Atomic Structure and Periodicity

1. Atomic Structure

Subatomic Particles

Atoms consist of three subatomic particles. Their properties define the behaviour of every element:

Property	Proton	Neutron	Electron
Symbol	$p^+$	$n^0$	$e^-$
Relative mass	$1$	$1$	$\approx 1/1836$
Actual mass (u)	$1.00728$	$1.00867$	$0.00055$
Charge	$+1$	$0$	$-1$
Location	Nucleus	Nucleus	Electron shells

Definition. The atomic number ($Z$) is the number of protons in the nucleus. It uniquely identifies an element.

Definition. The mass number ($A$) is the total number of protons and neutrons in the nucleus:

$$A = Z + N$$

where $N$ is the neutron number.

Definition. A nuclide is a specific atom characterised by its atomic number, mass number, and energy state, denoted as $\prescript{A}{}{Z}\mathrm{X}$.

Isotopes

Definition. Isotopes are atoms of the same element (same $Z$) with different numbers of neutrons (different $A$).

Isotopes have identical chemical properties (same electron configuration) but different physical properties (different mass, different nuclear stability).

Element	Isotope	$Z$	$A$	$N$	Natural Abundance
Hydrogen	$\mathrm{H}$ (protium)	$1$	$1$	$0$	$99.985\%$
Hydrogen	$\mathrm{D}$ (deuterium)	$1$	$2$	$1$	$0.015\%$
Hydrogen	$\mathrm{T}$ (tritium)	$1$	$3$	$2$	Trace (radioactive)
Carbon	$\mathrm{C}$-12	$6$	$12$	$6$	$98.89\%$
Carbon	$\mathrm{C}$-13	$6$	$13$	$7$	$1.11\%$
Carbon	$\mathrm{C}$-14	$6$	$14$	$8$	Trace (radioactive)
Chlorine	$\mathrm{Cl}$-35	$17$	$35$	$18$	$75.77\%$
Chlorine	$\mathrm{Cl}$-37	$17$	$37$	$20$	$24.23\%$

Relative Atomic Mass

Definition. The relative atomic mass ($A_r$) is the weighted average mass of an atom of an element relative to $1/12$ the mass of a carbon-12 atom, taking into account the natural abundances of all isotopes.

$$A_r = \sum_{i} (\mathrm{isotope mass})_i \times (\mathrm{fractional abundance})_i$$

Example — Chlorine

Chlorine has two naturally occurring isotopes: $\mathrm{Cl}$-35 ($75.77\%$, mass $\approx 34.97\mathrm{ u}$) and $\mathrm{Cl}$-37 ($24.23\%$, mass $\approx 36.97\mathrm{ u}$).

$$A_r = (34.97 \times 0.7577) + (36.97 \times 0.2423) = 26.50 + 8.96 = 35.46$$

The Nucleus

The nucleus is extremely small relative to the atom ($\approx 10^{-15}\mathrm{ m}$ diameter vs $\approx 10^{-10}\mathrm{ m}$ for the atom). It contains over $99.9\%$ of the atom's mass. Nuclear stability depends on the neutron-to-proton ratio:

• Light elements ($Z \lt 20$): stable when $N \approx Z$
• Heavier elements: stable when $N \gt Z$ (neutrons provide additional strong nuclear force to counteract electrostatic repulsion between protons)

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2. Quantum Model of the Atom

Evidence for Quantisation

The classical Rutherford-Bohr model failed to explain several observations:

1. Discrete emission line spectra (not continuous)
2. The stability of atoms (classically, orbiting electrons should radiate energy and spiral in)
3. The photoelectric effect

These required a quantum mechanical treatment where electron energy is quantised.

Electron Shells and Subshells

Electrons occupy shells (principal energy levels) labelled $n = 1, 2, 3, \ldots$

Each shell contains subshells, designated by the azimuthal quantum number $l$:

$n$	Subshells ($l$ values)	Maximum electrons
$1$	$1s$	$2$
$2$	$2s$, $2p$	$8$
$3$	$3s$, $3p$, $3d$	$18$
$4$	$4s$, $4p$, $4d$, $4f$	$32$

The maximum number of electrons in shell $n$ is $2n^2$.

Orbitals

Definition. An orbital is a region of space where there is a high probability ($\ge 90\%$) of finding an electron. Each orbital holds a maximum of two electrons with opposite spins.

Subshell	Number of orbitals	Max electrons	Orbital shape
$s$	$1$	$2$	Spherical
$p$	$3$	$6$	Dumbbell
$d$	$5$	$10$	Cloverleaf
$f$	$7$	$14$	Complex (multi-lobed)

Quantum Numbers

Each electron in an atom is described by four quantum numbers:

Quantum Number	Symbol	What it specifies	Allowed values
Principal	$n$	Energy level (shell)	$1, 2, 3, \ldots$
Azimuthal	$l$	Subshell shape	$0, 1, 2, \ldots, n-1$
Magnetic	$m_l$	Orbital orientation in space	$-l, -l+1, \ldots, 0, \ldots, l-1, l$
Spin	$m_s$	Electron spin direction	$+\frac{1}{2}$ or $-\frac{1}{2}$

The number of orbitals in a subshell is $2l + 1$.

Example — Quantum numbers for a $3p$ electron

For the $3p$ subshell: $n = 3$, $l = 1$, $m_l = -1, 0, +1$, $m_s = \pm\frac{1}{2}$

This gives three $p$-orbitals ($p_x$, $p_y$, $p_z$), each holding two electrons, for a total of six $3p$ electrons.

Electron Configuration Principles

Three rules govern how electrons fill orbitals:

1. Aufbau principle: Electrons fill orbitals starting from the lowest energy level upwards.
2. Pauli exclusion principle: No two electrons in the same atom can have identical sets of four quantum numbers. Each orbital holds a maximum of two electrons with opposite spins.
3. Hund's rule: Within a subshell, electrons occupy degenerate orbitals singly first, with parallel spins, before pairing up.

Filling Order

The filling order follows the $(n + l)$ rule: orbitals with a lower $(n + l)$ value fill first. When $(n + l)$ values are equal, the orbital with lower $n$ fills first.

$$1s \lt 2s \lt 2p \lt 3s \lt 3p \lt 4s \lt 3d \lt 4p \lt 5s \lt 4d \lt 5p \lt 6s \lt 4f \lt 5d \lt 6p \lt 7s \lt 5f \lt 6d \lt 7p$$

Writing Electron Configurations

Full notation — write every subshell explicitly:

$$\mathrm{Fe}: 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6$$

Noble gas core notation — replace the inner-shell electrons with the preceding noble gas symbol in brackets:

$$\mathrm{Fe}: [\mathrm{Ar}]\, 4s^2\, 3d^6$$

warning

Always write subshells in order of increasing $n$ first, then $l$ (i.e., $4s^2\, 3d^6$, not $3d^6\, 4s^2$). When writing configurations for ions, remove electrons from the highest $n$ value first: $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}]\, 3d^6$, not $[\mathrm{Ar}]\, 4s^2\, 3d^4$.

