Question

Solve this:

$\mathbf{11}\mathbf{.}\mathbf{}\mathbf{}Asolid{A}^{+}{B}^{-}hasNaCltypeclosepackedstructure.Iftheanionhasaradiusof241.5pm,whatshouldbetheidealradiusofthecation?Canacation{C}^{+}havingradiusof50pmbefittedintotetrahedralholeofthecrystal{A}^{+}{B}^{-}?$

Answer 1

Dear Student,

Limiting radius ratio = (Size of Cation ,A+)/(Size of anion B-)

In NaCl

For a cation to fit exactly an octahedral hole

r+/r-=0.414

r+ = 0.414 r-

= 0.414*241.5

=99.98 pm =100pm( aprox)

Therefore ideal radius of cation is =100pm( aprox)

Radius of of anion = 0.225x r ^- = 0.225 x241.5pm = 54.3pm
As the radius of the cation C⁺ (50pm) is smaller than the size of the tetrahedral hole, it can be placed into the tetrahedral void (but not exactly fit into it).