The interionic distance for caesium chloride crystal will be:

a

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\frac{a}{2}

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

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

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Solution

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

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Similar questions

Q. The interionic distance for cesium chloride crystal will be:

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The interionic distance for caesium chloride crystal will be:

The correct option is C √3a2The interionic distance for caesium chloride is between the Cs+ cation present at the body center and a Cl− anion present at the corner which is equal to half the length of diagonal = √3a2

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