Theoretical Considerations of the Energetics of Bond Formation Electrostatic attraction between atoms - the energetics of bond formation

We already know that Electropositive atoms form metallic bonds. Electronegative atoms form covalent bonds Electropositive and electronegative atoms form ionic bonds.

A chemical bond forms if, as a result, the energy of the bonded atoms is substantially lower than that of separate atoms so that the combination can be detected as a separate unit. If the lowest energy can be achieved by ion formation, then the bonding will be ionic. If the lowest energy can be achieved by electronsharing, then the bonding will be covalent. Hence, to come to any conclusion about the type of bonding present in a compound, we must consider the energy changes that accompany bond formation.

The Covalent bond What are the forces involved, and how do they change when two atoms come together? How does the energy change as a result? Let us consider two hydrogen atoms HA and HB coming towards each other. They will develop two types of forces: (a) Repulsive forces (i) repulsion between the electron of atom HA and the electron of atom HB (ii) repulsion between the nucleus of atom HA and the nucleus of atom HB.

(b) Attractive forces (i) attraction between electron of HA and the nucleus of HB (ii) attraction between electron of HB and the nucleus of HA. The net effect of the interplay of these forces changes as the two atoms come closer together. To begin with, the attractive forces dominate as HA and HB come closer to each other and increases till it reaches a maximum. Any further decrease in internuclear distance would lead to a decrease of attraction till the internuclear distance reaches a point where net attraction becomes zero beyond which a net repulsion is observed. When the two atoms are brought closer to each other under the influence of attraction the potential energy decreases. The potential energy reaches a minimum when the net attraction decrease to zero and the corresponding internuclear distance corresponds to the average bond length. (N.B. it is the average bond length since each bond keeps on increasing and decreasing in length due to bond vibration). The figure 1(a) shows the variations of the net interatomic force and figure 1(b) shows the potential energy variation with changing internuclear distance and are self explanatory. The energy difference between the energy of two 'free' hydrogen atoms and the potential energy minimum is known as the bond dissociation energy.

The Ionic Bond

The above description gives a general picture about the energetics of covalent bond formation where both atoms are electronegative and the resultant bonding involves a sharing of two electrons. If one atom is electropositive and the other electronegative, even then a potential energy minimum will be seen corresponding to covalent bond formation. Then why do we have the formation of an ionic bond insuch cases? If we look at the case of bonding between potassium (K) and fluorine (F), we can further elaborate on the puzzling situation. The figure 2 sums it up graphically. At large distance, the potential energy of a neutral (K + F) situation is much lower than the (K+ + F-) situations. This is because (ionisation energy of K - electron affinity of F) involves an increase in energy from a system containing isolated gaseous atoms of K and F. However, as the internuclear distance decreases, the coulombic attraction between K+ and F- gaseous ions starts increasing sharply. The potential energy curves for the ionic state then crosses the curve for the neutral state to reach a minimum value, which is much lower than that for the latter.

Fig. 2: The comparison of variation in potential energy of K and F in their neutral and ionic states with internuclear distance

Valence bond theory and molecular orbital theory - the two theories of bonding.

To understand the energetics of bonding on the basis of the quantum mechanical model of the atom, two different theories have been used using distinct approaches.

(a) The valence bond theory Here, to start with, we consider the orbitals of the bonding atoms as one electron wave functions and then go about combining them before calculating the electronic energy of the resultant 'bonds'. The essential feature in this approach is that to start with, we distinguish between electrons belonging to different atoms and correlate them accordingly.

According to quantum theory, the electrons in an atom or a molecule are indistinguishable since they constantly 'exchange' with each other. The valence bond theory takes care of it by using the concept of resonance where it implies that the actual "distribution" of electron is a 'resonance hybrid' of all the different possible "distributions" incorporating all possible types of correlations. The different "distributions" schemes are referred to as "limiting structures". Thus, for the H2 molecule we may consider four limiting structures corresponding to the correlations (i) Atom A - Electron A + Atom B - Electron B (covalent structure) (ii) Atom A - Electron B + Atom B - Electron A (covalent structure) (iii) Atom A - Electrons A and B (ionic structure) (iv) Atom B - Electrons A and B (ionic structure)

We have mentioned this, since the concept of resonance has been used extensively in explaining polar covalent bonds and bonding in molecules where the electrons are delocalized. Another special feature about the valence bond approach is that it implies an extra stabilization due to electron pairing. This is not a general principle.

