Chapter 3 · Classification of Elements and Periodicity in Properties

3.3 Periodic Trends in Properties

How size, ionisation enthalpy, electron gain enthalpy and electronegativity change across periods and down groups, why beryllium, nitrogen and fluorine break the pattern, and how the same trends explain valence, the odd behaviour of second-period elements, diagonal relationships and the acid–base nature of oxides.

Sodium and chlorine are neighbours in the same period, yet sodium gives up an electron so easily that it is stored under oil, while chlorine snatches electrons from almost anything. Caesium, at the bottom of group 1, is even more reactive than sodium; fluorine, at the top of group 17, is more reactive than chlorine. These are not separate facts to memorise. They follow from a few properties of atoms — their size, how strongly they hold their outer electrons, and how strongly they attract extra ones — and those properties change in a regular, periodic way across the table.

This section studies the four key periodic properties: atomic and ionic radius, ionisation enthalpy, electron gain enthalpy and electronegativity. Each trend is explained by two competing effects — the attraction of the nucleus, measured by the effective nuclear charge, and the distance of the outer electrons, set by the number of shells — together with the extra stability of filled and half-filled subshells from Section 2.4. Then we use the trends to explain valence, the anomalous behaviour of the second-period elements, diagonal relationships, and why oxides on the left of the table are basic and those on the right acidic.

Atomic and Ionic Radii

An atom has no sharp edge — its electron cloud fades away gradually (Section 2.3) — so "the radius of an atom" must be defined from the distance between nuclei in a real substance, which X-ray diffraction measures very accurately.

📘 Definition 3.5 — Covalent, metallic and van der Waals radii
  • The covalent radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond. In $\mathrm{Cl_{2}}$ the bond length is $198\ \mathrm{pm}$, so $r_{\mathrm{Cl}}=99\ \mathrm{pm}$.
  • The metallic radius is half the distance between the nuclei of neighbouring atoms in a metallic crystal. In copper metal the distance is $256\ \mathrm{pm}$, so $r_{\mathrm{Cu}}=128\ \mathrm{pm}$.
  • The van der Waals radius is half the distance between the nuclei of two identical non-bonded atoms in neighbouring molecules (or noble-gas atoms) in the solid. It is always larger than the covalent radius, because non-bonded atoms do not overlap.
"Atomic radius" in tables means the covalent radius for non-metals and the metallic radius for metals. Radii are quoted in picometres ($1\ \mathrm{pm}=10^{-12}\ \mathrm{m}$).

Two effects decide the size of an atom. Going across a period, electrons are added to the same shell while the nuclear charge rises by one each step. Electrons in the same shell screen each other only weakly, so the effective nuclear charge $Z_{\mathrm{eff}}$ — the net positive charge an outer electron actually feels after the inner electrons have shielded part of $Z$ — increases, and the outer electrons are pulled closer. Going down a group, each step adds a whole new shell further from the nucleus; the extra inner electrons shield the extra protons, so $Z_{\mathrm{eff}}$ changes little and the size grows.

📐 Rule 3.7 — Trends in atomic radius
  1. Across a period (left to right) atomic radius generally decreases, because $Z_{\mathrm{eff}}$ increases while the valence shell stays the same. Period 2: $\mathrm{Li}\ 152\gt\mathrm{Be}\ 111\gt\mathrm{B}\ 88\gt\mathrm{C}\ 77\gt\mathrm{N}\ 74\gt\mathrm{O}\ 66\gt\mathrm{F}\ 64\ \mathrm{pm}$.
  2. Down a group atomic radius increases, because the principal quantum number of the valence shell increases. Group 1: $\mathrm{Li}\ 152\lt\mathrm{Na}\ 186\lt\mathrm{K}\ 231\lt\mathrm{Rb}\ 244\lt\mathrm{Cs}\ 262\ \mathrm{pm}$.
  3. Noble gases are not compared with these values: only their van der Waals radii can be measured, and those are much larger.

When an atom loses electrons it shrinks; when it gains electrons it swells. A cation is smaller than its parent atom — it often loses its whole outer shell, and the remaining electrons are held by the same nuclear charge. An anion is larger — the added electrons repel each other and the existing ones, while the nuclear charge is unchanged. So $r(\mathrm{Na^{+}})=95\ \mathrm{pm}$ against $r(\mathrm{Na})=186\ \mathrm{pm}$, but $r(\mathrm{Cl^{-}})=181\ \mathrm{pm}$ against $r(\mathrm{Cl})=99\ \mathrm{pm}$. (Ionic radii in this course are Pauling's values.)

📘 Definition 3.6 — Isoelectronic species
Atoms and ions that have the same number of electrons are called isoelectronic species; for example $\mathrm{N^{3-}}$, $\mathrm{O^{2-}}$, $\mathrm{F^{-}}$, $\mathrm{Ne}$, $\mathrm{Na^{+}}$, $\mathrm{Mg^{2+}}$ and $\mathrm{Al^{3+}}$ all have $10$ electrons ($1s^{2}2s^{2}2p^{6}$). In an isoelectronic series the radius decreases as the nuclear charge $Z$ increases: the same electron cloud is pulled in more strongly. The larger the positive charge on the ion, the smaller it is; the larger the negative charge, the larger it is.
✏️ Example 1 — Radii from bond lengths
(a) The $\mathrm{C{-}C}$ bond length in diamond is $154\ \mathrm{pm}$ and the $\mathrm{H{-}H}$ bond length is $74\ \mathrm{pm}$. Find the covalent radii of carbon and hydrogen. (b) Using $r_{\mathrm{Cl}}=99\ \mathrm{pm}$, estimate the $\mathrm{H{-}Cl}$ bond length. The measured value is $127\ \mathrm{pm}$; suggest why the estimate is too large. (c) Arrange $\mathrm{Mg}$, $\mathrm{Na}$, $\mathrm{Si}$, $\mathrm{P}$ in order of increasing atomic radius, and $\mathrm{Br}$, $\mathrm{F}$, $\mathrm{I}$, $\mathrm{Cl}$ in order of decreasing radius.
(a) $r_{\mathrm{C}}=\tfrac12(154)=77\ \mathrm{pm}$; $r_{\mathrm{H}}=\tfrac12(74)=37\ \mathrm{pm}$.
(b) Sum of covalent radii: $37+99=136\ \mathrm{pm}$. The real bond is shorter because H and Cl differ in electronegativity: the polar bond has partial ionic character, which pulls the atoms closer than a pure covalent bond would.
(c) All four are in period 3, so radius falls left to right: $\mathrm{P}\ (110)\lt\mathrm{Si}\ (117)\lt\mathrm{Mg}\ (160)\lt\mathrm{Na}\ (186\ \mathrm{pm})$. Group 17, radius rises downwards: $\mathrm{I}\ (133)\gt\mathrm{Br}\ (114)\gt\mathrm{Cl}\ (99)\gt\mathrm{F}\ (64\ \mathrm{pm})$.
✏️ Example 2 — Isoelectronic ions
(a) Arrange $\mathrm{Mg^{2+}}$, $\mathrm{O^{2-}}$, $\mathrm{Na^{+}}$, $\mathrm{F^{-}}$, $\mathrm{N^{3-}}$ and $\mathrm{Al^{3+}}$ in order of increasing radius, and justify the order with the ratio $Z/(\text{number of electrons})$. (b) Name a cation and an anion isoelectronic with argon. (c) Why is $\mathrm{Fe^{3+}}$ smaller than $\mathrm{Fe^{2+}}$?
(a) count All have $10$ electrons. $Z$: N 7, O 8, F 9, Na 11, Mg 12, Al 13. $Z/e$: $0.7,\ 0.8,\ 0.9,\ 1.1,\ 1.2,\ 1.3$ — the electron cloud feels more pull per electron as $Z$ rises.
(a) order $\mathrm{Al^{3+}}\ (50)\lt\mathrm{Mg^{2+}}\ (65)\lt\mathrm{Na^{+}}\ (95)\lt\mathrm{F^{-}}\ (136)\lt\mathrm{O^{2-}}\ (140)\lt\mathrm{N^{3-}}\ (171\ \mathrm{pm})$.
(b) 18 electrons: e.g. $\mathrm{K^{+}}$ or $\mathrm{Ca^{2+}}$; $\mathrm{Cl^{-}}$ or $\mathrm{S^{2-}}$.
(c) Both have $Z=26$, but $\mathrm{Fe^{3+}}$ has $23$ electrons against $24$: the same nuclear charge holds fewer electrons, which repel each other less, so the ion is smaller.
Interactive 1 — Atoms, Ions and Isoelectronic Series to Scale

