufbau principle), each orbital holds at most two electrons of opposite spin (Pauli exclusion principle), and within a subshell electrons occupy separate orbitals before pairing (Hund's rule); half-filled and completely filled subshells are extra stable because of their symmetry and exchange energy.

Filling order:** $1s,2s,2p,3s,3p,4s,3d,4p,5s,\dots$

Worked example 1 — nitrogen ($Z=7$): $1s^{2}2s^{2}2p^3$, with three unpaired electrons, one in each 2p orbital.

Worked example 2 — the exceptions.

- Chromium ($Z=24$): $[\text{Ar}]3d^{5}4s^1$, not $3d^{4}4s^2$ — a half-filled 3d set
- Copper ($Z=29$): $[\text{Ar}]3d^{10}4s^1$, not $3d^{9}4s^2$ — a completely filled 3d set

Worked example 3 — ions. Iron ($Z=26$) is $[\text{Ar}]3d^{6}4s^2$. To form Fe${}^{3+}$, remove the two 4s electrons first, then one 3d electron:

$${\text{Fe}}^{3+}:[\text{Ar}]3d^5(5\text{ unpaired electrons})$$

An everyday example. Passengers boarding a bus with double seats usually take an empty seat each before anyone sits beside a stranger — the spirit of Hund's rule.

The substance. 4s fills before 3d, but 4s electrons are also the first to leave when a transition metal forms an ion.