Atomic Structure

The $4s$ orbital fills before $3d$. So when an iron atom loses two electrons, which orbital do they come from?

Not $3d$. Both come from $4s$, and ${\text{Fe}}^{2+}$ is $\left[\text{Ar}\right]3d^6$, not $\left[\text{Ar}\right]3d^{4}4s^2$.

The resolution is that "which is lower" depends on how many electrons are already there. In a bare potassium or calcium atom, $4s$ lies below $3d$, so it fills first. But once $3d$ begins to be occupied, the added nuclear charge pulls $3d$ down sharply — it is more compact and feels the nucleus more directly — and by the time you reach scandium, $3d$ is already below $4s$.

So the filling order and the ionisation order are governed by two different orderings of the same pair of orbitals, at two different points on the curve. Nothing is inconsistent; the energies simply moved.

This chapter is about that kind of detail. Main asks for the four quantum numbers and the order of filling. Advanced asks how many nodes an orbital has and of what kind, what the effective nuclear charge is, why chromium breaks the pattern, and what an unpaired-electron count implies for the magnetic moment.

1. Hydrogen-like species, and where Bohr stops

For any one-electron species, the Bohr results are exact:

$E_n=-\frac{13.6Z^2}{n^2}\text{eV},r_n=\frac{0.529n^2}{Z}\text{A˚},v_n=\frac{2.18\times 10^{6}Z}{n}{\text{m s}}^{-1}$

so ${\text{He}}^{+}$, ${\text{Li}}^{2+}$ and ${\text{Be}}^{3+}$ are all fair game. Transitions follow

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

and an atom excited to level $n$ can emit $\frac{n\left(n-1\right)}{2}$ distinct lines.

Bohr fails the moment a second electron appears, because electron-electron repulsion has no place in the model. It also cannot explain the fine structure of lines, the Zeeman effect, or why orbitals have shapes at all. Its survival in the syllabus is entirely due to the one-electron case being exactly right.

Illustration 1

Find the ionisation energy of ${\text{Li}}^{2+}$ and the wavelength of the transition $n=3\to 2$.

$IE=13.6\times 3^2=122.4$ eV

$\Delta E=13.6\times 9\left(\frac{1}{4}-\frac{1}{9}\right)=122.4\times 0.1389=17.0$ eV

$\lambda =\frac{1240}{17.0}=73$ nm

The $Z^2$ scaling is severe. Lithium's doubly charged ion binds its last electron nine times more strongly than hydrogen does, which is why removing it needs vacuum-ultraviolet photons.

2. Orbitals, nodes and radial distribution

An orbital is a wave function $\psi$; what has physical meaning is ${\psi }^2$, the probability density. Where $\psi$ passes through zero there is a node, and nodes come in two kinds:

$\text{radial (spherical) nodes}=n-l-1,\text{angular nodes (nodal planes)}=l$

$\text{total nodes}=n-1$

A $3p$ orbital therefore has $3-1-1=1$ radial node and $1$ nodal plane, two in all. A $3d$ orbital has no radial node and two nodal planes.

The radial distribution function $4\pi r^2{\psi }^2$ is what answers "where is the electron most likely to be found". For $1s$ its maximum sits at exactly the Bohr radius $a_0$ — the one place the quantum picture and Bohr's agree.

Illustration 2

Find the number of radial nodes, angular nodes and total nodes for $4d$ and $5f$.

$4d$: $n=4$, $l=2$. Radial $=4-2-1=1$; angular $=2$; total $=3$.

$5f$: $n=5$, $l=3$. Radial $=5-3-1=1$; angular $=3$; total $=4$.

The total is always $n-1$, whatever the subshell, which gives a one-second check on the two separate counts.

Illustration 3

Which has more radial nodes, $3s$ or $4d$? Which has more nodal planes?

$3s$: radial $=3-0-1=2$; nodal planes $=0$.

$4d$: radial $=4-2-1=1$; nodal planes $=2$.

So the answers point in opposite directions. Higher $l$ trades radial nodes for angular ones, which is why $s$ orbitals are the most "layered" and $f$ orbitals the most "lobed".

