Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,
[tex]\frac{r_{cation}}{r_{anion}}[/tex]
When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation, [tex]Na^+[/tex] = 102 pm
Radius of cation, [tex]Br^-[/tex] = 196 pm
[tex]\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520[/tex]
As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
