Chemistry - Atomic Structure and Bonding

The atomic number (proton number) $Z$ equals the number of protons in the nucleus.

The mass number $A$ equals the number of protons plus neutrons:

$A = Z + N$

Where $N$ is the number of neutrons.

Isotopes are atoms of the same element with the same number of protons but different numbers of Neutrons. They have identical chemical properties but different physical properties (e.g., different Masses).

Examples of isotopes:

• Carbon-12, Carbon-13, Carbon-14
• Hydrogen (protium), Deuterium, Tritium
• Chlorine-35, Chlorine-37

Chlorine has two isotopes: Cl-35 (75.8% abundance) and Cl-37 (24.2% abundance). Calculate the Relative atomic mass of chlorine.

$A_r = \frac{35 \times 75.8 + 37 \times 24.2}{100} = \frac{2653 + 895.4}{100} = \frac{3548.4}{100} = 35.48$

A mass spectrometer separates ions based on their mass-to-charge ratio ($m/z$). The stages are:

1. Ionisation: Atoms are ionised by electron bombardment to form positive ions
2. Acceleration: Ions are accelerated by an electric field
3. Deflection: Ions are deflected by a magnetic field (lighter ions are deflected more)
4. Detection: Ions hit a detector, producing a signal proportional to abundance

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

Each energy level contains subshells:

• $s$ subshell: holds up to 2 electrons
• $p$ subshell: holds up to 6 electrons
• $d$ subshell: holds up to 10 electrons
• $f$ subshell: holds up to 14 electrons

Electrons fill orbitals in order of increasing energy:

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

Each orbital can hold a maximum of 2 electrons with opposite spins.

When filling degenerate orbitals (orbitals of the same energy, such as the three $2p$ orbitals), Electrons occupy separate orbitals with parallel spins before pairing up.

Write the electron configuration of:

• Sodium ($Z = 11$): $1s^2\, 2s^2\, 2p^6\, 3s^1$
• Iron ($Z = 26$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6$
• Chlorine ($Z = 17$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5$

For transition metals, use the noble gas core notation:

• Iron: $[\mathrm{Ar}]\, 4s^2\, 3d^6$
• 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-full subshells are more stable)

• Periods: Horizontal rows (1 to 7). The period number equals the number of occupied electron shells.
• Groups: Vertical columns. Elements in the same group have the same number of valence electrons and similar chemical properties.

The atomic radius decreases across a period because the increasing nuclear charge pulls electrons Closer. It increases down a group because additional electron shells are added.

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

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

Trends in first ionisation energy:

• Increases across a period: Nuclear charge increases, electrons held more tightly
• Decreases down a group: Outer electrons are further from the nucleus and more shielded

Dips in ionisation energy occur at:

• Group 3 (e.g., Al): electron removed from $p$ subshell (higher energy than $s$)
• Group 6 (e.g., S): electron removed from a paired orbital (electron-electron repulsion)

Explain why the first ionisation energy of aluminium ($Z = 13$) is lower than that of magnesium ($Z = 12$).

Magnesium has electron configuration $1s^2\, 2s^2\, 2p^6\, 3s^2$. The electron is removed from the $3s$ subshell.

Aluminium has electron configuration $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^1$. The electron is removed From the $3p$ subshell.

The $3p$ subshell is at a slightly higher energy level than $3s$ So the $3p$ electron is less Tightly held and requires less energy to remove.

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

Ionic bonds form between metals (which lose electrons to form cations) and non-metals (which Gain electrons to form anions).

Ionic compounds form giant ionic lattices:

• Ions are arranged in a regular 3D pattern
• Each ion is surrounded by ions of opposite charge
• The electrostatic attraction between oppositely charged ions is the ionic bond
• There are no discrete molecules

Write the formula of magnesium oxide.

Magnesium is in Group 2: $\mathrm{Mg} \to \mathrm{Mg}^{2+} + 2e^-$

Oxygen is in Group 16: $\mathrm{O} + 2e^- \to \mathrm{O}^{2-}$

To balance charges: 1 $\mathrm{Mg}^{2+}$ ion balances 1 $\mathrm{O}^{2-}$ ion.