Each circle is drawn to scale from its radius in pm. Choose a set: the 10-electron series, the 18-electron series, or atoms beside their ions. Press ▶ Play to highlight the species one at a time; the bar chart shows the radius against the nuclear charge per electron, $Z/e$.

species—
radius—
Z / electrons—
electrons—

Ionisation Enthalpy

How tightly does an atom hold its outermost electron? We measure it by the energy needed to pull that electron right away from an isolated atom in the gas phase.

📘 Definition 3.7 — Ionisation enthalpy
The first ionisation enthalpy $\Delta_{i}H_{1}$ of an element is the enthalpy change when one mole of electrons is removed from one mole of isolated gaseous atoms in their ground state: $$\mathrm{X(g)\rightarrow X^{+}(g)+e^{-}}\qquad\Delta_{i}H_{1}$$ The second ionisation enthalpy $\Delta_{i}H_{2}$ refers to $\mathrm{X^{+}(g)\rightarrow X^{2+}(g)+e^{-}}$, and so on. Ionisation enthalpies are always positive (energy must be supplied), are measured in $\mathrm{kJ\,mol^{-1}}$, and successive values always increase: $\Delta_{i}H_{1}\lt\Delta_{i}H_{2}\lt\Delta_{i}H_{3}\ldots$, because each electron is removed from an increasingly positive ion. (The term ionisation energy is also used; the difference is negligible here.)

Ionisation enthalpy depends on the attraction between the nucleus and the electron being removed: it is larger when $Z_{\mathrm{eff}}$ is larger, when the electron is closer to the nucleus, and when the electron belongs to a particularly stable arrangement. The inner electrons shield (screen) the outer ones from the nucleus; s electrons penetrate closer to the nucleus than p electrons of the same shell, and so both shield better and are held more tightly.

📐 Rule 3.8 — Trends in ionisation enthalpy and their exceptions
  1. Across a period $\Delta_{i}H_{1}$ generally increases (higher $Z_{\mathrm{eff}}$, smaller atom), from a minimum at the alkali metal to a maximum at the noble gas.
  2. Down a group $\Delta_{i}H_{1}$ decreases: the outer electron is further away and better shielded. Group 1: $\mathrm{Li}\ 520\gt\mathrm{Na}\ 496\gt\mathrm{K}\ 419\gt\mathrm{Rb}\ 403\gt\mathrm{Cs}\ 376\ \mathrm{kJ\,mol^{-1}}$.
  3. Exceptions in a period: $\mathrm{Be}\gt\mathrm{B}$ and $\mathrm{Mg}\gt\mathrm{Al}$, because the electron removed from B or Al is a $p$ electron, which is shielded by the $ns^{2}$ pair and held less tightly than an $s$ electron; $\mathrm{N}\gt\mathrm{O}$ and $\mathrm{P}\gt\mathrm{S}$, because N and P have the extra-stable half-filled $p^{3}$, while in O and S one $p$ orbital holds a pair, and the repulsion between the paired electrons makes one of them easier to remove.
Period 2LiBeBCNOFNe
$\Delta_{i}H_{1}$ / $\mathrm{kJ\,mol^{-1}}$52090080110861402131416812081
Period 3NaMgAlSiPSClAr
$\Delta_{i}H_{1}$ / $\mathrm{kJ\,mol^{-1}}$4967385787871012100012511521

Values are rounded from spectroscopic ionisation energies ($1\ \mathrm{eV}$ per atom $=96.485\ \mathrm{kJ\,mol^{-1}}$); tables in different books differ by a few $\mathrm{kJ\,mol^{-1}}$ (e.g. Be is often quoted as $899$).

