Chemistry - Atomic Structure and Periodic Table

Subatomic Particles

Particle	Symbol	Relative Mass	Relative Charge	Location
Proton	$p$	1	$+1$	Nucleus
Neutron	$n$	1	$0$	Nucleus
Electron	$e^-$	$1/1836$ (negligible)	$-1$	Electron shells

Atomic Number and Mass Number

• Atomic number $Z$: the number of protons in the nucleus. This uniquely identifies the element.
• Mass number $A$: the total number of protons and neutrons.

$A = Z + N$

where $N$ is the number of neutrons.

In a neutral atom, the number of electrons equals the number of protons ($Z$).

---

Isotopes

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

Key Properties

• Same chemical properties (same electron configuration)
• Different physical properties (different mass)
• Different nuclear stability

Relative Atomic Mass

The relative atomic mass $A_r$ is the weighted average of the isotopic masses, accounting for their natural abundances:

$A_r = \sum (\mathrm{isotope mass} \times \mathrm{fractional abundance})$

Worked Example 1

Chlorine has two isotopes: $\mathrm{^{35}Cl}$ (75.8% abundance) and $\mathrm{^{37}Cl}$ (24.2% abundance). Calculate the relative atomic mass of chlorine.

Solution

$A_r = 35 \times 0.758 + 37 \times 0.242 = 26.53 + 8.954 = 35.48$

---

Electron Arrangements and Electronic Configuration

Shells and Subshells

Electrons occupy shells ($n = 1, 2, 3, \ldots$), each with a maximum capacity of $2n^2$ electrons.

Shell ($n$)	Max electrons	Subshells
1	2	$1s$
2	8	$2s$, $2p$
3	18	$3s$, $3p$, $3d$
4	32	$4s$, $4p$, $4d$, $4f$

Order of Filling (Aufbau Principle)

Electrons fill orbitals in order of increasing energy:

$1s \lt 2s \lt 2p \lt 3s \lt 3p \lt 4s \lt 3d \lt 4p \lt 5s \lt 4d \lt 5p$

Pauli Exclusion Principle

Each orbital holds a maximum of 2 electrons with opposite spins.

Hund's Rule

When filling degenerate orbitals (e.g., the three $2p$ orbitals), electrons occupy separate orbitals with parallel spins before pairing.

Worked Example 2

Write the electron configurations for the following elements:

Solution

• Sodium ($Z = 11$): $1s^2\, 2s^2\, 2p^6\, 3s^1$ or $[\mathrm{Ne}]\, 3s^1$
• Potassium ($Z = 19$): $[\mathrm{Ar}]\, 4s^1$
• Iron ($Z = 26$): $[\mathrm{Ar}]\, 4s^2\, 3d^6$
• Copper ($Z = 29$): $[\mathrm{Ar}]\, 4s^1\, 3d^{10}$ (exception: full $d$ subshell is more stable)
• Chromium ($Z = 24$): $[\mathrm{Ar}]\, 4s^1\, 3d^5$ (exception: half-filled $d$ subshell)

Important: Removing Electrons

When forming cations, $4s$ electrons are removed before $3d$ electrons, because once the $3d$ subshell is occupied, it sits at a lower energy than $4s$.

• $\mathrm{Fe}^{2+}$: $[\mathrm{Ar}]\, 3d^6$ (remove two $4s$ electrons)
• $\mathrm{Fe}^{3+}$: $[\mathrm{Ar}]\, 3d^5$ (remove two $4s$ and one $3d$)

---

The Periodic Table

The periodic table arranges elements in order of increasing atomic number. Elements in the same group (column) have similar chemical properties because they have the same number of valence electrons. Elements in the same period (row) have the same number of electron shells.

---

Periodic Trends

Trends Across a Period (Left to Right)

Property	Trend	Reason
Atomic radius	Decreases	Increasing nuclear charge pulls electrons closer
First ionisation energy	Generally increases	Electrons held more tightly by greater nuclear charge
Electronegativity	Increases	Greater attraction for bonding electrons
Metallic character	Decreases	Atoms are less willing to lose electrons

Trends Down a Group

Property	Trend	Reason
Atomic radius	Increases	Additional electron shells
First ionisation energy	Decreases	Outer electrons further from nucleus, more shielding
Electronegativity	Decreases	Less attraction for bonding electrons
Metallic character	Increases	Easier to lose outer electrons

---

Ionisation Energy

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms:

$\mathrm{X}(g) \to \mathrm{X}^+(g) + e^-$

Successive Ionisation Energies

Each subsequent ionisation energy is larger because the electron is removed from an increasingly positive ion. A large jump in successive ionisation energies indicates the removal of an electron from a new inner shell.

