Molecular Structure Calculations Answers

 

2. H3PO2 clearly shows no significant contribution from d orbitals. For example, no phosphorus hybrid is listed with d orbital character greater than d0.09. However, SF4 does show considerable d character in the hybridization. The strong d-character is in part caused the presence of four very electronegative fluorine atoms. For the central sulfur sp3d hybridization is expected, with four equivalent bonds to the fluorines. Note that the bond orders are not equivalent. The S-F bonds in the trigonal plane, S1-F3 and S1-F5, are stronger with a bond order of 0.872 than the axial bonds with a bond order of 0.638. This difference is because the hybridization on sulfur is different for the two types of bonds. Hybridization in the trigonal plane for the bonding pairs is s0.39p3d0.46, or about half as much d-character as expected. Hybridization for the sulfur for the axial bonds is s0.32p3d1.84, or about twice as much d character as expected. The penalty for more d character is a weaker bond, because the d-orbitals are higher in energy. These hybridizations differ from the expected. However, the average over the four bonds is about right, sp3d. SF4 is unusual in its agreement with the full extent of d-orbital character predicted from hybridization theory. Questions 3 and 4 deal with examples from the phosphorus oxyacids, which are more typical.

The hybridization for the lone-pair on sulfur is a different story. The lone pair is expected to also be in a sp3d hybrid. The calculated hybridization is, however, sp0.38. In other words, the lone pair is securely nestled in a low lying sulfur s orbital, with a little p character mixed in. The large amount of s-character is possible because the hybridization of the bonding orbitals has less s character than a pure sp3d hybrid, that is s0.39p3d0.46 or s0.32p3d1.84.

 

3. Some pi bonding is indicated in H3PO2 for the P=O bond between P(atom 1) and O(atom3 ) by the bond order of 1.670. Notice that the pi-bonding does not use d-orbital character on the phosphorus, which is expected since phosphorus is "expanding" its octet. For example, no hybrid is listed with d orbital character greater than d0.09. Also note that no pi-orbitals are listed explicitly. See SO2 or SO3 for examples of "Best Lewis Structures" with pi-orbitals. The pi-bonding acts primarily through the p-orbitals on phosphorus overlapping with the p-orbitals on oxygen. The hybridization of the P1-O3 bond is sp2.11, which is close to the hybridization that is expected for pi-bonded 2nd period elements of sp2. For examples of pi-bonding in 2nd period elements see ethylene, CH2=CH2, or ozone, O3 .

 

4. The following summary is taken from the printouts:

Molecule

Bond Angle

Bond Order

(expected)

Hybridization

Electronegativity

c (S)=2.5

H2S

93

0.956 (1)

s0.52p3

2.1

SF2

101

0.688 (1)

s0.24p3

4.0

SCl2

105

0.877 (1)

s0.23p3

3.0

SO2

118

1.645 (1.5)

s0.68p3

3.5

SO3

120

1.785 (1.333)

sp1.95

3.5

SO42-

109.5

1.424 (1)

sp3

3.5

On changing the attached atom from H to F to Cl, the bond angle opens up suggesting that there may be more s character in the sulfur hybrid. However, the hybridization shows the opposite is true. In fact the sulfur is getting even closer to pure p3 hybridization. The reason for the decrease in s character is that the sulfur is bonded to a more electronegative atom than it is in H2S. Bonding to a more electronegative atom increases the p character on the less electronegative atom (see Bent's rule and Question 1). The bond angles open up simply to accommodate the larger atoms, F and Cl.

For SO2 and SO3 pi bonding is important. Therefore, the sulfur needs to donate a p orbital to the pi system, leaving only two p orbitals left for the sigma-bond framework. In both SO2 and SO3 there are three pairs of electrons around the central sulfur in the sigma framework. With only two p orbitals left over, the sulfur must use s character to accommodate these three pairs. A similar argument applies to SO42-, in that there are four pairs of electrons around the central sulfur and so four hybrids are needed. Since four hybrids are needed, full sp3 hybridization does a good job of describing the contributions to the bonds on the central sulfur.

Some pi bonding is indicated in SO42- by the bond order of 1.42. Notice that the pi-bonding does not use d-orbital character on the sulfur. Significant d-character is expected since sulfur is "expanding" its octet. However, just as in Question 3, no d-orbital character is shown in the hybrids. The pi-bonding acts primarily through the p-orbitals on sulfur overlapping with the p-orbitals on oxygen.