3. The shapes of the orbitals

Shape is fixed by $l$ alone, and the shapes matter because bonding depends on how orbitals overlap in space.

$s$ orbitals are spherical, with no directional preference at all. Higher $s$ orbitals are spheres within spheres, separated by the radial nodes counted above.

$p$ orbitals are dumbbells aligned with the three axes. Each has exactly one nodal plane, passing through the nucleus perpendicular to its own axis — so $p_x$ has the $yz$ plane as its node, and the electron density there is exactly zero.

$d$ orbitals split into two families, and the distinction is the single most useful fact in the whole chapter:

Three of them — $d_{xy}$, $d_{yz}$ and $d_{zx}$ — have their lobes lying between the axes. The other two, $d_{x^2-y^2}$ and $d_{z^2}$, have lobes along the axes. In a free atom all five are degenerate, but surround the atom with ligands sitting on the axes and the two families are no longer equivalent — which is the whole origin of crystal field splitting.

The wave function also carries a sign, alternating from lobe to lobe. Two orbitals overlap constructively only where their signs match, which is why bonding and antibonding combinations exist at all.

Illustration 4

Identify the nodal planes of the $3d_{xy}$ orbital and state where its electron density is greatest.

Its lobes lie in the $xy$ plane, between the $x$ and $y$ axes.

Its two nodal planes are the $xz$ and $yz$ planes, since density vanishes wherever $x=0$ or $y=0$.

Maximum density lies along the lines $y=\pm x$ within the $xy$ plane.

Read the subscript literally. The function is proportional to the product $xy$, so it must vanish whenever either coordinate does — which locates both nodal planes without any further work.

Illustration 5

Which $d$ orbitals point directly at ligands placed on the six coordinate axes, and what follows?

$d_{x^2-y^2}$ and $d_{z^2}$ point straight at them; $d_{xy}$, $d_{yz}$ and $d_{zx}$ point between them.

Electrons in the first pair are therefore repelled more strongly and rise in energy.

This produces the octahedral splitting into a lower set of three and an upper set of two, which is the foundation of everything in coordination chemistry — colour, magnetism and geometry alike.

4. Quantum numbers, and what each one controls

Orbital angular momentum is $\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ — note that this is zero for every $s$ orbital, which Bohr's model could never accommodate, since it gave $nh/2\pi$ and never zero.

Capacities follow immediately: $2\left(2l+1\right)$ electrons per subshell and $2n^2$ per shell.

Illustration 6

How many electrons in an atom can have $n=4$ and $l=2$? How many can have $n=3$ and $m_l=+1$?

$n=4$, $l=2$ is the $4d$ subshell: $2\left(2\times 2+1\right)=10$ electrons.

$n=3$ with $m_l=+1$ requires $l\ge 1$, so $l=1$ or $2$: one orbital from each, two orbitals in all, holding $4$ electrons.

The second form is the harder one, because it cuts across subshells. Enumerate which $l$ values can produce the given $m_l$, count one orbital for each, and double.

5. Effective nuclear charge and Slater's rules

An outer electron does not feel the full nuclear charge; inner electrons screen it:

$Z_{eff}=Z-\sigma$

Slater's rules give $\sigma$ numerically. Group the configuration as $\left(1s\right)\left(2s,2p\right)\left(3s,3p\right)\left(3d\right)\left(4s,4p\right)\dots$ and for an electron in an $s$ or $p$ group:

For a $d$ or $f$ electron the rule is harsher: $0.35$ within the group and a full $1.00$ for everything inside, because $d$ orbitals penetrate poorly.

Illustration 7

Calculate the effective nuclear charge felt by a $3p$ electron in silicon ($Z=14$).

Configuration grouped: $\left(1s^2\right)\left(2s^{2}2p^6\right)\left(3s^{2}3p^2\right)$

$\sigma =0.35\times 3+0.85\times 8+1.00\times 2=1.05+6.80+2.00=9.85$

$Z_{eff}=14-9.85=4.15$

Only three of the fourteen protons are effectively felt. This is why the outer electrons of a large atom behave so much more like those of a light one than the raw nuclear charge would suggest.

