The interionic distance for cesium chloride crystal will be:

α

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\frac{α}{2}

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\frac{\sqrt{3} α}{2}

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\frac{2 α}{\sqrt{3}}

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Solution

The correct option is A $\frac{\sqrt{3} α}{2}$
The interionic distance for caesium chloride is between the $C s^{+}$ cation present at the body center and a $C l^{-}$ anion present at a the corner which is equal to half the length of diagonal $= \frac{\sqrt{3} a}{2} .$

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Similar questions

α

∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + α)^{2} & (1 + 2 α)^{2} & (1 + 3 α)^{2} (2 + α)^{2} & (2 + 2 α)^{2} & (2 + 3 α)^{2} (3 + α)^{2} & (3 + 2 α)^{2} & (3 + 3 α)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 648 α

\frac{\sqrt{3 a}}{2}

α^{t}

(α + p)^{m - 1} + (α + p)^{m - 2} (α + q) + (α + p)^{m - 3} (α + q)^{2} + . . . . . . . . (α + q)^{m - 1},

α \neq - q

p \neq q

(α + p)^{m - 1} + (α + p)^{m - 2} (α + q) + (α + p)^{m - 3} (α + q)^{2} + . . . . . . . . (α + q)^{m - 1},

4 n α = π

cot α \cdot cot 2 α \cdot cot 3 α \cdot . . . cot (2 n - 1) α

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