S3.1 The periodic table: classification of elements | IB DP Chemistry

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🎯What you need to be able to do

Identify the positions of metals, metalloids and non-metals, and recognise the s, p, d and f blocks.
Deduce the electron configuration of an atom up to \( Z = 36 \) from its position, and vice versa, and know the classifications alkali metals, halogens, transition elements and noble gases.
Explain the periodicity of atomic radius, ionic radius, ionization energy, electron affinity and electronegativity.
Describe and explain the reactions of group 1 metals with water and of group 17 elements with halide ions.
Deduce equations for the reactions with water of the oxides of group 1 and group 2 metals, of carbon and of sulfur, and relate these to acid rain and ocean acidification.
Deduce oxidation states of an atom in an ion or a compound, and explain why the oxidation state of an element is zero.
AHL Explain how discontinuities in first ionization energy across a period provide evidence for sublevels.
AHL Recognise the characteristic properties of transition elements, deduce the electron configurations of their ions, and explain their variable oxidation states.
AHL Apply the colour wheel to deduce the light absorbed and observed for a transition element complex.

📚The chemistry

How the table is organised

Periods are the horizontal rows, groups the vertical columns (numbered 1 to 18), and blocks the regions named after the sublevel being filled: s, p, d and f. Metals are on the left and centre, non-metals on the upper right, and the metalloids lie along the diagonal staircase between them — B, Si, Ge, As, Sb, Te.

The organisation is not arbitrary. Position is electron configuration:

the period number is the outer main energy level occupied by electrons;
elements in a group have the same number of valence electrons, which is why they behave alike chemically;
the block is the sublevel receiving the last electron.

So an element in period 3, group 16 has three occupied main levels and six valence electrons: \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^4 \) — sulfur. You should be able to run the deduction in both directions up to \( Z = 36 \). Four classifications must be known by name: alkali metals (group 1), halogens (group 17), noble gases (group 18) and transition elements (the d-block).

Periodicity: the five trends

Every trend on this list comes from the same two competing factors — effective nuclear charge and distance plus shielding — so learn the reasoning once and apply it five times.

Atomic radius
Across a period: decreases. Nuclear charge rises while electrons enter the same main level, so shielding is nearly constant and effective nuclear charge pulls the shell in.
Down a group: increases. An extra occupied main energy level, plus more shielding.

Ionic radius
Cations are smaller than their atoms (a whole shell lost, and fewer electrons for the same nuclear charge). Anions are larger (added electrons increase repulsion, with no extra protons). Across a period the cations shrink, then there is a jump upward at the first anion, then the anions shrink.

First ionization energy
Across: increases — smaller radius, higher effective nuclear charge, outer electron held more tightly.
Down: decreases — the outer electron is further out and better shielded. Full treatment in S1.3.

Electron affinity
The energy change when a gaseous atom gains an electron. Becomes more exothermic across a period (a stronger attraction for the incoming electron) and less exothermic down a group. First electron affinities are usually negative; the second is always positive, because you are forcing an electron onto an anion.

Electronegativity
The ability of an atom to attract the shared pair in a covalent bond. Increases across a period and decreases down a group, so fluorine is the highest and the bottom left is the lowest. Values are in the data booklet.

Metallic character
The mirror image: decreases across a period and increases down a group, because it tracks the ease of losing electrons.

Group 1 and group 17 reactivity

Group 1 metals become more reactive down the group. Reacting means losing the single outer electron; down the group the atomic radius increases and shielding increases, so the outer electron is held less tightly, ionization energy falls, and the electron is lost more readily. Each reacts with water to give a hydroxide and hydrogen:

\[ 2\mathrm{Na}(s) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2}(g) \]

The observations escalate down the group: lithium fizzes steadily, sodium melts into a ball and darts about, potassium ignites the hydrogen with a lilac flame, rubidium and caesium react explosively. The resulting solution is alkaline — hence the family name.

Group 17 elements become less reactive down the group. Reacting means gaining an electron; down the group the atom is larger and better shielded, so the incoming electron is attracted less strongly. This produces the displacement reactions:

\[ \mathrm{Cl_2}(aq) + 2\mathrm{KBr}(aq) \rightarrow 2\mathrm{KCl}(aq) + \mathrm{Br_2}(aq) \]

A more reactive halogen displaces a less reactive one from a solution of its halide: chlorine displaces bromide and iodide, bromine displaces iodide only, iodine displaces neither. The observation is the colour change — the solution turns orange (bromine) or brown (iodine). These are redox reactions, and they reappear in R3.2.

