IB SL

Structure 1: Models of the Particulate Nature of Matter

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Data booklet: You can use the IB Chemistry Data Booklet in the exam — all constants, the periodic table, and key equations are provided.

IB Chemistry SL — Atomic Structure & Periodicity

Complete Study Guide

Topics Covered

  1. Models of the Particulate Nature of Matter (S1.1)
  2. The Nuclear Atom — Isotopes & Mass Spectrometry (S1.2)
  3. Electron Configurations (S1.3)
  4. Counting Particles by Mass — The Mole (S1.4)
  5. Ideal Gases (S1.5)
  6. Periodic Trends — Atomic Radius, Ionisation Energy & Electronegativity

1. Models of the Particulate Nature of Matter (S1.1)

Chemistry begins with the idea that all matter is made of tiny, discrete particles. This section traces how our model of the atom has evolved from Dalton’s simple indivisible sphere to the quantum model used today. Understanding why each model changed is just as important as knowing the final picture — IB exams frequently ask you to evaluate the experimental evidence that drove each revision.

Dalton’s Atomic Model

John Dalton (1803) proposed the first modern atomic theory:

  • All matter is made of tiny, indivisible particles called atoms.
  • All atoms of the same element are identical in mass and properties.
  • Atoms of different elements have different masses and properties.
  • Atoms combine in whole-number ratios to form compounds.
  • In a chemical reaction, atoms are rearranged — they are never created or destroyed.

Dalton’s model explained the law of conservation of mass and the law of definite proportions. However, it could not explain the existence of subatomic particles or why elements emit specific colours of light.

Thomson’s Plum Pudding Model

J.J. Thomson (1897) discovered the electron through cathode ray tube experiments. He found that:

  • Cathode rays were deflected by electric and magnetic fields, proving they were charged particles.
  • The charge-to-mass ratio was the same regardless of the metal used for the cathode.

Thomson proposed a model in which the atom was a sphere of positive charge with electrons embedded throughout — like plums in a pudding. This was the first model to include subatomic particles, but it was soon overturned by Rutherford’s experiment.

Rutherford’s Gold-Foil Experiment (Nuclear Model)

Ernest Rutherford (1909–1911) fired a beam of positively charged alpha particles (α\alpha-particles, helium nuclei) at a very thin gold foil and observed the deflections using a detector screen.

Observations and conclusions:

ObservationConclusion
Most α\alpha-particles passed straight throughMost of the atom is empty space
A small fraction were deflected at large anglesThere is a small, dense, positively charged region — the nucleus
A tiny fraction bounced almost straight backThe nucleus is very small but contains most of the atom’s mass

This experiment disproved Thomson’s plum-pudding model (which predicted only tiny deflections from a diffuse positive charge). Rutherford’s nuclear model introduced a tiny, dense, positive nucleus surrounded by mostly empty space with electrons orbiting at large distances.

Exam trap — Rutherford’s experiment: The exam often asks you to explain which specific observation led to which specific conclusion. “Most particles passed through” proves empty space. “Some deflected at large angles / bounced back” proves a small dense positive nucleus. Do not just say “the nucleus exists” — link the observation to the conclusion explicitly.

The Bohr Model

Niels Bohr (1913) modified Rutherford’s model to explain atomic emission spectra (see Section 3). Bohr proposed:

  • Electrons orbit the nucleus in fixed energy levels (also called shells or principal quantum levels), labelled n=1,2,3,n = 1, 2, 3, \ldots
  • An electron in a specific energy level has a fixed energy — it does not radiate energy while in that orbit.
  • When an electron absorbs energy, it jumps to a higher energy level (excited state).
  • When an electron falls back to a lower energy level, it emits energy as a photon of light with a specific frequency.

The Bohr model successfully explained the line spectrum of hydrogen. However, it failed to accurately predict spectra of multi-electron atoms, and it could not explain why energy levels were stable.

The Quantum/Wave Model (Modern Atomic Model)

The modern model, developed by Schrödinger, Heisenberg, and others, treats electrons as waves rather than particles on defined orbits. Key ideas:

  • It is impossible to know both the exact position and momentum of an electron simultaneously (Heisenberg uncertainty principle).
  • Electrons occupy orbitals — regions of space where there is a high probability of finding an electron.
  • Orbitals are organised into sublevels (ss, pp, dd, ff) within each principal energy level.

