paper no.13 :applications of molecular symmetry and module...

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PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 33: Symmetry and chemical bonding part-IIIHybridization contd.

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Subject Chemistry

Paper No and Title 13, Applications of molecular symmetry and group theory

Module No and Title 33, Symmetry and chemical bonding: Part-III (Hybridization)

Module Tag CHE_P13_M33

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TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3. Hybridization in ABn molecules 3.1 σ-Bonding hybridization in ABn molecules 3.2 π-Bonding hybridization in ABn molecules 4. Summary

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1. Learning Outcomes

After studying this module, you shall be able to:

• Know more about scheme of hybridization. • Learn hybridization for σ-bonding in ABn molecule. • Find hybridization in PF5 molecule. • Know the type of hybridization for π-bonding in BCl3 molecule.

2. Introduction

You have already learnt about hybridization and how to deal with it in polyatomic molecules. We have discussed few examples for explaining the procedure for hybridization and finding the atomic orbitals involved in hybridization. In this module we will continue with hybridization.

3. Hybridization in ABn molecules

Let us take some more examples to explain the role of symmetry and group theory in explaining the hybridization in polyatomic molecules. In this module we will take hybridization involving π-bonding.

3.1 σ-Bonding hybridization in ABn molecules: Let us take PF5 molecule and work out the hybridization scheme for it. The structure of PF5 and P-F bond vectors are shown in figure 1.

P

F

F

F

F

F

C2,σv

C3

r1 r2

r3

r4

r5

Figure 1:Structureof PF5 and P-F bond vectors

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Take these five P-F bond vectors as the basis for finding ΓP-F . The effect of symmetry operations of D3h point group on five P-F bond vectors are shown in figure 2.

Figure 2: Effect of symmetry operations of D3h point group on five P-F vectors

The results for these symmetry operations are summarized in table below:

D3h E 2C3 3C2 σh 2S3 3σv

Unshifted vectors r 5 2 1 3 0 3

χ contributed

multiplication factor

x1 x1 x1 x1 x1 x1

ΓP-F

5 2 1 3 0 3

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ΓP-F can be reduced to irreducible representations by using standard reduction formula. The result is

given as:

ΓP-‐F = 2A1+ A2”+ E’

Five equivalent hybrid orbitals at central atom A must have the symmetries of the type, 2A1’, A2” E’. This points that two orbitals must be of 2A1’ symmetry, one orbital of A2” symmetry and two orbitals of E’ symmetry .Inspecting the character table of D3h point group gives information about the symmetries of various atomic orbitals as : D3h E 2C3 (z) 3C'2 h (xy) 2S3 3 v linear functions,

rotations

quadratic

functions

A'1 +1 +1 +1 +1 +1 +1 - x2+y2, z2 totally

symmetric s orbital

A'2 +1 +1 -1 +1 +1 -1 Rz -

E' +2 -1 0 +2 -1 0 (x, y) or( px ,py) (x2-y2, xy)or d x2-y2 ,dxy

atomic orbitals

A''1 +1 +1 +1 -1 -1 -1 - -

A''2 +1 +1 -1 -1 -1 +1 Z or pz atomic

orbital

-

E'' +2 -1 0 -2 +1 0 (Rx, Ry) (xz, yz) or dxz,dyz atomic

orbitals

The atomic orbitals of P corresponding to symmetry 2A1’+ A2”+ E’and that of E” symmetry are

summarized as in tabular form:

A1’ E’

A2” E”

s dz2

(px,py) (dx2-y2,dxy)

pz (dxz,dyz)

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Atomic orbitals (dxz,dyz) of symmetry E” will not be considered for mixing in forming hybrid orbitals as these are not the basis for any of the irreducible representation in ΓP-F .In order to form five equivalent hybrid orbitals on P atom we have to have two atomic orbitals of A1’ symmetry, one orbital of A2” symmetry and two orbitals of E’ symmetry. Based on this information we can list the possible combinations of atomic orbitals belonging to 2A1’+ A2”+ E’ symmetry in table 1.

