The ratio of densities if the same element undergoes $FCC$ and $HCP$ close packing is:

  • A
    $1$
  • B
    $4/6$
  • C
    $1.5$
  • D
    $\sqrt{2}/\sqrt{3}$

Explore More

Similar Questions

In an ionic crystalline solid,atoms of element $Y$ form an $hcp$ structure. The atoms of element $X$ occupy one-third of the tetrahedral voids. What is the formula of the compound?

How many tetrahedral voids are occupied in diamond? (in percentage)

What is the type of hole occupied if the limiting value of $\frac{r+}{r-}$ is in the range of $0.225$ to $0.414$?

$A$ compound is formed by elements $X$,$Y$,and $Z$. Atoms of $Z$ (anions) make $FCC$ lattice. Atoms of $X$ (cations) occupy all the octahedral voids. Atoms of $Y$ (cations) occupy $\frac{1}{3}$ rd of the tetrahedral voids. The formula of the compound is:

In a compound $XY$,the atoms of element $Y$ form an $fcc$ structure,while the atoms of element $X$ occupy $25\%$ of the tetrahedral voids. The molecular formula of the compound will be .......

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo