Atomic Structure and Periodicity

Example — Chlorine

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

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)

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 $l$:

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.

Quantum Numbers

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

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.

Writing Electron Configurations

Full notation — write every subshell explicitly:

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

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:

Deviations from the general trend across a period:

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 $Z$:

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:

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.

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.

Metallic and Non-Metallic Character

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

5. Group 1: Alkali Metals

Physical Properties

Chemical Properties

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

Reaction with Water

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.

Oxides

Alkali metals burn in oxygen to form oxides:

Hydroxides

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

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:

Uses

6. Group 17: Halogens

Physical Properties

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.

Displacement Reactions

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

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

Reaction with Alkali Metals

These are vigorous, exothermic reactions forming ionic halides.

Halide Ion Tests

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

Uses

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).

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

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.

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

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

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.

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.

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.

Coordination Number

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

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.

Spectrochemical series (increasing $\Delta$):

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

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 $1 : 4$.

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.

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.

Determining Molecular Formula from Isotope Peaks

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

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:

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 ($Z = 13$):

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:

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 $n = 3 \to n = 2$:

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

Energy of a Photon

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

Converting to electron-volts:

Number of Spectral Lines

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

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

12. Exam Practice

Question 1 (SL — 4 marks)

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

(b) 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 occurring sample. (2 marks)

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

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

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)

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)

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)

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)

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)

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)

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:

(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)

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.