Prediction of Shapes and Bond Angles The VSEPR or the valence shell electron pair repulsion theory was proposed initially by Sidgwick and Powell and was later developed by Gilespie.The electron pairs in the valence shell of a central atom repel to get as far away from each other to determine the bond angles and ultimately the shapes of molecules.Predicting shapes:Electrons around a central atom are two types.(a) Those involved in bonding with a neighbouring atom called bonding pair or BP.(b) Those not involved in bonding called lone pair or LPFor applying VSEPR theory, we donot consider the electrons involved in the formation of -bonds which are always present along with a -bond in a multiple bonding situation and donot move away from each other due to repulsion. The sum of these i.e. LP + BP = steric  number or SN.The steps involved in applying VSEPR theory to determine shapes of molecules are as follows. Step 1: Find out the steric number from the lewis structure.The steric number determines the basic arrangement of electron pairs as follows:

Steric number Arrangement of electron pairs Shape

2Linear

3 Trigonal planar

4 Tetrahedral

5 Trigonal bipyramid

6 Octahedral

7 Pentagonal bipyramid

 Step 2. Write down the VSEPR formula for the given molecule as ABmEn where m is the number of BP = number of connected neighbouring atoms (B) and n is the number of LP or non bonding electron pairs (E).Writing VSEPR formulas for molecules and identifying molecules having identical VSEPR formulas Write down the lewis structure. then number of LP = m and number of bonds or connected neighbours = n for the central atom.

Example 1 ClF3

Lewis structure Number of LP for Cl = 2 = nNumber of nearest neighbours of Cl = 3 = mVSEPR formula AB3E2

Note:m + 2n = 3 + 4 = 7 = number of valence electrons in Cl.

Example 2   SO2

Lewis structure Number of LP for s = 1 = nNumber of nearest neighbours of s = 2 = mVSEPR formula AB2E

Note:2m + 2n = 4 + 2 = 6 number of valence electrons in S.In the first case we have calculated the number of valence electrons as m + 2n but in the second case we have calculated it as 2m + 2n.This is because in ClF3 all the bonds are single bonds but in SO2 all the bonds are double bonds.If we remember that oxygen forms double bonds whereas halogens (e.g. F) forms single bonds we can identify different molecules having same VSEPR  formula.

 

The table above gives us an idea regarding the group number of the central atom corresponding to different VSEPR formulas and also tell us how the sets. (i) AB2, AB2E, AB2E2, AB2E3 (ii) AB3, AB3E, AB3E2 (iii) AB4, AB4E, AB4E2 (iv)AB5, AB5E and (v) AB6, AB6E will show variation of shapes within each set.

Step 3.Determine the shapes of the molecule with respect to ABm on the basis of the table. Since the lone pairs cannot be seen, the shape of molecule, will be different from the "arrangement of valence shell electrons", if lone pairs are present on the central atom.

Determining the shapes of molecules on the basis of VSEPR formula ABm En

Let us consider an example:XeF2 has VSEPR formula AB2 E3 with steric number 5. The arrangement of electrons will be trigonal bipyramid. But the shape will only be with respect to the bonded atoms since the lone pairs cannot be seen. The difference should become clear from the figure below.

Arrangement of electrons - trigonal bipyramidShape of molecule - linear

How do we know where to put the lone pairs?To have an answer we have to look at one of the important premises of the VSEPR theory regarding the relative magnitudes of repulsive forces due to LP and BP:'A lone pair (LP) repels stronger than a bond pair (BP)',This means LP - LP repulsion > LP - BP repulsion > BP - BP repulsion.We should however make a special note of the different cases of steric number 5, since the axial and the equatorial sites are not equivalent.