Oxides across a period

Metallic and non-metallic properties form a continuum, and the oxides show it clearly. Across period 3 the oxides run basic → amphoteric → acidic.

Group 1 and 2 oxides are basic and react with water to give alkaline solutions:
\( \mathrm{Na_2O}(s) + \mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) \)
\( \mathrm{MgO}(s) + \mathrm{H_2O}(l) \rightarrow \mathrm{Mg(OH)_2}(aq) \) (only sparingly soluble)
Aluminium oxide is amphoteric — it reacts with both acids and bases, sitting on the boundary.
Non-metal oxides are acidic and give acidic solutions:
\( \mathrm{CO_2}(g) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_2CO_3}(aq) \)
\( \mathrm{SO_2}(g) + \mathrm{H_2O}(l) \rightarrow \mathrm{H_2SO_3}(aq) \)
\( \mathrm{SO_3}(g) + \mathrm{H_2O}(l) \rightarrow \mathrm{H_2SO_4}(aq) \)

Two environmental consequences the syllabus names explicitly. Acid rain is caused by gaseous non-metal oxides — oxides of sulfur from burning sulfur-containing fossil fuels, and oxides of nitrogen formed in hot engines — dissolving in atmospheric water to give sulfurous, sulfuric and nitric acids. Ocean acidification is caused by rising atmospheric \( \mathrm{CO_2} \) dissolving in seawater to form carbonic acid, lowering the pH and making it harder for organisms to build carbonate shells.

Oxidation states

The oxidation state is the charge an atom would have if the compound were composed entirely of ions — that is, if every bonding pair were assigned completely to the more electronegative atom. It is a bookkeeping device, not a real charge, which is exactly how it differs from formal charge (S2.2).

Write it with the sign before the number: \( +2 \), \( -1 \). Working rules:

An uncombined element is zero — in \( \mathrm{O_2} \), \( \mathrm{Na} \) or \( \mathrm{P_4} \), the atoms are identical, so neither can be said to have taken the shared electrons.
A monatomic ion has an oxidation state equal to its charge.
Group 1 is +1, group 2 is +2, fluorine is always −1.
Hydrogen is +1, except in metal hydrides where it is −1 (\( \mathrm{NaH} \)).
Oxygen is −2, except in peroxides where it is −1 (\( \mathrm{H_2O_2} \)) and when bonded to fluorine.
The states sum to zero in a neutral compound, and to the charge in a polyatomic ion.

Naming uses Roman numerals for the oxidation state: iron(III) chloride, manganese(VII). Older generic names for oxyanions persist and are acceptable — \( \mathrm{NO_3^-} \) nitrate, \( \mathrm{NO_2^-} \) nitrite, \( \mathrm{SO_4^{2-}} \) sulfate, \( \mathrm{SO_3^{2-}} \) sulfite. The terms “oxidation number” and “oxidation state” are used interchangeably and either is accepted.

AHL Discontinuities as evidence for sublevels

The rise in first ionization energy across a period is interrupted twice, and the interruptions are evidence that main energy levels are divided into sublevels. Aluminium is lower than magnesium because its electron is removed from a 3p orbital rather than a 3s; sulfur is lower than phosphorus because its 3p⁴ configuration contains a paired orbital whose electrons repel. Full treatment, and the warning about “special stability” explanations, is in S1.3.

AHL Transition elements

A transition element is a d-block element with an incomplete d sublevel in at least one of its stable oxidation states. That definition is what excludes scandium and zinc: scandium’s only common ion, \( \mathrm{Sc^{3+}} \), is \( 3d^0 \), and zinc’s, \( \mathrm{Zn^{2+}} \), is \( 3d^{10} \) — so neither has a partially filled d sublevel, and neither shows the characteristic properties.

The characteristic properties, all traceable to the incomplete d sublevel:

Variable oxidation state. The 4s and 3d sublevels are close in energy, so successive ionization energies rise only gradually and several different numbers of electrons can be removed at comparable cost. Contrast group 2, where removing a third electron means breaking into an inner level.
High melting points and electrical conductivity — delocalized d-electrons, see S2.3.
Magnetic properties — unpaired d-electrons. Knowledge of the different types of magnetism is not assessed.
Catalytic activity — the ability to change oxidation state readily lets a transition element accept and release electrons during a reaction and be regenerated.
Coloured compounds and complex ion formation — below.