Formula: $\mathrm{MgO}$

A covalent bond is formed when two atoms share a pair of electrons. It occurs between non-metal Atoms.

Single bond: One shared pair of electrons, e.g., H-H

Double bond: Two shared pairs of electrons, e.g., O=O

Triple bond: Three shared pairs of electrons, e.g., N$\equiv$N

Dative (coordinate) bond: Both electrons in the shared pair come from the same atom, e.g., in The ammonium ion $\mathrm{NH}_4^+$

When two atoms with different electronegativities form a covalent bond, the bonding electrons are Pulled towards the more electronegative atom, creating a polar bond with a dipole.

• Non-polar covalent: $\Delta\mathrm{EN} \lt 0.5$ (e.g., H-H, Cl-Cl)
• Polar covalent: $0.5 \leqslant \Delta\mathrm{EN} \lt 1.7$ (e.g., H-Cl, H-O)
• Ionic: $\Delta\mathrm{EN} \geqslant 1.7$ (e.g., Na-Cl)

Use electronegativity values to predict the bond polarity of (a) $\mathrm{H_2O}$(b) $\mathrm{CCl_4}$ And (c) $\mathrm{KBr}$. Given: \mathrm{H} = 2.1$$\mathrm{O} = 3.5$$\mathrm{C} = 2.5$$\mathrm{Cl} = 3.0$$\mathrm{K} = 0.8$$\mathrm{Br} = 2.8.

The Valence Shell Electron Pair Repulsion theory predicts molecular shapes based on the idea that Electron pairs around a central atom repel each other and arrange themselves as far apart as Possible.

Predict the shape and bond angle of $\mathrm{NH}_3$.

Nitrogen has 5 valence electrons. Three are used in bonding with hydrogen, leaving one lone pair.

Total electron pairs = 4 (3 bonding + 1 lone pair)

The electron pair geometry is tetrahedral. With one lone pair, the molecular shape is trigonal Pyramidal.

The bond angle is approximately $107^\circ$ (less than $109.5^\circ$ due to lone pair repulsion).

Simple molecular (e.g., \mathrm{H}_2\mathrm{O}$$\mathrm{CO}_2$$\mathrm{I}_2):

• Low melting and boiling points (weak intermolecular forces between molecules)
• Do not conduct electricity
• gases or liquids at room temperature

Giant covalent (e.g., diamond, graphite, silicon dioxide):

• Very high melting and boiling points (strong covalent bonds throughout)
• Diamond: hard, insulator (all electrons in bonds)
• Graphite: soft (layers can slide), conducts electricity (delocalised electrons)

Explain why both diamond and $\mathrm{SiO_2}$ have very high melting points, but graphite has a lower (though still high) melting point and conducts electricity.

Intermolecular forces are weaker than intramolecular (covalent) bonds.

• Present between all molecules (including non-polar ones)
• Caused by instantaneous dipoles due to uneven electron distribution
• Strength increases with molecular size (more electrons) and shape (greater surface area contact)

Trend: larger molecules with more electrons have stronger van der Waals forces.

• Occur between polar molecules
• The positive end of one molecule attracts the negative end of another
• Stronger than van der Waals forces but weaker than hydrogen bonding

A special, strong type of dipole-dipole interaction that occurs when:

1. Hydrogen is covalently bonded to a highly electronegative atom (N, O, or F)
2. The hydrogen atom interacts with a lone pair on another N, O, or F atom

Conditions: H bonded to N, O, or F, and interacting with another N, O, or F.

Examples: \mathrm{H}_2\mathrm{O}$$\mathrm{NH}_3HF, DNA base pairing.

Explain why $\mathrm{H}_2\mathrm{O}$ has a higher boiling point than $\mathrm{H}_2\mathrm{S}$ Despite $\mathrm{H}_2\mathrm{S}$ having a larger molecular mass.

Both molecules have van der Waals forces, which are stronger for $\mathrm{H}_2\mathrm{S}$ (larger, More electrons).

However, $\mathrm{H}_2\mathrm{O}$ can form hydrogen bonds between molecules (H bonded to O), while $\mathrm{H}_2\mathrm{S}$ cannot (S is not electronegative enough).

