Q36 / 120 · JEE Advanced 2006 · Chemistry
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.
Correct answer: 216.5
Solution
For cubic lattice, $\rho=\frac{Z \times A}{N_A \times V}$ or $Z=\frac{\rho \times N_A \times V}{A}$
Given that,
$$
\begin{aligned}
& A=75 \mathrm{~g} / \mathrm{mol} \\
& V=(5 Å)^3=\left(5 \times 10^{-8}\right)^3 \mathrm{~cm}^3 \\
& =125 \times 10^{-24} \mathrm{~cm}^3 \\
& N_{\mathrm{A}}=6.02 \times 10^{23} \\
& \rho=2 \mathrm{gm} / \mathrm{cc} \\
& \therefore \quad Z=\frac{2 \times 6.02 \times 10^{23} \times 125 \times 10^{-24}}{75} \approx 2 \\
&
\end{aligned}
$$
The value of $Z$ represents that element must have body centred cubic (bcc) structure.
For bcc structure atomic radius $(r)=\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4} \times 5=2.165 Å$
$$
=216.5 \mathrm{pm}
$$
Hence, the correct answer is 216.5.