✏️ Example 3 — Explaining the anomalies
(a) Arrange Na, Mg, Al and Si in order of increasing first ionisation enthalpy and explain the position of aluminium. (b) Why is $\Delta_{i}H_{1}$ of nitrogen greater than that of oxygen, although oxygen has the higher nuclear charge? (c) Which has the larger $\Delta_{i}H_{2}$, sodium or magnesium? Why?
(a) $\mathrm{Na}\ (496)\lt\mathrm{Al}\ (578)\lt\mathrm{Mg}\ (738)\lt\mathrm{Si}\ (787)$. Aluminium ($[\mathrm{Ne}]\,3s^{2}3p^{1}$) loses a $3p$ electron, which is higher in energy and shielded by the $3s^{2}$ pair; magnesium must lose an electron from its filled, more penetrating $3s^{2}$.
(b) N is $2p^{3}$ — [↑][↑][↑], half-filled and extra stable. O is $2p^{4}$ — [↑↓][↑][↑]; the two electrons sharing one orbital repel each other, so removing one of them (which leaves the stable $2p^{3}$ of $\mathrm{O^{+}}$) costs less energy.
(c) $\mathrm{Na^{+}}$ is $[\mathrm{Ne}]$: its second electron must come from the filled $2p$ of an inner shell, much closer to the nucleus — $\Delta_{i}H_{2}(\mathrm{Na})=4562\ \mathrm{kJ\,mol^{-1}}$. $\mathrm{Mg^{+}}$ is $[\mathrm{Ne}]\,3s^{1}$ and loses a valence electron: $1451\ \mathrm{kJ\,mol^{-1}}$. Sodium has the larger value.
✏️ Example 4 — Successive ionisation enthalpies and energy per atom
(a) An element of period 3 has $\Delta_{i}H_{1}$ to $\Delta_{i}H_{4}$ equal to $578$, $1817$, $2745$ and $11\,577\ \mathrm{kJ\,mol^{-1}}$. Identify its group and the element. (b) How much energy is needed to convert $1.00\ \mathrm{g}$ of gaseous sodium atoms into $\mathrm{Na^{+}}$ ions ($M=22.99\ \mathrm{g\,mol^{-1}}$)? (c) What is the longest wavelength of light that can ionise a single gaseous sodium atom? ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $c=3.0\times10^{8}\ \mathrm{m\,s^{-1}}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$.)
(a) Ratios of neighbours: $1817/578=3.1$, $2745/1817=1.5$, $11\,577/2745=4.2$. The big jump comes after the third electron: three valence electrons, then a core electron. Group 13; in period 3 this is aluminium, $[\mathrm{Ne}]\,3s^{2}3p^{1}$.
(b) $n=1.00/22.99=0.04350\ \mathrm{mol}$; energy $=0.04350\times496=21.6\ \mathrm{kJ}$.
(c) Per atom: $E=\dfrac{496\times10^{3}}{6.022\times10^{23}}=8.24\times10^{-19}\ \mathrm{J}$. $\lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3.0\times10^{8}}{8.24\times10^{-19}}=2.41\times10^{-7}\ \mathrm{m}=241\ \mathrm{nm}$ (ultraviolet). Any longer wavelength carries too little energy per photon.

Electron Gain Enthalpy and Electronegativity

Ionisation enthalpy measures how hard it is to take an electron away. The opposite question — how much an atom "wants" an extra electron — is answered by the electron gain enthalpy.

📘 Definition 3.8 — Electron gain enthalpy
The electron gain enthalpy $\Delta_{eg}H$ of an element is the enthalpy change when one mole of electrons is added to one mole of isolated gaseous atoms: $$\mathrm{X(g)+e^{-}\rightarrow X^{-}(g)}\qquad\Delta_{eg}H$$ It is negative when energy is released (the anion is more stable than atom + electron, e.g. $\mathrm{Cl}$: $-349\ \mathrm{kJ\,mol^{-1}}$) and positive when energy must be supplied (e.g. the noble gases, whose extra electron would have to enter a new shell). Adding a second electron to an anion, as in $\mathrm{O^{-}(g)+e^{-}\rightarrow O^{2-}(g)}$, is always strongly endothermic (positive, several hundred $\mathrm{kJ\,mol^{-1}}$), because the electron is repelled by the negative ion.
📐 Rule 3.9 — Trends in electron gain enthalpy
  1. Across a period $\Delta_{eg}H$ generally becomes more negative (higher $Z_{\mathrm{eff}}$, smaller atom), reaching the most negative values at the halogens (group 17), which need just one electron to complete the noble-gas configuration.
  2. Down a group it generally becomes less negative, as the added electron enters a shell further from the nucleus. Group 17: $\mathrm{Cl}\ (-349)\lt\mathrm{F}\ (-328)\lt\mathrm{Br}\ (-325)\lt\mathrm{I}\ (-295\ \mathrm{kJ\,mol^{-1}})$, so chlorine — not fluorine — has the most negative value.
  3. Exceptions: (i) F is less negative than Cl, and O ($-141$) less negative than S ($-200$): in the very small $2p$ subshell the incoming electron is strongly repelled by the electrons already there, which outweighs the higher attraction. (ii) Noble gases, and Be, Mg (filled $ns^{2}$) have positive values; N (half-filled $2p^{3}$) has a value close to zero.

Ionisation enthalpy and electron gain enthalpy describe isolated atoms. In a molecule, what matters is how strongly an atom pulls the shared electrons of a bond towards itself.

📘 Definition 3.9 — Electronegativity
The electronegativity of an element is a qualitative measure of the ability of its atom, in a chemical compound, to attract shared electrons to itself. It is not a measurable energy but a relative number; on the most widely used scale, Linus Pauling's, fluorine is assigned $4.0$. Electronegativity increases across a period (Li $1.0$ to F $4.0$) and decreases down a group (F $4.0$, Cl $3.0$, Br $2.8$, I $2.5$). Fluorine is the most electronegative element, followed by oxygen ($3.5$); caesium ($0.7$) is among the least.

Electronegativity links directly to metallic character (Section 3.2): elements of low electronegativity readily give up electrons and are metallic; those of high electronegativity attract electrons and are non-metallic. The difference in electronegativity between two bonded atoms tells you how polar the bond is — a key idea of Chapter 4.

Interactive 2 — Periodic Trend Explorer

Choose a property and a series. The mini-table on the left is coloured as a heat map (darker = larger value; for electron gain enthalpy, darker = more negative); the bars on the right show the chosen period or group, with the anomalies outlined in red. Press ▶ Play for a tour of all four properties.

series—
largest—
smallest—
anomalies—
✏️ Example 5 — Electron gain enthalpy and electronegativity
(a) Arrange F, Cl, Br and I in order of increasingly negative electron gain enthalpy, and explain the position of fluorine. (b) Why is $\Delta_{eg}H$ of neon positive? (c) Which of the pairs has the more negative $\Delta_{eg}H$: (i) O or S; (ii) N or P; (iii) Li or Be? (d) Arrange C, N, O, F, Si in order of increasing electronegativity.
(a) $\mathrm{I}\ (-295)\gt\mathrm{Br}\ (-325)\gt\mathrm{F}\ (-328)\gt\mathrm{Cl}\ (-349)$, i.e. increasingly negative: I, Br, F, Cl. Fluorine's $2p$ subshell is so compact that the added electron suffers strong repulsion from the seven already there, so less energy is released than for chlorine, whose larger $3p$ subshell has more room.
(b) Neon is $1s^{2}2s^{2}2p^{6}$: an extra electron would have to enter the $3s$ orbital of a new, much higher-energy shell; energy must be supplied, so $\Delta_{eg}H\gt0$.
(c) (i) S ($-200$ vs $-141$): same small-size repulsion effect as F/Cl. (ii) P ($-72$; N is close to zero because its half-filled $2p^{3}$ is stable and small). (iii) Li ($-60$); Be's filled $2s^{2}$ makes the added electron go into $2p$, and its value is positive.
(d) Si $1.8$ $\lt$ C $2.5$ $\lt$ N $3.0$ $\lt$ O $3.5$ $\lt$ F $4.0$.