Exceptions to the Aufbau Principle

Half-filled and fully-filled $d$-subshells are more stable due to symmetry and exchange energy. This causes exceptions in chromium, copper, molybdenum, silver, and gold:

Element	Expected configuration	Actual configuration	Reason
$\mathrm{Cr}$	$[\mathrm{Ar}]\, 4s^2\, 3d^4$	$[\mathrm{Ar}]\, 4s^1\, 3d^5$	Half-filled $d$-subshell
$\mathrm{Cu}$	$[\mathrm{Ar}]\, 4s^2\, 3d^9$	$[\mathrm{Ar}]\, 4s^1\, 3d^{10}$	Fully-filled $d$-subshell
$\mathrm{Mo}$	$[\mathrm{Kr}]\, 5s^2\, 4d^4$	$[\mathrm{Kr}]\, 5s^1\, 4d^5$	Half-filled $d$-subshell
$\mathrm{Ag}$	$[\mathrm{Kr}]\, 5s^2\, 4d^9$	$[\mathrm{Kr}]\, 5s^1\, 4d^{10}$	Fully-filled $d$-subshell
$\mathrm{Au}$	$[\mathrm{Xe}]\, 6s^2\, 4f^{14}\, 5d^9$	$[\mathrm{Xe}]\, 6s^1\, 4f^{14}\, 5d^{10}$	Fully-filled $d$-subshell

Orbital Diagrams

Orbital diagrams represent electrons as arrows in boxes (one box per orbital). Up and down arrows represent $m_s = +\frac{1}{2}$ and $m_s = -\frac{1}{2}$.

For nitrogen ($1s^2\, 2s^2\, 2p^3$):

\begin`\{array}`{c} 1s\quad \boxed{\uparrow\downarrow} \\ 2s\quad \boxed{\uparrow\downarrow} \\ 2p\quad \boxed{\uparrow}\quad\boxed{\uparrow}\quad\boxed{\uparrow} \end`\{array}`

All three $2p$ electrons are unpaired with parallel spins, following Hund's rule.

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3. The Periodic Table

Structure

The periodic table arranges elements in order of increasing atomic number. The layout reflects the electron configurations of the elements.

Feature	Description
Periods	Horizontal rows; number = principal quantum number $n$ of the valence shell
Groups	Vertical columns; elements share similar valence electron configurations
Blocks	Regions corresponding to the subshell being filled

Block Structure

Block	Subshell being filled	Groups	Examples
$s$	$ns^1$ to $ns^2$	$1$, $2$	H, He, Li, Na, Mg, Ca
$p$	$np^1$ to $np^6$	$13$--$18$	B, C, N, O, F, Ne, Cl, Ar
$d$	$(n-1)d^1$ to $(n-1)d^{10}$	$3$--$12$	Sc, Ti, Fe, Cu, Zn
$f$	$(n-2)f^1$ to $(n-2)f^{14}$	Lanthanides/Actinides	Ce, Th, U

Group Numbering

The IB uses IUPAC group numbers $1$--$18$:

IUPAC Group	Common Name	Valence electrons
$1$	Alkali metals	$ns^1$
$2$	Alkaline earth metals	$ns^2$
$13$	Boron group	$ns^2\, np^1$
$14$	Carbon group	$ns^2\, np^2$
$15$	Nitrogen group	$ns^2\, np^3$
$16$	Oxygen group	$ns^2\, np^4$
$17$	Halogens	$ns^2\, np^5$
$18$	Noble gases	$ns^2\, np^6$ (except He: $1s^2$)

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4. Periodic Trends

Effective Nuclear Charge

Definition. The effective nuclear charge ($Z_{\mathrm{eff}}$) is the net positive charge experienced by an electron, after accounting for shielding by other electrons:

$$Z_{\mathrm{eff}} = Z - S$$

where $Z$ is the actual nuclear charge and $S$ is the shielding constant.

Shielding is the reduction in the attractive force between the nucleus and an electron due to repulsion by other electrons. Inner-shell electrons shield outer-shell electrons far more effectively than electrons in the same shell.

Atomic Radius

Trend	Explanation
Decreases across a period	$Z_{\mathrm{eff}}$ increases; electrons are pulled closer to the nucleus
Increases down a group	Additional electron shells increase the average distance from the nucleus

Definition. Atomic radius is half the distance between the nuclei of two bonded atoms of the same element.

For noble gases, atomic radius is taken as the van der Waals radius (half the distance between nuclei of adjacent atoms in the solid or liquid), which is significantly larger than covalent radii.

Ionic Radius

Trend	Explanation
Cations are smaller than their parent atoms	Fewer electron-electron repulsions; same $Z$ pulling fewer electrons
Anions are larger than their parent atoms	Increased electron-electron repulsion with the same $Z$
Ionic radius increases down a group	Additional shells
Across a period, ions decrease in size	Isoelectronic series: same number of electrons, increasing $Z$

Example — Isoelectronic series

$\mathrm{O}^{2-} \gt \mathrm{F}^- \gt \mathrm{Na}^+ \gt \mathrm{Mg}^{2+} \gt \mathrm{Al}^{3+}$

All have the neon configuration ($1s^2\, 2s^2\, 2p^6$, $10$ electrons). The nuclear charge increases from $Z = 8$ to $Z = 13$, so the radius decreases.

Ionization Energy

Definition. The first ionization energy ($IE_1$) is the minimum energy required to remove one mole of electrons from one mole of gaseous atoms:

$$\mathrm{X}(g) \to \mathrm{X}^+(g) + e^- \qquad \Delta H = IE_1$$

Trend	Explanation
Increases across a period	$Z_{\mathrm{eff}}$ increases; electrons held more tightly
Decreases down a group	Electrons are farther from the nucleus and more shielded

Deviations from the general trend across a period:

Deviation	Element pair	Explanation
Drop from Group $2$ to $13$	Be $\to$ B	Be: $2s^2$ (stable filled subshell); B: $2p^1$ (easier to remove)
Drop from Group $15$ to $16$	N $\to$ O	N: $2p^3$ (half-filled, stable); O: $2p^4$ (paired electron in $2p$ experiences repulsion)

Successive Ionization Energies

Each subsequent ionization energy is larger than the previous one because the remaining electrons experience less shielding from a shrinking electron cloud and are held by a constant $Z$:

$$IE_1 \lt IE_2 \lt IE_3 \lt \cdots$$

A large jump in successive ionization energies indicates the removal of an electron from a new inner shell. This reveals the electron configuration.

Example — Aluminium

For aluminium ($1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^1$):

• $IE_1 = 578\mathrm{ kJ/mol}$ (removes $3p$ electron)
• $IE_2 = 1817\mathrm{ kJ/mol}$ (removes $3s$ electron)
• $IE_3 = 2745\mathrm{ kJ/mol}$ (removes $3s$ electron)
• $IE_4 = 11577\mathrm{ kJ/mol}$ (removes $2p$ electron — large jump!)

