The periodic table and periodicity

The modern periodic table arranges elements in order of increasing atomic number ($Z$). Its layout is not arbitrary: it is built directly from electron configuration.

• A period is a horizontal row. The period number tells you the highest occupied principal energy level (outer shell). Period 3 elements have electrons up to the $n=3$ shell.
• A group is a vertical column. Elements in the same group have the same number of valence electrons, which is why they show similar chemistry. For the main groups, the group number (1, 2, 13-18) gives the number of outer electrons.
• A block is named after the subshell being filled by the outermost electron: the s-block (groups 1-2 and He), the p-block (groups 13-18), the d-block (transition metals) and the f-block (lanthanides and actinides).

So position encodes structure. Knowing an element is in period 3, group 16 immediately tells you its outer shell is $n=3$ and it has 6 valence electrons, i.e. $\text{[Ne]}\,3s^2\,3p^4$ (sulfur). The data booklet provides a full periodic table you may use in the exam.

The position of an element lets you write its electron configuration, and vice versa. Electrons fill subshells in order of increasing energy (the Aufbau order $1s,2s,2p,3s,3p,4s,3d,4p\ldots$).

• The block matches the subshell of the last electron: s-block elements end in $s^1$ or $s^2$, p-block in $p^1$ to $p^6$.
• The period number matches the largest $n$ in the configuration.
• For main-group elements the group number links to the count of outer electrons.

For example, chlorine (period 3, group 17) is $1s^2\,2s^2\,2p^6\,3s^2\,3p^5$, often shortened to $\text{[Ne]}\,3s^2\,3p^5$. Its 7 outer electrons place it in group 17.

Ions are formed by losing or gaining electrons to reach a stable (usually noble-gas) configuration. $\text{Na} \rightarrow \text{Na}^+$ loses one electron to give $\text{[Ne]}$; $\text{Cl} \rightarrow \text{Cl}^-$ gains one to give $\text{[Ar]}$. Recognising these patterns explains common ionic charges across the table.

Three key atomic properties vary predictably across the table because of three factors: nuclear charge, electron shielding and the number of occupied shells.

Atomic radius

• Across a period it *decreases*: nuclear charge rises while electrons enter the same shell, so the stronger pull contracts the atom.
• Down a group it *increases*: each element has an extra occupied shell.

First ionization energy (energy to remove the outer electron)

• Across a period it generally *increases* (greater nuclear charge, smaller atom holds electrons more tightly).
• Down a group it *decreases* (the outer electron is further out and better shielded).

Electronegativity (pull on a shared bonding pair)

• Across a period it *increases*; down a group it *decreases*. Fluorine is the most electronegative element.
• Metallic character shows the opposite trend to electronegativity: metals (left/bottom) lose electrons easily, non-metals (right/top) gain them. The data booklet lists radius, IE and electronegativity values.

Group 1 - the alkali metals (Li, Na, K, ...) are very reactive metals stored under oil.

• Reactivity increases down the group: the single outer electron is further from the nucleus and better shielded, so it is lost more easily.
• They react with water to give an alkaline hydroxide and hydrogen: $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$. The reaction becomes more vigorous from Li to K to Cs.

Group 17 - the halogens (F, Cl, Br, I) are reactive non-metals existing as diatomic molecules.

• Reactivity decreases down the group: a larger, more shielded atom attracts an extra electron less strongly.
• A more reactive halogen displaces a less reactive halide from solution, e.g. $$\text{Cl}_2\text{(aq)} + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)}$$ Here chlorine is reduced (it gains electrons) and bromide is oxidised. These displacement results confirm the reactivity order $\text{Cl} > \text{Br} > \text{I}$.

Whether an oxide is acidic, basic or amphoteric follows the metal/non-metal divide and links to position in the table.

• Metal oxides (left of the table) are generally basic. Soluble ones (group 1, group 2) react with water to give alkalis: $$\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$$ They neutralise acids.
• Non-metal oxides (right of the table) are generally acidic. They react with water to give acids: $$\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$$
• Some oxides near the metal/non-metal boundary are amphoteric (react with both acids and bases), the classic example being $\text{Al}_2\text{O}_3$.

Across period 3 the oxides shift smoothly from basic ($\text{Na}_2\text{O},\text{MgO}$) through amphoteric ($\text{Al}_2\text{O}_3$) to acidic ($\text{SiO}_2,\text{P}_4\text{O}_{10},\text{SO}_3,\text{Cl}_2\text{O}_7$).

• Acid deposition: oxides of sulfur and nitrogen ($\text{SO}_2$, $\text{NO}_x$) dissolve in atmospheric water to form acids, lowering rainwater pH and damaging ecosystems and buildings.

An oxidation state is a signed number that tracks electron control, assigned by treating every bond as if it were fully ionic. The main rules:

• Elements in their standard state = 0 (e.g. $\text{Na}$, $\text{O}_2$).
• Simple ion = its charge ($\text{Na}^+$ is $+1$).
• $\text{O}$ is usually $-2$ (but $-1$ in peroxides); $\text{H}$ is usually $+1$ (but $-1$ in metal hydrides).
• The oxidation states in a neutral compound sum to 0; in an ion they sum to the ion charge.

