Mr. Samuel,

For the requested space group you have to dispose 13 atoms Sms and Os with ratio 5:7.5, because of the main composition is SmO1.5. The space system is cubic and the transformation can be found here (Cubic space system, space group 215):

http://img.chem.ucl.ac.uk/sgp/large/215az1.htm.

You should place 5 full occupied positions to Sm atoms; and 7 full occupied positions to O atoms. The 8th position of O should be half-occupied (or 1/2). The total number of occupied positions should be kept (13). 'Half-occupied position' means '1/2 electron density', computing properties of such as system.

For example, if the oxidation states of your ions are:

Sm3+ and O2-

Full occupied position means:

Sm3+ with electronic configuration [Xe]6s24f3 (59 electrons)

O2- with electronic configuration [He]2s22p6 (10 electrons)

Half-occupied position means:

1/2O2- with preliminary unknown electronic configuration (most probably [He]2s22p1), but with 5 electrons around one isolated in the 3D space O-nucleus.

Thank You Prof. Ivanova for the help.

Israel