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Chemistry
The radius of Ag+ ion is 126 pm and that of I- ion is 216 pm. The coordination number of Ag+ ion is
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Q. The radius of $Ag^+$ ion is $126$ pm and that of $I^-$ ion is $216$ pm. The coordination number of $Ag^+$ ion is
The Solid State
A
$2$
0%
B
$4$
0%
C
$6$
50%
D
$8$
50%
Solution:
$\frac{r_{+}}{r_{-}} = \frac{126}{216} = 0.58$,
octahedral voids. Hence $C.N. = 6$.