For IB SL, you need to know the electron configuration model (shells and sublevels) — the full mathematical wave treatment is beyond the syllabus.

Key Definitions — Atoms, Elements, Compounds, Mixtures

TermDefinition
AtomThe smallest unit of an element that retains the chemical properties of that element
ElementA pure substance made of only one type of atom; cannot be broken down by chemical means
CompoundA pure substance made of two or more elements chemically bonded in fixed ratios
MixtureTwo or more substances physically combined; components retain their own properties and can be separated by physical means
IonAn atom or group of atoms that has gained or lost one or more electrons, giving it an overall charge

2. The Nuclear Atom — Subatomic Particles, Isotopes & Mass Spectrometry (S1.2)

Subatomic Particles

The nucleus contains protons and neutrons; electrons orbit outside the nucleus.

ParticleSymbolRelative ChargeRelative MassAbsolute Charge (C)Absolute Mass (kg)Location
Protonp+p^++1+111+1.60×1019+1.60 \times 10^{-19}1.673×10271.673 \times 10^{-27}Nucleus
Neutronn0n^00011001.675×10271.675 \times 10^{-27}Nucleus
Electronee^-1-1118360\frac{1}{1836} \approx 01.60×1019-1.60 \times 10^{-19}9.109×10319.109 \times 10^{-31}Shells (outside nucleus)

Key nuclear quantities:

  • Atomic number ZZ = number of protons (defines the element)
  • Mass number AA = number of protons + neutrons
  • Number of neutrons = AZA - Z
  • In a neutral atom: number of electrons = number of protons = ZZ
  • Notation: ZAX^A_Z\text{X} — e.g. 1123Na^{23}_{11}\text{Na} means sodium with A=23A=23, Z=11Z=11, so 11 protons, 12 neutrons, 11 electrons

Isotopes

Isotopes are atoms of the same element that have the same atomic number (ZZ) but different mass numbers (AA) — i.e., the same number of protons but different numbers of neutrons.

Examples:

  • Carbon: 612C^{12}_6\text{C} (6 protons, 6 neutrons) and 614C^{14}_6\text{C} (6 protons, 8 neutrons)
  • Chlorine: 1735Cl^{35}_{17}\text{Cl} (17 protons, 18 neutrons) and 1737Cl^{37}_{17}\text{Cl} (17 protons, 20 neutrons)
  • Hydrogen: 11H^1_1\text{H} (protium), 12H^2_1\text{H} (deuterium), 13H^3_1\text{H} (tritium)

Isotopes have identical chemical properties because they have the same number of electrons (and therefore the same electron configuration). They differ in physical properties (e.g. density, rate of diffusion) due to the difference in mass.

Exam trap — isotope properties: Students often confuse “same chemical properties” with “same physical properties”. Isotopes have the same chemical properties (same number of electrons / same electron configuration) but different physical properties (different masses). Do not state that isotopes are chemically different — they are not.

Relative Atomic Mass (ArA_r)

Because most elements exist as a mixture of isotopes, the relative atomic mass (ArA_r) is the weighted average mass of all naturally occurring isotopes of an element, relative to 112\frac{1}{12} of the mass of one atom of carbon-12.

Ar=(fractional abundance×mass number of isotope)A_r = \sum \left( \text{fractional abundance} \times \text{mass number of isotope} \right)

Worked Example 1 — Calculating ArA_r from isotope data

Chlorine has two naturally occurring isotopes:

  • 35Cl^{35}\text{Cl}: abundance 75.77%
  • 37Cl^{37}\text{Cl}: abundance 24.23%
Ar(Cl)=(0.7577×35)+(0.2423×37)A_r(\text{Cl}) = (0.7577 \times 35) + (0.2423 \times 37)Ar(Cl)=26.52+8.97=35.4935.5A_r(\text{Cl}) = 26.52 + 8.97 = 35.49 \approx 35.5

This matches the value you see on the periodic table. Note that ArA_r is never a whole number when two or more isotopes are present in significant amounts.

Common mistake: Using percentage instead of fractional abundance. Always divide percentages by 100 before substituting. Using 75.77 instead of 0.7577 gives an answer 100× too large.