The jump from $IE_3$ to $IE_4$ confirms that aluminium has three valence electrons.

Electron Affinity

Definition. Electron affinity ($EA$) is the enthalpy change when one mole of electrons is added to one mole of gaseous atoms:

$$\mathrm{X}(g) + e^- \to \mathrm{X}^-(g) \qquad \Delta H = EA$$

A more negative $EA$ indicates a greater tendency to accept an electron.

Trend	Explanation
Generally becomes more negative across a period	Increasing $Z_{\mathrm{eff}}$ attracts electrons more strongly
Generally becomes less negative down a group	Increased distance and shielding reduce the nuclear pull

Noble gases have positive (endothermic) electron affinities because the added electron enters a new, higher-energy subshell.

Electronegativity

Definition. Electronegativity is the ability of an atom to attract the shared pair of electrons in a covalent bond. The Pauling scale is the most common.

Trend	Explanation
Increases across a period	Increasing $Z_{\mathrm{eff}}$
Decreases down a group	Increased distance and shielding

Scale	Most electronegative	Least electronegative
Pauling	F ($3.98$)	Fr ($0.7$)

Metallic and Non-Metallic Character

Trend	Metallic character	Non-metallic character
Across a period	Decreases	Increases
Down a group	Increases	Decreases

Metallic character correlates with low ionization energy, low electronegativity, and large atomic radius. Non-metallic character correlates with high ionization energy, high electronegativity, and small atomic radius.

Summary Table of Periodic Trends

Property	Across a period (left to right)	Down a group (top to bottom)
Atomic radius	Decreases	Increases
Ionic radius	Decreases (isoelectronic)	Increases
Ionization energy	Increases	Decreases
Electron affinity	More negative	Less negative
Electronegativity	Increases	Decreases
Metallic character	Decreases	Increases
Non-metallic character	Increases	Decreases

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5. Group 1: Alkali Metals

Physical Properties

Property	Trend down the group
Melting point	Decreases (Cs is below room temp in some conditions)
Boiling point	Decreases
Density	Generally increases (Li, K anomalies)
Atomic radius	Increases
Softness	Increases (softer metals)

Chemical Properties

All alkali metals have the outer electron configuration $ns^1$. The single valence electron is easily lost, forming $\mathrm{M}^+$ ions.

Reaction with Water

$$2\mathrm{M}(s) + 2\mathrm{H}_2\mathrm{O}(l) \to 2\mathrm{MOH}(aq) + \mathrm{H}_2(g)$$

Reactivity increases down the group:

Metal	Observation
Li	Steady fizzing; moves on surface
Na	Rapid fizzing; melts into a ball; may ignite H$_2$
K	Ignites immediately with a lilac flame
Rb, Cs	Explosive reaction; often thrown from the water

Explanation of trend: Ionization energy decreases down the group. The valence electron is farther from the nucleus and more shielded, so less energy is required to remove it.

Oxides

Alkali metals burn in oxygen to form oxides:

Metal	Product with limited O$_2$	Product with excess O$_2$
Li	$\mathrm{Li}_2\mathrm{O}$ (oxide)	$\mathrm{Li}_2\mathrm{O}$
Na	$\mathrm{Na}_2\mathrm{O}$ (oxide)	$\mathrm{Na}_2\mathrm{O}_2$ (peroxide)
K	$\mathrm{K}_2\mathrm{O}_2$ (peroxide)	$\mathrm{KO}_2$ (superoxide)
Rb, Cs	Superoxides form readily	Superoxides

Hydroxides

All Group 1 hydroxides ($\mathrm{MOH}$) are strong bases and highly soluble in water:

$$\mathrm{MOH}(s) \to \mathrm{M}^+(aq) + \mathrm{OH}^-(aq)$$

Basicity increases down the group (solubility increases, so $[\mathrm{OH}^-]$ is higher).

Flame Tests

Alkali metal ions produce characteristic flame colours due to electron transitions:

Ion	Flame colour
$\mathrm{Li}^+$	Crimson red
$\mathrm{Na}^+$	Yellow
$\mathrm{K}^+$	Lilac (viewed through cobalt glass to filter Na)
$\mathrm{Rb}^+$	Red-violet
$\mathrm{Cs}^+$	Blue

Uses

Metal	Use
Li	Batteries, psychiatric medication (lithium carbonate)
Na	Street lamps (Na vapour), NaK coolant
K	Fertilisers ($\mathrm{KNO}_3$), potash

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6. Group 17: Halogens

Physical Properties

Property	Trend down the group
Melting point	Increases
Boiling point	Increases
Atomic radius	Increases
State at RT	F$_2$, Cl$_2$ (gas); Br$_2$ (liquid); I$_2$ (solid)
Colour	Pale yellow $\to$ yellow-green $\to$ red-brown $\to$ dark grey
Volatility	Decreases

The increase in melting and boiling points down the group is due to increasing London dispersion forces as the number of electrons (and therefore polarizability) increases.

Chemical Properties

All halogens have the outer electron configuration $ns^2\, np^5$. They gain one electron to form $\mathrm{X}^-$ ions, achieving a noble gas configuration.

Reactivity Trend

Reactivity decreases down the group. This is because atomic radius increases and $Z_{\mathrm{eff}}$ on the incoming electron decreases, so electron affinity becomes less favourable.

$$\mathrm{F}_2 \gt \mathrm{Cl}_2 \gt \mathrm{Br}_2 \gt \mathrm{I}_2$$

Displacement Reactions

A more reactive halogen displaces a less reactive halogen from its halide solution:

$$\mathrm{Cl}_2(aq) + 2\mathrm{KBr}(aq) \to 2\mathrm{KCl}(aq) + \mathrm{Br}_2(aq)$$ $$\mathrm{Br}_2(aq) + 2\mathrm{KI}(aq) \to 2\mathrm{KBr}(aq) + \mathrm{I}_2(aq)$$ $$\mathrm{Cl}_2(aq) + 2\mathrm{KI}(aq) \to 2\mathrm{KCl}(aq) + \mathrm{I}_2(aq)$$

But: $\mathrm{Br}_2$ cannot displace $\mathrm{Cl}^-$ and $\mathrm{I}_2$ cannot displace $\mathrm{Br}^-$ or $\mathrm{Cl}^-$.

Reaction with Alkali Metals

$$2\mathrm{M}(s) + \mathrm{X}_2(g) \to 2\mathrm{MX}(s)$$

These are vigorous, exothermic reactions forming ionic halides.

Halide Ion Tests

Halide	Reagent	Observation
$\mathrm{Cl}^-$	$\mathrm{AgNO}_3$(aq) + dilute HNO$_3$	White precipitate ($\mathrm{AgCl}$), soluble in dilute NH$_3$
$\mathrm{Br}^-$	$\mathrm{AgNO}_3$(aq) + dilute HNO$_3$	Cream precipitate ($\mathrm{AgBr}$), partially soluble in NH$_3$
$\mathrm{I}^-$	$\mathrm{AgNO}_3$(aq) + dilute HNO$_3$	Yellow precipitate ($\mathrm{AgI}$), insoluble in NH$_3$

Dilute HNO$_3$ is added first to remove any carbonate or hydroxide ions that would also form precipitates with $\mathrm{Ag}^+$.

