Question

9. A spinel is an oxide with the general formula A"B":04. The crystal structure is a close-packed array of oxide ions with the A and B metal ions occupying the holes in the lattice. If A occupies the tetrahedral holes and B the octahedral holes, the spinel is said to be normal, but if the A ions are in the octahedral holes and B in both kinds, the spinel is said to be inverse. When transition metal ions are involved, the ion (A or B) that has the greatest octahedral site preference energy (OSPE) will occupy the octahedral holes. OSPE = CFSE (octahedral) - CFSE (tetrahedral) Calculate the OSPE in terms of A, for the cations in each of the following spinels, and predict whether the spinel will be normal or inverse. All of the transition metal ions are high-spin. Remember that A. = 4/9 4. (a) CdTi20 (b) Co:04

Answer 1

E 0027 g 737: cari, Oh Here two ions are pringent Cd2+ + 3010 450 Berg cont in octahedral field toge egy So CFSE = 4-6 x 04+ 4 X016) - 010 cdht in the field eh tab so CFSE = 0445010 : Ti 37 7304450 cd2t in oct. field tugego . CASF = -(1x014) --04:40 2012+ in Tao field e!ti. CFSE = -(1x06) = -0,644 = -0.267 40 Son OSPE of 7237 = (-0.4 +0.267) 4. = -0.133 Aoi preference Here, Cd2+ has neither re ference for Tetrahedral hole non for octahedrail hole B. But 7:34 has higher. CASE in octahedral hole & la rist will remain in octahedral hole & cd27 will remain in tetrahedral hole. Catior is a normal Spinel.

© Co, 0, - Cô ca, 0, Colt 3d7 450 Coßt 306 450 cont in oct. field tog sege CESE = (5x04 +. 2.40.6) 4. -0.8 do Colt in Thi field 4 t 2 ² efse = 6 48.0.6 + 3x0.4) At = -1.2 At = -0.533 40 : Here coat in octahedral field has more higher CASE value than te tra thedsial field. so co2t will try to occupy the octahedral hole & cogly will have norm inverse Spinel Structure. OSFE of Co2+ = 60.8 +0.533 ). A. = -0.266 40 But in presence of weak of ligand fied 637 remains in low Spin in octahedral hole. SO CASE of cost in oct hole = (- 6x014) A. l togoogle) = - 214 A cost in to hole CFSE = (342) f 0.6X3 +0.4 x 3 ) 4, = -0+266 40

so ospe of cost in octahedral hole =(-204 +0.266) 4. = – 2013.1 So, ospe of cost is much more higher than the Ospe of cost i so, cost remains In octahedral hole & coat remains in tetrahedral hole. So, Coyoy has a normal Spinel Structure