(b) The molecular orbital theory Here we consider the orbitals as eigen states in atoms and combine them to form molecular orbitals before calculating the energies of the latter. Unlike atomic orbitals, the molecular orbitals belong to the molecule as a whole. After arranging the molecular orbitals in order of increasing energy, we apply Aufbau principle to fill them up and obtain a description of the electron distribution in the molecule.

(c) Theory and practice - the application of bonding theories for studying chemistry Both these theories have their advantages and disadvantages and some of the concepts we will be using are derived from either one or the other of the theories. Most concepts such as resonance, hybridization of atomic orbitals, orbital box diagrams for electronic configurations, number of bonds, descriptions of and bonds etc. are a direct consequence of the valence bond theory.

The molecular orbital theory on the other hand has given some approximately parallel concepts such as bond order instead of the number of bonds; and orbitals instead of and bonds; bonding, antibonding and nonbonding electrons instead of bond pairs and lone pairs; delocalised molecular orbitals instead of resonance, etc.

Representing Electron Distribution in Covalent Molecules The Lewis Octet Rule and the Lewis Dot Structure Lewis, in 1916, used his insight to explain the bonding in covalent molecules by identifying the essential features of a covalent bond. "A covalent bond is a pair of electrons shared between two atoms." He thought that, because the two electrons of a shared pair can attract the nuclei they lie between and would pull the two atoms together. The brilliance of Lewis lay in conjuring up this figure of 'two' electrons being shared on the basis of the knowledge about the number of valence electrons in an atom and about molecular formulas only. He had no idea about orbitals and the quantum mechanical model of the atom. Today, we know that in covalent bonding, two atoms share an orbital which because of Pauli's exclusion principle cannot have more than two electrons. Lewis had used an empirical rule called the octet rule to reach his conclusions. The Octet Rule and its Violations In electrovalent compounds, one atom loses electron whereas the other gains electrons until both the atoms have reached a noble gas configuration (a duplet in the valence shell for elements close to helium and an octet for all other elements). It was Lewis who suggested that atoms could also share electrons until they had reached a noble gas configuration and expressed this as the octet rule. Octet Rule: Atoms proceed as far as possible towards completing their octets by sharing electron pairs. (a) Duplet instead of octet When hydrogen forms a covalent bond, there is a duplet around the hydrogen atom. An octet is not possible here, since hydrogen has only one 1s orbital which cannot have more than two electrons.

The electron of hydrogen is represented by a dot. The valence electrons of Fluorine is represented by crosses 'X' (b) Expanded octet Elements belonging to the third and higher period have vacant low lying 'd' orbitals which can be used to accommodate electrons in excess of eight.

(c) Odd electron molecules In molecules such as NO2 where the total number of valence electrons is odd (N has 5 and two O have 12

total 17). Then at least one of the atoms will have only seven electrons. Note that none of the atoms can have nine electrons, i.e. more than an octet (unless in the case of expanded octets as mentioned above)

But not

(d) Incomplete octets When elements having less than four electrons form covalent bonds they do not have a complete octet. (The molecules they thus form are called electron deficient molecules.)

The Lewis dot structure and the information it conveys There is something incorrect in the way we have distinguished the electrons of different atoms by representing them either with dots or crosses. We know today, that electrons are indistinguishable and we cannot assign any one electron of a shared pair to a particular atom. Ideally, they should all be represented by dots (or crosses). This is precisely what we do in a Lewis structure of any molecule. In the process, we avoid assigning specific bonded electrons to specific atoms but convey most other essential information. (a) Connectivity The Lewis structure tells us which atoms in a molecule are connected with which other atom. e.g. from the Lewis structures of CO2 and N2O.