Uses

Halogen	Use
$\mathrm{F}_2$	Fluoridation of water, Teflon (PTFE) production
$\mathrm{Cl}_2$	Water purification, PVC, bleach ($\mathrm{NaClO}$)
$\mathrm{Br}_2$	Flame retardants, brominated compounds, photography
$\mathrm{I}_2$	Antiseptics, iodised salt, thyroid hormone synthesis

Interhalogens

Definition. Interhalogens are compounds formed between two different halogen atoms. The more electronegative halogen is the negative end of the molecule.

General formula: $\mathrm{XX}'_n$ where $n = 1, 3, 5, 7$ (depending on the size of the central halogen).

Example	Type	Structure
$\mathrm{ClF}$	Diatomic	Linear
$\mathrm{BrF}_3$	Triatomic	T-shaped
$\mathrm{IF}_5$	Pentaatomic	Square pyramidal
$\mathrm{IF}_7$	Heptaatomic	Pentagonal bipyramidal

Interhalogens are generally more reactive than the parent halogens because the bonds are polar.

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7. Group 18: Noble Gases

Properties

Noble gases have complete valence shells ($ns^2\, np^6$, except He which is $1s^2$), making them chemically inert under standard conditions. They exist as monatomic gases.

Element	Configuration	Boiling point (K)	First IE (kJ/mol)
He	$1s^2$	$4.2$	$2372$
Ne	$[\mathrm{He}]\, 2s^2\, 2p^6$	$27.1$	$2081$
Ar	$[\mathrm{Ne}]\, 3s^2\, 3p^6$	$87.3$	$1521$
Kr	$[\mathrm{Ar}]\, 4s^2\, 4p^6$	$119.8$	$1351$
Xe	$[\mathrm{Kr}]\, 5s^2\, 5p^6$	$165.0$	$1170$
Rn	$[\mathrm{Xe}]\, 6s^2\, 6p^6$	$211.0$	$1037$

Boiling Point Trend

Boiling points increase down the group because the number of electrons increases, leading to stronger London dispersion forces between atoms. The only intermolecular force in noble gases is London dispersion.

Reactivity

Under extreme conditions, the heavier noble gases can form compounds:

• Xenon forms $\mathrm{XeF}_2$, $\mathrm{XeF}_4$, $\mathrm{XeF}_6$, $\mathrm{XeO}_3$, $\mathrm{XeO}_4$
• Krypton forms $\mathrm{KrF}_2$ (extremely reactive)
• Argon forms very unstable compounds under extreme conditions

Xenon compounds exist because its ionization energy is low enough that highly electronegative fluorine and oxygen can remove or share electrons.

Uses

Noble Gas	Use
He	Balloons, cryogenics, helium-neon lasers, deep-sea diving gas mix
Ne	Neon signs (orange-red glow)
Ar	Inert atmosphere for welding, light bulbs
Kr	High-performance lighting, photography flash lamps
Xe	Xenon lamps (used in IMAX projectors, car headlights), ion propulsion

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8. Transition Metals (HL)

Definition

Definition. A transition metal is an element that has a partially filled $d$-subshell in either its atom or any of its common oxidation states.

This definition excludes scandium ($\mathrm{Sc}^{3+}$: $[\mathrm{Ar}]$) and zinc ($\mathrm{Zn}^{2+}$: $[\mathrm{Ar}]\, 3d^{10}$) as transition metals in their common oxidation states, though they are in the $d$-block.

Physical Properties

Property	Typical behaviour of transition metals
Melting/boiling points	High (strong metallic bonding from $d$-electrons)
Density	High
Hardness	Hard
Electrical conductivity	Good conductors
Malleability	Malleable and ductile

Variable Oxidation States

Transition metals can lose different numbers of $d$-electrons to form ions with different charges. This is because the $3d$ and $4s$ energy levels are close in energy.

Element	Common oxidation states
Ti	$+2, +3, +4$
V	$+2, +3, +4, +5$
Cr	$+2, +3, +6$
Mn	$+2, +3, +4, +6, +7$
Fe	$+2, +3$
Co	$+2, +3$
Cu	$+1, +2$

Trend: The maximum oxidation state increases across the period to manganese ($+7$) then decreases. Higher oxidation states become more stable with oxygen (oxoanions) than with water.

Complex Formation

Definition. A complex ion consists of a central metal ion surrounded by ligands coordinated via coordinate (dative covalent) bonds.

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}, \quad [\mathrm{Ag}(\mathrm{NH}_3)_2]^+, \quad [\mathrm{Fe}(\mathrm{CN})_6]^{4-}$$

Ligands

Definition. A ligand is a molecule or ion that can donate a lone pair of electrons to a central metal ion to form a coordinate bond.

Type of ligand	Examples	Denticity	Bonds donated
Monodentate	$\mathrm{H}_2\mathrm{O}$, $\mathrm{NH}_3$, $\mathrm{Cl}^-$, $\mathrm{CN}^-$	$1$	$1$
Bidentate	Ethylenediamine (en), oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$)	$2$	$2$
Hexadentate	EDTA ($\mathrm{EDTA}^{4-}$)	$6$	$6$

Coordination Number

Definition. The coordination number is the total number of coordinate bonds from ligands to the central metal ion.

Coordination number	Geometry	Example
$4$	Tetrahedral	$[\mathrm{CoCl}_4]^{2-}$
$4$	Square planar	$[\mathrm{Cu(NH}_3)_4]^{2+}$ (sometimes), $[\mathrm{Ni(CN)}_4]^{2-}$
$6$	Octahedral	$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, $[\mathrm{Fe(CN)}_6]^{4-}$

Colour of Transition Metal Complexes

Transition metal complexes are coloured because of $d$-$d$ electron transitions:

1. In an isolated atom/ion, all five $d$-orbitals are degenerate (same energy).
2. In a complex, ligands split the $d$-orbitals into groups of different energies ($d$-orbital splitting).
3. When white light passes through the complex, photons with energy matching the $\Delta E$ between split $d$-levels are absorbed.
4. The remaining light is transmitted, giving the complex its complementary colour.

\Delta E = hf = \frac`\{hc}`{\lambda}

Spectrochemical series (increasing $\Delta$):

$$\mathrm{I}^- \lt \mathrm{Br}^- \lt \mathrm{Cl}^- \lt \mathrm{F}^- \lt \mathrm{H}_2\mathrm{O} \lt \mathrm{NH}_3 \lt \mathrm{en} \lt \mathrm{CN}^- \lt \mathrm{CO}$$

Ligands that produce larger splitting are called strong-field ligands; those producing smaller splitting are weak-field ligands.

