The ionic radii of `A^(+)` and `B^(-)` ions are `0.98xx10^(-10)m` and `1.81xx10^(-10)m`.The coordination number of each ion in `AB` is :

Options

A. 4
B. 8
C. 2
D. 6

Answer

D

Explanation

Given, ionic radius of cation `(A^(+))=0.98xx10^(-10)m`
Ionic radius of anion `(B^(-))=1.81xx10^(-10)m`
`therefore" Coordination number of each ion in AB = ?"`
Now, we have
`"Radius ratio"=("Radius of cation")/("Radius of anion")=(0.98xx10^(-10)m)/(1.81xx10^(-10)m)`
`=0.541`
If radius ratio range is in between `0.441-0.732`, ion would have octahedral structure with coordination number `six'.