2.7.4: Expanded Octets and Molecular Orbitals (2023)

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In organic chemistry the octet rule is closer to a law;do not make more than four bonds to carbon! However, once you start dealing with atoms in the third row of the periodic table, the octet rule becomes a bit more of a guideline. Consider the common anion, hexfluorophospate, PF₆^-(Figure \(\PageIndex{1}\)), in which we draw the phosphorous atom with six bonds giving it an apparently electron count of 12.

Figure \(\PageIndex{1}\).The structure of the hexafluorophosphate anion.

The bonding in these types of systems is a bit more involved and tofully appreciate it we must consider the molecular orbital diagram for such amolecule. This will be done using the same techniques as were used in the previous section, but some adjustments will need to be made as we progress from the Lewis structure to the molecular orbital diagram.

Step 1 - Drawing the structure and determining the hybridization of the central atom

Exercise \(\PageIndex{1}\)

Using VSEPR, draw the three-dimensional structure of sulfur tetrafluoride(SF₄).

Answer

Sulfur has 6 valence electrons and fluorine has 7 giving a total of 34 electrons in this molecule.

To start, you would draw bonds between sulfur and each of the fluorine atoms which would use 8 of the 34 electrons.

As fluorine is more electronegative, the next step would be to give each fluorine 3 lone pair in order to fill their octets. This will use 26 of the remaining electrons. At this point you might be tempted to stop as the sulfur atom and all of the fluorine atoms have an octet.

The problem is that this structure only accounts for 32 of our 34 electrons. As fluorine isan element in the 2^ndrow, we will not add the missing electrons to any of the fluorine atoms. That leaves sulfur as the only option. Checking the formal charge on each of the atoms supports this placement.

Formal Charge = number of valence electrons - (2 x number of lone pair + number of bonds)

FC_F= 7 - (2(3) + 1) = 0

FC_S = 6 -(2(1) + 4) = 0

This makesthe sulfur AX₄E₁so the structure is going to be based on AX₅E₀which is trigonal bipyramidal.

Step 2 - Determine the point group of the molecule and assign the generator functions

Exercise \(\PageIndex{3}\)

What is the point group of sulfur hexafluoride?

Answer

SF₄ does not have an infinite rotation axis.

Thereone rotation axis passing through the lone pair which isC₂.

There are not 6 C₅ axes.

There are not 3 C₄ axes.

There are not 4 C₃ axes.

The are no perpendicular C₂axes.

There is not an S₄axis.

There is not a perpendicular mirror plane.

There are however two mirror planes.

The point group of SF₄is C_2v.

Once again we need the character table for the point group of SF₄.Shown below are two different representations for the character table for the C_2vpoint group.The first is the full character table. Although we ignored it in the methane example, this does include the d-orbital needed for hybridization. In theabbreviated character table the d-orbital has just been incorporated.

C_2v	E	C₂	σ_v (xz)	σ_v (yz)	Linear Functions	Quadratic Functions
A₁	+1	+1	+1	+1	z	x², y², z²
A₂	+1	+1	-1	-1	R_z	xy
B₁	+1	-1	+1	-1	x, R_y	xz
B₂	+1	-1	-1	+1	z, R_x	yz

Table \(\PageIndex{1}\): Full character table for the C_2vpoint group.

C_2v	Orbitals
A₁	s, p_z, d_z²
A₂
B₁	p_x
B₂	p_y

Table \(\PageIndex{2}\): Abbreviatedcharacter table for the C_2vpoint group.

Exercise \(\PageIndex{4}\)

What are the symmetry labels for the orbitals on sulfurused to make SF₄.

Answer

s - a₁

p_x- b₁

p_y- b₂

p_z- a₁

d_z²- a₁

Step 3- Making the LGOs

Before making the LGOs, we need to assignaxis labels to the molecule. While there are many ways to do this, the proper convention is to assign the principle axis to be the z-axis (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\). Assignment of the axes in SF₄.

As mentioned in the previous section, we can treat terminal halides using this method. To make the pictures less cluttered, the halides will be representedjust as the hydrogen atoms were and we will ignore the lone pair on the fluorine atoms. Based on the assigned axes above, the LGOscan be drawn to match the orbitals on the sulfur atom(Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\). LGOsin SF₄.

Step 3a- A few new twists

Before we go to step 4, we have a little bit of extra work to do. There are two features in this molecule that we have not dealt with before, the lone pair and the use of d-orbitals. Let's begin with the lone pair as this is something that could be encountered in molecules that do not break the octet rule. We know that this molecule has a lone pair on the sulfur atom. That lone pair has to be accounted for and must occupy one of the orbitals we considered as our basis functions. The question is, which orbital? The best option would be an orbital on the central atom that has no symmetry matched LGO. An example of this would be if all of the outer atoms in a molecule were sitting in the nodal plane of a p-orbital. We do not have that case in this example, so now we need to dig a little deeper. Our next best option is to consider all of the atomic orbitals of the central atom and ask which has the least amount of direct interaction with a LGO. The s-orbital on sulfur has interaction with all four fluorine atoms so it is not a good candidate. The p-orbitals all have interactions with two fluorine atoms, so they are slightly better options. In looking closer at the p-orbitals we see that the p_xorbital has very direct overlap with two fluorine atoms; the orbital is pointing directly at two atoms and both lobes interact. This is greater than the overlap of the p_yorbital in which the LGOdoes not lie directly on the y-axis. However, both lobes of the p_yorbital are interacting with the LGO. The p_z orbital is somewhat similar in that the LGOdoes not lie along the z-axis. However, for this orbital, only one lobe of the sulfur p_zis interacting with the LGO. So, given the options for the location of the lone pair, p_z> p_y> p_x> s, we must assign the p_z(a₁) orbital to be the one with the lone pair.You may have noticed that thed_z²was not considered in the previous discussion. That is because thed_z²orbital is not accessibleenergetically. It is too high in energy to participate in any meaningful bonding in this molecule. While, we draw an LGO that would match this orbital, we do not consider any possible bonding or antibonding interaction between the LGO and the d_z²orbital. This is going to become very critical in the next step.

