Q22 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Chemistry

Which of the following is correct in the crystal lattice given below?

  • A
    Nearest neighbour distance in $\mathrm{NaCl}=\frac{a}{\sqrt{2}}$
  • B
    Total volume of a unit cell in CCP $=(r \sqrt{2})^3$
  • C
    Packing fraction of $\mathrm{BCC}$ is more than that of a FCC unit cell
  • Correct answer
    D
    Nearest neighbour distance in $\mathrm{CsCl}=a \frac{\sqrt{3}}{2}$

Solution

(a) $\mathrm{NaCl}$ has fcc unit cell nearest neighbour distance in fcc unit cell $=a \sqrt{2}$ So, option (a) is incorrect. (b) Total volume of a unit cell in ccp $=(2 \sqrt{2} r)^3$ So, option (b) is incorrect. (c) Packing fraction of bcc unit cell $=68.04 \%$ Packing fraction of fcc unit cell $=74 \%$ So, packing fraction of fcc is greater than that of bcc unit cell. So, option (c) is also incorrect. (d) For a bcc, nearest neighbour distance $=a \sqrt{3} / 2$. Since, $\mathrm{CsCl}$ is a body centered cubic crystal, so the nearest neighbour distance will be equal to $a \sqrt{3} / 2$

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

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