Atomic structure

In a nutshell

Everything an element does chemically comes down to how its electrons are arranged, and that arrangement is set by the structure of the atom.

This subtopic covers the three sub-atomic particles, isotopes, how a time of flight mass spectrometer measures isotopic masses and abundances, how electrons fill shells and sub-shells, and how ionisation energies give direct evidence for that arrangement.

Assumed knowledge: GCSE atomic structure (protons, neutrons, electrons and the nuclear atom).

Core content

Fundamental particles

An atom is a tiny nucleus of protons and neutrons, surrounded by electrons arranged in shells.

You need the relative mass and relative charge of each particle:

Our picture of the atom has evolved over time as new evidence appeared: the indivisible sphere, then the "plum pudding" model (electrons in a ball of positive charge), then Rutherford's nuclear model (a dense, positive nucleus with most of the mass, from the alpha-particle scattering experiment), then Bohr's model (electrons in fixed energy levels). AQA only expects you to appreciate that the model changed as evidence grew, not to recall each experiment in detail.

Atomic number, mass number and isotopes

Every atom is labelled by two numbers, written around its symbol as $^{A}_{Z}\text{X}$:

• Atomic (proton) number, Z: the number of protons in the nucleus.
• Mass number, A: the total number of protons and neutrons in the nucleus.

From these you can count the particles in any atom or ion:

• protons = Z
• neutrons = A − Z
• electrons = Z (adjusted for the charge: add one electron per unit of negative charge, remove one per unit of positive charge).

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.

Because isotopes have the same number of electrons and the same electron configuration, they have the same chemical properties. Their different numbers of neutrons give them different masses, so their physical properties (such as density) differ slightly.

The time of flight (TOF) mass spectrometer

A mass spectrometer measures the mass and abundance of each ion in a sample, so it can identify elements, find relative isotopic masses and abundances, and measure relative molecular mass. The whole instrument is kept under a vacuum so the ions do not collide with air molecules.

There are four stages. Learn the order and what happens in each.

1. Ionisation. The sample is turned into positive ions (only ions can be accelerated and detected). Two methods are credited:

• Electron impact: the vaporised sample is bombarded with high-energy electrons from an electron gun, which knock off one electron from each particle to form a 1+ ion. Used for elements and small molecules. It can make larger molecules fragment. Equation: X(g) + e⁻ → X⁺(g) + 2e⁻
• Electrospray ionisation: the sample is dissolved in a volatile, polar solvent and pushed through a fine needle at high voltage, forming charged droplets. Each molecule gains a proton (H⁺) from the solvent to form MH⁺. Used for larger molecules; it does not cause fragmentation. Equation: M(g) + H⁺ → MH⁺(g)

2. Acceleration. The positive ions are accelerated by an electric field so that every ion gains the same kinetic energy.

3. Ion drift. The ions pass into a field-free flight tube. Since every ion has the same kinetic energy, lighter ions (smaller m/z) travel faster and reach the detector first; heavier ions (larger m/z) travel more slowly.

Still don't get it? · why lighter ions reach the detector first

Picture a sprinter and a heavyweight given exactly the same push of energy to start a race. The same energy launches the light sprinter to a high speed but barely gets the heavy one moving, so the sprinter crosses the line first.

Now the exam version. In the accelerator every ion is given the same kinetic energy, and kinetic energy is $KE = \frac{1}{2}mv^2$. If the energy is fixed and the mass $m$ is small, the velocity $v$ must be large; if $m$ is large, $v$ is small. So the lighter ion (smaller m/z) ends up moving faster and reaches the detector in a shorter time.

That is the whole logic AQA wants: same kinetic energy, so velocity depends on mass, so flight time depends on mass. The detector turns those different flight times into the m/z axis of the spectrum.

4. Detection. When a positive ion hits the detector it gains an electron. This flow of electrons produces a current, and the size of the current is proportional to the abundance of that ion. The computer converts flight times into m/z values and plots abundance against m/z.

The two equations behind the flight time are not something you must recall, but you may be asked to rearrange and use them:

$KE = \tfrac{1}{2}mv^2 \qquad\Rightarrow\qquad v = \sqrt{\frac{2\,KE}{m}}$

$t = \frac{d}{v} = d\sqrt{\frac{m}{2\,KE}}$

where $m$ is the mass of one ion in kilograms, $d$ is the flight-tube length in metres, and $t$ is the flight time in seconds.

