A binary compound (A^{+} B^{-}) has a zinc bl | vedclass

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

Question

(C) In a zinc blende structure,the anions $(B^{-})$ form a face-centered cubic $(FCC)$ lattice.Let the number of $B^{-}$ ions per unit cell be $N = 4$

Vedclass · Accepted Answer

$AB_2$

Vedclass · Answer

(C) In a zinc blende structure,the anions $(B^{-})$ form a face-centered cubic $(FCC)$ lattice.Let the number of $B^{-}$ ions per unit cell be $N = 4$.The number of tetrahedral voids in an $FCC$ lattice is $2N = 2 	imes 4 = 8$.The $A^{+}$ ions occupy $25\%$ of these tetrahedral voids.Number of $A^{+}$ ions $= 25\% 	ext{ of } 8 = 0.25 	imes 8 = 2$.The ratio of $A^{+} : B^{-}$ is $2 : 4$,which simplifies to $1 : 2$.Therefore,the formula of the compound is $AB_2$.

$A$ binary compound $(A^{+} B^{-})$ has a zinc blende structure where $B^{-}$ ions form the lattice and $A^{+}$ ions occupy $25\%$ of the tetrahedral voids. What is the formula of the compound?

Similar Questions

By increasing the pressure on $NaCl$ crystal,its coordination number will

How many "nearest" and "next nearest" neighbours,respectively,does potassium have in a $bcc$ lattice?

In the face-centred cubic $(FCC)$ lattice structure of gold, the closest distance between gold atoms is ('a' being the edge length of the cubic unit cell).

Calculate the volume occupied by all particles in a $bcc$ unit cell if the volume of the unit cell is $8.0 \times 10^{-23} \ cm^3$.

In a face centred cubic lattice,atom $A$ occupies the corner positions and atom $B$ occupies the face centre positions. If one atom of $B$ is missing from one of the face centred points,the formula of the compound is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module