Atomic Structure and Periodicity

Atomic structure and periodicity are like a lock and key, electron configuration determines where elements fit in the periodic table: Electron shell filling follows the Aufbau principle, explaining why the periodic table has its characteristic shape

Why it matters: Connecting atomic structure to periodicity enables prediction of chemical behavior across the entire table

The key insight: Electron shell filling follows the Aufbau principle, explaining why the periodic table has its characteristic shape

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

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 $\prescript{A}{Z}\mathrm{X}$.

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).

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.

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

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.

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

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.

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.

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

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

Reactivity increases down the group:

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.

Alkali metals burn in oxygen to form oxides:

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).

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

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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.

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 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.

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

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 $\mathrm{Cl}^-$.

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

These are vigorous, exothermic reactions forming ionic halides.

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

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).

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

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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.

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.

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.

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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.

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.

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.

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

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

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.

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

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.

::: :::caution $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{4+}$ 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.

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):

Homogeneous catalysis (catalyst in the same phase):

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

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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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.

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.

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

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:

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:

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

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

1. Ionization: Atoms or molecules are ionized ( 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.

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

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

• 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. ::: :::

Question 1: Calculating Relative Atomic Mass

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.

1. Writing half-equations without balancing charges or atoms. Always check electrons, hydrogen ions, and water molecules.

2. Forgetting to convert between units (e.g., $\text{cm}^3$ to $\text{dm}^3$) when calculating concentrations.

3. Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.

4. Confusing the terms ‘molar’ and ‘molecular’. Molar refers to per mole ($\text{mol}^{-1}$), while molecular refers to individual molecules.

• Stoichiometric Relationships — Mole concept and Avogadro’s number are foundational for understanding atomic structure calculations.
• Chemical Bonding — Electron configuration determines bonding behaviour, linking atomic structure to chemical properties.
• Periodicity — Trends in ionisation energy and electronegativity arise directly from the atomic structure covered here.
• Thermochemistry — Enthalpy changes in reactions depend on bond energies derived from atomic and molecular structure.