Valence, Anomalous Second-Period Elements and the Nature of Oxides

The valence of a representative element is the number of electrons it uses in bonding. It is usually equal to the number of valence electrons, or to eight minus that number, and so it repeats period after period — a property Mendeleev used to build his groups. The oxidation states of the elements (Chapter 7) follow the same pattern.

Group12131415161718
Valence electrons12345678
Valence12343, 52, 61, 70, 8
Hydride (period 3)$\mathrm{NaH}$$\mathrm{MgH_{2}}$$\mathrm{AlH_{3}}$$\mathrm{SiH_{4}}$$\mathrm{PH_{3}}$$\mathrm{H_{2}S}$$\mathrm{HCl}$—
Highest oxide (period 3)$\mathrm{Na_{2}O}$$\mathrm{MgO}$$\mathrm{Al_{2}O_{3}}$$\mathrm{SiO_{2}}$$\mathrm{P_{4}O_{10}}$$\mathrm{SO_{3}}$$\mathrm{Cl_{2}O_{7}}$—

The first element of each group, in period 2 (Li, Be, B, C, N, O, F), differs noticeably from the rest of its group. The reasons are its small size, large charge/radius ratio, high electronegativity and the absence of d orbitals in its valence shell. As a result, period-2 elements have a maximum covalence of $4$ (boron forms $\mathrm{[BF_{4}]^{-}}$, while aluminium forms $\mathrm{[AlF_{6}]^{3-}}$), and they readily form $p\pi$–$p\pi$ multiple bonds ($\mathrm{C{=}C}$, $\mathrm{C{\equiv}N}$, $\mathrm{N{\equiv}N}$, $\mathrm{O{=}O}$) that their heavier group members seldom form. Lithium and beryllium form compounds with marked covalent character, unlike the rest of groups 1 and 2.

📘 Definition 3.10 — Diagonal relationship
A diagonal relationship is the similarity in properties between an element of period 2 and the element diagonally below and to its right in period 3: $\mathrm{Li}$–$\mathrm{Mg}$, $\mathrm{Be}$–$\mathrm{Al}$, $\mathrm{B}$–$\mathrm{Si}$. Moving one step right raises electronegativity and $Z_{\mathrm{eff}}$, while moving one step down lowers them, so the two changes largely cancel and the diagonal pair have similar size, charge density and electronegativity. Examples: Li and Mg both burn in nitrogen to form nitrides ($\mathrm{Li_{3}N}$, $\mathrm{Mg_{3}N_{2}}$); Be and Al both form amphoteric oxides and covalent chlorides.

Periodic trends also explain chemical reactivity. On the far left, elements with very low ionisation enthalpy (the alkali metals) lose electrons easily and are strong reducing agents; on the far right, elements with very negative electron gain enthalpy and high electronegativity (the halogens) gain electrons easily and are strong oxidising agents. Reactivity is therefore highest at the two ends of a period and lowest in the middle; among metals it increases down a group, among non-metals it decreases.

📐 Rule 3.10 — Acid–base nature of oxides

Across a period the oxides change from strongly basic (left) through amphoteric to strongly acidic (right); down a group the basic character increases. Period 3:

$\mathrm{Na_{2}O}$, $\mathrm{MgO}$ (basic) $\ \rightarrow\ $ $\mathrm{Al_{2}O_{3}}$ (amphoteric) $\ \rightarrow\ $ $\mathrm{SiO_{2}}$, $\mathrm{P_{4}O_{10}}$, $\mathrm{SO_{3}}$, $\mathrm{Cl_{2}O_{7}}$ (acidic)

A basic oxide reacts with water to form a base or with acids to form salts; an acidic oxide gives an acid with water or reacts with bases; an amphoteric oxide reacts with both acids and bases. A few non-metal oxides, such as $\mathrm{CO}$, $\mathrm{NO}$ and $\mathrm{N_{2}O}$, are neutral.

✏️ Example 6 — Oxides, valence and diagonal pairs
(a) Write balanced equations for the reactions of $\mathrm{Na_{2}O}$ and $\mathrm{Cl_{2}O_{7}}$ with water, and say what each shows. (b) Write equations showing that $\mathrm{Al_{2}O_{3}}$ is amphoteric (with $\mathrm{HCl}$, and with $\mathrm{NaOH}$ giving $\mathrm{Na[Al(OH)_{4}]}$). (c) Using valence, predict the formulae of the compounds formed by (i) silicon and chlorine, (ii) aluminium and oxygen, (iii) magnesium and nitrogen. (d) Why can nitrogen not form $\mathrm{NCl_{5}}$, although phosphorus forms $\mathrm{PCl_{5}}$?
(a) $\mathrm{Na_{2}O(s)+H_{2}O(l)\rightarrow2NaOH(aq)}$ — a strong base: $\mathrm{Na_{2}O}$ is basic. $\mathrm{Cl_{2}O_{7}(l)+H_{2}O(l)\rightarrow2HClO_{4}(aq)}$ — perchloric acid, a very strong acid: $\mathrm{Cl_{2}O_{7}}$ is acidic.
(b) As a base: $\mathrm{Al_{2}O_{3}(s)+6HCl(aq)\rightarrow2AlCl_{3}(aq)+3H_{2}O(l)}$. As an acid: $\mathrm{Al_{2}O_{3}(s)+2NaOH(aq)+3H_{2}O(l)\rightarrow2Na[Al(OH)_{4}](aq)}$. Reacting with both makes it amphoteric. (Check of the second: Al $2=2$; Na $2=2$; O $3+2+3=8=2\times4$; H $2+6=8$ ✓.)
(c) Valences Si 4, Cl 1 → $\mathrm{SiCl_{4}}$; Al 3, O 2 → $\mathrm{Al_{2}O_{3}}$; Mg 2, N 3 → $\mathrm{Mg_{3}N_{2}}$.
(d) Nitrogen is a period-2 element: its valence shell ($n=2$) has only $2s$ and $2p$ orbitals — four orbitals in all — so its maximum covalence is $4$. Phosphorus ($n=3$) also has $3d$ orbitals available and can expand beyond eight electrons (Chapter 4).
📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

The s-block elements and the general trends of the p-block are taught and tested in school, but not in the year-end board exam. JEE Advanced lists both the s-block and the p-block. JEE Main and NEET list only the p-block — the general trends of groups 13–18 and the unique behaviour of the first element of each group; neither lists the s-block.

