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.