Mass Spectrometry

A mass spectrometer separates particles by their mass-to-charge ratio (m/zm/z). It is used to:

  1. Identify the isotopes present in an element and their relative abundances
  2. Determine the ArA_r of an element
  3. Identify unknown compounds (beyond SL scope)

How a mass spectrometer works (five stages):

StageProcessWhat happens
1. VaporisationSample is converted to a gasParticles are separated so they can be manipulated individually
2. IonisationParticles bombarded with high-energy electronsElectrons are knocked off atoms, forming positive ions (M+\text{M}^+)
3. AccelerationIons accelerated through an electric fieldAll ions gain the same kinetic energy; lighter ions travel faster
4. DeflectionIons passed through a magnetic fieldIons are deflected; lighter/more charged ions deflect more
5. DetectionIons strike a detectorA signal proportional to the number of ions is recorded at each m/zm/z ratio

Reading a mass spectrum:

  • The x-axis shows m/zm/z (for singly charged ions, this equals the mass number)
  • The y-axis shows relative abundance (%)
  • Each peak represents one isotope
  • The height of each peak is proportional to the natural abundance of that isotope

When peak heights (not percentages) are given, use this form to calculate ArA_r — it normalises the abundances automatically:

Ar=(m/z×relative abundance)relative abundanceA_r = \frac{\sum (m/z \times \text{relative abundance})}{\sum \text{relative abundance}}

IB Tip: You are expected to be able to read a mass spectrum and calculate ArA_r from it. In the exam, relative abundances may be given as peak heights rather than percentages — the calculation method is identical.


3. Electron Configurations (S1.3)

Energy Levels and Sublevels

Electrons in an atom occupy energy levels (principal quantum levels, n=1,2,3,4,n = 1, 2, 3, 4, \ldots). Within each energy level there are sublevels labelled ss, pp, dd, ff.

SublevelNumber of orbitalsMaximum electronsShape
ss12Spherical
pp36Dumbbell-shaped (3 orientations: pxp_x, pyp_y, pzp_z)
dd510Complex (5 orientations)
ff714Complex (7 orientations)

Each orbital can hold a maximum of 2 electrons (with opposite spins — Pauli exclusion principle).

Sublevels present in each principal energy level:

nnSublevels presentTotal electrons
11s1s2
22s2s, 2p2p8
33s3s, 3p3p, 3d3d18
44s4s, 4p4p, 4d4d, 4f4f32

The Three Rules for Filling Orbitals

1. Aufbau principle (“building up”): Electrons fill orbitals in order of increasing energy, starting with the lowest-energy sublevel.

The order of filling is:

1s2s2p3s3p4s3d4p5s4d5p1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \rightarrow 5s \rightarrow 4d \rightarrow 5p \ldots

Note that 4s4s fills before 3d3d because 4s4s is lower in energy than 3d3d for neutral atoms.

2. Hund’s rule: When electrons occupy orbitals of the same energy (degenerate orbitals), each orbital receives one electron before any orbital receives a second. All singly occupied orbitals have the same spin direction.

For example, in the 2p2p sublevel of nitrogen (2p32p^3), each of the three pp orbitals has one electron — none share an orbital yet.

3. Pauli exclusion principle: No two electrons in an atom can have the same set of four quantum numbers. In practice, this means each orbital holds a maximum of two electrons with opposite spins (represented as \uparrow\downarrow).

Writing Electron Configurations

Full configuration: List every sublevel and its electron count as a superscript.

Examples:

ElementZZFull electron configuration
Hydrogen11s11s^1
Helium21s21s^2
Lithium31s22s11s^2\, 2s^1
Carbon61s22s22p21s^2\, 2s^2\, 2p^2
Nitrogen71s22s22p31s^2\, 2s^2\, 2p^3
Oxygen81s22s22p41s^2\, 2s^2\, 2p^4
Neon101s22s22p61s^2\, 2s^2\, 2p^6
Sodium111s22s22p63s11s^2\, 2s^2\, 2p^6\, 3s^1
Magnesium121s22s22p63s21s^2\, 2s^2\, 2p^6\, 3s^2
Chlorine171s22s22p63s23p51s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5
Argon181s22s22p63s23p61s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6
Calcium201s22s22p63s23p64s21s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2
Iron261s22s22p63s23p64s23d61s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6

Condensed (noble gas core) notation: Replace the inner electrons with the symbol of the preceding noble gas in square brackets.

