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JEE Advanced · Chemistry · 15. Solid State

The edge length of unit cell of a metal having molecular weight \(75 \mathrm{~g} / \mathrm{mol}\) is 5 Å which crystallizes in cubic lattice. If the density is \(2 \mathrm{~g} / \mathrm{cc}\) then find the radius of metal atom, \(\left(N_A=6 \times 10^{23}\right)\). Give the answer in pm.

  1. A 216.5
  2. B 214.55
  3. C 341.55
  4. D 102.88
Verified Solution

Answer & Solution

Correct Answer

(A) 216.5

Step-by-step Solution

Detailed explanation

For cubic lattice, \(\rho=\frac{Z \times A}{N_A \times V}\) or \(Z=\frac{\rho \times N_A \times V}{A}\)
Given that,
\(A=75 \mathrm{~g} / \mathrm{mol} \)
\( V=(5 Å)^3=\left(5 \times 10^{-8}\right)^3 \mathrm{~cm}^3 \)
\( =125 \times 10^{-24} \mathrm{~cm}^3 \)
\( N_{\mathrm{A}}=6.02 \times 10^{23} \)
\( \rho=2 \mathrm{gm} / \mathrm{cc} \)
\( \therefore Z=\frac{2 \times 6.02 \times 10^{23} \times 125 \times 10^{-24}}{75} \approx 2\)
The value of \(Z\) represents that element must have body centred cubic (bcc) structure.
For bcc structure atomic radius \((r)=\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4} \times 5=2.165 Å\)
\(
=216.5 \mathrm{pm}
\)
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