We can conclude that the connectivities are: For CO2 O - C - O and not C - O - O and for N2O N - N - O and not N - O - N. (b) Formal charges In the example of N2O above, for example, we have shown that one of the nitrogen atoms in the given structure carries a formal positive charge and the other a formal negative charge. (c) Number of bonding and nonbonding electron pairs

Electrons involved in bonding area called bonding electron pairs, where as electrons which are not participating in bonding are called non bonding electron pairs or lone pairs. e.g. the oxygen atom in H2O has two lone pairs and two bond pairs.

(d) Number of bonds The number of bonds between any two atoms is obviously the number of bonded pairs of electrons between them. With the additional information that if there are more than one bonds between two atoms, the one of them is a bond and the others are bonds, the number of each type of bonds in a molecule can be determined.. For example, one can tell from the Lewis structure of CO2 given above, that there are a total of four bonds in the molecule, of which two are bonds and two are bonds.

Writing the Lewis Structures of Molecules HOW TO WRITE LEWIS STRUCTURE OF MOLECULES A systematic step by step approach to write Lewis structures is as follows: Deciding on Connectivity 1. Structural isomers have different connectivities. To decide on the connectivities in such cases, we must have experimental data. For example, C2H6O may have two possible connectivities.

and The first is a liquid alcohol and the second a gaseous ether. 2. Some connectivities are ruled out on the basis of octet (or duplet)considerations. For example, H2O cannot have H-H-O as the connectivity, since it would mean that the middle hydrogen exceeds its duplet. 3. When more than one oxygen atoms are present in a molecule, we normally do not have O-O

connectivity unless and until it is peroxide (or for e.g. ozone, O3). Thus, CH3CO2H cannot have H- -C-O-O-H, as the connectivity. 4. For oxoacids, the connectivity should show as many O-H bonds as is the basicity of the acid. For example H3PO3 (a dibasic acid) should have the connectivity

5. In some cases however, one has to write down the complete Lewis structure, in order to determine whether it would lead to a stable structure. For example, if you start with N-O-N as the connectivity, the

final structure would be something like which of course is improbable. Using Curved Arrows 1. The use of curved arrows to show the movement of electrons is very common, specially in organic chemistry. A curved arrow means that the pair of electrons from, which it originates proposedly moves to

the position (on an atom or between two atoms) it ends, e.g. means, that the resultant

structure would be . Note that in the process, the pair of bonding electrons have moved from between C and O on to O as a lone pair. Note: It may also denote the movement of electrons from one molecule to another, but it is never used to denote the movement of electron deficient species towards electrons, e.g.

(right) This cannot be shown as

(wrong) 2. In order to show the movement of a single electron, instead of a pair we use a half arrow, e.g.

Electron Book Keeping There are two types of electron counting to be done while writing a Lewis structure. 1. To check whether an octet is complete, incomplete or exceeded. 2. To find out how many of the electrons in the octet of an atom belong to the atom. For this we have to follow two simple rules, which are as follows: a) Both the electrons in a lone pair of an atom belong to the atom. b) Only one of the electrons of a bonded pair of an atom, belongs to it, the other belongs to the atom bonded to it. By counting the number of electrons which belong to a bonded atom, formal charge can be assigned to it, if necessary. All that needs to be done is to compare this number with the number of valence electrons in the neutral free atom. i) if they are the same, there is no formal charge. ii) if the former is greater by a number x, there is a formal charge of x-. iii) if the former is less by a number x, there is a formal charge of x+. Let us take some examples of different structures of the molecule CH2N2.

Structure 1. Carbon has 1 lone pair and 3 bond pairs. Total electrons in octet = 1 2 + 3 2 = 8 octet complete Number of electrons belonging to C = 1 2 + 3 1 = 5 which is, 5 - 4 = 1 more than the valence electrons of C (4) the formal charge is 1-. Middle nitrogen has 4 bond pairs. Total electrons in octet = 4 2 = 8 octet complete Electrons belonging to N = 4 1 = 4 which is, 4 - 5 = -1 or 1 less than valence electrons of N (5) the formal charge is 1+.