Table.1 Various possible combinations of atomic orbitals to give five equivalent hybrid orbitals for PF5 (D3h) Possible

ways

A1’ A2” E’

Hybridization

(i) s,(n+1)s pz (px,py) s2p3

(ii) s,(n+1)s pz ( dxy,dx2-y2) s2pd2

(iii) ndz2,(n+1) dz

2 pz (px,py) d2p3

(iv) ndz2,(n+1) dz

2 pz ( dxy,dx2-y2) d2pd2

(v) ns,ndz2 pz (px,py) sdp3

(vi) ns,ndz2 pz ( dxy,dx2-y2) sdpd2

For molecular structures with trigonal bipyramidal geometry (D3h point group, PF5) energy consideration make combinations (i) to (iv) of little importance as we use atomic orbitals from n and n+1 levels. Combination (v) is most probable and labeled as sdp3 .For molecules like MoCl5 both combination (v)and (vi ) are of equal importance. So for MoCl5 the hybrid orbital is: ψMo= a(sdp3) + b(d3sp) Where a ~ b i.e. (v) and (vi) combinations contribute equally to the hybrid orbitals of Mo. The d orbital in case of C4v is dx2-y2 in dsp3 hybridization, while in case of D3h the hybridization is also dsp3 but d orbital involved is dz

2. 3.2 π-Bonding hybridization in ABn molecules: In this section we will discuss the hybridization in ABn molecule involving π-bonding. For this let us take BCl3 molecule, which is trigonal planar molecule and belongs to D3h point group. We will follow the same procedure as discussed earlier in this module and

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earlier modules. Before discussing this let us define σ-bonding and π-bonding in a molecule. For this let us take the overlaps of two p-orbitals (i) side on overlap (ii) end on overlap. These overlaps are shown in figure.3

++ +

A A BB

π -bond

p type of AO on atoms A and B

nodal plane+- -

-

side on overlap

++ +

A B- -

end on on overlap+ - A B

σ− bond

Figure 3: Formation of σ- and π-bonds In ABn molecule where n=3 we take molecule BCl3 which belongs to D3h point group. We assume that boron has two available p-type atomic orbitals which are orthogonal to each other i.e. their nodal planes are mutually perpendicular (these may be 2px or 2py ) A.O on boron. If we want to form 2n number of hybrid orbitals then we must have 2n number of (p-type AO) on boron atom which match with 2n number of AOs of chlorine. Here we are concerned with LCAO of boron which will provide hybrid orbitals of right symmetry for matching with the atomic orbitals of chlorine. For getting irreducible representations we assign a pair of mutually perpendicular vectors to represent π-type AOs of chlorine instead on B atom. If we put all vectors on B atom it will be difficult to visualize the effect of symmetry operation on these .Each vector points towards the positive lobe of the orbital. In pair of vectors, one vector is perpendicular to molecular plane and second lies in the molecular plane but perpendicular to bond axis as shown in the figure 4. These vectors on chlorine will be used to develop the reducible representation which will be reduced to irreducible representations.

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Cl

Cl

Cl

r1

r2

r3

Fig.4 Orientation of Cl AO's available for π bonding

B

⊥

⊥

⊥

=

==

The ┴ and = shows perpendicular and parallel orbitals of π-bonding type. Six required hybrid orbitals,on boron atom ,can also be assigned six vectors and these vectors must match the orientation of six vectors on the chlorine atoms if π-bonds are to be formed between boron and chlorine. We will use the same procedure as described earlier to find ΓHybusing six vectors on boron atom and see the effect of symmetry operations of D3h point group on these vectors. This is difficult to visualize the effect of symmetry operations on these six vectors on B atom. So vectors on Cl will be considered, these will give the same results. It is clear if π-bond is to be formed six vectors on chlorine atoms must have same orientation as that of vectors on boron atom. It is easy to study the effect of symmetry operations on six separate vectors on chlorine atoms (i.e. two on each chlorine atom) rather than to study the effect of symmetry operations on six vectors present on boron atom (difficult to visualize). Another point which is to be noticed is that perpendicular vectors cannot be interchanged with vectors in plane by any of the symmetry operations of the D3h point group. Thus we have two distinct set of vectors which can form two distinct sets of reducible representation. One set of three perpendicular vectors and another set of three vectors in plane of the molecule and perpendicular to B-Cl bonds. Thus the problem can be solved easily by taking these two distinct and different set of basis vectors (three each). Thus ΓHyb can be written as: ΓHyb = ΓHyb(perpendicular )+ Γ Hyb(plane) If vector is shifted contribution to character is zero ie χ=0 and if vector is not shifted contribution is χ=+1, if vector is not shifted but direction of vector is reversed contribution to χ= -1. Following results are obtained by performing symmetry operations of D3h point group on these two sets of vectors. Here vectors represent as: perpendicular vectors represent py orbitals and planar vectors represent px orbitals. Effect of symmetry operations of D3h point group on perpendicular vectors is shown in figure 5. Similar effect of symmetry operations of D3h point group on planar vectors (px orbitas) can be worked out. Overall results of effects of symmetry operations are summarized in table2.