The s-block: groups 1 and 2. The alkali metals (Li, Na, K, Rb, Cs, Fr) have the outer configuration $ns^{1}$; the alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra) have $ns^{2}$. Their trends are the cleanest in the table (Interactive 3 below):

  • Size increases down each group, as a new shell is added each period. A group-2 atom is smaller than its group-1 neighbour, because of its extra nuclear charge.
  • Ionisation enthalpy falls down each group, so metallic character and reactivity increase: caesium reacts far more violently with water than lithium. Group 1 shows only the oxidation state $+1$; group 2 has higher ionisation enthalpies and shows only $+2$.
  • The hydration enthalpy of the ions falls as they get larger: $\mathrm{Li^{+}}\gt\mathrm{Na^{+}}\gt\mathrm{K^{+}}\gt\mathrm{Rb^{+}}\gt\mathrm{Cs^{+}}$. The very strongly hydrated $\mathrm{Li^{+}}$ ion makes lithium the strongest reducing agent of the group in water, although its ionisation enthalpy is the highest.
  • The oxides and hydroxides are basic, more strongly so down each group (Rule 3.10). Many of the metals and their salts colour a flame: Li crimson, Na yellow, K lilac, Ca brick red, Sr crimson, Ba apple green (Be and Mg give no colour).

Anomalous lithium and beryllium. The first member of each group is exceptionally small, with a high charge/radius ratio, so its compounds have marked covalent character. Lithium is harder and has a higher melting point than the other alkali metals; it burns in air to give mainly the normal oxide $\mathrm{Li_{2}O}$ (sodium gives the peroxide $\mathrm{Na_{2}O_{2}}$, potassium the superoxide $\mathrm{KO_{2}}$); it alone combines directly with nitrogen, forming $\mathrm{Li_{3}N}$; and $\mathrm{Li_{2}CO_{3}}$ decomposes on heating, whereas the other alkali metal carbonates do not. In these ways lithium resembles magnesium (Definition 3.10). Beryllium has a very small atom and high ionisation enthalpies, so its compounds are largely covalent and easily hydrolysed; with only four orbitals in its valence shell it never exceeds a coordination number of four (magnesium and the others can reach six); and its oxide and hydroxide are amphoteric, like those of aluminium.

The p-block: general trends in groups 13–18. The outer configuration is $ns^{2}np^{1}$ to $ns^{2}np^{6}$ (helium $1s^{2}$).

  • Oxidation states. The highest (group) oxidation state equals the number of valence electrons: $+3$ in group 13 up to $+7$ in group 17. Non-metals also show negative states (N $-3$, O $-2$, Cl $-1$).
  • Inert pair effect. In groups 13–15 a state two below the group state becomes more stable down the group, because the $ns^{2}$ pair of the heavy elements is reluctant to take part in bonding: Tl(I) is more stable than Tl(III), Pb(II) than Pb(IV) — so $\mathrm{PbO_{2}}$ is a strong oxidising agent — and Bi(III) than Bi(V).
  • Metallic character increases down each group: carbon and silicon are non-metals, germanium a metalloid, tin and lead metals. Non-metals sit at the top right of the block.
  • The first member is different (as in this section): small size, high electronegativity and no d orbitals give a maximum covalence of $4$ and strong $p\pi$–$p\pi$ multiple bonds — $\mathrm{N{\equiv}N}$ and $\mathrm{O{=}O}$ molecules, while phosphorus and sulphur form singly bonded $\mathrm{P_{4}}$ and $\mathrm{S_{8}}$.
  • Irregularities come from poor shielding by filled d and f subshells: gallium (135 pm) is slightly smaller than aluminium (143 pm), and thallium's ionisation enthalpy is higher than indium's. The noble gases (group 18) have the highest ionisation enthalpies of their periods, falling down the group — which is why xenon, the easiest to ionise of the stable ones, forms compounds such as $\mathrm{XeF_{2}}$ and $\mathrm{XeF_{4}}$.

Worked example 1. (a) Arrange $\mathrm{Li^{+}}$, $\mathrm{Na^{+}}$ and $\mathrm{K^{+}}$ in order of decreasing hydration enthalpy. (b) Lithium has the highest first ionisation enthalpy in group 1 ($520\ \mathrm{kJ\,mol^{-1}}$, against $419$ for potassium), yet it is the strongest reducing agent in water. Explain.
Solution. (a) $\mathrm{Li^{+}}\gt\mathrm{Na^{+}}\gt\mathrm{K^{+}}$: the smallest ion has the highest charge density and attracts water molecules most strongly. (b) Turning the solid metal into hydrated ions involves atomisation, ionisation and hydration. Lithium needs more energy to ionise, but the huge hydration enthalpy of the tiny $\mathrm{Li^{+}}$ ion more than repays it, so overall lithium gives up its electron in water most readily.

Worked example 2. (a) Carbon dioxide is a gas of separate $\mathrm{O{=}C{=}O}$ molecules, but silica, $\mathrm{SiO_{2}}$, is a hard solid. (b) $\mathrm{PbO_{2}}$ oxidises concentrated hydrochloric acid to chlorine, but $\mathrm{CO_{2}}$ and $\mathrm{SiO_{2}}$ do not. Explain both using p-block trends.
Solution. (a) The small carbon atom forms strong $p\pi$–$p\pi$ double bonds to oxygen, so each C is satisfied by two O atoms in a small molecule. The larger Si atom forms poor π bonds and instead makes four single Si–O bonds, linking every atom into a giant network. (b) The inert pair effect: for lead, the $+2$ state is more stable than $+4$, so Pb(IV) readily gains two electrons (reduced to $\mathrm{Pb^{2+}}$), oxidising $\mathrm{Cl^{-}}$ to $\mathrm{Cl_{2}}$. For carbon and silicon the group state $+4$ is the stable one.

Interactive 3 — Group Trends in the s- and p-Blocks (📘 school assessment)

Choose a group and a property. The bars run down the group; a red outline marks a step that breaks the expected trend, and the readouts give the change from the first to the last member. Press ▶ Play to tour the groups.

