The crystal structures of the compounds (\mathrm{Sr}{2} \mathrm{RuO}{4}) and (\mathrm{Sr}{2} \mathrm{IrO}{4}) have been analyzed at room temperature and at (10 \mathrm{~K}) with the neutron powder diffraction technique and the Rietveld method of profile fitting. (\mathrm{Sr}{2} \mathrm{RuO}{4}) crystallizes with the symmetry of space group (14 / \mathrm{mmm}) and room temperature lattice parameters (a=) (3.8730(3)) and (c=12.7323(9) \AA), while (\mathrm{Sr}{2} \mathrm{IrO}{4}) has the symmetry of space group (I 4_{1} /) acd and lattice parameters (a=5.4994(1)) and (c=25.7841(8) \AA). The unit cells of the two compounds are related to one another by the transformation matrix ( (1,-1,0 / 1,1,0 / 0), (0,2)). The main difference between the two structures is that the (M \mathrm{O}{6}) octahedra ( (M=\mathrm{Ru}, \mathrm{Ir}) ) have a regular, undistorted configuration in the ruthenium compound, while they are tilted by about (11^{\circ}) around the (c) axis of the unit cell in the iridium compound. The oxygen atoms of the (\mathrm{IrO}{2}) layers were found to be disordered over two sets of positions (x, x, \frac{1}{4}) with (x=\frac{1}{4} \delta). This means that the (\mathrm{IrO}{6}) octahedra assume two configurations and on a local level their relative orientation does not obey the symmetry requirements of space group 14 / /acd in all cases. The (M \mathrm{O}{6}) octahedra are elongated along the (c) axis, and this distortion is more pronounced in (\mathrm{Sr}{2} \mathrm{RuO}{4}) than in (\mathrm{Sr}{2} \mathrm{IrO}{4}). The coordination of the strontium atoms is ninefold in both compounds. Because of the tilting of the (\mathrm{IrO}_{6}) octahedra, however, the coordination polyhedra are different in the two cases. No phase transitions have been observed down to (10 \mathrm{~K}) in either compound. 1994 Academic Press. Inc.