In a solid,oxide ions are arranged in HCP. Ca | vedclass

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

Question

(C) Let the number of oxide ions $(O^{2-})$ in the $HCP$ lattice be $N = 6$.The number of $OHV$ (Octahedral Voids) is equal to the number of ions,so $

Vedclass · Accepted Answer

$ABO_3$

Vedclass · Answer

(C) Let the number of oxide ions $(O^{2-})$ in the $HCP$ lattice be $N = 6$.The number of $OHV$ (Octahedral Voids) is equal to the number of ions,so $OHV = 6$.The number of $THV$ (Tetrahedral Voids) is twice the number of ions,so $THV = 2 	imes 6 = 12$.Cations $A$ occupy $\frac{1}{6}$ of $THV = \frac{1}{6} 	imes 12 = 2$.Cations $B$ occupy $\frac{1}{3}$ of $OHV = \frac{1}{3} 	imes 6 = 2$.The ratio of atoms $A : B : O$ is $2 : 2 : 6$,which simplifies to $1 : 1 : 3$.Therefore,the empirical formula is $ABO_3$.

In a solid,oxide ions are arranged in $HCP$. Cations $A$ occupy one-sixth of $THV$ and cations $B$ occupy one-third of the $OHV$. The empirical formula of the compound is

Similar Questions

Consider the $bcc$ unit cells of the solid $1$ and $2$ with the position of atoms as shown below. The radius of atom $B$ is twice that of atom $A$. The unit cell edge length is $50\%$ more in solid $2$ than in $1$. What is the approximate packing efficiency in solid $2$? $........... \%$

Hexagonal close packed arrangement of ions is described as

Which of the following lattices has the highest packing efficiency: $(i)$ simple cubic,$(ii)$ body-centred cubic,and $(iii)$ hexagonal close-packed lattice?

In a close-packed lattice,how does the size of an octahedral void compare to that of a tetrahedral void?

If the distance between the corner and the center of a cube is $x$,then the tetrahedral void will be at a distance of .... from the corner of the cube.

Vedclass Test Series

Exam Paper Generator

Online Exam Module