VSEPR and HYBRID ORBITALS

Introduction

From a historical point of view the development of the localized bond and especially the octet structure with electron pairs in chemical bonds was of considerable importance. The obvious exception to the octet rule is hydrogen which requires two electrons to fill its shel1 but the majority of the representative elements have at least s and p orbitals available to form chemical bonds and thus account for the octet - that is two electrons either lone pairs or shared in each of the four atomic orbitals. Geometrical or symmetrical considerations may dictate that he atomic orbitals be "pure" s or p or they may be "mixed" or hybridized orbitals. Furthermore, for some compounds there may be less than eight electrons in the outermost shell (e.g. BF3) or, if there are d orbitals available, it may be possible to expand the valence shell to accommodate more than the classical Lewis octet. In this section use will briefly review the Lewis dot structure and the hybridization schemes and their relationship to the "valence shell electron pair repulsion" (VSEPR) idea. For most structures this should be a review of some of the ideas presented in freshman chemistry.

The VSEPR model, developed by Gillespie, is based on the idea that electron pairs i,1ll tend to move as far away from one another as possible, while still being attracted to a central nuclear charge. Of course, the latter may be shielded to some degree in inner-filled shells. If there are two pairs of electrons in the valence shell then they will tend to occupy regions with maximum charge density 180° from one another. Simple examples include BeF2 and BeH2. The molecules are linear and contain two pairs of bonding electrons and no non-bonding pairs (lone pairs) about the Be atom. In order to account for the bonding two sp hybrid orbitals are utilized which are made up of the atomic s and pz orbitals on Be (the z axis taken as the molecular axis). These orbitals contain pairs of electrons; one donated by the berillium and the other by the pendant atom. The px and py orbitals on Be remain unfilled.

If there are two pairs of electrons in the outermost shell, the molecular geometry will be trigonal planar and the bond angles will be 120 if the pendant atoms are identical. An example here is BCl3;. Here the three B-C1 bonds are made up of the three sp2 hybrid orbitals which contain the s, px and py orbit centered on the boron (the xy plane being the molecular plane) while the pz orbital is empty. If the molecule contains different atoms about the central atom, as in HBF2, then the molecular geometry will deviate somewhat from trigonal planar but this variation is slight.

For the case of four electron pairs in the outmost shell about a central atom, a tetrahedral geometry is predicted by the VSPR theory. The three classical examples, methane (CH4, ammonia, and water) illustrates the variety of molecular geometries (MG) that are associated with this electronic geometry (EG). In the case of methane both the MG and the EG are tetrahedral for ammonia, the EG is tetrahedral while the MG is trigonal pyramidal. In water the EG is again tetrahedral while the molecule is V-shaped. The localized bonding for these three molecules can be approximated by four sp³ hybrid orbitals made up of one s and three p orbitals. It is important to note that the number of atomic orbitals that go into a hybrid orbital scheme is equal to the number of resulting hybrid orbitals, whether or not the hybrid orbitals contain bonding or non-bonding electron pairs.

These rules apply to charge a species a w1ell as neutral molecules. For example, in the phosphate anion there are 8 electrons in the valence shell of phosphorous leading to a tetrahedral ES with three electron pairs bonded to the oxygen and one lone pair. This produces a trigonal pyramidal molecular geometry. For 5 electron pairs in the outmost shell two molecular geometries are possible. The first is the trigonal bipyramidal geometry as in PF5 . Here there are three equatorial and two axial fluorines. The expanded valence hell about the phosphorous contains 5 pairs of electrons and utilizes 5 dsp3 hybrid orbitals. The axial and equatorial orbitals are not equivalent. To a first approximation the axial P-F bonds are made up of two pd hybrid orbitals and the equatorial bonds made up of three sp2 hybrid orbitals.

The second molecular geometry for sp³ d electron pairs is the square pyramidal geometry which again utilizes orbitals but in the former case the d orbital was dz2 while here it is the dx2-y2 orbital. Some molecules are intermediate between these two geometries and others can interconvert from one form to another. This interconversion can be sufficiently rapid to obliterate the identity of a pendant atom as either axial or equatorial. An example here is PF5 in which the 5 fluorines appear identical on the NMR time scale.

For 6 electron pairs the geometry is octahedral and the bonding contains six equivalent sp3d2 hybrid orbitals. SF6 is a common example with all bonding pairs while BrF5 is a square pyramidal molecule with one lone pair.

