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$X^{- 2}$ ions adopt the following arrangement in which only tetrahedral voids are occupied by the ions of a metal “M”

If 0.01 moles of the smallest repeating units of the above arrangement are present in “800 ml” aqueous solution, the reduction potential of the metallic electrode is 0.44 V (S.R.P of metal electrode = 0.5 V) $(α = 1)$ . Find the number of tetrahedral voids occupied by metal ions in that arrangement?

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answer is 8.

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Detailed Solution

The crystal structure of $0.44 = 0.5 - \frac{0.059}{1}$ by $(\frac{x \times 0.01}{0.8})$

$x = 8$

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