Hydrogen bonding in $\mathrm{H}_2\mathrm{O}$ is much stronger than the van der Waals forces in $\mathrm{H}_2\mathrm{S}$Resulting in a higher boiling point for water.

In metallic bonding:

• Metal atoms lose their valence electrons to form positive ions (cations)
• The valence electrons are delocalised and form a “sea” of electrons
• The electrostatic attraction between the cations and the delocalised electrons is the metallic bond

An alloy is a mixture of two or more elements, at least one of which is a metal.

• Different-sized atoms disrupt the regular lattice
• Layers cannot slide as
• Alloys are harder and stronger than pure metals

Examples: steel (Fe + C), brass (Cu + Zn), bronze (Cu + Sn), solder (Sn + Pb)

Building blocks with rules: Atoms are like LEGO sets — protons determine the element, electrons determine the chemistry. The periodic table organizes elements by their electron configurations, revealing patterns in reactivity.

Why it matters: From semiconductors to pharmaceuticals, understanding atomic structure explains why elements behave the way they do and how to combine them.

The key insight: Ionisation energy increases across a period because nuclear charge increases while shielding stays roughly constant — electrons are held more tightly.

flowchart TD
    A[1_Atomic Structure And Bonding] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• For electron configuration questions, always use the correct order: $4s$ before $3d$ for filling, but $3d$ before $4s$ for ion formation.
• When explaining ionisation energy trends, mention nuclear charge, shielding, and atomic radius.
• In VSEPR questions, count total electron pairs (bonding + lone pairs) first, then determine the shape based on bonding pairs only.
• Remember that hydrogen bonding requires H bonded to N, O, or F specifically.
• When comparing boiling points, consider the type and strength of intermolecular forces, not just molecular mass.
• For ionic compounds, always state that they have high melting points due to strong electrostatic forces throughout the giant lattice.

Bond enthalpy (bond energy) is the average energy required to break one mole of a particular type of Bond in the gaseous state.

$\Delta H = \sum(\mathrm{Bonds broken}) - \sum(\mathrm{Bonds formed})$

• Bonds broken: endothermic (positive value)
• Bonds formed: exothermic (negative value)

Calculate the enthalpy change for the reaction: $\mathrm{H}_2(g) + \mathrm{Cl}_2(g) \to 2\mathrm{HCl}(g)$

Bonds broken:

• 1 $\times$ H-H = $436 \mathrm{ kJ/mol}$
• 1 $\times$ Cl-Cl = $243 \mathrm{ kJ/mol}$

Bonds formed:

• 2 $\times$ H-Cl = $2 \times 432 = 864 \mathrm{ kJ/mol}$

$\Delta H = (436 + 243) - 864 = 679 - 864 = -185 \mathrm{ kJ/mol}$

The reaction is exothermic.

Calculate the enthalpy of combustion of methane: $\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \to \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(g)$

Bonds broken:

• 4 $\times$ C-H = $4 \times 413 = 1652 \mathrm{ kJ/mol}$
• 2 $\times$ O=O = $2 \times 498 = 996 \mathrm{ kJ/mol}$

Bonds formed:

• 2 $\times$ C=O = $2 \times 745 = 1490 \mathrm{ kJ/mol}$
• 4 $\times$ O-H = $4 \times 463 = 1852 \mathrm{ kJ/mol}$

$\Delta H = (1652 + 996) - (1490 + 1852) = 2648 - 3342 = -694 \mathrm{ kJ/mol}$

The physical properties of ionic compounds can be explained by their giant ionic lattice structure:

High melting and boiling points:

The electrostatic attraction between oppositely charged ions acts throughout the entire lattice. A Large amount of energy is required to overcome these strong forces, resulting in high melting and Boiling points.

Electrical conductivity:

In the solid state, ions are held in fixed positions and cannot move, so ionic solids do not conduct Electricity. When molten or dissolved in water, ions are free to move and carry charge, allowing Conductivity.

Solubility in polar solvents:

Water molecules are polar and can attract ions from the lattice surface. The positive end of water Molecules (near $\mathrm{H}$) attracts anions, while the negative end (near $\mathrm{O}$) attracts Cations. If the hydration energy exceeds the lattice energy, the ionic compound dissolves.

