Atomic Structure

The three particles

Every atom is built from three kinds of particle. Two of them, protons and neutrons, sit together in a tiny, dense nucleus at the centre; the third, the electron, occupies the vast, mostly empty space around that nucleus, arranged in shells.

Naming the numbers: proton number and nucleon number

Two counts describe any atom completely, once you also know how many electrons it has.

• The proton (atomic) number, $Z$, is the number of protons. It is fixed for a given element — it is what makes the element that element — and it equals the number of electrons in a neutral atom, since a neutral atom has no net charge.
• The nucleon (mass) number, $A$, is the number of protons and neutrons together — i.e. everything in the nucleus.

From these two numbers alone:

because the nucleon number counts protons and neutrons together, and subtracting the protons leaves only the neutrons.

Extending to ions

A neutral atom has equal numbers of protons and electrons, so its overall charge is zero. An ion forms when electrons are removed or added. The nucleus, protons and neutrons both, is not touched. Many mistakes in this section come from forgetting this:

Forming an ion never changes the number of protons or neutrons. It only changes the number of electrons.

So for an ion of charge $q$ (positive $q$ means the ion has lost electrons; negative $q$ means it has gained them):

A $2+$ ion has two fewer electrons than the neutral atom ($q = +2$, so subtract $2$); a $1-$ ion has one more ($q = -1$, so subtracting a negative adds one). The number of protons and the number of neutrons are exactly what they are for the neutral atom, unaffected by $q$.

A clean demonstration

Take the nuclide $^{27}_{13}\text{Al}$, written in the notation shown in the figure above. Read the two numbers straight off it: nucleon number $A = 27$, proton number $Z = 13$.

Protons. Directly $Z = 13$.

Neutrons. $A - Z = 27 - 13 = 14$.

Electrons (neutral atom). Equal to $Z$, so $13$.

Now form the ion $\text{Al}^{3+}$. The charge is $q = +3$, meaning three electrons have been removed. Protons and neutrons are untouched by this — they are still $13$ and $14$. Only the electron count changes:

So $\text{Al}^{3+}$ has $13$ protons, $14$ neutrons and $10$ electrons — three fewer electrons than protons, giving the net $3+$ charge.

Now try an anion: $\text{O}^{2-}$, from oxygen, $Z = 8$. Protons $= 8$. For neutrons you would need the nucleon number of the isotope in question (commonly $16$, giving $8$ neutrons); take $^{16}_{8}\text{O}$ here, so neutrons $= 16 - 8 = 8$. The charge is $q = -2$, meaning two electrons have been added:

$\text{O}^{2-}$ has $8$ protons, $8$ neutrons and $10$ electrons.

Ions made of several atoms

Some ions contain more than one atom, for example the hydroxide ion $\text{OH}^-$ or the ammonium ion $\text{NH}_4^+$. These are called molecular ions. Count their particles atom by atom, then apply the charge once, to the whole ion:

• protons = the sum of the proton numbers of all the atoms;
• neutrons = the sum of $(A - Z)$ for all the atoms;
• electrons = total protons $- q$, where $q$ is the charge on the whole ion.

Demonstration: $^{16}\text{O}^{1}\text{H}^-$.

Protons: oxygen has $8$ and hydrogen has $1$, so $8 + 1 = 9$.

Neutrons: oxygen-16 has $16 - 8 = 8$ and hydrogen-1 has $1 - 1 = 0$, so $8 + 0 = 8$.

Electrons: the charge is $q = -1$, so $9 - (-1) = 10$.

One more useful fact: a proton and a neutron each have relative mass $1$ and an electron's mass is negligible. So the relative mass of an atom or ion is (very nearly) its total number of protons plus neutrons. Here, $9 + 8 = 17$, the relative formula mass of $^{16}\text{O}^{1}\text{H}^-$.