Note: In the cases of AB4E, AB3E2 and AB2E3, the lone pair is always assigned to the equatorial site.Hence, AB3E2 is T-shaped and not trigonal planar.No such confusion arises in the the case of AB4E2, since in the octahedral distribution, the axial and equatorial sites are equivalent. The two lone pairs should be trans to each other. Even if you put them trans to each other on two opposite equatorial sites the shape is still square planar

Steric number

Arrangement VSEPR formula

Shape Example

2Linear Bond angle 180o

AB2

Linear

CO2, NO2+, N2O, HgCl2,

OCS

3 Trigonal planar Bond angle 120o

AB3

Trigonal planar

BF3, NO3-, CO3

2-, SO3

AB2E

Bent

SO2, , NO2-

4 Tetrahedral Bond angle 109o 28'

AB4

Tetrahedral

CH4, CHCl3, NH4+,BH4

-, SO4

2-,PCl4+

AB3E

Trigonal pyramid

NH3,H3O+, PCl3,SO32-,

XeO3

AB2E2

Angular or V-shaped

H2O,SCl2,OF2,XeO2

5 Trigonal bipyramidBond angles 120o, 180o and 90o

AB5

Trigonal bipyramid

PCl5,PF3Cl2

AB4E

See-saw or distorted tetrahedral

SF4, IO2F2-

AB3E2

T-shaped

BrF3,ClF3

AB2E3

Linear

I3-,ICl2-, XeF2

6 Octahedral Bond angles 90o, 180o

AB6

Octahedral

SF6, PF6-

AB5E

Square pyramidal

BrF5, XeOF4

AB4E2

Square planar

ICl4+, XeF4

7Pentagonal bipyramidBond angles 90o, 180o and72o

AB7

Pentagonal bipyramid

IF7

AB6E

Distorted octahedral

XeF6

  Example 1.Using VSEPR theory assign the shapes of  the following molecules.(a) O3  (b) I3-  (c) NF3  (d) BF3 (e) ICl3(f) ICl2-  (g) ICl2+  (h) ICl4-  (i) ICl4+ (j) PF4

+

(k) OSF2 (l) O2SF2

Solution

Molecular formula

Lewis structure

VSEPR formula

Shape

(a) O3 AB2E Bent

(b) I3- AB2E3 Linear

(c) NF3 AB3E Trigonal pyramid

(d) BF3 AB3 Trigonal planar

(e) ICl3 AB3E2  T- shaped

(f) ICl2- AB2E3 Linear

(g) ICl2+ AB2E2 Angular or V-shaped

(h) ICl4+ AB4E2 Square planar

(i) ICl4- AB4E Distorted tetrahedral or see saw

(j) PF4+ AB4 Tetrahedral

(k) OSF2 AB3E Trigonal pyramid

(l) O2SF2 AB4 Tetrahedral

Predicting bond angles and explaining variations:It is clear that if the steric number of two molecules are the same, their bond angles should have similar values. For example, CH4, NH3 and H2O all have central atoms with SN = 4. The bond angles are as follows:H - C - H in CH4  109o28'  (AB4 molecule)H - N - H in NH3  107o  (AB3E  molecule)H - O - H in H2O  104.5o  (AB2E2 molecule)We can also explain the direction of variation using VSEPR theory (one being greater or less than the other). Lone pairs repel more than bond pairs. Hence, lone pairs will try to push the bond pairs together and reduce the angle between the bond pairs. The greater the number of lone pairs, the greater will be the reduction in bond angles. Thus we have the bond angles in the example above in the order.AB2E2 < AB3E < AB4

Note: The VSEPR theory cosiders only the number of BP and LP. It has no consideration for the size or electronegativity of the atoms concerned. Hence it cannot explain the difference in the bond angles of molecules having the same VSEPR formula.

Example:Explain the variation in bond angles for the following AB2E2 moleculesH2O (107o), H2S(92o), H2Te (90o).Answer: The actual explanation involves concepts of the valence bond theory and hybridisation along with ideas related to polarity of bonds and size of the central atoms.Since the electronegativities decrease as O > S >  Te. The polarity of bonds also

decrease as O - H > S - H > Te - H and the extent of  charge on the hydrogen

atom also decreases accordingly. Because of this  charge, the hydrogen atoms repel each other, against the direction of repulsion by lone pairs, and resists the decrease in bond angle in the case of H2O. This factor decreases as H2O > H2S > H2Te as the charge on the H atom decreases. Further the increase in size of the central atom from O to Te also 'increases' the distance between the H-atom. Hence the results.