NaCl is a FCC lattice where Na^{+} ions are a | vedclass

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

Question

(B) In a standard $NaCl$ crystal structure,$Na^{+}$ ions occupy the octahedral voids (edge centres and body centre) and $Cl^{-}$ ions occupy the $FCC$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) In a standard $NaCl$ crystal structure,$Na^{+}$ ions occupy the octahedral voids (edge centres and body centre) and $Cl^{-}$ ions occupy the $FCC$ lattice points (corners and face centres). However,based on the specific distribution provided in the question: $Na^{+}$ ions at corners $(8 	imes \frac{1}{8} = 1)$ and face centres $(6 	imes \frac{1}{2} = 3)$,total $Na^{+} = 1 + 3 = 4$. $Cl^{-}$ ions at edge centres $(12 	imes \frac{1}{4} = 3)$ and body centre $(1 	imes 1 = 1)$,total $Cl^{-} = 3 + 1 = 4$. Since there are $4$ $Na^{+}$ ions and $4$ $Cl^{-}$ ions per unit cell,the number of $NaCl$ formula units is $4$.

$NaCl$ is a $FCC$ lattice where $Na^{+}$ ions are at corner and face centre positions. Chloride ions are at edge centres and body centre positions. How many $NaCl$ formula units will be in a unit cell?

Similar Questions

What is the coordination number in a calcium fluoride $(CaF_2)$ type structure?

Which statement is correct for the antifluorite structure of $Na_2O$?

In a face-centered cubic $(FCC)$ unit cell,how many unit cells share an atom located at the face center?

$A$ crystal is formed by two elements $X$ and $Y$ in a cubic structure. $X$ atoms are at the corners of a cube,while $Y$ atoms are at the face centers. The formula of the compound will be:

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid $AX$ is given in the figure. If the radius of $X^{-}$ is $250 \ pm$, the radius of $A^{+}$ is: (in $pm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module