Single Choice

Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one eight of the tetrahedral holes occupied by one type of metal ion and one half of the octahedral holes occupied by another type of metal ion. Such a spinel is formed by $$Zn^{2+},\, Al^{3+},\, O^{2-}$$ with $$Zn^{2+}$$ in the tetrahedral holes. Give the formula of spinel :

AZnAl2O4
Correct Answer
BZn2AlO4
CZn2Al2O4
DNone of the above

Solution

The formula of spinel is $$\displaystyle ZnAl_2O_4$$
The general formula of spinels is $$\displaystyle AB_2O_4$$ or $$\displaystyle A[B_2]O_4$$ where square bracket represents the cations occupying octahedral sites.
The spinel structure consists of an fcc array of oxide ions in which the A cations occupy one-eighth of the tetrahedral holes and the B cations occupy the octahedral holes.
In inverse spinels, the cation distribution is $$\displaystyle B[AB]O_4 $$

SIMILAR QUESTIONS

Solid State

In a closest packed lattice, the number of octahedral sites as compared to tetrahedral ones will be:

Solid State

The ratio of close packed atoms to tetrahedral holes in cubic close packing is:

Solid State

In a closed packing, the ratio of numbers of tetrahedral holes to that of octahedral holes is given by:

Solid State

If the unit cell of a mineral has cubic packed (c.c.p.) array of oxygen atoms with $$m$$ fraction of octahedral holes occupied by aluminium ions and $$n$$ fraction of tetrahedral holes occupied by magnesium ions, m and n respectively are:

Solid State

An ionic compound has a unit cell consisting of $$A$$ ions at the corners of a cube and $$B$$ ions on the centres of the faces of the cube. The empirical formula for this compound would be:

Solid State

In an antifluorite structure, cations occupy:

Solid State

A crystal is made of particles X ,Y and Z. X forms fcc packing. Y occupies all the octahedral voids of X and Z occupies all the tetrahedral voids of X. If all the particles along one body diagonal are removed, then the formula of the crystal would be :

Solid State

The number of octahedral voids in a unit cell of cubic close packed structure is:

Solid State

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $$356$$ pm, then the radius (in pm) of carbon atom is:

Solid State

A cubic solid is made by atoms A forming closed packed arrangement, B occupying one/fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of the compound :-

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