Q22 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Chemistry

In a bcc lattice having the edge length of $200 \mathrm{pm}$, the cation has the radius of $70 \mathrm{pm}$. The radius ratio of $\frac{r^{+}}{r^{-}}$is (Given, $\sqrt{2}=1.4, \sqrt{3}=1.7$ and $\sqrt{6}=2.4$ )

  • Correct answer
    A
    0.7
  • B
    1
  • C
    0.4
  • D
    0.2

Solution

$\because$ For bcc lattice, radius ratio for $r^{+} / r^{-}$is 0.7 , i.e. radius ratio of cation by anion ranges from 0.732 to 1.0. Also, for bcc $r^{-}$ions are at corners, Thus, $4 \times r^{-}=\sqrt{3} a \quad(\because \sqrt{3} a=1.7 \times 200)$ (where, $a=$ edge length) and $\quad 4 \times r^{-}=1.7 \times 200$ $r^{-}=\frac{1.7 \times 200}{4}=85 \mathrm{pm}$ $\therefore \quad \frac{r^{+}}{r^{-}}=\frac{70}{85}=0.82$, which is close to 0.7.

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

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