Dips in Ionisation Energy Across a Period

Dip	Elements	Reason
Group 2 to 13	e.g. Mg to Al	$p$ electron is easier to remove than $s$ electron (higher energy subshell)
Group 15 to 16	e.g. P to S	Paired $p$ electron experiences electron-electron repulsion

Worked Example 3

The first five ionisation energies of an element are: 578, 1817, 2745, 11577, and 14842 kJ/mol. Identify the element and its group.

Solution

The large jump between the 3rd and 4th ionisation energies indicates the first three electrons are valence electrons and the fourth is from an inner shell. The element has three valence electrons, placing it in Group 13. With a first ionisation energy of 578 kJ/mol, this is aluminium.

---

Electronegativity

Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond.

• Fluorine is the most electronegative element (Pauling scale: 4.0)
• Electronegativity increases across a period and decreases down a group
• The difference in electronegativity ($\Delta\mathrm{EN}$) between bonded atoms determines bond character:
  • $\Delta\mathrm{EN} \lt 0.5$: non-polar covalent
  • $0.5 \leqslant \Delta\mathrm{EN} \lt 1.7$: polar covalent
  • $\Delta\mathrm{EN} \geqslant 1.7$: ionic

---

Group 1: Alkali Metals

Property	Trend down group	Explanation
Melting/boiling point	Decreases	Weaker metallic bonding as atomic radius increases
Density	Generally increases	Larger atoms, but structural effects dominate
Reactivity	Increases	Easier to lose the outer electron
Ionisation energy	Decreases	Outer electron further from nucleus
Flame test colour	Li: crimson	Na: yellow, K: lilac, Rb: red, Cs: blue

Common Reactions

Reaction with water (increasingly vigorous down the group):

$2\mathrm{Na} + 2\mathrm{H_2O} \to 2\mathrm{NaOH} + \mathrm{H_2}$

Reaction with oxygen:

$4\mathrm{Li} + \mathrm{O_2} \to 2\mathrm{Li_2O}$

$2\mathrm{Na} + \mathrm{O_2} \to \mathrm{Na_2O_2}$ (sodium peroxide)

---

Group 2: Alkaline Earth Metals

Property	Trend down group	Explanation
Melting/boiling point	Generally decreases	Weaker metallic bonding
Reactivity	Increases	Easier to lose two outer electrons
Solubility of hydroxides	Increases	Lattice energy decreases faster than hydration energy
Solubility of sulphates	Decreases	Lattice energy decreases slower than hydration energy

Common Reactions

Reaction with water:

$\mathrm{Ca} + 2\mathrm{H_2O} \to \mathrm{Ca(OH)_2} + \mathrm{H_2}$

Reaction with dilute acid:

$\mathrm{Mg} + 2\mathrm{HCl} \to \mathrm{MgCl_2} + \mathrm{H_2}$

Thermal Decomposition of Group 2 Compounds

The thermal stability of Group 2 nitrates and carbonates increases down the group:

$\mathrm{MgCO_3} \xrightarrow{\Delta} \mathrm{MgO} + \mathrm{CO_2}$

$\mathrm{CaCO_3} \xrightarrow{\Delta} \mathrm{CaO} + \mathrm{CO_2}$

$\mathrm{Mg(NO_3)_2} \xrightarrow{\Delta} \mathrm{MgO} + 2\mathrm{NO_2} + \tfrac{1}{2}\mathrm{O_2}$

---

Group 17: Halogens

Property	Trend down group	Explanation
Physical state	Gas to solid	Increasing van der Waals forces
Melting/boiling point	Increases	Stronger van der Waals forces
Reactivity	Decreases	Harder to gain an electron (decreased electronegativity)
Electronegativity	Decreases	Larger atomic radius, more shielding
Displacement ability	Decreases	$\mathrm{Cl_2}$ displaces $\mathrm{Br_2}$, not vice versa

Displacement Reactions

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

$\mathrm{Cl_2} + 2\mathrm{KBr} \to 2\mathrm{KCl} + \mathrm{Br_2}$

$\mathrm{Br_2} + 2\mathrm{KI} \to 2\mathrm{KBr} + \mathrm{I_2}$

$\mathrm{I_2}$ cannot displace $\mathrm{Br_2}$ or $\mathrm{Cl_2}$.

