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Standard XII

Chemistry

Examples of Ionic Compounds

In a normal s...

Question

In a normal spinel type structure, the oxide ions are arranged in ccp whereas $\frac{1}{8}$ tetrahedral holes are occupied by $Z n^{2 +}$ ions and 50% of octahedral holes are occupied by $F e^{3 +}$ ions. The formula of the compound is ?

$Z n_{2} F e_{2} O_{4}$

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$Z n F e_{2} O_{3}$

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$Z n_{2} F e_{2} O_{4}$

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$Z n_{2} F e_{2} O_{2}$

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Solution

The correct option is C
$Z n_{2} F e_{2} O_{4}$

Number of O - atoms per unit cell $= \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4$
Number of octahedral holes per unit cell $= 1 \times 4 = 4$
Number of $F e^{3 +}$ ions per unit cell $= \frac{50 \times 4}{100} = 2$
Number of $Z n^{2 +}$ ions per unit cell $= \frac{1}{8} \times 8 = 1$
Hence, formula : $Z n F e_{2} O_{4}$

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Similar questions

In a normal spinel type structure, the oxide ions are arranged in ccp whereas $\frac{1}{8}$ tetrahedral holes are occupied by $Z n^{2 +}$ ions and 50% of octahedral holes are occupied by $F e^{3 +}$ ions. The formula of the compound is ?

Q. In spinel structure, oxides ions form fcc lattice, whereas

\frac{1}{8}

th of tetrahedral holes are occupied by cations

A^{2 +}

and

\frac{1}{2}

of octahedral holes are occupied by cations

B^{3 +}

ions. The general formula of the compound having spinel structure is

Q. 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

Z n^{2 +}, A l^{3 +}, O^{2 -}

with

Z n^{2 +}

in the tetrahedral holes. Give the formula of spinel :

Q. Spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has 1/8 of the tetrahedral holes occupied by one type of metal ion and 1/2 of the octahedral holes occupied by another type of metal ion. Such a spine is formed by

Z n^{2 +}, A l^{3 +}

and

O^{2 -}

with

Z n^{2 +}

in the tetrahedral holes. If CCP arrangement of oxide ions remains undistorted in the presence of all the cations, formula of spinel and percentage of the packing fraction of crystal are respectively:

Q. A mixed oxide is made up of

Z n^{2 +} A l^{3 +}

and

O^{2 -}

Oxide ions are arranged in CCP layers one eight of the tetrahedral holes are occupied by

Z n^{2 +}

ions and half of the octahedral holes are occupied by

A l^{3 +}

ions. Find the formula of mixed oxide.

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In a normal spinel type structure, the oxide ions are arranged in ccp whereas 18 tetrahedral holes are occupied by Zn2+ ions and 50% of octahedral holes are occupied by Fe3+ ions. The formula of the compound is ?

The correct option is C Zn2Fe2O4 Number of O - atoms per unit cell =18×8+12×6=4 Number of octahedral holes per unit cell =1×4=4 Number of Fe3+ ions per unit cell =50×4100=2 Number of Zn2+ ions per unit cell =18×8=1 Hence, formula : ZnFe2O4

In a normal spinel type structure, the oxide ions are arranged in ccp whereas $\frac{1}{8}$ tetrahedral holes are occupied by $Z n^{2 +}$ ions and 50% of octahedral holes are occupied by $F e^{3 +}$ ions. The formula of the compound is ?