Structure 3. Classification of matter

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.

Translate representations without changing atom connectivity. For a skeletal formula, count every vertex and line end as carbon, add enough hydrogens to give carbon four bonds, and write heteroatoms explicitly. Then verify both the molecular formula and the connectivity.

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.

Matching atom totals alone cannot prove two drawings are the same compound: connectivity and, where relevant, stereochemistry must also agree. Do not reduce a molecular formula when the subscripts already have no common factor.

Identify the characteristic atoms and bonding pattern first, then give the functional-group name and relevant property context.

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.

Members of a homologous series share a functional-group pattern and general formula. Successive members differ by CH₂.

Recognize the series by its functional group and general formula, including alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, acids, ethers, amines, amides, esters, and halogenoalkanes.

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.

Melting and boiling points depend on chain length, branching, polarity, and intermolecular-force strength. Larger molecules often have stronger dispersion forces, while branching changes contact and packing.

Explain a comparison by naming the relevant structural difference and the resulting change in intermolecular attraction or packing.

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.

Apply the systematic sequence to saturated or mono-unsaturated compounds with up to six carbons and one functional-group type.

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.

Structural isomers have the same molecular formula but different atom connectivities. Types include straight-chain/branched, position, and functional-group isomers.

Classify primary, secondary, and tertiary alcohols, halogenoalkanes, and amines by the carbon or nitrogen environment attached to the functional group.

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.