Examples:

  • Sodium: [Ne]3s1[\text{Ne}]\, 3s^1
  • Sulfur: [Ne]3s23p4[\text{Ne}]\, 3s^2\, 3p^4
  • Calcium: [Ar]4s2[\text{Ar}]\, 4s^2
  • Iron: [Ar]4s23d6[\text{Ar}]\, 4s^2\, 3d^6

Worked Example 2 — Writing electron configurations

Write the full and condensed electron configurations for phosphorus (Z=15Z = 15).

Step 1: Total electrons = 15 (neutral atom).

Step 2: Fill in order: 1s22s22p63s23p31s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^3

Check: 2+2+6+2+3=152 + 2 + 6 + 2 + 3 = 15

Step 3: Condensed — the noble gas before P is neon (Z=10Z=10, configuration 1s22s22p61s^2\, 2s^2\, 2p^6):

[Ne]3s23p3[\text{Ne}]\, 3s^2\, 3p^3

Applying Hund’s rule to the 3p33p^3: The three 3p3p electrons each occupy a separate orbital with parallel spins before any pairing occurs:

3p:3p: \uparrow \quad \uparrow \quad \uparrow

Exceptions: Chromium and Copper

Two important exceptions to the Aufbau filling order arise from the extra stability of half-filled (d5d^5) and fully-filled (d10d^{10}) dd subshells.

ElementExpectedActualReason
Chromium (Z=24Z=24)[Ar]4s23d4[\text{Ar}]\, 4s^2\, 3d^4[Ar]4s13d5[\text{Ar}]\, 4s^1\, 3d^5Half-filled d5d^5 is extra stable
Copper (Z=29Z=29)[Ar]4s23d9[\text{Ar}]\, 4s^2\, 3d^9[Ar]4s13d10[\text{Ar}]\, 4s^1\, 3d^{10}Fully-filled d10d^{10} is extra stable

Exam trap — Cr and Cu exceptions: These are the two most commonly tested exceptions in IB Chemistry. If an exam question asks for the electron configuration of Cr\text{Cr} or Cu\text{Cu}, you must give the actual configuration, not the predicted one. For all other transition metals (Ti, V, Mn, Fe, Co, Ni, Zn), the regular Aufbau rule applies without exception.

Relating Electron Configuration to the Periodic Table

The periodic table is organised to reflect electron configurations:

  • ss-block (Groups 1–2): outermost electrons fill an ss sublevel
  • pp-block (Groups 13–18): outermost electrons fill a pp sublevel
  • dd-block (Groups 3–12, transition metals): outermost electrons fill a dd sublevel
  • ff-block (lanthanides and actinides): outermost electrons fill an ff sublevel

The period number gives the highest principal quantum level (nn) occupied by electrons in the ground state. The group number (for main-group elements) gives the number of valence electrons.

Shortcut for valence electrons:

  • Group 1: 1 valence electron (ns1ns^1)
  • Group 2: 2 valence electrons (ns2ns^2)
  • Groups 13–18: (group number − 10) valence electrons — e.g. Group 17 has 7 valence electrons (ns2np5ns^2\, np^5)
  • Noble gases (Group 18): 8 valence electrons (except He: 2)

The number of valence electrons determines bonding behaviour — this is why elements in the same group have similar chemistry.

Emission Spectra as Evidence for Discrete Energy Levels

When atoms are excited (by heat or electricity), electrons jump to higher energy levels. When they fall back to lower levels, they emit photons of light with specific frequencies. This produces a line emission spectrum — a series of discrete (separate) coloured lines on a dark background.

Key points:

  • The lines are discrete (not continuous), proving that electrons can only occupy specific energy levels.
  • Each element has a unique line spectrum — a “fingerprint” that can be used for identification (e.g. spectroscopy of stars).
  • The energy of the emitted photon equals the energy difference between the two levels: ΔE=hf\Delta E = hf (where hh is Planck’s constant and ff is frequency).
  • The visible lines in hydrogen’s spectrum (Balmer series) arise from transitions to the n=2n=2 level from n=3,4,5,6n=3, 4, 5, 6.

IB Tip: You do not need to calculate ΔE=hf\Delta E = hf at SL, but you must be able to explain why line spectra provide evidence for discrete energy levels. The key word is “discrete” — a continuous spectrum would be produced if electrons could have any energy.


