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Chemistry
The coordination number of Sr2+ ion in SrF2, if ionic radii are r (Si2+) = 113 pm and r (F-) = 136 pm; is
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Q. 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
The Solid State
A
6
21%
B
8
48%
C
12
18%
D
4
13%
Solution:
Radius ratio $ = \frac{r^+}{r^-} = \frac{113}{136}$
$ = 0.830 > 0.732$.
The coordination number of $Sr^{2+}$ is $8$.