(s) about crystal structures.… | JEE Advanced 2021 P1

(s) about crystal structures. (Avogadro constant NA=6.022×10^23 mol^-1.)

One or more options correct

Select the correct statement(s) about crystal structures. (Avogadro constant

N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}

A. In zinc blende ( $\mathrm{ZnS}$ ), $\mathrm{S^{2-}}$ ions form a ccp lattice and $\mathrm{Zn^{2+}}$ ions occupy all the tetrahedral voids
B. In a CsCl-type lattice in which cation and anion touch along the body diagonal and $r_+/r_-=0.732$ , the packing efficiency is about $73\%$
C. A metal of molar mass $27.0\ \mathrm{g\,mol^{-1}}$ that crystallises in an fcc lattice with edge length $405\ \mathrm{pm}$ has a density of about $2.70\ \mathrm{g\,cm^{-3}}$
D. In an fcc lattice, the packing efficiency is $74\%$ and there are $4$ octahedral and $8$ tetrahedral voids per unit cell

Show answer and solution

Answer: B, C, D

(A) is wrong: in zinc blende, the

\mathrm{S^{2-}}

ions are ccp and

\mathrm{Zn^{2+}}

ions occupy only half of the eight tetrahedral voids (four of the eight), as the formula

\mathrm{ZnS}

requires; filling all of them would give

\mathrm{Zn_2S}

. (B) is correct: at

r_+/r_-=0.732

the ions touch along the body diagonal,

a\sqrt3=2(r_++r_-)

, which for

r_-=1

gives

a=2

and anions touch along the edge. Packing efficiency

=\dfrac{\tfrac43\pi(1+0.732^3)}{8}=0.729

, that is

73\%

. (C) is correct:

\rho=\dfrac{4\times27.0}{6.022\times10^{23}\times(4.05\times10^{-8})^3}=2.70\ \mathrm{g\,cm^{-3}}

. (D) is correct: in fcc, packing efficiency

=\pi\sqrt2/6=74\%

, and there are

4

atoms,

4

octahedral and

8

tetrahedral voids per unit cell.

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