Why this belongs here

This tests whether you understand the charges and masses in the particle table (see "Protons, neutrons, electrons — and counting them"), not just remember them. A charged particle moving between two oppositely charged plates feels a force pulling it toward the plate of opposite sign. How much its path bends depends on the two properties you already know for each particle: its charge and its mass.

Applying that to the three particles

Run down the particle table with these two questions in mind.

• Proton ($+1$ charge, mass $1$): pulled toward the negative plate. It has real, "normal" mass, so it deflects, but only gently over the length of the plates.
• Neutron (no charge, mass $1$): not deflected at all — no charge means no force from the field, so a neutron beam travels straight through undisturbed, mass irrelevant.
• Electron ($-1$ charge, mass $1/1840$): pulled toward the positive plate — the opposite direction to the proton, because the charge has opposite sign. Because an electron is about $1840$ times lighter than a proton for the same size of force, it deflects far more sharply than the proton over the same distance.

Any two charged particles: compare charge ÷ mass

Questions sometimes use other particles, such as ions. For particles moving at the same velocity, the angle of deflection is proportional to

A bigger charge gives a bigger force, so more bending. A bigger mass resists the force, so less bending. The sign of the charge still decides the direction.

Demonstration. Compare a proton, $\text{p}^+$, with a deuterium ion, $^{2}\text{H}^+$ (one proton and one neutron, charge $+1$).

• Proton: $\dfrac{\text{charge}}{\text{mass}} = \dfrac{+1}{1} = 1$
• $^{2}\text{H}^+$: $\dfrac{\text{charge}}{\text{mass}} = \dfrac{+1}{2} = 0.5$

Both are positive, so both bend toward the negative plate. The proton has the larger ratio, so it bends more; the $^{2}\text{H}^+$ ion bends about half as much.

Defining isotope precisely

You have already met isotopes: magnesium-24, -25 and -26 in the first section are three isotopes of magnesium. The precise definition is:

Isotopes are atoms of the same element (same number of protons) that have different numbers of neutrons (and therefore different nucleon numbers).

Use both halves in an exam answer: same protons, different neutrons. Isotopes are written with the nuclide notation from the first section, for example $^{24}_{12}\text{Mg}$ and $^{26}_{12}\text{Mg}$.

From isotopes to a single number: relative atomic mass

A sample of a real element is a mixture of its isotopes in fixed natural proportions — for magnesium, roughly $79\%$ Mg-24, $10\%$ Mg-25 and $11\%$ Mg-26. The relative atomic mass, $A_r$, is a single weighted-average figure that represents this whole mixture, defined against carbon-12 as the standard:

The word weighted is the whole method: an isotope that makes up $92\%$ of a sample should count much more toward the average than one making up only $8\%$. So this is not a simple mean of the isotopic masses; each mass is weighted by how common that isotope is. (The AS Atoms, Molecules and Stoichiometry note defines these masses more formally.)

A clean demonstration with invented numbers

A sample of an imaginary element $Q$ contains two isotopes: $70\%$ with isotopic mass $60.0$, and $30\%$ with isotopic mass $62.0$. Find $A_r$.

Step 1 — multiply each mass by its abundance.

Step 2 — add the products.

Step 3 — divide by 100.

Sanity check. $A_r = 60.6$ sits between the two isotopic masses ($60.0$ and $62.0$), much closer to $60.0$ — exactly as expected, since the $60.0$ isotope is more than twice as abundant. If your answer ever falls outside the range of the isotopic masses you were given, you have made an arithmetic error: a weighted mean can never be more extreme than the values it is averaging.

What actually sets an atom's size

An atom has no sharp edge, so "atomic radius" is measured in practice, usually as half the distance between the nuclei of two neighbouring atoms. The trends in radius come from a tug-of-war between two things:

• the nuclear charge (the number of protons), which pulls the electrons inward: more protons, stronger pull;
• the number of occupied shells, which sets how far out the outer electrons are, together with shielding: electrons in inner shells repel the outer electrons and partly block the nucleus's pull on them.

Every trend below is this tug-of-war won by one side or the other.

