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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.

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  • $\begingroup$ more repulsion in axial bond I guess (one confront three equatorial in 90 degree)? $\endgroup$ – Rodriguez Jun 4 '16 at 3:38
  • $\begingroup$ I would love to answer, however I don’t know any pentacoordinated species that has $\mathrm{sp^3d}$ ‘hybridisation’. =C $\endgroup$ – Jan Jun 4 '16 at 11:50
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    $\begingroup$ $\ce{sp^{3}d}$ 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. $\endgroup$ – ron Jun 6 '16 at 21:50
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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.

phosphorus pentachloride, simplified structure
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.

Mesomeric structures of phosphorus pentachloride
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.

molecular orbitals forming the 4e3c bond.
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.

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

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    $\begingroup$ Why don't the equatorial bonds lengthen instead? $\endgroup$ – bon Jun 4 '16 at 7:39

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