outer configuration—
first → last—
trend breaks at—

Exercises

I
Atomic and ionic radii.
  • 1 Distinguish between covalent radius and van der Waals radius. Which is larger for chlorine?
  • 2 The $\mathrm{Br{-}Br}$ bond length is $228\ \mathrm{pm}$. Find the covalent radius of bromine.
  • 3 Explain why atomic radius decreases from Li to F.
  • 4 Arrange K, Mg, Na, Rb in order of increasing atomic radius.
  • 5 Why is a cation smaller and an anion larger than the parent atom?
  • 6 Arrange $\mathrm{S^{2-}}$, $\mathrm{Cl^{-}}$, $\mathrm{K^{+}}$, $\mathrm{Ca^{2+}}$ in order of decreasing radius.
  • 7 Which of the following are isoelectronic with $\mathrm{Ne}$: $\mathrm{Na^{+}}$, $\mathrm{Cl^{-}}$, $\mathrm{O^{2-}}$, $\mathrm{Mg^{2+}}$, $\mathrm{K^{+}}$, $\mathrm{F^{-}}$?
  • 8 By what percentage is the radius of $\mathrm{Mg^{2+}}$ ($65\ \mathrm{pm}$) smaller than that of the Mg atom ($160\ \mathrm{pm}$)?
1 Covalent radius: half the distance between nuclei of two bonded identical atoms. Van der Waals radius: half the distance between nuclei of non-bonded atoms in neighbouring molecules. The van der Waals radius is larger.
2 $228/2=114\ \mathrm{pm}$.
3 Electrons enter the same shell ($n=2$) while $Z$ rises from 3 to 9; electrons in one shell shield each other poorly, so $Z_{\mathrm{eff}}$ increases and pulls the cloud in.
4 $\mathrm{Mg}\ (160)\lt\mathrm{Na}\ (186)\lt\mathrm{K}\ (231)\lt\mathrm{Rb}\ (244\ \mathrm{pm})$.
5 A cation has fewer electrons (often losing its whole outer shell) held by the same nuclear charge; an anion has extra electrons that increase repulsion, with no extra nuclear charge.
6 All have $18$ electrons; radius falls as $Z$ rises: $\mathrm{S^{2-}}\ (184)\gt\mathrm{Cl^{-}}\ (181)\gt\mathrm{K^{+}}\ (133)\gt\mathrm{Ca^{2+}}\ (99\ \mathrm{pm})$.
7 $\mathrm{Na^{+}}$, $\mathrm{O^{2-}}$, $\mathrm{Mg^{2+}}$, $\mathrm{F^{-}}$ ($10$ electrons). $\mathrm{Cl^{-}}$ and $\mathrm{K^{+}}$ have $18$.
8 $\dfrac{160-65}{160}\times100=59\%$ smaller (the ion is $0.41$ times the size of the atom).
II
Ionisation enthalpy. Use the values in this section where needed.
  • 1 Define first ionisation enthalpy and write the equation for the first ionisation of magnesium.
  • 2 Why are successive ionisation enthalpies of an element always larger than the previous one?
  • 3 Arrange B, Be, C, O, N in order of increasing first ionisation enthalpy.
  • 4 Explain why $\Delta_{i}H_{1}$ of phosphorus is greater than that of sulphur.
  • 5 Why does $\Delta_{i}H_{1}$ decrease from Li to Cs?
  • 6 An element has $\Delta_{i}H_{1}=738$, $\Delta_{i}H_{2}=1451$ and $\Delta_{i}H_{3}=7733\ \mathrm{kJ\,mol^{-1}}$. To which group does it belong? Explain.
  • 7 Calculate the energy needed to convert $1.00\ \mathrm{g}$ of gaseous magnesium atoms into $\mathrm{Mg^{2+}}$ ions ($M=24.31\ \mathrm{g\,mol^{-1}}$).
  • 8 Find the energy per atom (in J) needed to ionise a potassium atom ($\Delta_{i}H_{1}=419\ \mathrm{kJ\,mol^{-1}}$) and the longest wavelength of light that can do it.
1 The enthalpy change for removing one mole of the most loosely held electrons from one mole of isolated gaseous atoms: $\mathrm{Mg(g)\rightarrow Mg^{+}(g)+e^{-}}$, $\Delta_{i}H_{1}=738\ \mathrm{kJ\,mol^{-1}}$.
2 After each removal the same nuclear charge holds fewer electrons in a smaller, positively charged ion, so the next electron is held more strongly.
3 $\mathrm{B}\ (801)\lt\mathrm{Be}\ (900)\lt\mathrm{C}\ (1086)\lt\mathrm{O}\ (1314)\lt\mathrm{N}\ (1402)$.
4 P is $3p^{3}$ (half-filled, extra stable); S is $3p^{4}$, and the paired electron is repelled by its partner and is removed more easily.
5 Each step adds a shell, so the $ns^{1}$ electron is further from the nucleus and more shielded; the increase in $Z$ is largely cancelled by shielding.
6 $1451/738\approx2.0$ but $7733/1451\approx5.3$: the big jump comes after two electrons, so there are two valence electrons — group 2 (this is magnesium).
7 $n=1.00/24.31=0.04114\ \mathrm{mol}$; energy $=0.04114\times(738+1451)=0.04114\times2189=90.0\ \mathrm{kJ}$.
8 $E=\dfrac{419\times10^{3}}{6.022\times10^{23}}=6.96\times10^{-19}\ \mathrm{J}$; $\lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3.0\times10^{8}}{6.96\times10^{-19}}=2.86\times10^{-7}\ \mathrm{m}=286\ \mathrm{nm}$.
III
Electron gain enthalpy and electronegativity.
  • 1 Define electron gain enthalpy. What does a negative value mean?
  • 2 Which has the most negative electron gain enthalpy of all the elements? Why not fluorine?
  • 3 Why is the second electron gain enthalpy of oxygen positive?
  • 4 Arrange O, S, F, Cl in order of increasingly negative electron gain enthalpy.
  • 5 Why do Be and Mg have positive electron gain enthalpies, while Li and Na have negative ones?
  • 6 How does electronegativity differ from electron gain enthalpy?
  • 7 Arrange Na, Cl, Mg, P, Si in order of increasing electronegativity.
  • 8 Find the electronegativity difference in $\mathrm{NaCl}$ and in $\mathrm{C{-}H}$ (Na $0.9$, Cl $3.0$, C $2.5$, H $2.1$). Which bond is more polar?
1 The enthalpy change when one mole of electrons is added to one mole of isolated gaseous atoms, $\mathrm{X(g)+e^{-}\rightarrow X^{-}(g)}$. Negative: energy is released, the anion is more stable.
2 Chlorine ($-349\ \mathrm{kJ\,mol^{-1}}$). Fluorine ($-328$) is smaller: the added electron is crowded into the compact $2p$ subshell and strongly repelled by the electrons already there.
3 The second electron is added to an ion that is already negative ($\mathrm{O^{-}}$) and is repelled; energy must be supplied.
4 $\mathrm{O}\ (-141)$, $\mathrm{S}\ (-200)$, $\mathrm{F}\ (-328)$, $\mathrm{Cl}\ (-349)$.