4. Counting Particles by Mass — The Mole (S1.4)

Atoms are too small to count individually. Chemists use the mole as a counting unit — one mole of any substance contains the same number of particles as there are atoms in exactly 12 g of carbon-12.

Avogadro’s Constant

NA=6.022×1023 mol1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}

One mole of any substance contains 6.022×10236.022 \times 10^{23} particles (atoms, molecules, ions, etc.).

Molar Mass

The molar mass (MM) of a substance is the mass of one mole of that substance, expressed in g mol1\text{g mol}^{-1}. Numerically, it equals the ArA_r or MrM_r (relative formula mass) in grams.

  • M(C)=12.01g mol1M(\text{C}) = 12.01\, \text{g mol}^{-1}
  • M(H2O)=2(1.01)+16.00=18.02g mol1M(\text{H}_2\text{O}) = 2(1.01) + 16.00 = 18.02\, \text{g mol}^{-1}
  • M(NaCl)=22.99+35.45=58.44g mol1M(\text{NaCl}) = 22.99 + 35.45 = 58.44\, \text{g mol}^{-1}

The Mole Equation

n=mMn = \frac{m}{M}

Where:

  • nn = amount of substance (mol)
  • mm = mass (g)
  • MM = molar mass (g mol1\text{g mol}^{-1})

Rearrangements: m=n×Mm = n \times M and M=mnM = \frac{m}{n}

Worked Example 3 — Mole calculations

Part (a): How many moles are in 9.80 g of sulfuric acid, H2SO4\text{H}_2\text{SO}_4?

Step 1 — Find molar mass: M(H2SO4)=2(1.01)+32.07+4(16.00)=2.02+32.07+64.00=98.09g mol1M(\text{H}_2\text{SO}_4) = 2(1.01) + 32.07 + 4(16.00) = 2.02 + 32.07 + 64.00 = 98.09\, \text{g mol}^{-1}

Step 2 — Apply n=m/Mn = m/M: n=9.8098.09=0.09990.100moln = \frac{9.80}{98.09} = 0.0999 \approx 0.100\, \text{mol}

Part (b): What mass of sodium chloride (NaCl\text{NaCl}) contains 1.20×10241.20 \times 10^{24} formula units?

Step 1 — Find moles: n=1.20×10246.022×1023=1.992.00moln = \frac{1.20 \times 10^{24}}{6.022 \times 10^{23}} = 1.99 \approx 2.00\, \text{mol}

Step 2 — Find mass: m=n×M=2.00×58.44=116.9117gm = n \times M = 2.00 \times 58.44 = 116.9 \approx 117\, \text{g}

Relative Formula Mass

  • Relative formula mass (MrM_r): sum of ArA_r values for all atoms in the formula unit — used for ionic compounds and molecules alike (e.g. Mr(H2O)=2(1.01)+16.00=18.02M_r(\text{H}_2\text{O}) = 2(1.01) + 16.00 = 18.02)
  • Both ArA_r and MrM_r are dimensionless (they are ratios relative to 112\frac{1}{12} of the mass of 12C^{12}\text{C}, not absolute masses)

Empirical and Molecular Formulae

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule.

  • Molecular formula is always a whole-number multiple of the empirical formula.
  • The multiple is found from: multiple=Mr(molecular)Mr(empirical)\text{multiple} = \frac{M_r(\text{molecular})}{M_r(\text{empirical})}

Worked Example 4 — Empirical and molecular formula from percentage composition

A compound has the following percentage composition by mass: C 40.00%, H 6.72%, O 53.28%. Its molar mass is Mr=90.08g mol1M_r = 90.08\, \text{g mol}^{-1}. Find its empirical and molecular formulae.

Step 1: Assume 100 g of compound — percentages become masses in grams:

  • m(C)=40.00gm(\text{C}) = 40.00\, \text{g}, m(H)=6.72gm(\text{H}) = 6.72\, \text{g}, m(O)=53.28gm(\text{O}) = 53.28\, \text{g}

Step 2: Convert to moles: n(C)=40.0012.01=3.331n(H)=6.721.008=6.667n(O)=53.2816.00=3.330n(\text{C}) = \frac{40.00}{12.01} = 3.331 \quad n(\text{H}) = \frac{6.72}{1.008} = 6.667 \quad n(\text{O}) = \frac{53.28}{16.00} = 3.330

