Molecular geometry

Predicting The Geometry of Molecules

The ideal shapes of molecules, which are predicted based on electron pairs and lone pairs of electrons are mentioned in the table below:

The following procedure uses VSEPR theory to determine the geometry of the molecules:

1. Write the Lewis structure of the molecule or polyatomic ion.

2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.

3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral

4. Use the number of lone pairs to determine the molecular structure. If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position.

5. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.

For example, BCl₃ has three electron pairs and no lone pairs of electrons. Thus these three electron pairs will arrange themselves in a trigonal planar geometry as shown below. The bond angle between each B-Cl bond is 120^oC.

Related Topics:

• Vsepr Theory
• Molecular Orbital Theory
• Energy Level Diagram

Also Read:

Some Solved Examples

Example 1: In which of the following molecules/ions are all the bonds not equal?

1)SF₄

2)SiF₄
3)XeF₄
4)BF₄⁻

Solution

In SF₄ Hybridisation is sp³ d and the shape is see-saw but all bonds are not equal. The length of the Axial bond is longer than the equatorial bond.

But, in SiF_4, XeF₄ and BF₄⁻ all bonds are equal.

seesaw with the bond angles equal to 89⁰ and 177⁰ instead of the expected angles of 90⁰and 180⁰ respectively.

SiF₄: sp³ hybridisation, and tetrahedral geometry.

XeF₄ : Sp³d² the shape is square planar instead of octahedral due to the presence of two lone pairs of electrons on the Xe atom.

BF₄^- : sp³ hybridization and tetrahedral geometry

Example 2: The maximum number of 90⁰ angles between bond- pair, bond pair of electrons is observed in

1) dsp³ hybridisation
2) sp³d hybridisation
3) dsp² hybridisation
4) sp³d² hybridisation

Solution

Hybridization | Structure | Number of 90⁰ Angle

dsp³ square Pyramidal 5

sp³d Trigonal Bipyramidal 6

dsp² Square Planar 4

sp³d² Octahedral 12

dsp² hybridisation (four 90^oangles between bond pair and bond pair) & sp³d or dsp³ hybridisation
(six 90^∘ angles between bond pair and bond pair)

sp³d² hybridisation (twelve 90⁰ angle between bond pair and bond pair)

Example 3: In which of the following pairs the two species are not isostructural?

1)CO₃^2- and NO₃^-
2) PCl₄⁺and SiCl₄
3)PF₅ and BrF₅
4) AlF₆^3-and SF₆

Solution
PCl₄⁺ and SiCl₄ ⇒ both tetrahedral
PF₅⇒ trigonal bipyramidal
BrF₅⇒BrF₅ square pyramidal
AIF₆³⁻ AlF₆³⁻AlF₆^3-and SF₆ SF₆ both are octahedral
CO₃²⁻ CO₃^2- and NO₃^- NO₃⁻both are trigonal planar.

Hence, the answer is the option (3).

Example 4: The pair of species having identical shapes for molecules of both species is

1)CF₄,SF₄ CF₄,SF₄
2)XeF₂,CO₂ XeF₂,CO₂
3)BF₃,PCl₃ BF₃,PCl₃
4)PF₅,IF₅ PF₅,IF₅

Solution

Shapes of the molecule:

CF₄ Tetrahedral

SF₄ See saw-shaped

XeF₂ Linear

CO₂ Linear

BF₃ Triangular planar

PCl₃ Pyramidal

PF₅ Triangular bipyramidal

IF₅ Square pyramidal

Hence, the answer is the option (2).

Example 5: Which of the following statements is correct for ClF₃ ?

1)All bond lengths Cl-F are identical.

2)All bonds are identical

3)Equatorial bonds are larger than axial bonds.

4) Equatorial bonds are smaller than axial bonds.

Solution

ClF₃ has 5 electron groups/pairs around the central Cl atom ( Three bonds and two lone pairs). With 5 electron groups around the central atom, the molecule will adopt a trigonal bipyramidal shape.

F atoms occupy the axial positions and one F occupies the equatorial positions of trigonal bipyramidal arrangement. Usually, the bond length between the Cl - F axial is more than the Cl - F equatorial to reduce the repulsions between the lone pair - lone pair, bond pair - lone pair, and bond pair-bond pair repulsions, according to VSEPR theory.

Hence, the option number (4) is the correct option.

Example 6: Which among the following molecules is (a) involved in sp³d hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom ? [JEE Main 2025]

1) PF₅

2) XeF₄

3) SF₄

4) XeF₂

Solution:

(a) Hybridisation $={\mathrm{sp}}^3\mathrm{d}$
(b) All bonds are not identical
(c) No lone pair on central atom

(a) ${\mathrm{XeF}}_4:-{\mathrm{sp}}^3{\mathrm{d}}^2$ Hybridisation

(b) 2 lone pairs on central atom

(a) Hybridisation $={\mathrm{sp}}^3\mathrm{d}$
(b) All bonds are not identical
(c) 1 lone pair on central atom

(a) Hybridisation $={\mathrm{sp}}^3\mathrm{d}$
(b) All bonds are identical
(c) 3 lone pairs on the central atom

From all the given molecules, SF₄ molecule is involved in sp³d hybridization, has different bond lengths, and has a lone pair of electrons on the central atom

Hence, the correct answer is option (3).

Practice More Questions From the Link Given Below:

Summary

The common shapes are linear, bent, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. For example, CO₂ is linear, while NH₃ has a trigonal pyramidal geometry due to a lone pair at nitrogen. Molecular geometry largely determines such things as polarity, reactivity, physical state, color, magnetism, and biological activity.

Also Check:

• NCERT Exemplar Class 11th Chemistry Solutions
• NCERT Exemplar Class 12th Chemistry Solutions
• NCERT Exemplar Solutions for All Subjects

NCERT Chemistry Notes:

• NCERT Notes Class 11th Chemistry
• NCERT Notes Class 12th Chemistry
• NCERT Notes For All Subjects