Across a period: radius shrinks

Moving left to right across a period, each element has one more proton, but the new electron goes into the same outer shell. No new inner shell is added, so shielding stays about the same. The nuclear charge rises while shielding stays similar, so the outer electrons are attracted more strongly and pulled closer: the atom shrinks.

Down a group: radius grows

Moving down a group, each element has one more occupied shell. That shell is further from the nucleus, and it is shielded by all the shells inside it. The extra distance and extra shielding outweigh the rise in nuclear charge, so the atom gets bigger, even though it has more protons.

Ionic radius: the same tug-of-war, now with a changed electron count

Forming an ion changes the number of electrons but never the number of protons, so the same nuclear charge now has a different number of electrons to hold.

• Cations (positive ions) are smaller than their atoms. A metal atom usually loses its whole outer shell (Na, 2,8,1 → Na⁺, 2,8). The ion has one fewer shell, and the same nuclear charge now pulls on fewer electrons, so they are held more tightly.
• Anions (negative ions) are larger than their atoms. Adding electrons to the outer shell increases the repulsion between electrons, which pushes them further apart. The nuclear charge is unchanged but now has more electrons to hold, so each is held less tightly.

Isoelectronic species: radius with the electron count fixed

The cleanest way to see the role of nuclear charge on its own is to hold the electron count constant and vary only the number of protons. Species with identical electron counts are called isoelectronic. Take $\text{Na}^+$, $\text{Ne}$ and $\text{F}^-$: all three have exactly $10$ electrons, but

With the electron count fixed, whichever species has the most protons pulls that fixed cloud in hardest — so it is the smallest. Fewer protons means a weaker pull on the same ten electrons, so the radius grows:

Beyond $2, 8, 8$: shells split into sub-shells

The IGCSE picture of shells holding $2$, then $8$, then $8$ electrons is correct but incomplete. AS Chemistry looks inside each shell.

Each shell is labelled by a principal quantum number, $n = 1, 2, 3, \ldots$, counting outward from the nucleus. Within a shell, electrons occupy one or more sub-shells, labelled $\text{s}, \text{p}, \text{d}$ (and $\text{f}$, not needed at this level). Shell $n$ contains $n$ types of sub-shell: shell $1$ has only an s sub-shell; shell $2$ has s and p; shell $3$ has s, p and d.

Within a sub-shell, electrons occupy orbitals. An orbital is a region of space that can hold at most two electrons, and two electrons in the same orbital must have opposite spins. (In the next section you draw each orbital as a box and each electron as an arrow, pointing up or down for its spin.)

The order of increasing energy

Electrons fill the lowest-energy sub-shells first (the Aufbau principle, from the German for "building up"). Within a shell, energy rises $\text{s} < \text{p} < \text{d}$. Between shells, energy usually rises with $n$, but there is one exception you must know:

The bold step is the exception: the 4s sub-shell is slightly lower in energy than 3d, even though $n = 4 > 3$. So 4s fills before 3d when you build up an atom. (The next section shows that 4s also empties before 3d when a transition-metal atom forms an ion.)

When every electron is in the lowest-energy arrangement available, following this order, the atom is in its ground state: its lowest energy state. If an electron is given energy and moves up to a higher level, the atom is no longer in its ground state. In this course every configuration you write is the ground state.

What s and p orbitals actually look like

• An s orbital is a sphere centred on the nucleus: the electron is equally likely to be found in every direction. There is one s orbital in each s sub-shell. A 2s orbital is the same shape as a 1s orbital, only bigger.
• A p orbital has a dumbbell shape: two lobes on opposite sides of the nucleus. There are three p orbitals in each p sub-shell, at right angles to each other, along the $x$, $y$ and $z$ axes. They are labelled $\text{p}_x$, $\text{p}_y$ and $\text{p}_z$.