Complex ion	Colour observed	Colour absorbed
$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	Blue	Orange/red
$[\mathrm{Cu}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+}$	Deep blue	Yellow/orange
$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	Pink	Green
$[\mathrm{CoCl}_4]^{2-}$	Blue	Yellow/orange

warning

A substance is colourless if either: (a) it has no $d$-electrons (e.g., $\mathrm{Sc}^{3+}$, $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{4+}$ has $d^0$), or (b) it has a full $d$-subshell (e.g., $\mathrm{Zn}^{2+}$, $[\mathrm{Cu}(\mathrm{NH}_3)_4]^+$ has $d^{10}$). In both cases, there are no $d$-$d$ transitions possible.

Catalytic Properties

Transition metals are effective catalysts because they can adopt variable oxidation states and form intermediate complexes, providing alternative reaction pathways with lower activation energies.

Heterogeneous catalysis (catalyst in a different phase):

Catalyst	Reaction
Fe	Haber process: $\mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3$
V$_2$O$_5$	Contact process: $2\mathrm{SO}_2 + \mathrm{O}_2 \to 2\mathrm{SO}_3$
Ni	Hydrogenation of alkenes
Pt/Pd	Catalytic converters (oxidation of CO and hydrocarbons, reduction of NO$_x$)

Homogeneous catalysis (catalyst in the same phase):

Catalyst	Reaction
$\mathrm{Fe}^{2+}/\mathrm{Fe}^{3+}$	Fenton's reagent (oxidation of organic pollutants)
$\mathrm{Mn}^{2+}$	Decomposition of $\mathrm{H}_2\mathrm{O}_2$

Magnetic Properties

Transition metals and their complexes can be paramagnetic (attracted to a magnetic field) or diamagnetic (weakly repelled).

Property	Condition	Example
Paramagnetic	Unpaired $d$-electrons present	$[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ ($d^5$)
Diamagnetic	All $d$-electrons paired	$[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($d^{10}$)

The number of unpaired electrons determines the magnetic moment (measured in Bohr magnetons, $\mu_B$):

$$\mu \approx \sqrt{n(n + 2)}\ \mu_B$$

where $n$ is the number of unpaired electrons.

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9. Spectral Evidence for Atomic Structure

Emission Spectra

When atoms absorb energy (e.g., from heat or electricity), electrons are excited to higher energy levels. When they fall back to lower levels, they emit photons with energies corresponding to the energy differences:

\Delta E = E_{\mathrm{higher}} - E_{\mathrm{lower}} = h\nu = \frac`\{hc}`{\lambda}

where:

• $h = 6.626 \times 10^{-34}\mathrm{ J}\cdot\mathrm{s}$ (Planck's constant)
• $c = 3.00 \times 10^8\mathrm{ m/s}$ (speed of light)
• $\lambda$ = wavelength of emitted light
• $\nu$ = frequency

Each element produces a unique line emission spectrum — a series of discrete lines at specific wavelengths. This is the basis of flame tests and spectroscopic analysis.

Absorption Spectra

When white light passes through a cool gas, the gas absorbs photons at wavelengths corresponding to the energy differences between its electron levels. The transmitted light shows dark lines at these wavelengths on a continuous spectrum.

Hydrogen Spectral Series

The hydrogen emission spectrum shows several series, each corresponding to transitions to a specific lower energy level:

Series	Final $n$	Spectral region	Wavelength range
Lyman	$1$	Ultraviolet	$< 400\mathrm{ nm}$
Balmer	$2$	Visible	$400$--$700\mathrm{ nm}$
Paschen	$3$	Infrared	$> 700\mathrm{ nm}$
Brackett	$4$	Infrared	$> 700\mathrm{ nm}$

The energy levels of hydrogen are given by:

$$E_n = -\frac{13.6\mathrm{ eV}}{n^2} = -\frac{2.18 \times 10^{-18}\mathrm{ J}}{n^2}$$

For the Balmer series (transitions to $n = 2$), the first four lines correspond to:

Transition	$\lambda$ (nm)	Colour
$3 \to 2$	$656$	Red
$4 \to 2$	$486$	Cyan
$5 \to 2$	$434$	Blue
$6 \to 2$	$410$	Violet

Convergence

Lines in each series converge at the series limit (the ionization energy). As $n$ increases, the energy levels get closer together and transitions approach a continuum:

$$E_{\infty} = 0\mathrm{ J} \quad (\mathrm{ionization threshold})$$

Significance

The existence of discrete spectral lines is direct evidence that electron energy is quantised. The classical model predicted a continuous spectrum, which is never observed for individual atoms.

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10. Mass Spectrometry

Principle

Mass spectrometry measures the mass-to-charge ratio ($m/z$) of ions. The general process:

1. Ionization: Atoms or molecules are ionized (typically by electron impact — high-energy electrons knock an electron off the sample, forming positive ions).
2. Acceleration: Ions are accelerated by an electric field. All ions receive the same kinetic energy: $\frac{1}{2}mv^2 = zV$ where $V$ is the accelerating voltage and $z$ is the charge on the ion.
3. Deflection: Ions pass through a magnetic field and are deflected. Lighter ions (or more highly charged ions) are deflected more: $r = \frac{\sqrt{2mV}}{zB}$ where $r$ is the radius of curvature and $B$ is the magnetic field strength.
4. Detection: A detector records the abundance of ions at each $m/z$ value, producing a mass spectrum.

Interpreting Mass Spectra

A mass spectrum plots relative abundance (y-axis) against $m/z$ (x-axis).

Isotopic Abundance

For a single element, the mass spectrum shows peaks at each isotope's mass, with heights proportional to natural abundance.

Example — Boron

Boron has two isotopes: $\mathrm{B}$-10 ($19.9\%$) and $\mathrm{B}$-11 ($80.1\%$).

The mass spectrum shows peaks at $m/z = 10$ and $m/z = 11$ with relative heights in the ratio $19.9 : 80.1$, approximately $1 : 4$.

$$A_r = (10 \times 0.199) + (11 \times 0.801) = 1.99 + 8.81 = 10.81$$

Molecular Ion

The molecular ion peak ($\mathrm{M}^+$) corresponds to the intact molecule with one electron removed. Its $m/z$ value gives the molecular mass.

$$\mathrm{CH}_4 + e^- \to \mathrm{CH}_4^{+\bullet} + 2e^-$$

The molecular ion peak for $\mathrm{CH}_4$ appears at $m/z = 16$.

Fragmentation

After ionization, the molecular ion often breaks apart into smaller fragments. The fragmentation pattern is characteristic of the molecule and can be used to identify it.

