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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Chemistry
Chapter
Solid State

JEE Advanced 2006 · Chemistry · Solid State

The edge length of unit cell of a metal having molecular weight 75...

The edge length of unit cell of a metal having molecular weight 75 g/mol75 \mathrm{~g} / \mathrm{mol} is 5 Å which crystallizes in cubic lattice. If the density is 2 g/ccthenfindtheradiusofmetalatom,(NA=6×1023)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

0

Step-by-step solution

For cubic lattice, ρ=Z×ANA×Vor\rho=\frac{Z \times A}{N_A \times V} orZ=\frac{\rho \times N_A \times V}{A}Giventhat,Given that,A=75 g/molV=(5A˚)3=(5×10−8)3 cm3=125×10−24 cm3NA=6.02×1023ρ=2gm/cc∴Z=2×6.02×1023×125×10−2475≈2\begin{aligned}& A=75 \mathrm{~g}/ \mathrm{mol}\\ & V=(5 Å)^3=\left(5 \times10^{-8}\right)^3 \mathrm{~cm}^3 \\ & =125 \times10^{-24} \mathrm{~cm}^3 \\ & N_{\mathrm{A}}=6.02 \times10^{23} \\ & \rho=2 \mathrm{gm}/ \mathrm{cc}\\ & \therefore \quad Z=\frac{2 \times 6.02 \times 10^{23} \times 125 \times 10^{-24}}{75} \approx2 \\ & \end{aligned}ThevalueofThe value ofZrepresentsthatelementmusthavebodycentredcubic(bcc)structure.Forbccstructureatomicradiusrepresents 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=216.5\mathrm{pm}$
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