Hi friends.
Can you give structural positions of Sm2O3 with P-43n as the space group
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.
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