$A$ compound is formed by two elements $M$ and $N$. Element $N$ forms a hexagonal closed pack $(HCP)$ array with $2/3$ of the octahedral holes occupied by $M$. The formula of the compound is $....$
- A$M_4 N_3$
- B$M_2 N_3$
- C$M_3 N_2$
- D$M_3 N_4$
Explore More
Similar Questions
In a crystal lattice,$C$ anions form a cubic close-packed $(CCP)$ structure. Cations $(A)$ occupy $50\%$ of the tetrahedral voids and cations $(B)$ occupy $50\%$ of the octahedral voids. Determine the molecular formula of the compound.
Medium
View Solution$A$ solid has $CCP$ arrangement having atoms $A, B, C$. If $A$ atoms are present at face centres,$B$ at corners and $C$ atoms occupy $50\%$ tetrahedral voids,then the molecular formula of the solid will be:
Medium
View SolutionIf $Z$ is the number of atoms in the unit cell that represents the closest packing sequence $ABCABC .....,$ the number of tetrahedral voids in the unit cell is
Medium
View Solution$A$ compound is formed by elements $P$ and $Q$. Atoms of $Q$ are in $ccp$ arrangement. If $P$ occupies all the tetrahedral voids,what is the formula of the compound?
Medium
View SolutionWhat fraction of one edge-centred octahedral void lies in one unit cell of $fcc$?
MediumNEET 2023
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