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Cl

Cl

Cl

B⊥

E

Cl

Cl

Cl No vector or py changed

Cl

Cl

ClC3or S3

All vectors py shifted

C2

C3

Cl

Cl

Cl

C2

Direction one vector (pyorbital here) reversed

σh

Cl

Cl

Cl

Direction of three vectors reverse

Cl

Cl

Cl

σv

One vector remains unshifted

Figure 5: Effect of symmetry operations of D3h point group on perpendicular vectors or py

Table 2: Effect of symmetry operations on vectors (px,py orbitals)

D3h E 2C3 3C2 σh 2S3 3σv

Unshifted vectors(┴) 3 0 -1 -3 0 1

ΓHyb(┴) 3 0 -1 -3 0 1

Unshifted vectors( In plane) 3 0 -1 3 0 -1

ΓHyb(plane) 3 0 -1 3 0 -1

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Therefore ΓHyb = ΓHyb(┴)+ ΓHyb(plane)and their

characters are = ( 3 0 -1 -3 0 1) and (3 0 -1 3 0 -1 ) 6 0 -2 0 0 0

Using standard reduction formula we can reduce these reducible representations as:

ΓHyb(┴) =A2”+E”

ΓHyb(plane)= A2’+E’

Symmetries of various atomic orbitals on boron can be obtained from D3h character table as done earlier and these are summarized in table 3:

Table 3: Symmetries of atomic orbitals on boron for π-bonding

Out of plane (┴) In plane

A2” E” A2’ E’

pz (dxz.dyz) None (px,py) (dx2-y2,dxy)

For formation of three perpendicular π-‐bonds between atomic orbitals on boron and atomic orbitals on each chlorine, the orbitals on boron must be of A2” and E” symmetry. The pz orbital has A2” symmetry and pair of (dxz, dyz) orbitals have E’’ symmetry.Therefore, for formation three π-‐bonds ⊥ to plane of the molecule one should have one atomic orbital (pz) having A2” +and a pair of (dxz, dyz) orbitals having E’’ i.e. pd2 type of hybridisation is there for ⊥ π-‐bonding in BCl3 . For formation of three π-‐bonds in plane of the molecules we must have atomic orbitals of boron with symmetries A2’ and E’. There is no atomic orbital on boron of A2’ symmetry but there are pair of orbitals (px, py) or (dx2-‐y2, dxy) which have right symmetry E’. Thus in the plane of molecule no three equivalent π-‐bonds involving (px, py) set or (dx2-‐y2, dxy) set are formed. It does not mean that no π-‐bonds are formed in the plane of the molecule. It only means that two π-‐bonds are formed in plane involving three chlorine atoms i.e. two π-‐bonds are shared equally between three chlorines in plane of BCl3 molecule.

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4. Summary

• More about scheme of hybridization has been discussed. • Hybridization for σ-bonding in ABn molecule i.e., hybridization in PF5

molecule has been discussed in detail. • The type of hybridization for π-bonding in BCl3 molecule involving

perpendicular py orbitas and in plane px orbitals vectors has been discussed in detail.

paper no.13 :applications of molecular symmetry and module...

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