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.

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.

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.

Stereoisomers have the same constitution but different spatial arrangements. Cis-trans isomerism requires restricted rotation, while enantiomers are non-superimposable mirror images caused by a chiral carbon.

Check the spatial arrangement, not just connectivity, and identify whether the case is a non-cyclic alkene/cycloalkane cis-trans pair or a chiral mirror-image pair.

For alkene cis–trans isomerism, each C of the C=C must carry two different groups; restricted rotation alone is not enough. For chirality, verify four different substituents on one tetrahedral carbon and test whether the mirror-image pair can be superimposed.

In a substituted C3 or C4 cycloalkane, restricted ring geometry gives cis when relevant substituents are on the same side of the ring and trans when they are on opposite sides. For a chiral centre, a solid wedge points out of the page and a dashed wedge behind it. A pair of enantiomers rotates plane-polarized light in opposite directions; an equal racemic mixture has no net rotation. E/Z nomenclature is outside this syllabus scope.

The molecular-ion peak provides a mass constraint, while fragmentation peaks reveal structural features of the organic molecule. Use the supplied fragment data rather than assuming an operational mechanism not given.

For the commonly supplied singly charged positive ions, z = +1 so the numerical m/z value can be read as the ion mass; state or check that assumption if charge is not specified. Build a candidate grid: first match the molecular ion to Mr, then ask whether each proposed structure can conserve atoms while producing every supplied diagnostic fragment. A fragment supports a substructure, but cannot by itself prove the full connectivity.

Separate two jobs: the molecular-ion peak constrains relative molecular mass, while a fragment peak constrains a possible substructure. The tallest base peak is the most abundant detected ion and need not be M⁺; accept a candidate only if its formula can account for the supplied fragment masses.

IR absorptions identify bond types through characteristic wavenumbers. Use the functional-group region and the supplied data-booklet values to match absorptions with structural features.

Treat an absorption as evidence for a bond or functional group, then check whether the proposed structure accounts for all decisive peaks.

A broad O–H absorption and a strong C=O absorption together support a carboxylic acid more strongly than either peak alone. Use the data-booklet range, then check both presence and absence of decisive absorptions; IR identifies bonds and groups, not a unique whole structure by itself.

Greenhouse-gas link: an IR-active vibration must change the molecule's dipole moment, allowing it to absorb matching outgoing infrared radiation. A molecule can be non-polar overall yet have IR-active vibrations; $\ce{CO2}$ is the key example. Absorption at characteristic wavenumbers supports the presence of particular vibrating bonds, but greenhouse effect also depends on concentration, absorption bands and atmospheric lifetime, not one peak alone.

Use all three features together to constrain the structure; do not infer the whole molecule from one signal alone.

Ethanol gives three proton environments with an expected integration ratio 3:2:1 for CH₃, CH₂ and OH, though the OH shift can vary. Normalize integrations to a whole-number ratio and match shifts to environments before assembling fragments; signal count alone is insufficient.

Protons share one signal only when they are chemically equivalent in the molecular environment; use symmetry or a substitution test rather than visual closeness. Integration gives relative signal area, so normalize ratios rather than treating raw values as absolute proton counts. Chemical-shift ranges can overlap, so use shift together with integration, signal count and structure.

Identify the relevant neighbouring proton set, predict n + 1, then compare with the actual pattern and cross-check shift/integration. Treat n + 1 as the introductory local rule: exchangeable protons may not show expected coupling, non-equivalent neighbour sets can give more complex patterns, and overlapping peaks can hide multiplicity. Do not force such evidence into a simple singlet/doublet/triplet/quartet label.

An ethyl fragment commonly gives a three-H triplet next to CH₂ and a two-H quartet next to CH₃. Apply the n + 1 pattern only to relevant neighbouring, non-equivalent hydrogens, then confirm the assignment with integration and chemical shift.

Use each technique as an independent constraint, then reject any candidate that contradicts one of the supplied data sets.

Use an elimination workflow: molecular mass and formula limit atom totals, IR requires or excludes functional groups, and NMR fixes hydrogen environments and neighbours. Write each constraint beside a candidate and reject it immediately when one spectrum conflicts; agreement with a single striking peak is not enough.