For a main-group element, the highest oxidation state often equals the group number (e.g. sulfur reaches $+6$ in $\text{SO}_3$; chlorine reaches $+7$ in $\text{Cl}_2\text{O}_7$).

• In naming, the oxidation state is shown with a Roman numeral: iron(II) chloride is $\text{FeCl}_2$, iron(III) chloride is $\text{FeCl}_3$.

Worked check for sulfur in $\text{SO}_4^{2-}$: oxygen contributes $4 \times (-2) = -8$, the total must be $-2$, so $S = -2 - (-8) = +6$.

First ionization energy generally rises across a period, but there are two small dips in periods 2 and 3 that reveal the subshell structure of the atom.

Dip 1: group 2 to group 13 (e.g. Be -> B, Mg -> Al)

• In group 13 the electron removed is from a higher-energy p subshell, which is slightly further out and shielded by the full s subshell. It is therefore easier to remove, so IE *falls* slightly.

Dip 2: group 15 to group 16 (e.g. N -> O, P -> S)

• Group 15 has a stable half-filled p subshell ($p^3$, one electron in each p orbital). In group 16 ($p^4$) two electrons must pair in one orbital, and electron-electron repulsion makes one easier to remove, so IE *falls* slightly.

These discontinuities are strong evidence that subshells (s, p) exist and that full and half-full subshells carry extra stability.

• The overall pattern of successive ionization energies (big jumps when you break into a new inner shell) is separate evidence for the existence of shells themselves.

A transition element is best defined as a d-block metal that forms at least one ion in which the d subshell is only partly occupied (period 4: Sc-Zn). By this test, Sc (only $\text{Sc}^{3+}$, $d^0$) and Zn (only $\text{Zn}^{2+}$, $d^{10}$) sit in the d-block but are usually not counted as typical transition metals.

Because their d subshell is incomplete, these metals share a cluster of related behaviours, all traceable to the closely spaced 3d and 4s energy levels:

• they take up more than one stable oxidation state (e.g. iron forms $+2$ and $+3$);
• they act as catalysts, either as the metal or as compounds (e.g. Fe in the Haber process, $\text{MnO}_2$, $\text{V}_2\text{O}_5$), because switching oxidation state opens alternative reaction pathways;
• they build complex ions, as ligands donate lone pairs into vacant d orbitals;
• and they form coloured compounds, a consequence of the partly filled d orbitals.
• Configuration note: the 4s subshell fills before 3d but empties first when ions form, so $\text{Fe}$ is $\text{[Ar]}\,3d^6\,4s^2$ while $\text{Fe}^{2+}$ is $\text{[Ar]}\,3d^6$ (not $3d^4\,4s^2$).

A defining feature of transition metals is the ability to exist in several stable oxidation states, because the 3d and 4s electrons are similar in energy and can be removed in varying numbers without a large energy cost.

• The maximum oxidation state for the early 3d metals rises to a peak at manganese: $\text{Mn}$ can reach $+7$ (as in $\text{MnO}_4^-$, manganate(VII)).
• All first-row transition metals show a common $+2$ state (loss of the two 4s electrons), and most also show $+3$.
• Examples: iron as $\text{Fe}^{2+}$ (iron(II), often pale green) and $\text{Fe}^{3+}$ (iron(III), yellow-brown); copper as $\text{Cu}^+$ and $\text{Cu}^{2+}$ (blue).

This behaviour underlies their role as catalysts and as redox-active species. For instance $\text{MnO}_4^-$ is a strong oxidising agent precisely because Mn can drop from $+7$ to a lower, more stable state.

• Use Roman numerals to name the state: $\text{Cu}_2\text{O}$ is copper(I) oxide, $\text{CuO}$ is copper(II) oxide. Remember to deduce the metal's state from oxygen ($-2$) and overall neutrality.

Most transition-metal compounds are coloured, unlike the colourless compounds of s-block and most p-block metals. The colour comes from the d electrons.

Why complexes form

• A ligand (e.g. $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$, $\text{CN}^-$) has a lone pair that it donates into vacant orbitals of the central metal ion, forming coordinate (dative) bonds and a complex ion such as $[\text{Cu(H}_2\text{O})_6]^{2+}$.

Why they are coloured

• When ligands surround the ion, the five d orbitals split into two energy levels separated by a gap $\Delta E$.
• A d electron can absorb a photon of visible light whose energy matches $\Delta E$, jumping to the higher level (a d-d transition).
• The colour we see is the complement of the colour absorbed (e.g. $[\text{Cu(H}_2\text{O})_6]^{2+}$ absorbs red/orange and looks blue).

Factors changing $\Delta E$ - and so the colour - include the identity of the metal, its oxidation state, the type of ligand and the number/geometry of ligands. A full d subshell ($d^{10}$, e.g. $\text{Zn}^{2+}$) has no possible d-d transition, so its compounds are colourless.