:

The terminal nitrogen also has one lone pair and three bond pairs as with the carbon. The octet is complete and 5 electrons belong to this nitrogen. The number is equal to the number of valence electrons of nitrogen and so there will be no formal charges. The final assignment will be

Structure 2. All octets are complete. Terminal carbon owns 4 electrons and has no charge. Middle nitrogen owns 4 electrons and has 1+ charge. Terminal nitrogen owns 6 electrons and has 1- charge.

Final assignment is

Structure 3. Octet of terminal carbon is incomplete. Terminal carbon owns 3 electrons and has 1+ charge Middle nitrogen owns 5 electrons and has no charge Terminal nitrogen owns 6 electrons and has 1- charge.

Final assignment is Identifying Coordinate Covalent Bonds A coordinate covalent bond is, as the name suggests, a covalent bond. The only difference is, the bond was not formed by the two atoms contributing one electron each. Instead, one of the atoms has contributed both the electrons. While counting the electrons belonging to a bonded atom, however, the history of how the bond was formed does not matter. There the rules are simple, the two atoms equally share the two electrons of the bond. The donor atom in a coordinate bond ends up owning one less electron and the acceptor atom ends up owning one extra electron. Example 1. NH3 and BF3 combine to form a coordinate covalent bond, where N is the donor atom and B is the acceptor atom as shown below.

Thus N owns one electron less and has a positive charge and B owns one extra electron and has a negative charge. We may identify this situation as a coordinate bond between N and B and depict it as shown below.

Note that both these representations are equally valid. Example 2. HNO3 can be shown as

:

:

You should also note that, if the acceptor atom carried a positive charge, before coordinate bond formation, then in the final situation it should become neutral. The only clue for coordinate bond formation comes from the appearance of a positive charge on the donor atom. Example 3. Consider the reaction of NH3 with H+, where N is the donor atom and H+ is the acceptor atom as shown below.

Though all the four N-H bonds are equivalent (showing why we insist that a coordinate bond is not different from a covalent bond), the positive charge on N divulges the information that a coordinate bond has been formed. Some thumb rules, regarding when C, N and O carry a charge and when are they neutral in the bonded state.

Lone pairs Bond pairs

for neutral carbon for neutral nitrogen for neutral oxygen

none 1 2

4 3 2

for C+ for N+ for O+

none none 1

3 4 3

for C-

for N- for O-

1 2 3

3 2 1

C-, neutral N and O+ are isoelectronic, i.e they own the same number of electrons (which is 5) and they have identical bonding environment (1 lone pair and 3 bond pairs). Other similar examples are as follows: Neutral C and N+ No lone pair + 4 bond pairs Neutral N and O+ 1 lone pair + 3 bond pairs Neutral O and N- 2 lone pairs + 2 bond pairs Writing the complete Lewis structure and finding the steric number of an atom To write a complete Lewis structure, these five steps are followed: 1. Write down the connectivity. 2. Give each atom its due share of electrons. 3. Pair up all the electrons. 4. Check and correct octets. 5. Assign formal charges. The steric number (SN) of an atom is given by SN = number of lone pairs + number of bonds. If there is only one bond between two atoms, it is a bond. If there are multiple bonds between two atoms, then one of them is a bond and the others are bonds. Example 1. CO Step 1. (connectivity) C - O

Step 2. (give electrons) (carbon has 4 electrons and oxygen has 6)

Step 3. (pair up)

Step 4. (check and correct octet) Carbon has a sextet. To correct this, borrow a lone pair from neighbouring oxygen to form a new C - O bond as shown.

Step 5. (assign formal charges) Note that both C and O possess 5 electrons each and there is one electron extra in C and one less in O. Hence the final assignment is

Further, since both C and O have one lone pair and one bond each, the steric number is 1 + 1 = 2 in both the cases. Example 2. O3

Step 1. O - O - O

Step 2.

Step 3.

Step 4. Note that the central oxygen has more than an octet (10 electrons). Correct this by transferring a pair of its bonding electrons to the corresponding terminal oxygen as shown below.