Fragment $m/z$	Likely species	Common origin
$15$	$\mathrm{CH}_3^+$	Loss of H from $\mathrm{CH}_4^{+\bullet}$
$29$	$\mathrm{C}_2\mathrm{H}_5^+$ or $\mathrm{CHO}^+$	Ethanol, aldehydes
$43$	$\mathrm{C}_3\mathrm{H}_7^+$ or $\mathrm{CH}_3\mathrm{CO}^+$	Ketones, propanol
$77$	$\mathrm{C}_6\mathrm{H}_5^+$	Benzene ring

Determining Molecular Formula from Isotope Peaks

For molecules containing chlorine or bromine, the isotope patterns are distinctive:

Element	Isotopes	Approximate ratio
Cl	$\mathrm{Cl}$-35, $\mathrm{Cl}$-37	$3 : 1$
Br	$\mathrm{Br}$-79, $\mathrm{Br}$-81	$1 : 1$

A molecule with one chlorine atom shows an $\mathrm{M}$ and $\mathrm{M}+2$ peak in a $3:1$ ratio. A molecule with one bromine atom shows an $\mathrm{M}$ and $\mathrm{M}+2$ peak in a $1:1$ ratio.

Example — Chlorobenzene

Chlorobenzene ($\mathrm{C}_6\mathrm{H}_5\mathrm{Cl}$) shows:

• $\mathrm{M}^+$ at $m/z = 112$ ($\mathrm{C}_6\mathrm{H}_5{}^{35}\mathrm{Cl}$)
• $\mathrm{M}+2$ at $m/z = 114$ ($\mathrm{C}_6\mathrm{H}_5{}^{37}\mathrm{Cl}$)
• Ratio of peak heights: approximately $3:1$

High-Resolution Mass Spectrometry

High-resolution MS can determine exact masses to several decimal places, distinguishing between molecules with the same nominal mass but different molecular formulas:

Species	Exact mass (u)
$\mathrm{C}_2\mathrm{H}_4\mathrm{O}$	$44.0262$
$\mathrm{CO}_2$	$43.9898$
$\mathrm{N}_2\mathrm{O}$	$44.0011$

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11. HL Extensions

Slater's Rules for Effective Nuclear Charge

Slater's rules provide a systematic way to estimate the shielding constant $S$ for an electron in a many-electron atom.

Rules

1. Write the electron configuration in groups: $(1s)(2s, 2p)(3s, 3p)(3d)(4s, 4p)(4d)(4f)(5s, 5p) \ldots$

2. Electrons in groups to the right of the electron of interest contribute $0$ to $S$.

3. Other electrons in the same group contribute:

   • For $ns$ or $np$ electrons: each other electron contributes $0.35$ (except $1s$, where the other electron contributes $0.30$)
   • For $nd$ or $nf$ electrons: each other electron contributes $0.35$

4. Electrons in the $n-1$ shell contribute:

   • $0.85$ each (for $s$ and $p$ electrons in the $n$ shell)
   • $1.00$ each (for $d$ and $f$ electrons in the $n$ shell)

5. Electrons in shells $n-2$ or lower contribute $1.00$ each.

Example — $Z_{\mathrm{eff}}$ for a $3p$ electron in chlorine ($Z = 17$)

Configuration: $(1s)^2(2s, 2p)^8(3s, 3p)^7$

For a $3p$ electron:

• Same $(3s, 3p)$ group: $6$ other electrons $\times 0.35 = 2.10$
• $n - 1$ shell $(2s, 2p)^8$: $8 \times 0.85 = 6.80$
• $n - 2$ shell $(1s)^2$: $2 \times 1.00 = 2.00$

$S = 2.10 + 6.80 + 2.00 = 10.90$

$Z\_{\mathrm{eff}} = 17 - 10.90 = 6.10$

Example — $Z_{\mathrm{eff}}$ for a $3d$ electron in scandium ($Z = 21$)

Configuration: $(1s)^2(2s, 2p)^8(3s, 3p)^8(3d)^1(4s)^2$

For the $3d$ electron:

• Same $(3d)$ group: $0$ other electrons $\times 0.35 = 0$
• $n - 1$ shell: $(3s, 3p)^8$ each contributes $1.00$ (for $d$ electrons, the rule is different) = $8 \times 1.00 = 8.00$
• Shells $n - 2$ and lower: $(2s, 2p)^8(1s)^2$ = $10 \times 1.00 = 10.00$

$S = 0 + 8.00 + 10.00 = 18.00$

$Z\_{\mathrm{eff}} = 21 - 18.00 = 3.00$

The low $Z_{\mathrm{eff}}$ on the $3d$ electron explains why the $4s$ orbital fills before $3d$ — the $4s$ electron experiences a higher effective nuclear charge.

Successive Ionization Energy Graphs and Electron Configuration

Plotting $\log(IE)$ against ionization number reveals jumps that correspond to the removal of electrons from inner shells.

For aluminium ($Z = 13$):

$$\underbrace{IE_1, IE_2, IE_3}_{\mathrm{valence } 3s^2\, 3p^1} \ll IE_4, IE_5, \ldots, IE_{11} \ll IE_{12}, IE_{13}$$

Ionization number	Electron removed	Approximate IE (kJ/mol)	Shell
$1$	$3p^1$	$578$	$n = 3$
$2$	$3s^1$	$1817$	$n = 3$
$3$	$3s^1$	$2745$	$n = 3$
$4$	$2p^1$	$11577$	$n = 2$ (jump!)
$\ldots$
$11$	$2s^1$	$6521$	$n = 2$
$12$	$1s^1$	$18560$	$n = 1$ (jump!)
$13$	$1s^1$	$21600$	$n = 1$

The jumps reveal:

• $3$ valence electrons ($n = 3$)
• $8$ electrons in $n = 2$
• $2$ electrons in $n = 1$

Energy Level Transitions and Spectral Lines

Energy-Wavelength Calculations

The Rydberg equation for hydrogen gives the wavelength of any spectral line:

$$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$

where $R_H = 1.097 \times 10^7\mathrm{ m}^{-1}$ is the Rydberg constant, $n_i$ is the initial energy level, and $n_f$ is the final energy level ($n_i \gt n_f$ for emission).

Example — Wavelength of the first Balmer line

For the transition $n = 3 \to n = 2$:

$$\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 \times 10^7 \times 0.1389 = 1.524 \times 10^6\mathrm{ m}^{-1}$$$$\lambda = \frac{1}{1.524 \times 10^6} = 6.56 \times 10^{-7}\mathrm{ m} = 656\mathrm{ nm}$$

This corresponds to the red line in the Balmer series (H$\alpha$).

Energy of a Photon

E = h\nu = \frac`\{hc}`{\lambda}

For the $n = 3 \to n = 2$ transition:

$$E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{6.56 \times 10^{-7}} = 3.03 \times 10^{-19}\mathrm{ J}$$

Converting to electron-volts:

$$E = \frac{3.03 \times 10^{-19}}{1.602 \times 10^{-19}} = 1.89\mathrm{ eV}$$

Number of Spectral Lines

The number of possible spectral lines from energy level $n$ down to the ground state is:

$$N = \frac{n(n - 1)}{2}$$

For $n = 4$: $N = 6$ spectral lines.