Electron configurations of the ions: remove the 4s electrons first. \( \mathrm{Fe} \) is \( [\mathrm{Ar}]\,4s^2\,3d^6 \), so \( \mathrm{Fe^{2+}} \) is \( [\mathrm{Ar}]\,3d^6 \) and \( \mathrm{Fe^{3+}} \) is \( [\mathrm{Ar}]\,3d^5 \). Remember also the two neutral-atom exceptions, \( \mathrm{Cr} \) and \( \mathrm{Cu} \).

AHL Complexes and colour

A complex ion forms when ligands — species with a lone pair, such as \( \mathrm{H_2O} \), \( \mathrm{NH_3} \), \( \mathrm{Cl^-} \), \( \mathrm{CN^-} \) — donate that pair to a transition element cation, forming coordination bonds (S2.2). To deduce the overall charge, add the charge on the metal ion to the charges on the ligands: \( \mathrm{[Cu(H_2O)_6]^{2+}} \) is 2+ because water is neutral, but \( \mathrm{[CuCl_4]^{2-}} \) is 2− because \( +2 + 4(-1) = -2 \).

Why they are coloured. Ligands cause the five d orbitals, which are degenerate in the free ion, to split into groups of slightly different energy. An electron can then be promoted between the split d orbitals by absorbing a photon of exactly the right energy, and that energy corresponds to a wavelength in the visible region. The light we see is the light not absorbed, so the colour observed is complementary to the colour absorbed.

Use the colour wheel in the data booklet: the absorbed and observed colours sit opposite one another. A solution that appears blue is absorbing orange; a solution that appears green is absorbing red. Convert between colour, wavelength and frequency with \( c = \lambda f \), also in the booklet. Splitting patterns and their relation to coordination number are explicitly not required.

This is also the basis of colorimetry: because the absorbance of a coloured complex is proportional to concentration, a calibration curve of known standards lets you determine an unknown concentration — a technique worth remembering for the IA.

✏️Worked example

(a) Deduce the oxidation state of the named element in each: Cr in \( \mathrm{Cr_2O_7^{2-}} \), S in \( \mathrm{H_2SO_4} \), H in \( \mathrm{CaH_2} \), O in \( \mathrm{Na_2O_2} \), and Mn in \( \mathrm{MnO_4^{-}} \).
(b) Explain why the atomic radius decreases from Na to Cl but increases from F to I.
(c) Write an equation for the reaction of potassium with water and state two observations. Explain why potassium reacts more vigorously than sodium.
(d) AHL A solution of \( \mathrm{[Cu(H_2O)_6]^{2+}} \) appears pale blue. State the colour of light absorbed, and explain why the ion is coloured at all while a solution of \( \mathrm{Zn^{2+}} \) is colourless.

(a) Apply the rules and let the sum condition do the work.

\( \mathrm{Cr_2O_7^{2-}} \): oxygen is −2, so \( 2x + 7(-2) = -2 \), giving \( 2x = 12 \), \( x = \mathbf{+6} \).
\( \mathrm{H_2SO_4} \): hydrogen +1, oxygen −2, neutral compound, so \( 2(+1) + x + 4(-2) = 0 \), giving \( x = \mathbf{+6} \).
\( \mathrm{CaH_2} \): calcium is group 2, so +2; this is a metal hydride, so hydrogen is \( \mathbf{-1} \). Check: \( +2 + 2(-1) = 0 \).
\( \mathrm{Na_2O_2} \): sodium is +1, so \( 2(+1) + 2x = 0 \) gives \( x = \mathbf{-1} \) — a peroxide, the standard exception.
\( \mathrm{MnO_4^{-}} \): \( x + 4(-2) = -1 \), so \( x = \mathbf{+7} \).

(b) Two different comparisons.

Na to Cl (across period 3): the number of protons increases from 11 to 17, but the outer electrons are all being added to the same main energy level (n = 3), so the shielding by inner shells stays essentially constant. The effective nuclear charge therefore increases, pulling the outer shell in more strongly, and the atomic radius decreases.

F to I (down group 17): each successive element has an additional occupied main energy level, so the outer electrons are further from the nucleus, and there are more inner shells shielding them. Nuclear charge increases too, but distance and shielding dominate, so the radius increases.

(c)

\[ 2\mathrm{K}(s) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{KOH}(aq) + \mathrm{H_2}(g) \]

Observations (any two): the metal floats and melts into a ball; it moves rapidly on the surface; there is effervescence as hydrogen is released; the hydrogen ignites with a lilac flame; the metal disappears; the resulting solution turns universal indicator purple.