Two ways of writing the same fact

An electron configuration records how many electrons are in each sub-shell, using superscripts for the count: $1\text{s}^2$ means "two electrons in the $1\text{s}$ sub-shell". You need both of these formats:

• Full configuration: every sub-shell written out. For iron ($Z = 26$): $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6\,4\text{s}^2$
• Shorthand (noble-gas core) configuration: the inner electrons are replaced by the symbol of the noble gas with exactly that configuration, in square brackets: $[\text{Ar}]\,3\text{d}^6\,4\text{s}^2$ This works because argon ($Z=18$) is $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6$.

Both describe the same electrons. The shorthand is just quicker to write. You will see 3d written before 4s (as in the syllabus, and here) and 4s written before 3d (the order they fill in); both describe the same electrons.

A clean demonstration

Write the full electron configuration of phosphorus, $\text{P}$, $Z = 15$.

Step 1. $15$ electrons to place.

Step 2. Fill in energy order, tracking the running total:

Step 3. Only $15 - 12 = 3$ electrons remain, and $3\text{p}$ can hold up to $6$ — so it is only partly filled:

Check the total: $2+2+6+2+3 = 15$. ✓ Shorthand form, using neon ($Z=10$, $1\text{s}^2 2\text{s}^2 2\text{p}^6$) as the core: $[\text{Ne}]\,3\text{s}^2 3\text{p}^3$.

For an ion, change the electron count first. The sulfide ion $\text{S}^{2-}$ has $16 + 2 = 18$ electrons, so it fills exactly as far as argon: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6$. For main-group ions this always works. Transition-metal ions need one extra rule, shown later in this section.

Electrons in boxes, and Hund's rule

The electrons-in-boxes notation draws one box per orbital and one arrow per electron. The arrow points up or down to show the electron's spin, and two electrons in one box point opposite ways. This notation shows something the superscripts hide: which orbitals in a sub-shell are occupied, and whether their electrons are paired.

That is decided by Hund's rule: in a sub-shell with more than one orbital (p or d), electrons go into separate orbitals singly, with the same spin, before any pairing occurs. The reason is inter-electron repulsion: two electrons in the same orbital repel each other strongly. Spreading out into empty orbitals first keeps them apart, which is the lower-energy arrangement, and the ground state is always the lowest-energy arrangement.

Two atoms that break the filling order: chromium and copper

Following the filling order exactly would give chromium ($Z = 24$) $[\text{Ar}]\,3\text{d}^4\,4\text{s}^2$ and copper ($Z = 29$) $[\text{Ar}]\,3\text{d}^9\,4\text{s}^2$. The real ground states are different: in each, one electron moves from 4s into 3d.

The 4s and 3d sub-shells are very close in energy, so a small energy saving is enough to change the arrangement. In chromium, $3\text{d}^5\,4\text{s}^1$ has six unpaired electrons in six separate orbitals, so no two electrons share an orbital and repulsion is less. In copper, the 3d sub-shell becomes completely full. In both cases the new arrangement has the lower energy, so it is the ground state. These are the only two exceptions between hydrogen and krypton, so learn them. The syllabus example for iron, which follows the normal order, is drawn first in the figure for comparison.

Transition-metal ions: 4s empties before 3d

4s is lower in energy than 3d while an atom is being filled, so it fills first. But once there are electrons in 3d, the energies change: the 4s electrons are now the highest-energy (outermost) electrons. So when a transition-metal atom loses electrons to form an ion, it loses its 4s electrons first, even though they were added before the 3d electrons.

Take iron, $\text{[Ar]}\,3\text{d}^6 4\text{s}^2$. Forming $\text{Fe}^{2+}$ removes both 4s electrons: $\text{[Ar]}\,3\text{d}^6$. Forming $\text{Fe}^{3+}$ removes one more, from 3d, because 4s is already empty: $\text{[Ar]}\,3\text{d}^5$. It is not $\text{[Ar]}\,3\text{d}^6 4\text{s}^1$: that answer removes the last electron you wrote down instead of the highest-energy electron.