A^{+}B^{-} shows rock salt type arrangement. | vedclass

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

Question

(B) In a rock salt $(NaCl)$ type structure,$A^{+}$ ions occupy octahedral voids and $B^{-}$ ions form a $FCC$ lattice.$1$. The number of $A^{+}$ and $

Vedclass · Accepted Answer

Number of nearest cations from a given cation are $8$

Vedclass · Answer

(B) In a rock salt $(NaCl)$ type structure,$A^{+}$ ions occupy octahedral voids and $B^{-}$ ions form a $FCC$ lattice.$1$. The number of $A^{+}$ and $B^{-}$ ions per unit cell is $4$ each,so option $C$ is correct.$2$. In $FCC$,$A^{+}$ ions form an $FCC$ lattice (octahedral voids in $FCC$ form an $FCC$ lattice),so option $D$ is correct.$3$. The distance between nearest cations in an $FCC$ lattice is $\frac{a}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \ \mathring{A}$,so option $A$ is correct.$4$. In a rock salt structure,the nearest cations to a given cation are located at the corners of the unit cell relative to the face-centered position,which are $12$ in number (next-nearest neighbors),not $8$. Therefore,option $B$ is incorrect.

$A^{+}B^{-}$ shows rock salt type arrangement. Edge length of unit cell is $6 \ \mathring{A}$. If radius of $A^{+}$ is $1 \ \mathring{A}$,then the incorrect option is:

Similar Questions

In the calcium fluoride structure,the coordination number of the cation and anion are respectively:

In a cubic crystal structure,'$W$' atoms are at the corners of the cube,'$O$' atoms are at the edges,and '$Na$' atoms are at the body center. Determine the formula of this crystal.

How many particles per unit cell are present in $BCC$ structure?

Which type of cubic lattice has the maximum number of atoms per unit cell?

Calculate the number of atoms per unit cell in a face-centered cubic $(fcc)$ unit cell.

Vedclass Test Series

Exam Paper Generator

Online Exam Module