The FCC unit cell of a compound contains ions | vedclass

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

Question

(C) In an $FCC$ unit cell,the number of atoms at the corners is $8 	imes \frac{1}{8} = 1$.Since ions of $A$ are at the corners,the number of $A$ atom

Vedclass · Accepted Answer

$AB_3$

Vedclass · Answer

(C) In an $FCC$ unit cell,the number of atoms at the corners is $8 	imes \frac{1}{8} = 1$.Since ions of $A$ are at the corners,the number of $A$ atoms per unit cell is $1$.In an $FCC$ unit cell,the number of atoms at the face centres is $6 	imes \frac{1}{2} = 3$.Since ions of $B$ are at the face centres,the number of $B$ atoms per unit cell is $3$.Therefore,the ratio of $A:B$ is $1:3$.The formula of the compound is $AB_3$.

The $FCC$ unit cell of a compound contains ions of $A$ at the corners and ions of $B$ at the centre of each face. What is the formula of the compound?

Similar Questions

The $8:8$ type packing is present in which of the following?

Coordination number of $Zn^{2+}$ in $ZnS$ (zinc blende) is

The solid $AB$ has a $NaCl$ type structure. If the radius of the $A^{+}$ ion is $120 \, pm$, then the maximum radius of the $B^{-}$ ion will be ........... $pm$.

In a $ccp$ lattice of $X$ and $Y$,$X$ atoms are present at the corners while $Y$ atoms are at face centers. What would be the formula of the compound if one of the $X$ atoms from a corner is replaced by a $Z$ atom?

The crystal structure of $NaCl$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module