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CHEMISTRY

 

MODULE 1

 

1.3    Shapes of Molecules

Ionic bonds formed by electrostatic attraction between ions are non-directional in space. When covalent bonds are formed between atoms they occur by the overlapping of atomic orbitals. Some of these orbitals point in particular directions in space (p, d and f). It therefore follows that the bonds formed will have preferred directions in space and the molecules will have definite shapes.

 

Explanation in terms of electron pair repulsion theory for the shapes of molecules containing up to four pairs of electrons around the central atom such as BeCl2, BF3, CH4, NH3, H2O, and CO2. (Questions will not be set on hybridisation of orbitals). The departure of the bond angles in NH3 and H2O from the predicted tetrahedral, explained in terms of the increasing repulsion between bonding pair-bonding pair, lone pair-bonding pair and lone pair-lone pair electrons.

When you have finished this section you should be able to:

 

·         Apply the electron-pair-repulsion theory to molecules involving two, three and four pairs of bonding electrons.

·         Extend the electron-pair-repulsion theory to include non-bonding pairs (lone pairs) of electrons.

·         Use the following terms correctly to describe the shapes of molecules - linear, bent or V-shaped, trigonal planar, tetrahedral, trigonal pyramidal, trigonal bipyramidal, octahedral.

 

 

The shapes of covalent molecules

A covalent molecule will have a shape which is determined by the angles between the bonds joining the atoms together.  A simple theory to account for the shapes of molecules was put forward by Sidgwick and Powell in 1940.  It is called the VSEPR  (Valence Shell Electron Pair Repulsion) theory.  Note that the term valence shell is often used for what we call the outer-shell electrons.

The basic idea of the theory is that charge clouds (i.e. electron pairs) in the outer –shell of an atom in a molecule will arrange themselves to stay as far away from each other as possible. This will minimise the total energy of electrostatic repulsion between them.

With the help of this theory it is possible to predict

(i)      The general geometric shape of a molecule.

(ii)     The departure of interbond angles from the values for regular

          geometric shapes.

 

 

Linear molecules

Exercise 1

(a)         How many pairs of bonding electrons are there in a molecule of beryllium hydride, BeH2?

 

(b)    If the electron pairs repel each other so that they occupy regions of space as far apart as possible, what shape must the molecules have?

 

 

Molecules formed from the bonding of two atoms such as HCl, Cl2, HI and Br2 must be linear as it will always be possible to connect the nuclei of the two atoms by a straight line.

Molecules of the type

                                                A         B          A

 

having just two electron pairs around the central atom (an electron deficient compound) will also be linear. A linear arrangement of atoms ( with a bond angle of 180o ) puts the two electron clouds as far apart as possible. e.g. Cl-Be-Cl

 

You can apply the same principle to molecules containing multiple bonds. Electron pairs in multiple bonds are assumed to occupy the position of one electron pair in a single bond.

 

 

Exercise 2

What shape would you expect for the following molecules?

Explain your answer.

(a) CO2     (b) HCN

 


Trigonal (triangular) planar molecules

 

Exercise 3

(a)     How many pairs of bonding electrons are there in a molecule of boron trifluoride, BF3?

 

(b)    By considering repulsion of electron pairs, what shape do you expect for BF3?

 

 

 

When there are three pairs of electrons around the central

Atom (or their equivalent), the bonds lie in the same plane at an angle of 120o.  The arrangement is described as trigonal planar.

                                 A

                       

 

                                                B                      A

 


                       

                                  A

 

Exercise 4   

The molecule of methanal (formaldehyde) is trigonal planar, and the molecule of ethene is based on a trigonal planar arrangement around each carbon atom, as shown below.

 

                                               H                                                      H                  H

                                       120o       C=O                                                    C=C

                                               H                                                      H                  H

                                                methanal                                                   ethene

 

Explain why the bonds around each carbon atom are in a trigonal planar arrangement.

 

 


Tetrahedral molecules

Both the shapes we have considered so far are planar and are therefore easy to draw on paper.  For this reason, dot-and-cross diagrams for linear and trigonal planar molecules can correspond also to their actual shapes.  This is not the case for tetrahedral molecules, which are three-dimensional, as you can see in the next exercise.

 

 

Exercise 5

(a)         How many pairs of bonding electrons are there in a molecule of methane, CH4?

 

(b)         By considering repulsion of electron pairs, what shape do you expect for the CH4 molecule?

 


(c)         By considering bond angles, show that electron pair repulsion would not give the square planar shape of the dot-and-cross diagram below.

(d)    State whether you expect the following molecules and ions to have an identical shape or a very similar shape to the CH4 molecule.  Give reasons.

 

(i) NH4+     (ii) SiCl4

 

 

When there is a complete octet in the valence shell with four electron pairs around the central atom the molecule adopts a tetrahedral structure with bond angles of 109.5o.

The 3-dimensional tetrahedral shape can be represented as shown :

                                                            B

 

 


                                           109½ o

                                                            A

                                                                              B

                                          B

                                                                           B

It is most important for you to be able to visualise tetrahedral bonding and to be able to draw it adequately, because it forms the basis of so many molecular shapes, especially in organic chemistry.

Molecules with lone pairs

So far we have considered only molecules where all the outer shell electrons are used in bonding. However, many molecules have some electrons , called non-bonding pairs or lone pairs, which are not shared between two atoms. These electrons also affect the shape of the molecule by electron pair repulsion.

 

 

 

Exercise 6

(a)         Draw a dot-and-cross  diagram of the ammonia molecule, and count the number of pairs of electrons surrounding the nitrogen atom.

(b)         In which directions (from the N atom) would you expect to find the electron pairs (or regions of greatest charge density)?

(c)         What shape is outlined by the four atoms in the molecule?

 

 

 


The tetrahedron is not quite regular, because the four electron pairs are not identical. The lone pair, since it is not shared, remains closer to the N atom than the bonding pairs. The repulsion between non-bonding pairs and bonding pairs is greater, which pushes the hydrogen atoms closer together. This gives a slightly smaller H-N-H bond angle than the tetrahedral angle of 109½ o.

 

 


The shape of the ammonia molecule is described as trigonal pyramidal.

When describing the shape only the positions of the bonded atoms are considered (even thought the non-bonded pairs help to determine the overall shape).

 

 

 

The strength of the repulsion between electron pairs decreases in the order:

lone-pair / lone-pair  > bonded-pair / lone-pair  > bonded-pair / bonded-pair

        strongest                            strong                           least strong        

                                                                              

 


Two non-bonding pairs

 

Exercise 7

(a)         Draw a dot-and-cross diagram of the water molecule.

(b)         Draw the shape of a water molecule, showing how it fits into a tetrahedron.

(c)         Use your answer to the previous exercise to predict approximately the H-O-H bond angle.

 

The shape of the water molecule is described as bent or V-shaped.

 

SUMMARY

 

 

 

 

 

Molecule

Electron pairs around central atom

 

 

 

Number of lone pairs

 

 

Dot-and-cross diagram

 

 

 

 

Shape

 

 

 

 

Bond angle

BeCl2

 

2

 

0

 

 

 

Linear

 

BF3

 

3

 

0

 

 

 

 

Trigonal planar

 

 

CH4

 

 

 

 

4

 

 

0

 

 

 

Tetrahedral

 

NH3

 

4

 

0

 

 

 

 

Trigonal pyramidal

 

H2O

 

4

 

2

 

 

 

 

Bent or V-shaped

 

CO2

4

(two bonding regions)

 

0

 

 

 

 

Linear