A face-centred cubic (FCC) unit cell is made | vedclass

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

Question

(D) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.Initially,the number of atoms of $A$ at corners $= 8 	imes \frac{1}{8} = 1$.The

Vedclass · Accepted Answer

$AB_4$

Vedclass · Answer

(D) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.Initially,the number of atoms of $A$ at corners $= 8 	imes \frac{1}{8} = 1$.The number of atoms of $B$ at face centres $= 6 	imes \frac{1}{2} = 3$.One body diagonal passes through $2$ corners and the body centre. However,in an $FCC$ lattice,there are no atoms at the body centre.When atoms along one body diagonal are removed,$2$ corner atoms are removed.Remaining atoms of $A = 1 - (2 	imes \frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}$.Since no face-centred atoms lie on the body diagonal,the number of $B$ atoms remains $3$.The ratio of $A:B = \frac{3}{4} : 3 = 3 : 12 = 1 : 4$.Therefore,the empirical formula is $AB_4$.

$A$ face-centred cubic $(FCC)$ unit cell is made up of two types of atoms $A$ and $B$,in which $A$ occupies the corner positions and $B$ occupies the face centres. If atoms along one of the body diagonals are removed,what is the empirical formula of the remaining solid?

Similar Questions

Explain the calculation of packing efficiency in $hcp$ and $ccp$ structures with the help of a diagram.

An ionic compound is formed by elements $X$ and $Y$. Element $Y$ forms a face-centered cubic $(FCC)$ lattice, and element $X$ occupies one-third of the tetrahedral voids. What is the formula of the compound?

What is the total number of spheres that surround the tetrahedral hole in a close-packed three-dimensional structure?

$A$ mineral consists of a cubic close-packed structure formed by $O^{2-}$ ions,where half the octahedral voids are occupied by $Al^{3+}$ and one-eighth of the tetrahedral voids are occupied by $Mn^{2+}$. The chemical formula of the mineral is

The edge of unit cell of $FCC$ $Xe$ crystal is $620 \ pm$. The radius of $Xe$ atom is $..........$ $pm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module