If the radius of ${Br}^{-}$ ion is $0.182 nm$ , how large can a cation be fit in its tetrahedral holes?

$0.0753 nm$

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$0.092 nm$

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$0.125 pm$

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$0.554 pm$

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Solution

The correct option is A
$0.0753 nm$

Explanation:
Calculating the size of the cation:
Given,
Radius of ${Br}^{-}$ ion = $0.182 nm$
According to the radius ratio rule, for a tetrahedral void, it should be $= 0.225 - 0.414$ (maximum)
To fit into the tetrahedral void the cation size can be calculated as;
$0.414 equals r to the power of plus over r to the power of minus sign rightwards double arrow 0.414 equals fraction numerator r to the power of plus sign over denominator 0.182 end fraction comes from r to the power of plus sign equals 0.414 cross times 0.182 space space space space space space space equals 0.0753 space nm$
Hence, If the radius of ${Br}^{-}$ ion is $0.182 nm$ , the cation size that can be fit in its tetrahedral holes is calculated as $0.0753 nm$ .

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