CDS Chemistry · Chemical Bonding

Molecular Shapes and Bond Counting

How hybridisation sets a molecule's shape, from linear HCN to trigonal bipyramidal PCl₅, and how to count sigma and pi bonds.

All Chemical Bonding notes

Why this matters

Three CDS questions, two of them in 2019 (I): a match of four molecules to their shapes, and the hybridisation of PCl₅ (HARD). The third counts the single bonds in cyclohexane.

Concept 1 of 2: Hybridisation and molecular shape

Electron pairs around a central atom push each other as far apart as possible. Count the bonding pairs and lone pairs around the centre, and the shape follows. Lone pairs take up room but are not seen in the shape, so ammonia is a pyramid, not a tetrahedron.

Definition

Count the electron groups (bonds + lone pairs) on the central atom:

2 groups — sp — linear: BeCl₂, CO₂, HCN.
3 groups — sp² — trigonal planar: BF₃, HCHO.
4 groups — sp³ — tetrahedral: CH₄, CH₃F. With one lone pair: trigonal pyramidal (NH₃). With two: bent (H₂O).
5 groups — sp³d (dsp³) — trigonal bipyramidal: PCl₅.
6 groups — sp³d² — octahedral: SF₆.

Molecule	Hybridisation	Shape
HCN	sp	Linear
HCHO	sp²	Trigonal planar
CH₃F	sp³	Tetrahedral
NH₃	sp³ (one lone pair)	Trigonal pyramidal
PCl₅	sp³d (dsp³)	Trigonal bipyramidalQ
SF₆	sp³d²	Octahedral

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — General Knowledge · Q51Moderate

Example 1 · Chemical Bonding · Bond Counting and Molecular Structure

Match List-I with List-II and select the correct answer using the code given below the Lists :

List-I (Compound/Molecule)	List-II (Shape of Molecule)
A. $\mathrm{CH_3F}$	1. Trigonal planar
B. $\mathrm{HCHO}$	2. Tetrahedral
C. $\mathrm{HCN}$	3. Trigonal pyramidal
D. $\mathrm{NH_3}$	4. Linear

Code :

Ammonia is a pyramid, not a tetrahedron

NH₃ has four electron groups (sp³), but one is a lone pair, so the atoms form a trigonal pyramid. CH₄ and CH₃F, with no lone pair, are tetrahedral.

Five groups need a d orbital

PCl₅'s five bonds need five hybrid orbitals: one s, three p and one d, so sp³d (dsp³). sp³ gives only four.

Drill 1 more on hybridisation and molecular shape

Concept 2 of 2: Counting sigma and pi bonds

Every bond between two atoms has exactly one sigma bond. A double bond adds one pi bond, a triple bond adds two. So count the bonds in the structure, then split each into its sigma and pi parts.

Definition

The count:

Single bond = 1 σ. Double bond = 1 σ + 1 π. Triple bond = 1 σ + 2 π.
A saturated compound (only single bonds) has no pi bonds; every bond is a single (sigma) bond.
Total bonds in a hydrocarbon = C–C bonds + C–H bonds.

Sigma and pi bonds

\text{single} = 1\sigma \qquad \text{double} = 1\sigma + 1\pi \qquad \text{triple} = 1\sigma + 2\pi

Worked example

How many sigma and pi bonds are there in ethene, C₂H₄ (CH₂=CH₂)?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (I) 2022 — General Knowledge · Q12Moderate

Example 2 · Chemical Bonding · Bond Counting and Molecular Structure

The number of saturated and unsaturated bonds in cyclohexane are :

Count the C–H bonds too

A ring of six carbons has 6 C–C bonds, but a hydrocarbon's total also includes every C–H bond. Leaving them out undercounts badly.

Saturated means zero pi bonds

Saturated compounds, cycloalkanes included, have only single bonds. A ring is not a double bond.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Counting sigma and pi bonds
Sigma and pi bonds
$\text{single} = 1\sigma \qquad \text{double} = 1\sigma + 1\pi \qquad \text{triple} = 1\sigma + 2\pi$

Reference tables (1)

Hybridisation and molecular shape6 rows

Molecule	Hybridisation	Shape
HCN	sp	Linear
HCHO	sp²	Trigonal planar
CH₃F	sp³	Tetrahedral
NH₃	sp³ (one lone pair)	Trigonal pyramidal
PCl₅	sp³d (dsp³)	Trigonal bipyramidalQ
SF₆	sp³d²	Octahedral

Watch out for (4)

Ammonia is a pyramid, not a tetrahedron
→ Hybridisation and molecular shape
Five groups need a d orbital
→ Hybridisation and molecular shape
Count the C–H bonds too
→ Counting sigma and pi bonds
Saturated means zero pi bonds
→ Counting sigma and pi bonds

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