5 Be and Mg have filled $ns^{2}$ subshells; an extra electron must enter the higher-energy $np$ subshell. Li and Na ($ns^{1}$) can accept the electron into the half-filled $ns$ orbital.
6 Electron gain enthalpy is a measurable energy for an isolated gaseous atom; electronegativity is a relative, dimensionless measure of an atom's pull on shared electrons in a compound.
7 $\mathrm{Na}\ (0.9)\lt\mathrm{Mg}\ (1.2)\lt\mathrm{Si}\ (1.8)\lt\mathrm{P}\ (2.1)\lt\mathrm{Cl}\ (3.0)$.
8 NaCl: $3.0-0.9=2.1$; C–H: $2.5-2.1=0.4$. NaCl is far more polar — in fact ionic.
IV
Valence, anomalies, oxides and reactivity.
  • 1 Predict the formulae of the compounds of (a) calcium and chlorine, (b) aluminium and sulphur, (c) silicon and oxygen.
  • 2 Name the three diagonal pairs and give one property shared by Be and Al.
  • 3 Give two reasons for the anomalous behaviour of the second-period elements.
  • 4 Classify as acidic, basic, amphoteric or neutral: $\mathrm{MgO}$, $\mathrm{SO_{3}}$, $\mathrm{Al_{2}O_{3}}$, $\mathrm{CO}$, $\mathrm{K_{2}O}$, $\mathrm{P_{4}O_{10}}$.
  • 5 Write a balanced equation for the reaction of $\mathrm{SO_{3}}$ with water and of $\mathrm{CaO}$ with water.
  • 6 Why is reactivity highest at the two ends of a period?
  • 7 Which is the stronger reducing agent, sodium or potassium? Which is the stronger oxidising agent, fluorine or iodine?
  • 8 Applied. A water-treatment engineer must choose between $\mathrm{CaO}$ and $\mathrm{SO_{2}}$ to raise the pH of acidic mine drainage. Which, and why?
1 (a) Ca 2, Cl 1: $\mathrm{CaCl_{2}}$; (b) Al 3, S 2: $\mathrm{Al_{2}S_{3}}$; (c) Si 4, O 2: $\mathrm{SiO_{2}}$.
2 Li–Mg, Be–Al, B–Si. Be and Al both form amphoteric oxides (and covalent chlorides).
3 Very small size and high electronegativity (large charge/radius ratio); no d orbitals in the valence shell, so maximum covalence 4. They also form $p\pi$–$p\pi$ multiple bonds readily.
4 $\mathrm{MgO}$ basic; $\mathrm{SO_{3}}$ acidic; $\mathrm{Al_{2}O_{3}}$ amphoteric; $\mathrm{CO}$ neutral; $\mathrm{K_{2}O}$ basic; $\mathrm{P_{4}O_{10}}$ acidic.
5 $\mathrm{SO_{3}(g)+H_{2}O(l)\rightarrow H_{2}SO_{4}(aq)}$; $\mathrm{CaO(s)+H_{2}O(l)\rightarrow Ca(OH)_{2}(aq)}$.
6 On the left, very low ionisation enthalpies make metals lose electrons easily; on the right, very negative electron gain enthalpies and high electronegativities make non-metals gain electrons easily. Elements in the middle do neither readily.
7 Potassium (lower $\Delta_{i}H_{1}$: $419$ vs $496$). Fluorine (highest electronegativity, smallest atom; iodine holds extra electrons least strongly).
8 $\mathrm{CaO}$: a basic oxide that forms $\mathrm{Ca(OH)_{2}}$ and neutralises acid. $\mathrm{SO_{2}}$ is an acidic oxide and would lower the pH further.
V
📘 School assessment only (not in the year-end board exam). The s-block and p-block trends.
  • 1 Arrange Li, Na, K, Rb and Cs in order of (a) increasing atomic radius, (b) decreasing first ionisation enthalpy, (c) decreasing hydration enthalpy of the $\mathrm{M^{+}}$ ion.
  • 2 Give two ways in which lithium differs from the other alkali metals, and one way in which it resembles magnesium.
  • 3 Explain why beryllium chloride is largely covalent and why beryllium never shows a coordination number above four.
  • 4 Use the inert pair effect to explain why $\mathrm{PbCl_{2}}$ is more stable than $\mathrm{PbCl_{4}}$, and why thallium commonly forms $\mathrm{Tl^{+}}$ while aluminium forms only $\mathrm{Al^{3+}}$.
  • 5 Nitrogen exists as $\mathrm{N_{2}}$ molecules, but phosphorus as $\mathrm{P_{4}}$ units with single bonds. Explain.
1 (a) Li < Na < K < Rb < Cs (a new shell each step). (b) Li > Na > K > Rb > Cs ($520$, $496$, $419$, $403$, $376\ \mathrm{kJ\,mol^{-1}}$). (c) $\mathrm{Li^{+}}$ > $\mathrm{Na^{+}}$ > $\mathrm{K^{+}}$ > $\mathrm{Rb^{+}}$ > $\mathrm{Cs^{+}}$ — the smaller the ion, the more strongly it attracts water molecules.
2 Any two: lithium is harder with a higher melting point; it burns in air to the normal oxide $\mathrm{Li_{2}O}$ (others give peroxides or superoxides); it combines directly with nitrogen to form $\mathrm{Li_{3}N}$; $\mathrm{Li_{2}CO_{3}}$ decomposes on heating to $\mathrm{Li_{2}O}$ and $\mathrm{CO_{2}}$; its compounds are more covalent. Like magnesium it forms a nitride (Mg gives $\mathrm{Mg_{3}N_{2}}$), and its carbonate decomposes on heating — the diagonal relationship.
3 $\mathrm{Be^{2+}}$ would be tiny with a very high charge/radius ratio, and beryllium's ionisation enthalpies are high, so it shares electrons instead of forming ions: $\mathrm{BeCl_{2}}$ is covalent. Its valence shell ($n=2$) has only four orbitals ($2s$, three $2p$), so at most four electron pairs can be accepted: coordination number $4$, e.g. $\mathrm{[Be(OH)_{4}]^{2-}}$.
4 In the heavy elements of groups 13–15 the $ns^{2}$ pair is held tightly and is reluctant to take part in bonding, so the state two below the group state becomes the more stable one: Pb(II) rather than Pb(IV), Tl(I) rather than Tl(III). In light Al the $3s$ electrons are used readily, so only $+3$ is found.
5 Small nitrogen atoms come close enough for strong $p\pi$–$p\pi$ overlap, so $\mathrm{N{\equiv}N}$ triple bonds form. The larger phosphorus atoms overlap poorly sideways, so phosphorus forms three single bonds to three other atoms instead, giving tetrahedral $\mathrm{P_{4}}$ units.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The correct order of radii is (A) $\mathrm{N^{3-}\lt O^{2-}\lt F^{-}\lt Na^{+}}$ (B) $\mathrm{Na^{+}\lt F^{-}\lt O^{2-}\lt N^{3-}}$ (C) $\mathrm{F^{-}\lt Na^{+}\lt O^{2-}\lt N^{3-}}$ (D) $\mathrm{Na^{+}\lt N^{3-}\lt O^{2-}\lt F^{-}}$

