3.1 The periodic table

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.

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.

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.

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.

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.

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.