Interpreting a mass spectrum and relative atomic mass

A mass spectrum of an element shows one peak per isotope: its position gives the m/z (the relative isotopic mass for a 1+ ion) and its height gives the relative abundance.

The relative atomic mass is the weighted mean mass of the isotopes:

$A_r = \frac{\sum(\text{isotopic mass} \times \%\text{ abundance})}{100}$

If the data are given as relative abundances rather than percentages, divide by the total relative abundance instead of by 100.

The result is worked through in full in Worked examples below.

Relative molecular mass from a spectrum

Mass spectrometry also measures relative molecular mass, M_r:

• With electron impact, the peak at the highest m/z is the molecular ion (the whole molecule minus one electron). Its m/z equals M_r. Smaller fragment peaks appear at lower m/z.
• With electrospray, the molecule is detected as MH⁺, so M_r = (m/z of the peak) − 1.

Electron configuration

Electrons occupy principal energy levels (shells) numbered 1, 2, 3, 4..., which are divided into sub-shells labelled s, p and d (and f). Each sub-shell is made of orbitals, and each orbital holds a maximum of 2 electrons of opposite spin.

Sub-shells fill in order of increasing energy. The one twist is that 4s fills before 3d (4s is lower in energy):

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p

Two rules govern how electrons go in:

• fill the lowest-energy sub-shell first;
• within a sub-shell, put one electron in each orbital before pairing any up (electrons repel, so they stay unpaired where they can).

Write a configuration as sub-shell labels with the electron count as a superscript. For example:

• oxygen (Z = 8): 1s² 2s² 2p⁴
• calcium (Z = 20): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
• vanadium (Z = 23): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³

The periodic table is divided into blocks by which sub-shell the outer electron is in: the s block (Groups 1 and 2), the p block (Groups 3 to 0), and the d block (the transition metals in the centre).

Chromium, copper and ions

Two atoms up to Z = 36 break the simple pattern. Chromium and copper each move one electron from 4s into 3d, because a half-filled or full 3d sub-shell with a singly occupied 4s is more stable:

• chromium (Z = 24): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
• copper (Z = 29): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰

When atoms form ions:

• positive ions: electrons are removed from the outermost shell first. For d-block atoms this means the 4s electrons are lost before the 3d. Example: Fe is [Ar] 4s² 3d⁶, so Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵.
• negative ions: electrons are added to the outer sub-shell. Example: O is 1s² 2s² 2p⁴, so O²⁻ is 1s² 2s² 2p⁶.

Ionisation energies

The first ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, to form one mole of gaseous 1+ ions.

Every first-ionisation-energy equation has the same shape (all species gaseous), even for an element that would not normally form a 1+ ion:

Mg(g) → Mg⁺(g) + e⁻

The second ionisation energy is the energy needed to remove one electron from each ion in one mole of gaseous 1+ ions, to form one mole of gaseous 2+ ions:

Mg⁺(g) → Mg²⁺(g) + e⁻

Three factors control how hard an electron is to remove:

1. Nuclear charge: more protons means a stronger attraction on the electron, so a higher ionisation energy.
2. Atomic radius / distance: the further the outer electron is from the nucleus, the weaker the attraction, so a lower ionisation energy.
3. Shielding: inner shells of electrons repel the outer electron and reduce the pull of the nucleus, lowering the ionisation energy.

Successive ionisation energies: evidence for shells

Removing electrons one after another from the same atom gives successive ionisation energies. Each one is larger than the last, because once an electron is gone the ion is more positive, so it holds the remaining electrons more tightly.

The size of the jumps reveals the shell structure. A large jump happens when the next electron comes from a shell closer to the nucleus, with less shielding.

Magnesium shows two small values, then a large jump to the third: the first two electrons come from the outer (n = 3) shell, and the third from the n = 2 shell underneath. Two electrons before the big jump means magnesium is in Group 2. In general, the number of electrons removed before the first big jump equals the group number.

First ionisation energy across a period and down a group

These trends are the evidence AQA asks you to explain for sub-shells (across Period 3) and for shells (down Group 2).

Down Group 2 (Be to Ba), first ionisation energy decreases. Each element down the group has its outer electrons in a shell further from the nucleus with more shielding by inner shells. Even though nuclear charge increases, the greater distance and shielding win, so the attraction on the outer electron is weaker and it is easier to remove. This falling trend is evidence that electrons sit in shells at increasing distances.

