30 points
>
Time management is very much important in IIT JAM. The eduncle test series for IIT JAM Mathematical Statistics helped me a lot in this portion. I am very thankful to the test series I bought from eduncle.
Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Chemistry (CY)
What is spinel structure?
1 Answer(s)
Answer Now
- 0 Likes
- 1 Comments
- 0 Shares
-
![comment-profile-img]()
>
Do You Want Better RANK in Your Exam?
Start Your Preparations with Eduncle’s FREE Study Material
- Updated Syllabus, Paper Pattern & Full Exam Details
- Sample Theory of Most Important Topic
- Model Test Paper with Detailed Solutions
- Last 5 Years Question Papers & Answers
Sign Up to Download FREE Study Material Worth Rs. 500/-
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.
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.