Step 4- Constructing the MO diagram

In looking at the orbital energies from the previous section, we see that the 2s orbital of fluorine is MUCH lower in energy than the 2p of fluorine or the 3s and 3p of sulfur. It will not interact and does not need to be considered for this diagram. The 2p orbitals of fluorine fall between the 3s and 3p orbitals of sulfur. We will be bringing together a total of eightorbitals. Why only eight? Let's consider each of the combinations above(Figure \(\PageIndex{3}\)).

Remember, we are ignoring the extra lone pairs on fluorine so that leaves a total of 12 orbitals that will not be shown in our diagram.
The s-orbital (a₁) from sulfur and the corresponding LGO will make a bonding and antibondingset.
The p_xorbital (b₁) orbital from sulfur and the correspondingLGO will make a bonding and antibonding set.
The p_yorbital (b₂) orbital from sulfur and the corresponding LGOwill make a bonding and antibonding set.
The p_zorbital (a₁) has been assigned as the orbital holding the lone pair so we ignore the LGO it could interact with. Note that this does not mean that the fluorine orbitals do not contribute to this orbital, but for the purpose ofconstructing the MO diagram we will ignore any contribution.
The d_z² orbital (a₁) of sulfur is not energetically accessible so we do not include it, but we do consider the LGOthat would interact with it.

In sketching the MO diagram, the s-orbital of sulfur is the lowest energy orbital depicted and as such will give the lowest energy bonding MO and likely the highest energy antibonding MO. We will have two a₁orbitals that will be labeled as non-bonding (nb), one for the p_zorbital with the lone pair, and one for the LGO that would match the d_z² orbital. As we are considering the orbitals to be localized on specific atoms, we will not change their energy from that of the atomic orbital from which they are derived. That leaves the b₁and b₂orbitals that will make a bonding and antibonding set. At first glance, you might expect these to be degenerate as they both are derivedfrom p-orbitals on sulfur and the corresponding LGOs. However, upon examining the pictures above,(Figure \(\PageIndex{3}\)), it might be expected that the b₁bonding orbital might be slightly lower in energy than the b₂since there is a more direct overlap between the sulfur orbital and the LGO. For now we will treat the orbitals as being degenerate. This would lead to a MO diagram as pictured below(Figure \(\PageIndex{4}\)).

(Figure \(\PageIndex{4}\)). MO diagram for SF₄.

Overall, this would give a bond order\( = \frac{6-0}{2}\ = 3\nonumber\) which is quite different from the Lewis structure. The bonding in structures of p-block elements with 'expanded' octets is not as straightforward as the Lewis structure might suggest. There are filled bonding orbitals that involve more than two atoms. There are also significant ionic contributions to what is typically drawn as covalent bonds.

How did we do?

Our MO diagram was generated using a pictorial method. Using computational methods it is possible to derive a more accurate MO diagram for this molecule. Show below are the orbitals and their energies as calculated by WebMO(Table\(\PageIndex{3}\)). Overall our MO diagram looks pretty good. We see a VERY slight difference in the energy of the b₁and b₂ bonding orbitals. This difference is more pronounced in the antibonding orbitals. Based on theABIMABTBIBprinciple, it is not surprising that we see a greater difference in the antibonding orbitals. So our approximation of the b₁and b₂orbitals as being degenerate is not unreasonable for the bonding MOs but not accurate for the antibonding.

You may also notice that all of the fluorine contributions are from p-orbitals. The 2s orbital of fluorine is much too low in energy to participate in these interactions. So the only orbital involved in bonding with the sulfur would be the p-orbital pointing directly at the sulfur atom. This justifies our treatment of the fluorine as being hydrogen-like. Again, we are ignoring any possibleπ-bonding interactions in this molecule.

We also estimated that the a₁(nb) would not change in energy much from the atomic orbitals they were derived from. The energy for the a₁(nb) derived from the sulfur p_zis -12.049 eV which compares very favorably to the value of -12.0 eV for the 3p orbital of sulfur found in the previous page. For the a₁(nb) LGO, the energy was calculated to be -17.853 eV which is slightly higher than the -18.7 eV listed for the 2p orbitals of fluorine. Again, while not completely accurate, our pictorial method did a fairly reasonable job of generating the MO diagram.

(Table\(\PageIndex{3}\)).The MOs for SF₄(ignoring the lone pairs) as calculated and displayed byWebMO. The data for SF₄was imported into WebMOfrom the NIST Webbook as a computed 3-D structure and calculations were performed usingB3LYP-6-31G(d).