Across Period 3 (Na to Ar), first ionisation energy generally increases. Electrons are added to the same shell, so shielding stays similar, but the nuclear charge increases and the atoms get smaller, so the outer electron is held more strongly.

Two dips break the rise, and they are evidence for sub-shells and orbitals:

• Magnesium to aluminium (drop). Aluminium's outer electron is in a 3p sub-shell, which is higher in energy and is shielded by the 3s electrons, whereas magnesium's outer electron is in 3s. So aluminium's outer electron is easier to remove, and its first ionisation energy is lower. This shows the 3s and 3p sub-shells are at different energies.
• Phosphorus to sulfur (drop). In sulfur, the fourth 3p electron must pair up in a 3p orbital that already holds one electron. The repulsion between the two paired electrons in that orbital makes one easier to remove, so sulfur's first ionisation energy is lower than phosphorus's (where all three 3p electrons are unpaired).

Still don't get it? · why sulfur's first ionisation energy dips below phosphorus's

Imagine seats on a bus. Everyone prefers an empty double-seat to sitting next to a stranger. Phosphorus has three 3p "double-seats" with one person in each, all comfortable. Sulfur has one more electron, and it is forced to share a seat, where the two shove against each other.

Now the exam version. Phosphorus is 3p³, with one electron in each of the three 3p orbitals, none paired. Sulfur is 3p⁴: the fourth electron must go into an orbital that already has one, so the two electrons are paired in the same 3p orbital. These two negative electrons repel each other, which makes one of them easier to pull away. That extra repulsion lowers sulfur's first ionisation energy below phosphorus's, even though sulfur has one more proton.

The marks live in the detail: say the electron being removed is from a 3p orbital that now holds a pair, and that the dip is due to repulsion between those paired electrons. "Half-full shells are more stable" earns nothing.

Worked examples

Example 1: counting particles in an atom and an ion.

For ⁵⁶Fe³⁺ (iron, Z = 26):

$\text{protons} = Z = 26$ $\text{neutrons} = A - Z = 56 - 26 = 30$ $\text{electrons} = 26 - 3 = 23 \ \ (\text{3 electrons lost to give the 3+ charge})$

For ¹⁶O²⁻ (oxygen, Z = 8): protons = 8, neutrons = 16 − 8 = 8, electrons = 8 + 2 = 10.

Example 2: relative atomic mass from a mass spectrum.

Using the magnesium spectrum above (²⁴Mg 79.0 %, ²⁵Mg 10.0 %, ²⁶Mg 11.0 %):

$A_r = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}$ $A_r = \frac{1896 + 250 + 286}{100} = \frac{2432}{100}$ $A_r = 24.3$

The abundances are given to 3 significant figures, so the answer is quoted to 3 significant figures: 24.3 (no units, because it is a relative mass).

Example 3: percentage abundances from a known relative atomic mass.

Chlorine has two isotopes, ³⁵Cl and ³⁷Cl, and A_r = 35.5. Let the fraction of ³⁵Cl be $x$, so the fraction of ³⁷Cl is $(1-x)$:

$35x + 37(1 - x) = 35.5$ $35x + 37 - 37x = 35.5$ $-2x = -1.5$ $x = 0.75$

So the sample is 75 % ³⁵Cl and 25 % ³⁷Cl.

Example 4: time of flight.

A ²⁴Mg⁺ ion is given a kinetic energy of 2.50 × 10⁻¹⁶ J and travels down a flight tube of length 1.00 m. Find its flight time. (L = 6.022 × 10²³ mol⁻¹.)

First, the mass of one ion, converted to kilograms:

$m = \frac{24}{6.022 \times 10^{23}} = 3.985 \times 10^{-23}\ \text{g} = 3.985 \times 10^{-26}\ \text{kg}$

Then the velocity from $v = \sqrt{2\,KE/m}$:

$v = \sqrt{\frac{2 \times (2.50 \times 10^{-16})}{3.985 \times 10^{-26}}} = \sqrt{1.255 \times 10^{10}} = 1.120 \times 10^{5}\ \text{m s}^{-1}$

Then the time from $t = d/v$:

$t = \frac{1.00}{1.120 \times 10^{5}} = 8.93 \times 10^{-6}\ \text{s}$

The single most common error is leaving the mass in grams: the kinetic energy is in joules, so the mass must be in kilograms.

