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

What is the coordination number of a particle in $fcc$ structure?

In a face-centered cubic lattice,a unit cell face is shared equally by how many unit cells?

In an ionic crystal,ion $A$ is present at the corners of the cubic unit cell,and ion $B$ is present at the centers of the faces. What is the formula of this ionic compound?

Find the radius of an atom in $fcc$ unit cell having edge length $393 \ pm$. (in $pm$)

Identify the type of unit cell that has particles at the centre of each face in addition to the particles at eight corners of a cube?

Vedclass Test Series

Exam Paper Generator

Online Exam Module