Step 5. The central oxygen owns only 5 electrons. This implies that there is a positive charge on it. The right terminal oxygen owns 7 electrons. This implies that there is a negative charge on it. Final assignment

The Steric number of a) the left terminal oxygen is: 2 lone pairs + 1 = 3 b) the central oxygen is: 1 lone pair + 2 = 3 c) the right terminal oxygen is: 3 lone pairs + 1 = 4 Example 3. SO2 Here, since S and O belong to the same group and have the same number of valence electrons, one of the

possible structures would be

However, since S belongs to the 3rd period and can undergo valence shell expansion, the octet restriction need not be applied here and

is also a valid structure. Example 4. NO2

+ Step 1. O - N - O (not N - O - O, which implies a peroxide)

Step 2.

Since it is and not NO2, we must remove one electron either from N or from O.

Step 3.

Step 4. Correct the incomplete octet of the right terminal oxygen of the first case.

(which is the same situation as in the other case) Step 5. The central nitrogen owns only 4 electrons. This implies that there is a positive charge on it.

Final assignment Steric number for oxygen atoms is 2 lone pairs + 1 = 3 Steric number for nitrogen is zero lone pair + 2 = 2

The general formula for a open chain hydrocarbon with no bond (saturated hydrocarbon) is CnH2n + 2. The number of hydrogen atoms decreases in units of two for every - bond. Thus the general formula for an alkene is CnH2n and for an alkyne it is CnH2n-2. In case of cyclic compounds, each ring means a decrease of two hydrogen atoms from the formula of open chain saturated hydrocarbon) Thus an index is formulated called hydrogen deficiency index (HDI) of a hydrocarbon having formula CxHy as

HDI = where HDI = (number of bonds) + (number of rings) For example in C6H6 (benzene)

HDI = In benzene we have 3 bonds and 1 ring. In case hetero atoms are present, simply remove the atoms to get back to the parent hydrocarbon formula (i) Remove halogen and replace by a hydrogen atom. (ii) Remove oxygen of sulphur - no further adjustment is necessary. (iii) Remove nitrogen or phosphorus and remove a hydrogen atom along with it. (In the same manner if you decrease one carbon atom to step down in the homologous series, remove one CH2) For example,

Structural formula Molecular formula Formula after removal of hetero atom

C2H4O C2H4 C2H4

C2H3Cl C2H4

C2H4

C2H5N C2H4 C2H4

Molecules with more than One Lewis Structures: Resonance and Conjugation

In the previous section the structure of O3 was written as . It could also be written as

. The natural question that comes to our mind is, which one is correct? Ozone is not any unique case. The concept of resonance is a very important component of the valence bond theory of bonding and is actually related to the more general phenomenon of delocalization of electrons in a molecule.

The concept of resonance: Let us look at the actual electron distribution in ozone. Both the oxygen-oxygen bonds are exactly equivalent and each of the terminal oxygen atoms carries 1/2 unit of negative formal charge. Using the concept of resonance, therefore, each of the structures is a hypothetical limiting structure and the actual electron distribution is considered as a resonance hybrid of all possible limiting structures. Conventionally double headed arrows are used to show resonance involving limiting structures.

Thus for O3 we write

For SO2 we write , etc. When do we have to use the concept of resonance?

The simple answer is, whenever there is a multiple bond, resonance becomes important. For example,

Each description is used to explain the various aspects of reactivities of carbonyl compounds. A more significant use of resonance is to describe the long range delocalization of electrons (as we have observed in ozone) leading to an extra stability in molecules. Such a delocalization can take place when there is 'conjugation' in the molecule.

Molecules with more than One Lewis Structures: Resonance and Conjugation Identifying conjugated electrons in a molecule Conjugated electrons can get delocalised. Hence, such electrons must either be non-bonding electrons or loosely bound -bonding electrons. There are cases where even -bonding electrons participate in resonance - referred as hyperconjugation. For our convenience we will refer to such electrons as mobile electrons. In order to get delocalised, there should be some vacant or vacatable orbital in another part of the molecule where the mobile electrons may move to. The condition for conjugation is that there must be a gap of one bond between the mobile electrons, and the vacant and vacatable orbital. Example 1

Example 2

The resonance can be shown as

Example 3 If there is a gap of more than one bond, then there will be no conjugation.

