Q3 / 75 · VITEEE 2021 · 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 $2 \mathrm{~g}$ / cc then find the radius of metal atom. $\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}\right)$. Give the answer in $\mathrm{pm}$.

  • Correct answer
    A
    $217 \mathrm{pm}$
  • B
    $210 \mathrm{pm}$
  • C
    $220 \mathrm{pm}$
  • D
    $205 \mathrm{pm}$

Solution

$$ \begin{aligned} & \text { } \rho=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{V}} \\ & \mathrm{Z}=\frac{\rho \mathrm{N}_{\mathrm{A}} \mathrm{V}}{\mathrm{M}}=\frac{2 \times 6 \times 10^{23} \times\left(5 \times 10^{-8}\right)^3}{75} \\ & \mathrm{Z}=2, \text { which represents } b c c \text { structure } \\ & \therefore \quad r=\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4} \times 5=2.165 Å=216.5 \mathrm{pm} \approx \\ & 217 \mathrm{pm} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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