Brittleness:

When a force is applied, layers of ions shift. Ions of the same charge come adjacent to each other, And the repulsive forces cause the crystal to fracture along a cleavage plane.

Lattice energy is the energy released when one mole of an ionic compound is formed from its gaseous Ions.

Factors affecting lattice energy:

• Ionic charge: Higher charge leads to stronger attraction and larger lattice energy
• Ionic radius: Smaller ions can get closer together, increasing lattice energy

Valence electrons are the electrons in the outermost shell of an atom. They determine the chemical Properties of an element.

Transition metals have the following characteristic properties:

• Variable oxidation states (e.g., Fe: +2 and +3; Mn: +2, +4, +7)
• Formation of coloured compounds (due to d-d electron transitions)
• Catalytic activity (e.g., Fe in Haber process, V$_2$O$_5$ in Contact process)
• Formation of complex ions (e.g., $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$)

Write the electron configuration of $\mathrm{Fe}^{3+}$.

Fe ($Z = 26$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6$

$\mathrm{Fe}^{3+}$: Remove 3 electrons. Since $4s$ electrons are lost before $3d$:

$\mathrm{Fe}^{3+}$: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^5$

Note that $\mathrm{Fe}^{3+}$ has a half-filled $3d$ subshell ($3d^5$), which contributes to its Relative stability compared to $\mathrm{Fe}^{2+}$ ($3d^6$).

Period 3 elements ($\mathrm{Na}$ to $\mathrm{Ar}$) show clear trends that are frequently examined:

Atomic radius decreases across the period because increasing nuclear charge pulls electrons closer.

• \mathrm{Na}$$\mathrm{Mg}$$\mathrm{Al}: Metallic bonding, increasing strength (more delocalised electrons)
• $\mathrm{Si}$: Giant covalent structure, very high melting point
• \mathrm{P}$$\mathrm{S}$$\mathrm{Cl}$$\mathrm{Ar}: Simple molecular, weak van der Waals forces

• \mathrm{Na}$$\mathrm{Mg}$$\mathrm{Al}: Good conductors (metallic bonding with delocalised electrons)
• $\mathrm{Si}$: Semiconductor (conductivity increases with temperature)
• \mathrm{P}$$\mathrm{S}$$\mathrm{Cl}$$\mathrm{Ar}: Non-conductors (no mobile charge carriers)

Write the electron configuration of $\mathrm{S^{2-}}$ and explain why it has the same configuration as argon.

Predict the shape and bond angle of $\mathrm{SF_4}$.

Explain why propanone ($\mathrm{CH_3COCH_3}$B.p. $56^\circ\mathrm{C}$) has a higher boiling point than propane ($\mathrm{CH_3CH_2CH_3}$B.p. $-42^\circ\mathrm{C}$), but a lower boiling point than propan-1-ol ($\mathrm{CH_3CH_2CH_2OH}$B.p. $97^\circ\mathrm{C}$).

Hybridisation is the concept of mixing atomic orbitals to form new hybrid orbitals that are Equivalent in energy and suitable for bonding.

Determine the hybridisation of the central atom in $\mathrm{SF}_4$.

Sulphur has 6 valence electrons. In $\mathrm{SF}_4$Four are used for bonding with fluorine, Leaving one lone pair. Total electron pairs = 5 (4 bonding + 1 lone pair).

Hybridisation: $sp^3d$ (trigonal bipyramidal electron pair geometry, with the lone pair in an Equatorial position, giving a “see-saw” molecular shape).

Molecular orbital theory describes bonding in terms of the combination of atomic orbitals to form Molecular orbitals that extend over the entire molecule.

Bonding orbitals: Lower in energy than the original atomic orbitals; stabilise the molecule.

Antibonding orbitals: Higher in energy than the original atomic orbitals; destabilise the Molecule.

Bond order:

$\mathrm{Bond order} = \frac{1}{2}(\mathrm{bonding electrons} - \mathrm{antibonding electrons})$

• Bond order = 1: single bond
• Bond order = 2: double bond
• Bond order = 1.5: intermediate (e.g., $\mathrm{O}_2^-$)
• Bond order = 0: no bond (molecule does not exist, e.g., $\mathrm{He}_2$)

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• Chemistry - Atomic Structure and Bonding