Example 5: deducing the group from successive ionisation energies.

An element has these successive ionisation energies (kJ mol⁻¹):

The first big jump is between the 3rd and 4th ionisation energies. So three electrons are removed fairly easily (from the outer shell) before the fourth must come from a shell closer to the nucleus. Three outer electrons means the element is in Group 3.

Common exam mistakes

• Omitting state symbols in ionisation-energy equations. Every species is gaseous, so (g) is required on the atom and the ion; it is a separate mark and is lost without it.
• Defining first ionisation energy without the word gaseous, or writing about "an element" or "an atom" instead of one mole of gaseous atoms. The gaseous state and the "per mole" (or "one electron from each atom") are both needed.
• Showing the ionisation forming a 2+ ion (for example Mg(g) → Mg²⁺(g) + 2e⁻) when asked for the first ionisation energy, which must form only a 1+ ion.
• Explaining a trend with a vague phrase like "bigger nucleus" or "greater effective nuclear charge". AQA wants more protons / greater nuclear charge for the attraction, and marks "effective nuclear charge" as neutral.
• For the Mg-to-Al dip, forgetting to say the electron is removed from a 3p sub-shell that is higher in energy / shielded by 3s.
• For the P-to-S dip, writing "repulsion between electrons" without saying the two electrons are paired in the same 3p orbital, or wrongly claiming a "more stable half-full sub-shell". The credit is for the paired-electron repulsion.
• Writing the electron configuration of the atom when the ion is asked for, and (for d-block ions) removing 3d electrons before 4s. Remove 4s before 3d.
• Forgetting the chromium and copper anomalies (4s¹ 3d⁵ and 4s¹ 3d¹⁰).
• In a time of flight calculation, leaving the mass in grams instead of converting to kilograms.
• Quoting a calculated A_r to the wrong precision. Report it to the number of significant figures of the abundance data, not more.
• Giving a mass-spectrum peak as the element symbol alone when the m/z value (or the species with its charge and mass number, for example ²⁴Mg⁺) is wanted.

Key definitions

• Atomic (proton) number (Z): the number of protons in the nucleus of an atom.
• Mass number (A): the total number of protons and neutrons in the nucleus of an atom.
• Isotopes: atoms of the same element with the same number of protons but different numbers of neutrons.
• Relative isotopic mass: the mass of an atom of an isotope relative to 1/12 of the mass of an atom of carbon-12.
• Relative atomic mass (A_r): the weighted mean mass of an atom of an element relative to 1/12 of the mass of an atom of carbon-12.
• Relative molecular mass (M_r): the weighted mean mass of a molecule relative to 1/12 of the mass of an atom of carbon-12.
• First ionisation energy: the energy needed to remove one electron from each atom in one mole of gaseous atoms, to form one mole of gaseous 1+ ions.
• Second ionisation energy: the energy needed to remove one electron from each ion in one mole of gaseous 1+ ions, to form one mole of gaseous 2+ ions.

Specification

• I can state the relative mass and relative charge of a proton, neutron and electron, and describe the atom as a nucleus surrounded by electrons.
• I can appreciate that the model of atomic structure has changed over time as new evidence appeared.
• I can use mass number, atomic number and charge to work out the numbers of protons, neutrons and electrons in any atom or ion.
• I can explain the existence of isotopes and why they have the same chemical properties.
• I can describe the five stages of a simple TOF mass spectrometer: ionisation, acceleration to constant kinetic energy, ion drift, detection and data analysis.
• I can write equations for electron impact and electrospray ionisation.
• I can interpret a simple mass spectrum of an element and calculate relative atomic mass from isotopic abundances (mononuclear ions).
• I can use mass spectrometry to determine a relative molecular mass.
• I can write the electron configuration of atoms and ions up to Z = 36 using s, p and d sub-shells, including the Cr and Cu anomalies.
• I can define first and successive ionisation energies and write equations for them with state symbols.
• I can explain how first and successive ionisation energies in Period 3 (Na to Ar) and Group 2 (Be to Ba) give evidence for electron arrangement in sub-shells and shells.

Related notes

• Amount of substance
• Bonding
• Periodicity