In a structure of mixed oxide,oxide ions are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In a $CCP$ structure,the number of oxide ions $(O^{2-})$ per unit cell is $4$. Number of tetrahedral voids $(TV)$ = $2 	imes 4 = 8$. Number of oc

Vedclass · Accepted Answer

$A_4B_5O_{10}$

Vedclass · Answer

(A) In a $CCP$ structure,the number of oxide ions $(O^{2-})$ per unit cell is $4$. Number of tetrahedral voids $(TV)$ = $2 	imes 4 = 8$. Number of octahedral voids $(OV)$ = $4$. Number of $A^{2+}$ ions = $\frac{1}{5} 	imes 8 = \frac{8}{5}$. Number of $B^{3+}$ ions = $\frac{1}{2} 	imes 4 = 2$. The ratio of $A : B : O$ is $\frac{8}{5} : 2 : 4$. Multiplying the ratio by $5$,we get $8 : 10 : 20$. Simplifying the ratio by dividing by $2$,we get $4 : 5 : 10$. Thus,the formula of the oxide is $A_4B_5O_{10}$.

In a structure of mixed oxide,oxide ions are in $CCP$. One-fifth of tetrahedral voids are occupied by divalent ion $(A^{2+})$ while half of the octahedral voids are occupied by trivalent ion $(B^{3+})$. The formula of the oxide is:

Similar Questions

Assertion : In close packing of spheres,a tetrahedral void is surrounded by four spheres whereas an octahedral void is surrounded by six spheres.
Reason : $A$ tetrahedral void has a tetrahedral shape whereas an octahedral void has an octahedral shape.

Calculate the void volume of a simple cubic unit cell if the volume of the unit cell is $5.5 \times 10^{-22} \ cm^3$.

In a face-centred cubic $(FCC)$ lattice,where are the tetrahedral voids located?

If $Z$ is the number of atoms in the unit cell that represents the closest packing sequence $ABCABC .....,$ the number of tetrahedral voids in the unit cell is

What is the number of tetrahedral voids present in $0.5 \text{ mole}$ of a compound forming an $hcp$ structure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module