1.

If the radius of the bromide ion is 0.182 nm, how large a cation can fit in each of the tetrahedral holes ?

Answer»


Solution :For a tetrahedral hole, `(r_(+))/(r_(-))= 0.225`
`:. r_(+)` (radius of tetrahedral hole) = `0.225 xx r_(-) =0.225 xx 0.182 NM= 4.01 xx 10^(-2)` nm
Therefore, CATION having radius of `4.01 xx 10^(-2)` nm will just FIT in the tetrahedral hole.


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