Yash posted an Question
April 18, 2020 • 21:27 pm 30 points
  • IIT JAM
  • Chemistry (CY)

What is spinel structure?

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    Lingareddy 1 Best Answer

    The spinels have the general chemical formula AB2X4.      Where:     AII = a divalent cation like Mg, Cr, Mn, Fe, Co, Ni, Cu, Zn, Cd, Sn      BIII = a trivalent cation like Al, Ga, In, Ti, V, Cr, Mn, Fe, Fe, Co, Ni      X = O, S, Se etc. Examples of Normal Spinels: MgAl2O4 (known as spinel), Mn3O4, ZnFe2O4, FeCr2O4 (chromite) etc.

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    A spinel unit cell is made up of 8 FCC cells. The anions (usually oxide ions: O2-) occupy the FCC lattice points. The divalent AII cations occupy 1/8th  of the tetrahedral voids, whereas the trivalent BIII cations occupy one half (1/2) of octahedral voids. Thus a normal spinel can be represented as: (AII)tet(BIII)2octO4  Note: We know that in one FCC lattice unit cell, the effective numbers of atoms (or ions) occupying the lattice points is 4. At the same time, the effective number of  tetrahedral voids (holes) = 8 and that of octahedral voids = 4. That means, in a normal spinel, there are 8 x 4 = 32 anions occupying the lattice points of 8 FCC unit cells. Whereas, the number of divalent AII cations occupying tetrahedral voids is 8 x 1/8 x 8 = 8 and the number of trivalent BIII ions occupying the octahedral voids = 8 x 1/2 x 4 = 16. i.e. The ratio of AII : BIII : O2- = 8 : 16 : 32 = 1:2:4 which confirms with the formula of normal spinels.

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