Why are axial bonds are longer than equatorial bond in sp3d hybridized atoms?

Why are axial bonds are longer than equatorial bond in case of $\mathrm{sp^3d}$ hybridization? I have done some research but I can't seem to find the answer.

• more repulsion in axial bond I guess (one confront three equatorial in 90 degree)? – Rodriguez Jun 4 '16 at 3:38
• I would love to answer, however I don’t know any pentacoordinated species that has $\mathrm{sp^3d}$ ‘hybridisation’. =C – Jan Jun 4 '16 at 11:50
• $\ce{sp^{3}d}$ hybridization is not involved here. See this earlier answer for an explanation as to the hybridization\bonding involved and why the axial bonds are longer. – ron Jun 6 '16 at 21:50

2 Answers

You are asking for $$\mathrm{sp^3d}$$ hybridisation, but I do not know of a case where $$\mathrm{sp^3d}$$ hybridisation actually happens. Either it does not make sense to discuss hybridisation at all (the iron in pentacarbonyliron) or the hybridisation is actually not $$\mathrm{sp^3d}$$ but $$\mathrm{sp^2 + p}$$ (the phosphorus in $$\ce{PCl5}$$). I shall discuss the latter, because I believe it could be the one you are on about.

Phosphorus pentachloride is often drawn like in figure 1 below for simplicity. This depiction implies five identical covalent 2-electron-2-centre bonds. However, that does not agree with the octet rule.

Figure 1: Simplified structure of $$\ce{PCl5}$$.

Instead, one can draw a set of two mesomeric structures, each one conforming with the octet rule — see figure 2.

Figure 2: Mesomeric structures of $$\ce{PCl5}$$ conforming to the octet rule.

This already hints us towards the answer: Rather than assuming two bonds which are equal to the other three bonds we need to consider three ‘classical’ 2-electron-2-centre bonds to the equatorial chlorines and one 4-electron-3-centre bond to the two axial chlorines. This bond’s order is $$0.5$$ rather than $$1$$. Typically, the lower the bond order the weaker a bond and therefore the greater the bond length is — the experiment is in fine agreement with theory.

But no bonding discussion is truly complete without an orbital consideration. Check out the three collinear p-orbitals of phosphorus and the two chlorines, that together form three molecular orbitals labelled $$\Psi_1$$ to $$\Psi_3$$ in figure 3.

Figure 3: Representation of the three molecular orbitals that form the 4-electron-3-centre bond.

As you can see, the lowest MO $$\Psi_1$$ is bonding with respect to both $$\ce{P-Cl}$$ bonds. The highest MO $$\Psi_3$$ is antibonding with respect to both. And the middle one is bonding with respect to $$\ce{Cl-Cl}$$ but nonbonding if we add the phosphorus atom. We need to fill in four electrons into these three orbitals ($$\ce{PCl3}$$ has one lone pair, and we are effectively using the bond of a $$\ce{Cl-Cl}$$ molecule as our second electron pair). Thus, the bonding and the nonbonding orbitals are filled. Bond order can now be calculated by:

$$\text{bond order} = \frac{(\text{electrons in bonding orbitals}) - (\text{electrons in antibonding orbitals})}{\text{number of bonds}\times 2}\\= \frac{2-0}{2 \times 2} = 0.5$$

Again, a lower bond order typically correlates with a greater bond length.

• Please, consider partial charge interactions in bond order determination. – permeakra Jun 6 '16 at 22:47

I think it's because the equatorial bonds lie on the same plane and so would be of equivalent length but the axial bonds experience repulsion from the equatorial bonds and as a result try to move away as far as possible so as to minimize repulsion. Thus, the axial bonds would be longer. Hope this helps :)

• Why don't the equatorial bonds lengthen instead? – bon Jun 4 '16 at 7:39