 Solution

The ideal eclipsed ion has D4h molecular point group symmetry. Put the Mo–Mo axis along $z$, with the metal atoms at $(0,0,\pm c)$, and orient the four chloride positions around each metal along $\pm x,\pm y$.

There is a $C_4(z)$ axis, perpendicular twofold axes, a horizontal plane through the midpoint, vertical mirror planes and an inversion centre at the bond midpoint. The eclipsed arrangement makes the two chloride squares coincide in projection along $z$. These labels refer to the ideal molecular geometry, rather than small crystal-packing distortions.