In an $fcc$ crystal lattice,atoms $A$ are at the corners of the cube and atoms $B$ are at the face centers. If one atom is removed from one of the face centers,find the formula of the compound.
- A$A_2B_5$
- B$A_4B_5$
- C$A_8B_5$
- D$A_2B_3$
Explore More
Similar Questions
The close packing that represents the $ABC ABC......$ order is:
Medium
View SolutionThe number of tetrahedral voids present in $0.5 \ mole$ of $hcp$ crystal structure is:
In a face-centred cubic lattice,atom $A$ occupies the corner positions and atom $B$ occupies the face-centred positions. If one atom of $B$ is missing from one of the face-centred points,the formula of the compound is
Medium
View Solution$A$ compound is formed from elements $X$ and $Y$. The atoms of $Y$ (anions) form a $ccp$ lattice. The atoms of $X$ (cations) occupy half of the octahedral voids and half of tetrahedral voids. What is the formula of the compound?
The number of tetrahedral voids in the unit cell of a face-centered cubic $(fcc)$ lattice of similar atoms is:
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo