In a closed packed structure of mixed oxides, it is found that lattice has O2- (oxide ion) and half of octahedral voids are occupied by trivalent cations (A3+) and one-eighth of tetrahedral voids are occupied by divalent cations (B2+). Thus, mixed oxide is

Question

In a closed packed structure of mixed oxides, it is found that lattice has O2- (oxide ion) and half of octahedral voids are occupied by trivalent cations (A3+) and one-eighth of tetrahedral voids are occupied by divalent cations (B2+). Thus, mixed oxide is (A) A2BO3 (B) A2BO4 (C) AB2O4 (D) ABO3

Tardigrade Admin · Accepted Answer

In closed packed structure lattice.
Number of octahedral voids

= 1

Number of tetrahedral voids

= 2

Charge on

(A_{2} B)^{x +} = (2 \times 3 + 2 \times 1) = + 8

Thus, number of

O

-atoms

= \frac{8}{2} = 4

(to balance charge)
Lattice is

A_{2} B O_{4}

.

Q. In a closed packed structure of mixed oxides, it is found that lattice has $O^{2 -}$ (oxide ion) and half of octahedral voids are occupied by trivalent cations $(A^{3 +})$ and one-eighth of tetrahedral voids are occupied by divalent cations $(B^{2 +})$ . Thus, mixed oxide is

$A_{2} B O_{3}$

$A_{2} B O_{4}$

$A B_{2} O_{4}$

$A B O_{3}$

In closed packed structure lattice.
Number of octahedral voids $= 1$
Number of tetrahedral voids $= 2$

Charge on $(A_{2} B)^{x +} = (2 \times 3 + 2 \times 1) = + 8$
Thus, number of $O$ -atoms $= \frac{8}{2} = 4$ (to balance charge)
Lattice is $A_{2} B O_{4}$ .

Q. In a closed packed structure of mixed oxides, it is found that lattice has O2− (oxide ion) and half of octahedral voids are occupied by trivalent cations (A3+) and one-eighth of tetrahedral voids are occupied by divalent cations (B2+). Thus, mixed oxide is

A2​BO3​

A2​BO4​

AB2​O4​

ABO3​