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12. Exam Practice

Question 1 (SL — 4 marks)

(a) Define the term relative atomic mass. (2 marks)

(b) Naturally occurring boron consists of two isotopes, $\mathrm{B}$-10 and $\mathrm{B}$-11. The relative atomic mass of boron is $10.81$. Calculate the percentage abundance of $\mathrm{B}$-10. (2 marks)

Markscheme:

(a) The weighted mean mass of an atom of an element relative to $1/12$ the mass of an atom of carbon-12, based on the abundance of isotopes in a naturally occurring sample. (2 marks)

(b) Let $x$ = fractional abundance of $\mathrm{B}$-10, so $(1 - x)$ = fractional abundance of $\mathrm{B}$-11.

$$10.81 = 10x + 11(1 - x)$$ $$10.81 = 10x + 11 - 11x$$ $$10.81 = 11 - x$$ $$x = 0.19$$

Percentage abundance of $\mathrm{B}$-10 = $19\%$. (1 mark for setup, 1 mark for answer)

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Question 2 (SL — 3 marks)

Explain why the first ionization energy of sodium is lower than that of magnesium, but the first ionization energy of magnesium is lower than that of aluminium.

Markscheme:

Na ($1s^2\, 2s^2\, 2p^6\, 3s^1$) to Mg ($1s^2\, 2s^2\, 2p^6\, 3s^2$): $Z_{\mathrm{eff}}$ increases across the period, so the $3s$ electrons in Mg are held more tightly. (1 mark)

Mg ($3s^2$) to Al ($3s^2\, 3p^1$): the electron removed from Al is a $3p$ electron, which is at a higher energy level than the $3s$ electrons of Mg and is partially shielded by the $3s$ electrons. (1 mark)

Also: Mg has a stable filled $3s$ subshell configuration. (1 mark)

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Question 3 (SL — 4 marks)

(a) State the electron configuration of $\mathrm{Fe}^{2+}$ using noble gas notation. (1 mark)

(b) Describe the trend in atomic radius across Period 3 and explain this trend in terms of effective nuclear charge. (3 marks)

Markscheme:

(a) $[\mathrm{Ar}]\, 3d^6$ (1 mark; note: electrons are removed from $4s$ before $3d$)

(b) Atomic radius decreases from Na to Ar. (1 mark)

Across a period, the nuclear charge ($Z$) increases by one proton per element, but the additional electrons enter the same shell and provide only partial shielding. (1 mark)

Therefore $Z_{\mathrm{eff}}$ increases across the period, pulling the electron cloud closer to the nucleus and decreasing the atomic radius. (1 mark)

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Question 4 (SL — 3 marks)

A sample of chlorine gas is analysed by mass spectrometry. Describe and explain the appearance of the mass spectrum.

Markscheme:

Two peaks are observed at $m/z = 35$ and $m/z = 37$. (1 mark)

The heights of the peaks are in the approximate ratio $3:1$, reflecting the natural abundances of $\mathrm{Cl}$-35 ($75.77\%$) and $\mathrm{Cl}$-37 ($24.23\%$). (1 mark)

The sample is diatomic ($\mathrm{Cl}_2$), so additional peaks appear at $m/z = 70$ ($^{35}\mathrm{Cl}$--$^{35}\mathrm{Cl}$), $m/z = 72$ ($^{35}\mathrm{Cl}$--$^{37}\mathrm{Cl}$), and $m/z = 74$ ($^{37}\mathrm{Cl}$--$^{37}\mathrm{Cl}$) in the ratio $9:6:1$. (1 mark)

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Question 5 (HL — 5 marks)

(a) State the four quantum numbers for each of the valence electrons in a ground-state oxygen atom. (3 marks)

(b) Explain why the fourth ionization energy of beryllium is much larger than the third. (2 marks)

Markscheme:

(a) Oxygen: $1s^2\, 2s^2\, 2p^4$. The six valence electrons (in $n = 2$):

$2s$ electrons: $(2, 0, 0, +\frac{1}{2})$ and $(2, 0, 0, -\frac{1}{2})$

$2p$ electrons: $(2, 1, -1, +\frac{1}{2})$, $(2, 1, 0, +\frac{1}{2})$, $(2, 1, +1, +\frac{1}{2})$, $(2, 1, -1, -\frac{1}{2})$ (1 mark for $n, l, m_l$; 1 mark for $m_s$ showing Hund's rule with first three $2p$ electrons having parallel spins; 1 mark for the fourth being paired)

(b) Beryllium: $1s^2\, 2s^2$. $IE_1$, $IE_2$, $IE_3$ remove the two $2s$ electrons and one $1s$ electron. $IE_4$ removes the remaining $1s$ electron. (1 mark)

The $1s$ electron is much closer to the nucleus and experiences far less shielding, so it requires much more energy to remove. This is an inner shell electron. (1 mark)

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Question 6 (HL — 6 marks)

(a) Explain what is meant by the term ligand and give one example of a bidentate ligand. (2 marks)

(b) The complex $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is yellow but $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ is blue. Explain this difference. (2 marks)

(c) Explain why transition metals often show catalytic activity, using the Haber process as an example. (2 marks)

Markscheme:

(a) A ligand is a molecule or ion that can donate a lone pair of electrons to a central metal ion via a coordinate bond. (1 mark) Example of bidentate ligand: ethylenediamine (en) or oxalate ion ($\mathrm{C}_2\mathrm{O}_4^{2-}$). (1 mark)

(b) $\mathrm{NH}_3$ is a stronger-field ligand than $\mathrm{H}_2\mathrm{O}$ on the spectrochemical series, so it causes greater $d$-orbital splitting ($\Delta$) in $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$. (1 mark) A larger $\Delta$ means higher-energy photons are absorbed, so the complementary colour transmitted is different (yellow vs blue). (1 mark)

(c) Transition metals have variable oxidation states, allowing them to form intermediate compounds with reactants. This provides an alternative reaction pathway with a lower activation energy. (1 mark) In the Haber process, iron catalyses the reaction by adsorbing $\mathrm{N}_2$ and $\mathrm{H}_2$ onto its surface, weakening the $\mathrm{N}\equiv\mathrm{N}$ triple bond and facilitating the formation of $\mathrm{NH}_3$. (1 mark)

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Question 7 (HL — 4 marks)

The successive ionization energies of an element X are shown below (in kJ/mol):

$IE_1 = 577$, $IE_2 = 1816$, $IE_3 = 2744$, $IE_4 = 11577$, $IE_5 = 14842$, $IE_6 = 18376$

(a) Identify element X and explain your reasoning. (2 marks)

(b) Write the electron configuration of X$^{2+}$. (1 mark)

(c) Explain why $IE_4$ is significantly larger than $IE_3$. (1 mark)

Markscheme:

(a) The element is aluminium. (1 mark) There is a large jump between $IE_3$ and $IE_4$, indicating that the first three electrons are removed from the valence shell and the fourth electron is from an inner shell. This is consistent with Group 13, and aluminium is the element in Period 3, Group 13. (1 mark)

(b) $[\mathrm{Ne}]\, 3s^1$ (1 mark; removing two electrons from the $3s^2\, 3p^1$ configuration)

(c) $IE_4$ removes an electron from the $n = 2$ shell, which is closer to the nucleus and experiences much less shielding. The effective nuclear charge on inner-shell electrons is much higher. (1 mark)

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Question 8 (HL — 4 marks)

Calculate the wavelength of radiation emitted when an electron in a hydrogen atom transitions from $n = 5$ to $n = 2$. Identify the spectral series and the region of the electromagnetic spectrum.