Step 3: Divide by the smallest value (3.330): C:3.3313.3301H:6.6673.3302O:3.3303.330=1\text{C}: \frac{3.331}{3.330} \approx 1 \quad \text{H}: \frac{6.667}{3.330} \approx 2 \quad \text{O}: \frac{3.330}{3.330} = 1

Empirical formula: CH2O\text{CH}_2\text{O}Mr(empirical)=12+2+16=30g mol1M_r(\text{empirical}) = 12 + 2 + 16 = 30\, \text{g mol}^{-1}

Step 4: Find the multiple: multiple=90.0830=3.0033\text{multiple} = \frac{90.08}{30} = 3.003 \approx 3

Molecular formula: C3H6O3\text{C}_3\text{H}_6\text{O}_3

Exam trap — empirical formula ratios: After dividing by the smallest mole value, you may get a ratio like 1 : 1.5 : 2. Do not round 1.5 to 2 — multiply all values by 2 to get whole numbers (1 : 1.5 : 2 becomes 2 : 3 : 4). Only round to the nearest integer if the value is within 0.1 of a whole number.


5. Ideal Gases (S1.5)

An ideal gas is a hypothetical gas whose molecules:

  • Have negligible volume (treated as point masses)
  • Have no intermolecular forces between them
  • Collide elastically (kinetic energy is conserved)

Real gases approximate ideal behaviour at low pressure and high temperature.

The Ideal Gas Law

PV=nRTPV = nRT

Where:

  • PP = pressure (Pa or kPa — be consistent with RR)
  • VV = volume (m3\text{m}^3 or L — be consistent with RR)
  • nn = amount of substance (mol)
  • RR = gas constant =8.314J mol1K1= 8.314\, \text{J mol}^{-1}\text{K}^{-1} (use with Pa and m3\text{m}^3)
  • TT = temperature (K) — always convert: T(K)=T(°C)+273.15T(\text{K}) = T(°\text{C}) + 273.15

Molar Volume at STP

Standard Temperature and Pressure (STP): T=273KT = 273\, \text{K} (0°C), P=100kPaP = 100\, \text{kPa}

At STP, one mole of any ideal gas occupies a volume of 22.7 L (22.7×103m322.7 \times 10^{-3}\, \text{m}^3).

Vm=RTP=8.314×273100000=0.02270m3=22.7L mol1V_m = \frac{RT}{P} = \frac{8.314 \times 273}{100\,000} = 0.02270\, \text{m}^3 = 22.7\, \text{L mol}^{-1}

Exam trap — STP definition: The IB (from 2016 onwards) defines STP as 273 K and 100 kPa (not 101.3 kPa). This gives Vm=22.7L mol1V_m = 22.7\, \text{L mol}^{-1}, not 22.4L mol122.4\, \text{L mol}^{-1} (which was the old definition at 101.3 kPa). Using 22.4 in an IB exam will cost you marks.

Worked Example 5 — Ideal gas calculation

Calculate the volume occupied by 3.20 g of oxygen gas (O2\text{O}_2) at 298 K and 101 kPa. Give your answer in litres.

Step 1: Find moles: M(O2)=2×16.00=32.00g mol1M(\text{O}_2) = 2 \times 16.00 = 32.00\, \text{g mol}^{-1}

n=3.2032.00=0.100moln = \frac{3.20}{32.00} = 0.100\, \text{mol}

Step 2: Apply PV=nRTPV = nRT. Use R=8.314J mol1K1R = 8.314\, \text{J mol}^{-1}\text{K}^{-1}, PP in Pa:

V=nRTP=0.100×8.314×298101000=247.8101000=2.45×103m3V = \frac{nRT}{P} = \frac{0.100 \times 8.314 \times 298}{101\,000} = \frac{247.8}{101\,000} = 2.45 \times 10^{-3}\, \text{m}^3

Step 3: Convert to litres (1m3=1000L1\, \text{m}^3 = 1000\, \text{L}):

V=2.45×103×1000=2.45LV = 2.45 \times 10^{-3} \times 1000 = 2.45\, \text{L}


Although the IB 2025 syllabus places periodic trends formally in Structure 3, they are commonly examined alongside Structure 1 content and are essential for understanding bonding (Structure 2). This section covers the three most important trends.

Atomic Radius

The atomic radius is defined as half the distance between the nuclei of two identical atoms in a covalent bond (covalent radius).