Potassium reacts more vigorously because reaction requires the loss of the single outer electron. Potassium’s outer electron is in the fourth main energy level rather than the third, so it is further from the nucleus and more shielded by inner shells. Its first ionization energy is therefore lower, the electron is lost more readily, and the reaction is faster and more exothermic.

(d) The ion appears blue, so it is transmitting blue light and absorbing the complementary colour, which the colour wheel gives as orange (roughly 600–620 nm).

It is coloured because \( \mathrm{Cu^{2+}} \) is \( [\mathrm{Ar}]\,3d^9 \) — a partially filled d sublevel. The six water ligands split the five d orbitals into two sets of slightly different energy, and an electron is promoted from the lower set to the higher by absorbing a photon whose energy matches the gap. Because that gap corresponds to visible light, a colour is removed from white light and the complementary colour is seen.

\( \mathrm{Zn^{2+}} \) is \( [\mathrm{Ar}]\,3d^{10} \) — the d sublevel is completely full. There is no vacancy for an electron to be promoted into, so no visible light is absorbed and the solution is colourless. This is also why zinc is not classified as a transition element.

Check it. For every oxidation state, substitute back and confirm the total: in \( \mathrm{Cr_2O_7^{2-}} \), \( 2(+6) + 7(-2) = 12 - 14 = -2 \), matching the ion’s charge. Sanity-check magnitudes too — an oxidation state greater than the group number is impossible for a main-group element, so if you obtain +8 for sulfur, the arithmetic is wrong. For (d), check the pairing on the colour wheel by reversing it: if the observed colour and the absorbed colour are not opposite, you have read the wheel the wrong way round.

Writing oxidation states like charges, and vice versa. An oxidation state is written sign first: \( +2 \), \( -1 \). An ionic charge is written number first, as a superscript: \( \mathrm{Cu^{2+}} \), \( \mathrm{Cl^{-}} \). Examiners do penalise the swap, because it hides a real confusion: the copper in \( \mathrm{[Cu(H_2O)_6]^{2+}} \) has oxidation state +2, but the complex has charge 2+ and the copper does not carry all of it. The second trap on this page is forgetting that the oxygen exception applies to peroxides — assume −2 in \( \mathrm{Na_2O_2} \) and you get an impossible +1 for sodium.

📝Practise

Work through these on paper, then reveal the answer.

1. An element is in period 4, group 16. Deduce its full electron configuration, its block, and the formula and charge of the ion it most commonly forms.

Period 4 means the outer main energy level is n = 4; group 16 means six valence electrons, arranged \( 4s^2\,4p^4 \). Filling in Aufbau order and remembering that 3d is filled before 4p in period 4: \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^4 \). That is \( Z = 34 \), selenium. The last electron enters a p sublevel, so it is in the p-block. Gaining two electrons completes the 4p sublevel and gives the krypton configuration, so the common ion is \( \mathrm{Se^{2-}} \), the selenide ion — exactly as for oxygen and sulfur above it.

2. Explain the trend in first ionization energy across period 3 from sodium to argon, and account for the two points at which the trend is broken.

General trend: increases. Across the period the nuclear charge rises from 11 to 18, while electrons are added to the same main energy level, so shielding by inner shells is essentially unchanged. The effective nuclear charge increases and the atomic radius falls, so the outer electron is held more tightly and more energy is needed to remove it. First break: Al is lower than Mg. Magnesium loses a 3s electron, aluminium a 3p electron. The 3p sublevel is higher in energy and is slightly shielded by the filled 3s orbital, so less energy is needed despite the higher nuclear charge. Second break: S is lower than P. Phosphorus is \( 3p^3 \) with one electron in each 3p orbital; sulfur is \( 3p^4 \), so one orbital holds a pair whose electrons repel, raising the energy of the one removed. Both breaks are evidence that main energy levels are divided into sublevels, and that orbitals within a sublevel fill singly first.

3. Chlorine water is added to separate solutions of potassium bromide and potassium iodide, and then bromine water is added to potassium chloride. Predict what happens in each case and write ionic equations where a reaction occurs.

Chlorine + potassium bromide: reaction occurs; the solution turns orange as bromine is formed. \( \mathrm{Cl_2}(aq) + 2\mathrm{Br^{-}}(aq) \rightarrow 2\mathrm{Cl^{-}}(aq) + \mathrm{Br_2}(aq) \). Chlorine + potassium iodide: reaction occurs; the solution turns brown as iodine is formed, and a dark solid may appear. \( \mathrm{Cl_2}(aq) + 2\mathrm{I^{-}}(aq) \rightarrow 2\mathrm{Cl^{-}}(aq) + \mathrm{I_2}(aq) \). Bromine + potassium chloride: no reaction — no colour change. The pattern is that a more reactive halogen displaces a less reactive one from its halide, and reactivity decreases down group 17. Chlorine has the smallest atomic radius of the three and the least shielding, so it attracts an additional electron most strongly and is the strongest oxidising agent; bromine cannot take an electron from chloride because chlorine holds it more tightly.