Idea All four have $10$ electrons; radius decreases as $Z$ increases: N (7) largest, Na (11) smallest.
Answer (B) ($95\lt136\lt140\lt171\ \mathrm{pm}$).

E2. JEE Main The first ionisation enthalpies of Na, Mg and Si are $496$, $738$ and $787\ \mathrm{kJ\,mol^{-1}}$. The value for Al is closest to (A) $575$ (B) $760$ (C) $800$ (D) $1000\ \mathrm{kJ\,mol^{-1}}$

Idea Al loses a $3p$ electron, shielded by $3s^{2}$, so its value lies below Mg's, not between Mg and Si.
Answer (A) (actual $578\ \mathrm{kJ\,mol^{-1}}$).

E3. Board · 3 marks Explain why (i) the first ionisation enthalpy of nitrogen is higher than that of oxygen; (ii) chlorine has a more negative electron gain enthalpy than fluorine; (iii) $\mathrm{Na^{+}}$ is smaller than $\mathrm{Na}$.

1 mark N has the extra-stable half-filled $2p^{3}$; in O ($2p^{4}$) one electron is paired and repulsion makes it easier to remove.
1 mark F is very small: an added electron is strongly repelled by the electrons in the compact $2p$ subshell; in the larger $3p$ of Cl this repulsion is less.
1 mark $\mathrm{Na^{+}}$ has lost its only $3s$ electron (the whole third shell), and its $10$ electrons are held by $11$ protons, so it is much smaller ($95$ vs $186\ \mathrm{pm}$).

E4. JEE Main Numerical answer. An element of period 3 has successive ionisation enthalpies $787$, $1577$, $3232$, $4356$ and $16\,091\ \mathrm{kJ\,mol^{-1}}$. What is its group number?

Jumps $1577/787=2.0$, $3232/1577=2.0$, $4356/3232=1.3$, $16\,091/4356=3.7$. The large jump comes after the fourth electron: four valence electrons ($3s^{2}3p^{2}$, silicon).
Answer 14.

E5. JEE Advanced One or more options correct. (A) The first ionisation enthalpy of Be is greater than that of B. (B) Electronegativity decreases from F to I. (C) $\mathrm{Al_{2}O_{3}}$ reacts with both $\mathrm{HCl}$ and $\mathrm{NaOH}$. (D) Of all the elements, fluorine has the most negative electron gain enthalpy.

(A) True ($900$ vs $801$): B loses a $2p$ electron.
(B) True ($4.0$, $3.0$, $2.8$, $2.5$).
(C) True: it is amphoteric.
(D) False: chlorine ($-349$) is more negative than fluorine ($-328\ \mathrm{kJ\,mol^{-1}}$).
Answer (A), (B), (C).

E6. Board · Case study The oxides of the period-3 elements are $\mathrm{Na_{2}O}$, $\mathrm{MgO}$, $\mathrm{Al_{2}O_{3}}$, $\mathrm{SiO_{2}}$, $\mathrm{P_{4}O_{10}}$, $\mathrm{SO_{3}}$ and $\mathrm{Cl_{2}O_{7}}$. (i) How does the acid–base nature of these oxides change across the period, and why? (2 marks) (ii) Name the amphoteric oxide. (1 mark) (iii) Write the equation for the reaction of $\mathrm{Na_{2}O}$ with water. (1 mark)

(i) From strongly basic ($\mathrm{Na_{2}O}$, $\mathrm{MgO}$) through amphoteric ($\mathrm{Al_{2}O_{3}}$) to strongly acidic ($\mathrm{SiO_{2}}$ to $\mathrm{Cl_{2}O_{7}}$). Metallic character falls and electronegativity rises across the period: metal oxides are ionic and give $\mathrm{OH^{-}}$ with water, non-metal oxides are covalent and give acids.
(ii) Aluminium oxide, $\mathrm{Al_{2}O_{3}}$.
(iii) $\mathrm{Na_{2}O(s)+H_{2}O(l)\rightarrow2NaOH(aq)}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why is lithium — and increasingly sodium — the metal of choice for rechargeable batteries in phones, electric cars and grid storage?

Why it matters. Lithium-ion batteries power billions of devices and most electric vehicles. Lithium is mined mainly in Australia, Chile, Argentina and China, and demand is rising so fast that companies in China, Europe and India are developing sodium-ion batteries, using a cheaper and far more abundant element from the same group.

What to do.

  1. Use the group-1 data in this section (atomic radius, $\Delta_{i}H_{1}$, electronegativity) to explain why the alkali metals give up an electron so readily.
  2. A battery stores more energy per kilogram if its ions are light. Compare the molar masses of Li ($6.94$), Na ($22.99$) and K ($39.10\ \mathrm{g\,mol^{-1}}$): how many moles of each metal are there in $1.00\ \mathrm{kg}$?
  3. In a battery the ions move in and out of solid electrodes. Using the ionic radii $\mathrm{Li^{+}}\ 60$, $\mathrm{Na^{+}}\ 95$, $\mathrm{K^{+}}\ 133\ \mathrm{pm}$ (Pauling), explain why a small ion is an advantage, and estimate by how much the volume of $\mathrm{Na^{+}}$ exceeds that of $\mathrm{Li^{+}}$.
  4. Find out (from an energy agency report or a recent review) one advantage and one drawback of sodium-ion batteries compared with lithium-ion.

Tools: Rule 3.7 and Rule 3.8, Interactive 2 (group 1), and a reliable energy or materials-science source.

Hand in: A short table of the group-1 data and your calculations, and a half-page recommendation on where sodium-ion batteries make most sense.

Going further. Lithium has the most negative standard electrode potential of all the elements, although its ionisation enthalpy is the highest in group 1. Find out what this has to do with the very large hydration enthalpy of the small $\mathrm{Li^{+}}$ ion (Chapter 7 and Class 12).

Instructor note: moles in 1.00 kg: Li $1000/6.94=144$, Na $1000/22.99=43.5$, K $1000/39.10=25.6\ \mathrm{mol}$ — lithium stores about 3.3 times as many charge carriers per kilogram as sodium. Volume ratio $(95/60)^{3}=3.97\approx4$. Sodium-ion: cheaper, abundant, safer at low temperature, no cobalt; drawbacks: lower energy density, heavier cells.

🧪 Self-check quiz

0/5 correct

Q1. Which species is the smallest?

Q2. The first ionisation enthalpy of oxygen is lower than that of nitrogen because

Q3. Which element has the most negative electron gain enthalpy?

Q4. Going down group 17, electronegativity

Q5. Which oxide is amphoteric?