The density of sodium chloride at 25°C is 2.163 x 10³ kg m^-3. When X-rays from a palladium target having a wavelength of 58.1 pm are rsd the (200) reflection of sodium chloride occurs at an angle of 5.90°. How many Na⁺ and Cl^- ions are there in a unit cell? (NaCl = 58.443)

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4 Na⁺, 4 Cl^-

4 Na⁺, 2 Cl^-

2 Na⁺, 4Cl^-

3 Na⁺, 1 Cl^-

answer is C.

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Detailed Solution

We have, $nλ = 2 dsin θ$
For $n = 1, d_{200} = \frac{λ}{2 \sin θ} = \frac{58 .1 (pm)}{2 \sin 5 {.9}^{\circ}} = 282 pm$
Now, $d_{200} = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}}$
$282 (pm) = \frac{a}{\sqrt{2^{2} + 0^{2} + 0^{2}}}; a = 564 pm$
Thus, using the equation
$\begin{array}{l} ρ = \frac{zM}{NV} = \frac{zM}{{Na}^{3}} \\ ∴ = \frac{ρ {Na}^{3}}{M} = \frac{2 .163 \times 10^{3} \times (6 .022 \times 10^{23}) \times {(564 \times 10^{- 12})}^{3}}{58 .443 \times 10^{- 3}} = 3 .999 \times 4 \end{array}$
Thus, a unit cell contains 4 Na⁺ and 4Cl^- ions as expected.