The -electrons in the C = C cannot get delocalised to the C = O site. Example 4 If there are no gaps, then too there will be no delocalisation or conjugation.

has a vacatable orbital on the central carbon, if we write as follows:

But since there is no gap of any bond, we cannot write resonance structures (as suggested below). There is no conjugation possible here.

(No conjugation - no resonance) How significant is a resonance structure?

We may write two different limiting structures for

Which one of the limiting structures, I or II, is more significant? The general rule is that if a limiting structure denotes an unstable (hypothetical) molecule, it is not significant. On these grounds we may say that structure I showing a positive charge on an electronegative oxygen atom is not significant. Let us study some of the rules which will tell us how to distinguish between possible and impossible resonance structures, and also between those that are significant and there which are not. Applying rules of resonance 1. Possible and impossible structures a) The number of paired and unpaired electrons must be the same in all the limiting structures.

b) Resonance involves only the movement of electrons for a given nuclear arrangement, i.e., the relative position of the nuclei cannot differ from one limiting structure to another (distinction from tautomerism) c) Resonance implies the delocalization of electrons which is possible when the molecule is planar. For example, benzene is planar and there is electron delocalization because of resonance.

But there may be no resonance involved, though all the conditions required for conjugation may 'apparently' exist. For example:

does not show resonance since it is not planar, but tub shaped as shown below.

d) Structures should not imply the expansion of octets for elements of the second period (or duplets for

elements of the first period). For example, we have already seen that cannot be a contributing structure. 2. Important and unimportant structures The actual electron distribution is a resonance hybrid of that represented by the individual contributing structures. But all the structures are not equally important. The following set of rules help to choose the ones which are more important or significant for explaining the properties of molecules.

a) Contributing structures in which there is no change in the number of and -bonds respectively are called isovalent structures. If the numbers are different they are called heterovalent structures. Other factors being equal isovalent resonance structures are more significant than heterovalent structures. For example,

b) Generation of isolated charges in a structure reduces its significance. This is another reason why the second structure shown for butadiene above becomes so insignificant. Structures with more than two isolated charges have insignificant contribution. c) Heterovalent resonance may give rise to significant structures if it increases the negative charge (electron density) on the more electronegative atom. We have already seen an example of this in the example of resonance in but-2-enal. d) Structures where like charges are present close to each other are normally not significant. For example, in

The structure II has insignificant contribution because: i) There are more than two isolated charges ii) Like charges are crowded close to each other 3. Resonance energy The stabilization of a molecule because of resonance or delocalization is called resonance or delocalisation energy. This energy becomes large when: a) The contributing resonance structures are equivalent. For example, as in

.

b) Larger number of contributing structures of roughly comparable energies is possible. For example, ClO3

- has greater resonance energy than ClO2-

Three contributing structures

Two contributing structures c) There are (4n + 2) conjugated -electrons in a planar cyclic compound. Here n may have values 0,1,2,

etc. For n = 1, we have the well known case of benzene where the extra resonance stabilization is related to aromaticity. Aromaticity and aromatic stabilization may be present in compounds other than benzene and also for other values of n. Some other such examples of molecules and ions showing aromatic stabilization are given below.

Cyclooctatetraene is not aromatic because it has 8 electrons which does not fit the (4n + 2) rule.

The VSEPR Theory 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 LP For 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

2

Linear

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 = n Number of nearest neighbours of Cl = 3 = m VSEPR 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 = n Number of nearest neighbours of s = 2 = m VSEPR 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 bipyramid Shape 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

2 Linear 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,SO3

2-, XeO3

AB2E2

Angular or V-shaped

H2O,SCl2,OF2,XeO2

5 Trigonal bipyramid Bond 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

7 Pentagonal bipyramid Bond angles 90o, 180o and72o

AB7

Pentagonal bipyramid

IF7

AB6E

Distorted octahedral

XeF6

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.