JEE CHEMISTRY MCQ: SOLID STATE

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Analyzing +4 | -1

Distance between nearest cation and anion in an ionic solid is

\frac{\sqrt{3} a}{4}

then ionic solid may be:

View answer and explanation

Correct option: D

All

Option D (All) is correct because in ZnS,

Na_2O

, and

CaF_2

structures, the distance between the nearest cation and anion can be

\frac{\sqrt{3} a}{4}

when the cation occupies a tetrahedral void.

More Questions from SOLID STATE

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{CaF}_2

{F}^{-}

\mathrm{d}_{\mathrm{e}}

\mathrm{d}_{\mathrm{fd}}, \mathrm{d}_{\mathrm{bd}}

\mathrm{AB}_2

\mathrm{A}