Hybridization

Let us consider a linear X-Y-X molecule with electron pair bonds made up of an

S and Pz orbital centered on Y and appropriate orbitals on X. Now S and PZ

are two orbitals of the full orthonormal set available to Y. That is

< s/Pz > = 0

< s/s > = 1

I

< p/p > = 1

By symmetry the two X-Y bonds are to be equivalent and should be directed 1800 away from each other. In other words, 4f!e wish to construct two sp hybrid orbits. As a starting point let us take the appropriate hybrid orbitals as 1inear combinations of the s and PZ atomic orbitals. Since we start with two atomic orbitals we expect two hybrid orbitals

w1 = as + bPZ

w2 = cs + dPZ

where a, b, c and d are constants to be determined by the chemical and physical consideration of the problem:

1. they must be normalized

2. they must be orthogonal

3. they must contain equal amounts of s and p character

The first condition leads to:

< w1/w1 > = 1 = < (as + bpz)/ (as + bpz) >

= a2(1) + 2ab(0) + b2(1)

= a2 + b2 = 1 '(1)

Similarly:

w2/w2 > = 1

leads to:

c2 + d2 = 1 (2)

by orthogonality:

< w1/w2 > = 0

ac + bd = 0 j (3)

The condition that the amount of s characterizing the two wavefunctions be equal leads to:

a2 = c2 (4)

and finally the condition that the amount of llpz???? character in the two wavefunctions be equal leads to:

I

b2 = d2 (5)

Equations (1), (2), (4) and (5) lead to the result that the absolute value of all the coefficients is square root (SR) of 1/2. Equation (3) 's the one that determines the signs of the coefficients. There are several possibilities but the usual choice is to take the sign of the s orbital o be positive and then either b or d is negative to satisfy (3).

I

The final result is: W1 = SR(1/2)² + SR(1/2)P

W2 = SR(1/2)S - SR(1/2)Pz

2

~2 = 2 s - pz

The first wavefunction has the following interpretation. Take one half of an s orbital and add to one half of a Pz orbital. The result is a large positive loop pointing upward and a small residual negative loop pointing downward. It is the upward loop that forms the two electron bonds with X. The second wavefunction results from the addition of one half of an s orbital with a positive phase factor everywhere with one half of a pz orbital with the positive loop down and the negative loop up. The negative sign in front of pz changed the phase relationship. The result of the addition is a positive loop pointing downward. This second sp hybrid orbital has the correct phase relationship or forming the second electron pair bond. It should be emphasized that the proper combination of two orthonormal atomic wavefunctions on one atomic center can be constructed and results in two orthonormal hybridized orbitals which are equivalent in

their atomic orbit composition but are directly differently in space.

For a trigonal planar XY3 molecule with sp2 hybrid orbitals it is convenient to choose the two p orbitals to be px and pV and to choose one of the X-Y bonds to lie along the positive X axis. The first of the three hybrid orbitals is:

W1 = SR(1/3) + SR(2/3)Px

The interpretation of this wavefunction is to take one third of the atomic s orbital with a positive factor and acid i t to 2/3 of an atomic px orbital oriented in the usual way (with its positive loop along the positive x axis). The resultant hybrid orbital has a large positive loop alone the positive x axis.

The wavefunctions 2 and 3 make equal angles to both the x and y axes. Therefore the amount of px character and py character in each of these must be the same and the total amount of px and py character must equal 1.

Orthogonality requires that the sign of Py be different on W2 and W3.

W2 = SR(1/3)S + SR(1/6)Px + SR(1/2)Py

W3 = SR(1/3)S - SR(1/6)Px - SR(1/2)Py

W2 has a positive loop directed along a vector making an angle of 120 with the positive x axis. Although the analogy is not strictly correct, it may be helpful to visualize this result as a combination of the positive upward thrust of px, pulled a little to the left by the negative sign on the small contribution from px, all reinforced by the positive sign of s above the x axis. The positive sign of s below the s axis causes a cancellation of electron density in that region due to the interference effects with the negative sign of py and to a lesser extent px. In a similar way the constructive and destructive interference alone may be used to describe W3.

For the case of a sn3 hybrid orbital scheme it is convenient to orient the coordinate system at the center of a cube with x, y and z axes coming out of the face centers. If one of the y-x bonds is along! the (111) direction, that is the direction defined by doing one unit along x, one unit along y and one unit along z then the position of the other three y-x bonds are also defined: (-1, -1, 1), (1, -1, -1) and (1, -1, -1).