3.1 The periodic table

Use the table's row, column, and block together; do not substitute period number for group or block identity.

Use bromine as a three-coordinate check: it lies in period 4, group 17 and the p block, so its outer shell is n = 4 with a p-subshell being filled. Block describes the subshell pattern, period the highest occupied main level, and group the repeating valence pattern—three related but different labels.

Read the configuration in both directions: position predicts the outer pattern, and the outer pattern identifies the position.

The configuration 1s²2s²2p⁶3s²3p⁵ ends at n = 3 and p⁵, placing the element in period 3, group 17 and the p block. Reverse the reasoning by using a table position to predict the outer configuration, then check that the total electron count matches the atomic number.

Explain a trend with effective nuclear charge, shielding, shell, distance, and attraction; a direction alone is not a complete explanation.

Across period 3, nuclear charge rises while added electrons enter the same main shell, so effective attraction generally increases, radius falls and first ionization energy rises. For ions, compare electron count and charge as well as position; an isoelectronic species with more protons is smaller.

Electron affinity needs a sign check. Under the enthalpy-change convention, a more favourable first electron gain is more negative: it generally becomes more negative across a period as nuclear attraction increases, and less negative down a group as distance and shielding increase. Sublevel energy and electron repulsion cause exceptions, so compare the stated data rather than forcing every element into a smooth trend.

Group 1 becomes more metallic down the group, while Group 17 becomes less non-metallic down the group. These trends help predict displacement and reaction outcomes.

Use the reactivity order to decide whether a Group 1 metal reacts with water or whether a halogen displaces a halide ion, then write and explain the observation or equation.

Chlorine displaces bromide because Cl₂ is the stronger oxidizing agent: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂; bromine cannot reverse that reaction. For Group 1 with water, use the downward decrease in ionization energy to explain faster electron loss, then balance metal + water → hydroxide + H₂.

Use balanced equations as evidence for the classification: Na₂O + H₂O → 2NaOH and SO₃ + H₂O → H₂SO₄ are representative basic and acidic cases. The bonding/electronegativity trend explains why the character changes across the period, but it does not guarantee that every oxide reacts readily with water.

Al₂O₃ is the useful boundary case: it is amphoteric, so it can react with an acid such as HCl and with a strong base such as NaOH. Do not label an oxide from the element's position alone—check the stated reaction and distinguish a water reaction from acid–base behaviour in another medium.

Environmental link: sulfur oxides dissolve and can be oxidized to acids that increase $\ce{H+}$ in rainwater, causing acid rain. Atmospheric $\ce{CO2}$ dissolves in seawater and participates in $\ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-}$, increasing $\ce{H+}$ and lowering ocean pH. These are acidification mechanisms; do not treat every non-metal oxide as reacting with water in exactly the same way.

An oxidation state is the charge an atom would have if bonding electrons were assigned according to the ionic convention. It is not necessarily the physical charge on an atom in a covalent compound.

Use known oxidation-state rules and the overall charge to solve for the unknown state in compounds and ions.

In MnO₄⁻, four O atoms contribute −8, so Mn must be +7 to give the overall −1 charge. Write the charge-sum equation explicitly and remember that +7 is an oxidation-state assignment, not a claim that manganese exists as a free Mn⁷⁺ ion in permanganate.

Discontinuities in the general first-ionization-energy trend provide evidence for sublevel energies and electron pairing. A higher-energy p electron can be easier to remove than an s electron, and paired-electron repulsion can lower IE.

Name the sublevel and pairing evidence, not just the direction of the graph change.

The drop from Mg to Al occurs because Al loses a higher-energy 3p electron after Mg loses 3s; the drop from N to O reflects pairing repulsion in one 2p orbital. These local electronic effects explain exceptions without overturning the overall across-period rise.

A transition element has an atom or at least one stable/common ion with a partially filled d subshell. Test the actual configurations: occupying the d block is not sufficient by itself, so species whose relevant atom and common ions are d⁰ or d¹⁰ do not meet the definition merely because of table position. Then connect each claimed characteristic property to electronic or complex-ion evidence.

Check the d subshell in the atom or a common ion: a species with an incomplete d subshell can show unpaired-electron magnetism, variable oxidation states, complex formation and colour. Do not infer every property from the label alone; connect colour to d-level splitting and catalysis to accessible oxidation states or adsorption pathways.

First-row transition elements can show variable oxidation states because successive ionization energies for outer s and d electrons are relatively close. Forming an ion removes 4s electrons before 3d electrons.

Write the neutral configuration, remove the required 4s electrons first, then remove 3d electrons and count the remaining configuration.

Start Fe as [Ar]4s²3d⁶. Fe²⁺ loses the two 4s electrons to give [Ar]3d⁶; Fe³⁺ loses one more 3d electron to give [Ar]3d⁵. The written filling order of the neutral atom does not change the rule that 4s electrons are removed first.

Light can be absorbed to promote an electron between split d-orbitals. The observed colour is complementary to the colour absorbed.

Use the colour wheel to map absorbed colour to observed complementary colour, and explain the absorption through d-orbital splitting and promotion.

Worked calculation: for absorbed light of wavelength $600\,\mathrm{nm}=600\times10^{-9}\,\mathrm{m}$, $f=c/\lambda=(3.00\times10^8\,\mathrm{m\,s^{-1}})/(600\times10^{-9}\,\mathrm{m})=5.00\times10^{14}\,\mathrm{s^{-1}}$. Use the colour wheel separately: absorption in the orange region means the observed colour is the complementary blue region. Frequency and wavelength describe the absorbed radiation, not the colour label by themselves.

If a complex absorbs orange light, use the colour wheel to predict the observed complementary blue. The absorbed photon promotes a d electron between ligand-split levels; the observed colour is transmitted or reflected light, not the colour of the absorbed radiation.

The simple d–d model requires an electron in a lower split d level and an available higher d level. A d⁰ or d¹⁰ ion therefore has no d–d transition in this model. The energy gap—and hence absorbed wavelength—also depends on the metal ion, oxidation state, ligand and geometry, so an exact shade cannot be predicted without the coordination environment.