Illustration 8

Compare the effective nuclear charge on a $4s$ and a $3d$ electron in scandium ($Z=21$).

For $4s$: $\sigma =0.35\times 0+0.85\times 9+1.00\times 10=7.65+10=17.65$, giving $Z_{eff}=3.35$.

For $3d$: everything inside counts fully, $\sigma =1.00\times 18=18$, giving $Z_{eff}=3.00$.

Two electrons in the same atom feel measurably different charges. Screening, not the bare nuclear charge, is what determines chemical behaviour.

6. Aufbau, and the two orderings

Filling follows the $\left(n+l\right)$ rule: lower $\left(n+l\right)$ fills first, and where two are equal, the lower $n$ wins. That is why $4s$ ($n+l=4$) precedes $3d$ ($n+l=5$).

But as the hook showed, once $3d$ is occupied it drops below $4s$. Two consequences follow, and both are examined:

Electrons are removed from the highest $n$ first, so transition metals always lose $4s$ before $3d$. ${\text{Fe}}^{2+}$ is $3d^6$; ${\text{Ti}}^{3+}$ is $3d^1$; ${\text{Zn}}^{2+}$ is $3d^{10}$.

No transition-metal cation ever retains an $s$ electron in its ground state, which makes writing ionic configurations mechanical once the rule is remembered.

Illustration 9

Write the ground-state configurations of $\text{Fe}$, ${\text{Fe}}^{2+}$ and ${\text{Fe}}^{3+}$, and count the unpaired electrons in each.

$\text{Fe}$: $\left[\text{Ar}\right]3d^{6}4s^2$ — the $3d^6$ has $4$ unpaired.

${\text{Fe}}^{2+}$: $\left[\text{Ar}\right]3d^6$ — still $4$ unpaired.

${\text{Fe}}^{3+}$: $\left[\text{Ar}\right]3d^5$ — $5$ unpaired, a half-filled set.

The three-plus ion is the more stable of the two, which is why iron(III) salts are so common. A half-filled $d$ subshell carries extra exchange stabilisation, as the next section explains.

7. Exceptions, exchange energy and magnetic moment

Chromium is $\left[\text{Ar}\right]3d^{5}4s^1$ and copper is $\left[\text{Ar}\right]3d^{10}4s^1$, not the $3d^{4}4s^2$ and $3d^{9}4s^2$ that naive filling predicts. Two effects combine.

Exchange energy. Electrons of parallel spin in degenerate orbitals can exchange places, and every such pair lowers the energy. The number of exchange pairs for $n$ parallel electrons is $\frac{n\left(n-1\right)}{2}$, so going from $d^4$ to $d^5$ raises the count from $6$ to $10$ — a gain of four pairs, more than enough to pay for promoting an $s$ electron.

Symmetrical distribution. Half-filled and fully filled subshells are spherically symmetric, which lowers the repulsion energy.

The number of unpaired electrons is measurable, through the spin-only magnetic moment:

$\mu =\sqrt{n\left(n+2\right)}\text{BM}$

Illustration 10

Calculate the spin-only magnetic moments of ${\text{Mn}}^{2+}$, ${\text{Ni}}^{2+}$ and ${\text{Zn}}^{2+}$.

${\text{Mn}}^{2+}$ is $3d^5$, $n=5$: $\mu =\sqrt{35}=5.92$ BM

${\text{Ni}}^{2+}$ is $3d^8$, $n=2$: $\mu =\sqrt{8}=2.83$ BM

${\text{Zn}}^{2+}$ is $3d^{10}$, $n=0$: $\mu =0$, diamagnetic

A measured moment identifies the ion. Going backwards from $\mu$ to $n$ and then to the configuration is a standard Advanced question, and $5.92$ BM is the unmistakable signature of a $d^5$ high-spin ion.

Illustration 11

An ion has a magnetic moment of $3.87$ BM. Find the number of unpaired electrons and suggest a $3d$ ion.

