Question

The coordination number of $Sr^{2+}$ ion in $SrF_2$, if ionic radii are r $(Si^{2+})$ = 113 pm and r $(F^-)$ = 136 pm; is

• A. 6
• B. 8
• C. 12
• D. 4

Answer 1

Answer: (B)

8

Answer 2

Radius ratio $= \frac{r^+}{r^-} = \frac{113}{136}$ $= 0.830 > 0.732$. The coordination number of $Sr^{2+}$ is $8$.