# Dipole moment of ferrocene conformations

Which conformation of ferrocene has a zero dipole moment?

I thought that the answer should be the staggered conformation because it would have the cyclopentadienyl groups oriented away from each other. However, the answer given is the eclipsed form. I can’t understand why that happens.

• Now that's a cryptic question indeed. You are right in that the staggered conformation has zero dipole moment because of symmetry. So does the eclipsed conformation. So does every other conformation as well. – Ivan Neretin Oct 15 '17 at 18:40
• I would also argue both the eclipsed and staggered form have zero dipole-moment due to symmetry, by the argument that, if ferrocene is aligned along the z axis, then for any dipole $v = (v_x, v_y, v_z)$ there is a dipole $v' = (v_x, v_y, -v_z)$ so that the net dipole moment cancels. Maybe it is about polarizability? – logical x 2 Oct 15 '17 at 19:23

To find out whether a compound can have a nonzero dipole moment, consider its symmetry. A dipole moment is a vector which must be transformed onto itself by every symmetry operation present in the molecule’s point group.

It should be easy to see that te eclipsed conformation has five planes of symmetry perpendicular to the cyclopentadienyl rings ($5\sigma_\mathrm v$). Another sixth plane of symmetry is sandwiched between the two rings and passes through the iron atom ($\sigma_\mathrm h$). The only vector that will be transformed onto itself by all these planes of symmetry is the null vector $\vec{0}$. Therefore, this conformation has zero dipole moment.

The staggered conformation also has five planes of symmetry perpendicular to the ring planes as the eclipsed conformation ($5\sigma_\mathrm v$). It also features a centre of symmetry in the iron atom ($i$) as well as improper rotation ($S_{10}$). The inversion centre alone is enough to determine that only the null vector $\vec 0$ will be transformed upon itself. Therefore, this form also has no dipole moment.

Both you and the book are correct.