$\sqrt{n\left(n+2\right)}=3.87\Rightarrow n\left(n+2\right)=15\Rightarrow n=3$

Three unpaired electrons in a $3d$ set means $d^3$ or $d^7$: ${\text{Cr}}^{3+}$ or ${\text{Co}}^{2+}$.

The moment alone cannot distinguish them. Colour, coordination geometry or oxidation-state chemistry is needed to settle which, which is exactly why the question usually supplies one more clue.

8. Uncertainty, matter waves and the end of the orbit

$\Delta x\cdot \Delta p\ge \frac{h}{4\pi },\lambda =\frac{h}{mv}$

The two together destroy the idea of an orbit. If an electron's position within an atom is known to about $10^{-10}$ m, its momentum is uncertain by enough to give a velocity uncertainty of order $10^6$ m s${}^{-1}$ — comparable with the speed itself. A trajectory is therefore not merely unknown but meaningless.

Orbit and orbital are different objects: an orbit is a definite path, an orbital is a region in which the probability of finding the electron is high.

Illustration 12

An electron is confined to a region of $1$ Å. Estimate the minimum uncertainty in its velocity.

$\Delta v\ge \frac{h}{4\pi m\Delta x}=\frac{6.63\times 10^{-34}}{4\pi \left(9.11\times 10^{-31}\right)\left(10^{-10}\right)}$

$\Delta v\ge 5.8\times 10^5$ m s${}^{-1}$

Compare the Bohr velocity of $2.18\times 10^6$ m s${}^{-1}$. The uncertainty is a quarter of the speed itself, which is precisely why the electron cannot be said to follow a path.

Illustration 13

Repeat the estimate for a cricket ball of $150$ g localised to $1$ mm.

$\Delta v\ge \frac{6.63\times 10^{-34}}{4\pi \left(0.15\right)\left(10^{-3}\right)}=3.5\times 10^{-31}$ m s${}^{-1}$

Utterly unmeasurable. The principle applies to the ball exactly as it does to the electron; it simply produces no observable consequence, which is why classical mechanics survives everywhere except inside atoms.

Summary

• $4s$ fills before $3d$ but $3d$ lies lower once occupied, so $4s$ empties first: ${\text{Fe}}^{2+}$ is $3d^6$.
• Hydrogen-like: $E_n=-\frac{13.6Z^2}{n^2}$ eV, $r_n=\frac{0.529n^2}{Z}$ Å; level $n$ emits $\frac{n\left(n-1\right)}{2}$ lines.
• Bohr fails the instant a second electron appears, since it contains no electron-electron repulsion.
• Radial nodes $=n-l-1$; nodal planes $=l$; total $=n-1$ always.
• $4\pi r^2{\psi }^2$ peaks at $a_0$ for $1s$ — the one point where Bohr and quantum mechanics agree.
• $d_{xy}$, $d_{yz}$, $d_{zx}$ have lobes between the axes; $d_{x^2-y^2}$ and $d_{z^2}$ lie along them — the origin of crystal field splitting.
• Orbital angular momentum is $\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$, and is zero for every $s$ orbital.
• Capacities: $2\left(2l+1\right)$ per subshell, $2n^2$ per shell.
• $Z_{eff}=Z-\sigma$ by Slater: $0.35$ same group, $0.85$ for $\left(n-1\right)$, $1.00$ deeper — but $1.00$ for everything inside a $d$ or $f$ electron.
• Filling by the $\left(n+l\right)$ rule; ionisation always removes the highest $n$ first, so cations of transition metals keep no $s$ electrons.
• Exchange pairs $=\frac{n\left(n-1\right)}{2}$ for parallel electrons; $d^4\to d^5$ gains four pairs, which is why chromium is $3d^{5}4s^1$.
• Half-filled and fully filled subshells are also spherically symmetric, lowering repulsion.
• $\mu =\sqrt{n\left(n+2\right)}$ BM: $5.92$ BM is the signature of five unpaired electrons.
• $\Delta x\Delta p\ge \frac{h}{4\pi }$ makes a velocity uncertainty comparable to the speed itself for a confined electron — an orbital is not an orbit.