Halogen Reactions with Sodium Hydroxide

Cold, dilute $\mathrm{NaOH}$:

$\mathrm{Cl_2} + 2\mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{NaClO} + \mathrm{H_2O}$

This disproportionation reaction produces sodium chloride and sodium chlorate(I).

Worked Example 4

Explain why bromine is a liquid at room temperature while chlorine is a gas.

Solution

Bromine atoms are larger than chlorine atoms and have more electrons. This results in stronger instantaneous dipole-induced dipole (London dispersion) forces between bromine molecules. These stronger intermolecular forces require more energy to overcome, giving bromine a higher boiling point ($59^\circ\mathrm{C}$) compared to chlorine ($-34^\circ\mathrm{C}$).

---

Group 18: Noble Gases

Noble gases are inert because they have a full outer shell of electrons ($ns^2\, np^6$, except helium which has $1s^2$). They have very high ionisation energies and very low electronegativities.

Property	Trend down group	Explanation
Boiling point	Increases	Stronger van der Waals forces
Density	Increases	Larger atomic mass
First ionisation energy	Decreases	Outer electrons further from nucleus

Uses

Noble Gas	Use	Reason
Helium	Balloons, airships	Low density, non-flammable
Neon	Advertising signs	Emits red-orange light when ionised
Argon	Welding atmosphere, light bulbs	Unreactive, prevents oxidation
Krypton	High-performance light bulbs	Higher density, reduces evaporation

---

Transition Metals (Basic)

Transition metals are d-block elements that form at least one stable ion with an incomplete d subshell. Key properties:

• Variable oxidation states (e.g., $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$)
• Formation of coloured compounds (due to d-d electron transitions)
• Catalytic activity (e.g., iron in the Haber process, nickel in hydrogenation)
• Formation of complex ions

---

Common Pitfalls

• Writing $4s^2\, 3d^6$ instead of $3d^6\, 4s^2$ when writing the full configuration. The $4s$ fills before $3d$, so it is written first. However, $4s$ electrons are removed first when forming cations.
• Confusing atomic number with mass number. Atomic number $Z$ counts protons; mass number $A$ counts protons plus neutrons.
• Assuming that electronegativity differences cleanly divide bonds into ionic and covalent. The boundary at $\Delta\mathrm{EN} = 1.7$ is approximate.
• Forgetting that the first ionisation energy of an element is the energy to remove the outermost electron, not any electron.
• Confusing Group 2 hydroxide solubility trends (increases down group) with Group 2 sulphate solubility trends (decreases down group).

---

Summary Table

Topic	Key Point
Atomic number $Z$	Number of protons
Mass number $A$	$Z + N$
Isotopes	Same $Z$, different $N$
Aufbau principle	Fill orbitals in order of increasing energy
Pauli exclusion	Max 2 electrons per orbital
Hund's rule	Fill degenerate orbitals singly first
Ionisation energy trend	Increases across period, decreases down group
Electronegativity trend	Increases across period, decreases down group
Group 1 trend	Reactivity increases down group
Group 17 trend	Reactivity decreases down group

---

Problem Set

Problem 1: The first four ionisation energies of an element are 738, 1451, 7733, and 10540 kJ/mol. To which group does this element belong?

If you get this wrong, revise: Successive Ionisation Energies

Solution

The large jump between the 2nd and 3rd ionisation energies indicates the first two electrons are valence electrons. The element has two valence electrons, placing it in Group 2.

Problem 2: Write the electron configuration of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$. Which ion is relatively more stable and why?

If you get this wrong, revise: Electron Configuration — Removing Electrons

Solution

Fe ($Z = 26$): $[\mathrm{Ar}]\, 4s^2\, 3d^6$

$\mathrm{Fe}^{2+}$: $[\mathrm{Ar}]\, 3d^6$ (remove two $4s$ electrons)

$\mathrm{Fe}^{3+}$: $[\mathrm{Ar}]\, 3d^5$ (remove two $4s$ and one $3d$)

$\mathrm{Fe}^{3+}$ is relatively stable due to its half-filled $3d^5$ configuration.

