The volume occupied by a single $CsCl$ ion pair in a crystal is $7.014 \times 10^{-23} \ cm^{3}$. The smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance is equal to the length of the side of the cube corresponding to the volume of one $CsCl$ ion pair. The smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance is nearly:

  • A
    $4.3 \ \mathring{A}$
  • B
    $4.5 \ \mathring{A}$
  • C
    $4.4 \ \mathring{A}$
  • D
    None of the above

Explore More

Similar Questions

An element with density $2.8 \ g \ cm^{-3}$ forms an $fcc$ unit cell having an edge length of $4 \times 10^{-8} \ cm$. Calculate the molar mass of the element.

Calculate the volume of the unit cell for an element having a molar mass of $56 \ g \ mol^{-1}$ that forms $bcc$ unit cells. $\left[\rho \cdot N_{A} = 4.8 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$

In an $NaCl$ crystal, the radii of $Na^{+}$ and $Cl^{-}$ ions are $95 \, pm$ and $180 \, pm$ respectively. The edge length of its unit cell is ........... $pm$.

Which formula is used to calculate edge length $(a)$ in a $bcc$ structure,where $r$ is the radius of the atom?

Aluminum crystallizes in a cubic close-packed structure. If its metallic radius is $125 \ pm$, what is the side length of the unit cell?

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