Structure 3. Classification of matter

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.

An empirical formula is a ratio, not necessarily the complete molecule: hydrogen peroxide has molecular formula H₂O₂ but empirical formula HO. Reduce all subscripts by their greatest common factor; if no common factor exists, the molecular and empirical formula are identical.

Identify the complete bonding pattern: an aldehyde has a terminal –CHO carbonyl, a ketone has C=O between carbons, and an ester contains –C(=O)–O–. Do not label every O–H as an alcohol or every C–N as an amine without checking the neighbouring carbonyl.

Saturation describes carbon-carbon bonding: a saturated compound has only C-C single bonds, while an unsaturated compound contains at least one C=C or C≡C bond. A carbonyl C=O does not by itself make the carbon skeleton unsaturated. Identify saturation separately from identifying hydroxyl, carbonyl, carboxyl or other functional groups.

Moving from one member to the next adds CH₂, so molar mass and dispersion forces change gradually while the shared functional group gives similar reaction patterns. Use both the functional group and general formula: formula alone can overlap with another structural family.

Straight-chain pentane has a larger contact surface and higher boiling point than more highly branched isomers of the same formula; lengthening a series usually strengthens dispersion forces. For different functional groups, include polarity and hydrogen bonding before attributing the trend to size alone.

For CH₃CH(OH)CH(CH₃)CH₃, choose the four-carbon chain containing –OH, number from the end that gives –OH the lower locant, and name 3-methylbutan-2-ol. The principal suffix controls numbering before a substituent does; check locants, punctuation and retained unsaturation at the end.

C₄H₁₀O can represent different carbon skeletons, different –OH positions, or an ether instead of an alcohol. Draw each connectivity once, then compare molecular formulae. Rotating or redrawing one connectivity does not create a new structural isomer.

For alcohols and halogenoalkanes, classify the carbon carrying the functional group; for amines, classify the nitrogen by how many carbon groups are bonded to it. Do not use the position number alone: butan-2-ol is secondary because its OH-bearing carbon is attached to two other carbons.