Mathematical Applications

1. Identify and Count Atoms: Identify each element based on the sphere's colour or size (e.g., black = Carbon, white = Hydrogen, blue = Nitrogen). Count the number of spheres to determine the molecular formula (e.g., 1 Nitrogen and 3 Hydrogens = $NH_3$).
2. Determine Connectivity: Look at the "sticks" (bonds) or where spheres touch (in space-filling models) to find the central atom.
3. Sketch the Displayed Formula:
   • Write the symbol for the central atom in the middle.
   • Arrange the symbols for the surrounding atoms around it.
   • Replace every 3D "stick" with a single solid 2D line ($-$) to represent a covalent bond.
4. Sketch the Dot-and-Cross Diagram:
   • Draw the outer electron shell of the central atom overlapping with the shells of the outer atoms.
   • Place a shared pair (one dot and one cross) in each overlap area for every bond identified in the 3D model.
   • Crucial Step: Add any lone pairs (non-bonding electrons) to the central atom. These are often not shown in 3D models but must be added to ensure the central atom has a full outer shell (or matches its Group number).

Worked Example: Sketching Ammonia ($NH_3$) from a Ball-and-Stick Model

• 3D Observation: A central blue sphere with 3 white spheres attached by 3 sticks in a pyramidal shape.
• 2D Displayed Formula: Draw '$N$' with 3 lines connecting to 3 '$H$' atoms.
• 2D Dot-and-Cross: Draw $N$ with 5 outer electrons total. 3 are shared with $H$ atoms (3 pairs of dot-and-cross). The remaining 2 must be drawn as a lone pair on the $N$ atom.

Comparing Chemical Models

Different models highlight distinct features, but every model has limitations. You must evaluate these explicitly in the exam.

Model Type	What it shows clearly	Limitations (Negative features)
Displayed formula (2D)	Connectivity of atoms and all bonds as solid lines.	Does not show 3D shape, bond angles, or relative sizes of atoms.
Dot-and-cross diagram (2D)	Outer-shell electron arrangement and the source of shared electrons.	Does not show the 3D physical arrangement or relative sizes.
Ball-and-stick model (3D)	3D geometry, exact bond angles, and spatial arrangement.	Misleadingly depicts bonds as physical sticks; shows non-existent empty gaps.
Space-filling model (3D)	Relative atomic radii and how tightly atoms are packed.	Internal bonds and the underlying structure are completely hidden.

Simple Molecular Geometries (3D Shapes)

• Methane ($CH_4$): Forms a tetrahedral shape with uniform bond angles of $109.5^\circ$.
• Ammonia ($NH_3$): Forms a pyramidal shape.
• Water ($H_2O$): Forms a V-shaped (or "bent") molecule.

Representing Giant Lattices

• A giant ionic lattice (e.g., $NaCl$) is a 3D repeating pattern of alternating ions.
• In $NaCl$, each ion has a 6:6 coordination: every $Na^+$ ion is surrounded by six $Cl^-$ ions, and vice versa.
• To sketch in 2D: Draw a cross-section showing alternating charges (e.g., $+ - + -$) arranged in a regular square grid.

Visualising Giant Covalent Structures

• Diamond: A 3D giant covalent lattice where every carbon atom is covalently bonded to four others in a rigid tetrahedral arrangement. This makes diamond exceptionally hard.
• Graphite: 2D layers of hexagonal rings stacked into a 3D structure. Each carbon is bonded to three others within its layer, leaving delocalised electrons.
• To sketch Graphite: Use solid lines for covalent bonds within layers and dashed lines for the weak intermolecular forces between layers.
• Fullerenes: 3D molecules like Buckminsterfullerene ($C_{60}$), which is a hollow sphere of 20 hexagons and 12 pentagons.

Worked Example: Deducing Empirical Formula from a Lattice

A 3D ball-and-stick model of a metallic halide shows a repeating unit. Counting the spheres reveals for every two metal ions ($M^{2+}$), there are four halide ions ($X^-$). Deduce the empirical formula.

1. Identify ratio: $M:X = 2:4$.
2. Simplify: $1:2$.
3. Formula: $MX_2$.

Worked Example: Calculating SA:V Ratio

A cubic nanoparticle has a side length of $20\text{ nm}$. Calculate its surface area to volume ratio.

1. Surface Area: $6 \times (20 \times 20) = 2400\text{ nm}^2$.
2. Volume: $20 \times 20 \times 20 = 8000\text{ nm}^3$.
3. Ratio: $2400 : 8000 = 0.3\text{ nm}^{-1}$.