VSEPR and Shapes of Molecules: PECTAA Class 11 Chemistry, Punjab Board

CH4, NH3 and H2O each have four electron pairs around the central atom. Yet one is a tetrahedron, one is a pyramid and one is bent. The difference is the lone pairs. Learn to count them and every shape question becomes routine.

Chapter 3 of the PECTAA Chemistry 11 book is Chemical Bonding. Section 3.8 covers the valence shell electron pair repulsion (VSEPR) model and how it predicts the shapes of molecules. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

The idea. The book credits the model to Sidgwick and Powell (1940). Electron pairs around a central atom push each other away, so they settle as far apart as possible. That arrangement fixes the geometry.

The book's main points, in short.

1. Both lone pairs and bond pairs decide the geometry.
2. Pairs sit at maximum distance from each other.
3. A lone pair takes more space than a bond pair, because only one nucleus holds it. Repulsion order: lp–lp > lp–bp > bp–bp.
4. A double or triple bond counts as one region, even though it holds more charge.
5. Highly electronegative atoms on the central atom weaken bond pair repulsions, while the lone pair repels more strongly.

Two different words. The electron pair geometry comes from the total number of pairs (bond + lone). The shape describes only the positions of the atoms, so lone pairs are left out.

Counting pairs (the book's NH3 method).

• Valence electrons on N = 5. Electrons shared by three H atoms = 3. Total = 8 electrons = 4 pairs.
• Bond pairs = 3, so lone pairs = 4 − 3 = 1.

The main cases.

Key points

• Lone pairs squeeze angles. 109.5° (CH4) falls to 107° (NH3, one lone pair) and 104.5° (H2O, two lone pairs).
• Electronegative neighbours squeeze further. In NF3 the book gives 102.5°, smaller than in NH3.
• Multiple bonds count once. SO3 has three S=O regions and is trigonal planar. SO2 has two S=O regions and one lone pair, so it is bent.

Worked examples

Example 1. Shape of H2O.

1. O has 6 valence electrons. Two H atoms share 2. Total 8 electrons = 4 pairs.
2. Bond pairs = 2, so lone pairs = 2.
3. Four pairs give a tetrahedral electron pair geometry. Leaving out the two lone pairs, the shape is angular.
4. Two lone pairs push hardest, so the angle drops to 104.5°.

Example 2. Why I3− is linear.

1. The central I atom has five electron pairs: two bond pairs and three lone pairs.
2. Five pairs give a trigonal bipyramidal arrangement. The three lone pairs take the triangle in the middle.
3. The two outer I atoms sit above and below, 180° apart. So the ion is linear.

Example 3. Why XeF4 is square planar.

1. Xe has six pairs: four bond pairs and two lone pairs.
2. Six pairs give an octahedral arrangement. The two lone pairs sit opposite each other, 180° apart.
3. The four F atoms lie in one square, with 90° angles.

Common mistakes in the exam

1. Mixing geometry and shape. H2O has a tetrahedral electron pair geometry but an angular shape.
2. Forgetting lone pairs. NH3 is not trigonal planar. Its lone pair makes it a pyramid.
3. Counting a double bond as two regions. SO2 has three regions, not five.
4. Writing the repulsion order backwards. Lone pair–lone pair is the strongest, bond pair–bond pair the weakest.
5. Calling I3− bent or XeF4 tetrahedral. Lone pairs placed to minimise repulsion give linear I3− and square planar XeF4.

Exam technique

• Write the four steps: choose the central atom, count all pairs, name the electron pair geometry, then give the shape without lone pairs.
• Show the pair count for N or O in one line, as in the NH3 method above.
• Quote the angle with the reason (lone pair repulsion) for full marks.
• Check your board's notice for the topics in your paper.

Practice questions

1. Predict the shape of SiH4 and its bond angle. Answer: Four bond pairs, no lone pair: tetrahedral, 109.5°.
2. How many lone pairs are on P in PH3? Answer: P has 5 valence electrons, H atoms add 3, so 4 pairs. Three are bond pairs, so one lone pair. The shape is trigonal pyramidal.
3. Predict the shape of H2Se. Answer: Se has 6 valence electrons, two H add 2, so 4 pairs with 2 lone pairs. The shape is angular.
4. Why is the angle in NF3 smaller than in NH3? Answer: The F atoms are highly electronegative. N–F bond pairs lie closer to F and repel each other less, while the lone pair on N repels more strongly.
5. Name the shape of BeCl2 and its angle. Answer: Linear, 180°.

Ready to practise this? Continue in 1st Year Chemistry.