Trend across a period (left → right): decreases

As you move across a period, the nuclear charge ZZ increases (more protons added), but the electrons are added to the same energy level (same principal quantum number nn). The increased nuclear charge pulls all the electrons closer to the nucleus, so the atomic radius decreases.

Trend down a group (top → bottom): increases

As you move down a group, each successive element has electrons in an additional energy level (higher nn). The outermost electrons are further from the nucleus, so the atomic radius increases. Additionally, electrons in inner shells shield the outer electrons from the full nuclear charge (shielding effect).

Periodic trends summary table:

PropertyAcross period (→)Down group (↓)Reason
Atomic radiusDecreasesIncreasesNuclear charge vs. new energy levels
1st ionisation energyGenerally increases*DecreasesEffective nuclear charge vs. shielding
ElectronegativityIncreasesDecreasesNuclear attraction on bonding electrons

*With exceptions: see Section 6.2

Increasing nuclear charge →↓ Increasing shellsAtomicRadiusIE₁Electro-negativitysmallerlargerlowerhigherlowerhigher

Periodic trends: atomic radius, ionisation energy and electronegativity change directionally across periods and down groups.

First Ionisation Energy (IE₁)

The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms in the ground state:

X(g)X+(g)+eΔH=IE1\text{X}(g) \rightarrow \text{X}^+(g) + e^- \quad \Delta H = \text{IE}_1

IE₁ is always endothermic (energy must be supplied to overcome the attraction between the electron and nucleus).

Trend across a period: general increase

As nuclear charge increases across a period (electrons in the same shell), the effective nuclear charge experienced by the outer electrons increases. The outer electrons are held more tightly, so more energy is needed to remove them.

Anomalies across Period 2 and Period 3:

There are two important dips in the general increasing trend:

Anomaly 1 — Group 2 to Group 13 (e.g. Be → B, Mg → Al):

IE₁(B) < IE₁(Be), even though B has higher ZZ.

  • Beryllium: outer electron is in 2s2s (1s22s21s^2\, 2s^2)
  • Boron: outer electron is in 2p2p (1s22s22p11s^2\, 2s^2\, 2p^1)
  • The 2p2p sublevel is higher in energy and further from the nucleus than 2s2s, and is also partially shielded by the filled 2s2s electrons. Therefore the 2p2p electron in B is easier to remove than the 2s2s electron in Be.

Anomaly 2 — Group 15 to Group 16 (e.g. N → O, P → S):

IE₁(O) < IE₁(N), even though O has higher ZZ.

  • Nitrogen: 2p32p^3 — each 2p2p orbital has one electron (Hund’s rule)
  • Oxygen: 2p42p^4 — one 2p2p orbital must hold two electrons (paired)
  • Electron–electron repulsion between the paired electrons in oxygen makes it easier to remove one electron from O than from N.

Exam trap — IE anomalies: These two anomalies (B < Be and O < N) are among the most frequently tested points in IB Chemistry. You must be able to explain them using sublevel energies and electron pairing repulsion. Simply saying “because of the electron configuration” earns no marks — you must explain the specific reason (higher sublevel / electron repulsion from pairing).

Trend down a group: decreases

Down a group, the outer electron is in successively higher energy levels (greater nn), further from the nucleus. Additionally, there are more inner electron shells providing greater shielding from the nuclear charge. Both factors reduce the effective nuclear charge felt by the outer electron, so less energy is needed to remove it.

Electronegativity

Electronegativity is a measure of the ability of an atom in a covalent bond to attract the shared pair of electrons towards itself. It is measured on the Pauling scale, where fluorine (the most electronegative element) is assigned a value of 4.0.

Trend across a period: increases

Moving across a period, nuclear charge increases while the atomic radius decreases (electrons are closer to the nucleus). The nucleus exerts a stronger attraction on the bonding electrons, so electronegativity increases.

Trend down a group: decreases

Moving down a group, the atomic radius increases and the shielding effect increases. The nucleus has less effective attraction over the bonding electrons, so electronegativity decreases.

IB Tip: Noble gases are not typically assigned electronegativity values because they do not generally form covalent bonds. Caesium (Cs) has the lowest electronegativity of commonly discussed elements; fluorine (F) has the highest. Electronegativity differences are used to determine bond polarity and bond type — you will use this concept extensively in Structure 2 (Bonding & Structure).


Practice Questions

Test your understanding with these IB-style multiple choice questions.