Markscheme:

$$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{25}\right)$$ $$= 1.097 \times 10^7 \times \frac{21}{100} = 2.304 \times 10^6\mathrm{ m}^{-1}$$ $$\lambda = \frac{1}{2.304 \times 10^6} = 4.34 \times 10^{-7}\mathrm{ m} = 434\mathrm{ nm}$$

(2 marks for correct substitution and calculation)

This is part of the Balmer series (transitions to $n = 2$) and falls in the visible region of the electromagnetic spectrum (blue-violet). (2 marks)

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IB Exam Tip

When answering "explain" questions about periodic trends, always reference effective nuclear charge and shielding. The marking scheme expects these terms. A two-mark explanation requires the trend statement AND the reasoning.

Common Mistake

When writing electron configurations for transition metal ions, always remove electrons from the $ns$ orbital first (highest principal quantum number), NOT from the $(n-1)d$ orbital. So $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}]\, 3d^5$, not $[\mathrm{Ar}]\, 4s^2\, 3d^3$.

Common Mistake

Do not confuse atomic radius trends with ionic radius trends. When comparing ionic radii within an isoelectronic series, the ion with the largest nuclear charge has the smallest radius. For example, $\mathrm{Na}^+$ is smaller than $\mathrm{F}^-$ even though $\mathrm{Na}$ has a larger atomic radius than $\mathrm{F}$.

danger

Common Pitfalls

• Confusing first ionisation energy with electronegativity: First ionisation energy is the energy required to REMOVE the outermost electron from a gaseous atom. Electronegativity is the ability of an atom to ATTRACT electrons in a covalent bond. Both generally increase across a period, but they measure fundamentally different properties and have different periodic trends down a group.

• Misunderstanding why ionisation energy decreases down a group: Ionisation energy decreases down a group because the outermost electron is in a higher energy shell, FURTHER from the nucleus and more shielded by inner electrons. The increased distance and shielding outweigh the increased nuclear charge. Students often mention only one factor when both are needed.

• Confusing periodic trends across a period: Across a period, atomic radius DECREASES (increasing nuclear charge pulls electrons closer), first ionisation energy generally INCREASES, electronegativity INCREASES, and metallic character DECREASES. Students frequently get one or more of these trends backwards.

• Misidentifying exceptions in ionisation energy trends: The general increase in first ionisation energy across a period has dips at Group 13 (e.g., boron) and Group 16 (e.g., oxygen). Group 13 dips because the p1 electron is in a higher energy p-subshell. Group 16 dips because the p4 electron is paired with another electron in the same orbital, creating repulsion. These exceptions are frequently tested in IB exams.

Practice Problems

Question 1: Calculating Relative Atomic Mass

Naturally occurring boron consists of two isotopes: $\mathrm{B}$-10 ($19.9\%$ abundance, mass $10.01\mathrm{ u}$) and $\mathrm{B}$-11 ($80.1\%$ abundance, mass $11.01\mathrm{ u}$). Calculate the relative atomic mass of boron.

Answer

$A_r = (10.01 \times 0.199) + (11.01 \times 0.801) = 1.992 + 8.819 = 10.81$

The relative atomic mass of boron is $10.81\mathrm{ u}$.

Question 2: Electron Configuration and Quantum Numbers

(a) Write the electron configuration of $\mathrm{Cr}$ ($Z = 24$) using noble gas notation.

(b) State the four quantum numbers for the last electron added to chromium.

Answer

(a) Chromium is an exception to the Aufbau principle. A half-filled $d$-subshell is more stable:

$\mathrm{Cr}: [\mathrm{Ar}]\, 4s^1\, 3d^5$

(b) The last electron enters the $3d$ subshell:

• Principal quantum number: $n = 3$
• Azimuthal quantum number: $l = 2$ (for $d$-orbital)
• Magnetic quantum number: $m_l = +2$ (one of $-2, -1, 0, +1, +2$)
• Spin quantum number: $m_s = +\frac{1}{2}$ (Hund's rule: first five electrons have parallel spins)

Question 3: Periodic Trends

Explain why the first ionization energy of aluminium is lower than that of magnesium, but the first ionization energy of sulfur is lower than that of phosphorus.

Answer

Aluminium vs Magnesium: Mg has the electron configuration $[\mathrm{Ne}]\, 3s^2$ with a stable, filled $3s$ subshell. Al has $[\mathrm{Ne}]\, 3s^2\, 3p^1$. The $3p$ electron in Al is at a higher energy level than the $3s$ electrons of Mg and is partially shielded by the $3s$ electrons, so it requires less energy to remove.

Sulfur vs Phosphorus: P has the configuration $[\mathrm{Ne}]\, 3s^2\, 3p^3$ with a stable half-filled $3p$ subshell. S has $[\mathrm{Ne}]\, 3s^2\, 3p^4$, where the fourth $3p$ electron is paired with another electron in the same orbital. The paired electrons experience mutual repulsion, making the paired electron easier to remove.

Question 4: Isoelectronic Series

Arrange the following ions in order of increasing ionic radius and explain your reasoning: $\mathrm{O}^{2-}$, $\mathrm{F}^-$, $\mathrm{Na}^+$, $\mathrm{Mg}^{2+}$, $\mathrm{Al}^{3+}$.

Answer

All five species are isoelectronic with the neon configuration ($1s^2\, 2s^2\, 2p^6$, 10 electrons).

$\mathrm{Al}^{3+} \lt \mathrm{Mg}^{2+} \lt \mathrm{Na}^+ \lt \mathrm{F}^- \lt \mathrm{O}^{2-}$

All have the same number of electrons, but the nuclear charge increases from $\mathrm{O}$ ($Z = 8$) to $\mathrm{Al}$ ($Z = 13$). A higher nuclear charge pulls the electron cloud closer to the nucleus, resulting in a smaller ionic radius.

Question 5: Spectral Line Calculation

Calculate the wavelength of the photon emitted when an electron in a hydrogen atom transitions from $n = 4$ to $n = 2$. Use the Rydberg equation with $R_H = 1.097 \times 10^7\mathrm{ m}^{-1}$.

Answer

$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{16}\right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \left(\frac{4 - 1}{16}\right) = 1.097 \times 10^7 \times 0.1875 = 2.057 \times 10^6\mathrm{ m}^{-1}$

$\lambda = \frac{1}{2.057 \times 10^6} = 4.86 \times 10^{-7}\mathrm{ m} = 486\mathrm{ nm}$

This corresponds to the cyan line in the Balmer series (visible region).

For the A-Level treatment of this topic, see Atomic Structure & Periodicity.