Problem 3: Neon has three naturally occurring isotopes: $\mathrm{^{20}Ne}$ (90.48%), $\mathrm{^{21}Ne}$ (0.27%), and $\mathrm{^{22}Ne}$ (9.25%). Calculate the relative atomic mass.

If you get this wrong, revise: Isotopes and Relative Atomic Mass

Solution

$A_r = 20 \times 0.9048 + 21 \times 0.0027 + 22 \times 0.0925 = 18.096 + 0.0567 + 2.035 = 20.19$

Problem 4: Explain why the first ionisation energy of sodium is much lower than that of neon.

If you get this wrong, revise: Ionisation Energy and Electron Configuration

Solution

Sodium ($Z = 11$) has the configuration $[\mathrm{Ne}]\, 3s^1$. The single $3s$ electron is in a new shell, further from the nucleus and well shielded by the inner 10 electrons. It is relatively easy to remove.

Neon ($Z = 10$) has a stable noble gas configuration ($1s^2\, 2s^2\, 2p^6$). Removing an electron requires breaking this stable full outer shell, requiring much more energy.

Problem 5: Predict the products when chlorine gas is bubbled into a solution of potassium iodide.

If you get this wrong, revise: Group 17 — Displacement Reactions

Solution

Chlorine is more reactive than iodine, so it displaces iodine:

$\mathrm{Cl_2} + 2\mathrm{KI} \to 2\mathrm{KCl} + \mathrm{I_2}$

The solution turns brown as iodine is liberated. Adding starch indicator produces a blue-black colour.

Problem 6: State and explain the trend in reactivity of the Group 1 metals with water.

If you get this wrong, revise: Group 1 — Properties and Trends

Solution

Reactivity increases down Group 1. As you descend, the atomic radius increases and the outer electron is further from the nucleus with more electron shielding. This means the outer electron is less strongly attracted to the nucleus and is more easily lost, resulting in more vigorous reactions with water.

Problem 7: Explain why the thermal stability of Group 2 carbonates increases down the group.

If you get this wrong, revise: Group 2 — Thermal Decomposition

Solution

As you descend Group 2, the ionic radius of the metal cation increases. The larger cation has a lower charge density and polarises the carbonate anion less. The $\mathrm{CO_3^{2-}}$ ion is therefore less distorted and more thermally stable, requiring more energy to decompose. So $\mathrm{MgCO_3}$ decomposes readily on heating while $\mathrm{BaCO_3}$ is much more resistant to thermal decomposition.

Problem 8: Explain why the solubility of Group 2 hydroxides increases down the group while the solubility of Group 2 sulphates decreases.

If you get this wrong, revise: Group 2 — Solubility Trends

Solution

For hydroxides, as you descend Group 2, the lattice energy decreases (larger ions) but the hydration energy decreases more slowly. Since the hydration energy becomes more significant relative to lattice energy, dissolution becomes more favourable, so solubility increases.

For sulphates, the opposite occurs: the lattice energy (which is large due to the doubly-charged $\mathrm{SO_4^{2-}}$ ion) decreases more slowly than the hydration energy as ionic size increases. This makes dissolution less favourable down the group, so solubility decreases.

Problem 9: The first seven ionisation energies of an element are: 789, 1577, 3232, 4356, 16091, 19784, and 23793 kJ/mol. Identify the group of this element.

If you get this wrong, revise: Successive Ionisation Energies

Solution

Looking at the jumps: 789 to 1577 (small), 1577 to 3232 (moderate), 3232 to 4356 (moderate), 4356 to 16091 (very large).

The large jump between the 4th and 5th ionisation energies means the first four electrons are valence electrons. The element has four valence electrons, placing it in Group 14.

Problem 10: Use electronegativity values to classify the bond in hydrogen chloride ($\mathrm{HCl}$). (EN of H $= 2.1$, EN of Cl $= 3.0$)

If you get this wrong, revise: Electronegativity

Solution

$\Delta\mathrm{EN} = 3.0 - 2.1 = 0.9$

Since $0.5 \leqslant \Delta\mathrm{EN} \lt 1.7$, the bond in $\mathrm{HCl}$ is polar covalent.

The shared electron pair is displaced towards the more electronegative chlorine atom, creating a dipole with a partial negative charge on Cl and a partial positive charge on H.