4. Write equations for the reactions with water of Na₂O, MgO, CO₂ and SO₃, and state whether the resulting solution is acidic or alkaline in each case. Relate two of them to a named environmental issue.

\( \mathrm{Na_2O}(s) + \mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) \) — alkaline, strongly so. \( \mathrm{MgO}(s) + \mathrm{H_2O}(l) \rightarrow \mathrm{Mg(OH)_2}(aq) \) — alkaline, but only weakly, since magnesium hydroxide is sparingly soluble. \( \mathrm{CO_2}(g) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_2CO_3}(aq) \) — acidic (weakly). \( \mathrm{SO_3}(g) + \mathrm{H_2O}(l) \rightarrow \mathrm{H_2SO_4}(aq) \) — acidic, strongly so. The pattern is that metal oxides are basic and non-metal oxides are acidic, with aluminium oxide amphoteric on the boundary. Environmental links: \( \mathrm{SO_3} \) (and \( \mathrm{SO_2} \)), released when sulfur-containing fossil fuels are burned, dissolves in atmospheric water to produce acid rain, which damages buildings, forests and freshwater ecosystems. Rising atmospheric \( \mathrm{CO_2} \) dissolving in seawater causes ocean acidification, lowering the pH and hindering the formation of carbonate shells and coral skeletons.

5. AHL Explain why iron forms both \( \mathrm{Fe^{2+}} \) and \( \mathrm{Fe^{3+}} \) while calcium forms only \( \mathrm{Ca^{2+}} \). Give the electron configuration of each ion.

Calcium is \( [\mathrm{Ar}]\,4s^2 \). Removing the two 4s electrons gives \( \mathrm{Ca^{2+}} \), \( [\mathrm{Ar}] \). A third electron would have to come from the filled 3p sublevel of an inner main energy level, which is much closer to the nucleus and far less shielded, so the third ionization energy is enormously higher — there is a huge jump in the successive values, and no ordinary chemical reaction supplies that much energy. Iron is \( [\mathrm{Ar}]\,4s^2\,3d^6 \). The 4s and 3d sublevels are close in energy, so successive ionization energies rise only gradually. Removing the two 4s electrons gives \( \mathrm{Fe^{2+}} \), \( [\mathrm{Ar}]\,3d^6 \); removing one more 3d electron costs comparatively little extra and gives \( \mathrm{Fe^{3+}} \), \( [\mathrm{Ar}]\,3d^5 \). The absence of a large jump between the second and third ionization energies is precisely why transition elements show variable oxidation states and calcium does not.

6. AHL A complex ion has the formula \( \mathrm{[Fe(CN)_6]^{3-}} \) and its solution appears yellow. (a) Deduce the oxidation state of iron. (b) State the type of bond between iron and the cyanide ligands. (c) State the colour of light absorbed and explain the origin of the colour.

(a) Each cyanide ligand carries a charge of −1, so \( x + 6(-1) = -3 \), giving iron an oxidation state of +3. (b) The ligands each donate a lone pair to the metal ion, so the bonds are coordination (dative covalent) bonds — both electrons of the shared pair originate from the ligand. Cyanide is acting as a Lewis base and the iron(III) ion as a Lewis acid (R3.4). (c) A yellow solution is transmitting yellow, so it is absorbing the complementary colour, which the colour wheel gives as violet/blue (roughly 400–430 nm). The origin: \( \mathrm{Fe^{3+}} \) is \( [\mathrm{Ar}]\,3d^5 \), a partially filled d sublevel. The six ligands split the five d orbitals into sets of slightly different energy, and an electron absorbs a photon whose energy exactly matches that gap and is promoted from the lower set to the higher. Because the gap happens to correspond to visible light, a colour is removed from white light and the complementary colour is what we see.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The Royal Society of Chemistry interactive periodic table — every trend on this page as real plotted data, which is far more convincing than being told the shape of the graph.
PhET — simulations of the group 1 and group 17 reactions, useful precisely because the real experiments with rubidium and caesium cannot be done in a school laboratory.
Your data booklet — the colour wheel, the electronegativity table and the periodic table itself. If you can find all three inside ten seconds, this sub-topic gets substantially easier.