Question 1

Rutherford’s gold-foil experiment provided evidence for which of the following?

A. Electrons are present in discrete energy levels.

B. The atom contains a small, dense, positively charged nucleus.

C. Atoms are mostly empty space with electrons embedded in positive charge.

D. Neutrons and protons have approximately equal masses.


Question 2

Which of the following is the correct electron configuration for chromium (Z=24Z = 24)?

A. [Ar]4s23d4[\text{Ar}]\, 4s^2\, 3d^4

B. [Ar]4s23d2[\text{Ar}]\, 4s^2\, 3d^{2}

C. [Ar]4s13d5[\text{Ar}]\, 4s^1\, 3d^5

D. [Ar]4s03d6[\text{Ar}]\, 4s^0\, 3d^6


Question 3

An element has two isotopes: 63X^{63}\text{X} with 69.2% natural abundance and 65X^{65}\text{X} with 30.8% natural abundance. What is the relative atomic mass of this element?

A. 63.5

B. 64.0

C. 63.6

D. 64.4


Question 4

Which of the following correctly explains why the first ionisation energy of oxygen is lower than that of nitrogen?

A. Oxygen has a higher nuclear charge than nitrogen.

B. Oxygen has a larger atomic radius than nitrogen.

C. The paired electron in oxygen’s 2p2p sublevel experiences greater electron–electron repulsion, making it easier to remove.

D. Oxygen’s 2p2p sublevel is at a lower energy than nitrogen’s.


Question 5

A sample of gas occupies 4.50×103m34.50 \times 10^{-3}\, \text{m}^3 at a temperature of 300 K and a pressure of 1.20×105Pa1.20 \times 10^5\, \text{Pa}. What is the amount of substance (in mol) present in the sample?

A. 0.180 mol

B. 0.216 mol

C. 0.270 mol

D. 1.80 mol


Show Answers
  1. B — The key evidence from the gold-foil experiment was that a small fraction of α\alpha-particles were deflected at large angles or bounced back, proving the existence of a small, dense, positively charged nucleus. A refers to evidence from line emission spectra (Bohr model). C describes Thomson’s plum-pudding model, which Rutherford’s experiment disproved. D relates to mass spectrometry, not the gold-foil experiment.

  2. C[Ar]4s13d5[\text{Ar}]\, 4s^1\, 3d^5. Chromium is one of two common exceptions to the Aufbau rule. A half-filled 3d53d^5 subshell (combined with a 4s14s^1 rather than 4s24s^2) is especially stable due to exchange energy. The predicted configuration (A) would be [Ar]4s23d4[\text{Ar}]\, 4s^2\, 3d^4, but this is not observed in practice.

  3. CAr=(0.692×63)+(0.308×65)=43.60+20.02=63.6263.6A_r = (0.692 \times 63) + (0.308 \times 65) = 43.60 + 20.02 = 63.62 \approx 63.6. This element is copper (Cu\text{Cu}). A (63.5) is the textbook value rounded differently; B and D result from arithmetic errors in the weighted average.

  4. C — In nitrogen (2p32p^3), each 2p2p orbital contains one electron (Hund’s rule). In oxygen (2p42p^4), one 2p2p orbital must contain two paired electrons. The repulsion between the two electrons in the same orbital reduces the energy needed to remove one of them, lowering IE₁ relative to nitrogen. A is true but explains the general increase, not the anomaly. B is wrong — oxygen is actually smaller than nitrogen (higher nuclear charge). D is wrong — subshell energy is determined by sublevel type, not by which element occupies it.

  5. B — Using PV=nRTPV = nRT: n=PVRT=1.20×105×4.50×1038.314×300=5402494.2=0.21650.216moln = \frac{PV}{RT} = \frac{1.20 \times 10^5 \times 4.50 \times 10^{-3}}{8.314 \times 300} = \frac{540}{2494.2} = 0.2165 \approx 0.216\, \text{mol}. A common error is using temperature in °C rather than K, which gives a very different (incorrect) answer. Always convert to Kelvin first.

Next Topic

The concepts in this guide — especially electron configurations and electronegativity — are directly applied in the next topic:

Bonding & Structure (Structure 2) — ionic bonding, covalent bonding, VSEPR geometry, intermolecular forces, and the bonding triangle all build directly on electron configurations and periodic trends covered here.