A^{2+},B^{2+} and C^{-} form an ionic complex | vedclass

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(C) The dissociation of the complex $A_{x-2}[B(C)_x]_2$ is given by: $A_{x-2}[B(C)_x]_2 \longrightarrow (x-2) A^{2+} + 2[B(C)_x]^{-(x-2)}$ Total numbe

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$5$

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(C) The dissociation of the complex $A_{x-2}[B(C)_x]_2$ is given by: $A_{x-2}[B(C)_x]_2 \longrightarrow (x-2) A^{2+} + 2[B(C)_x]^{-(x-2)}$ Total number of ions produced per formula unit,$n = (x-2) + 2 = x$. The van't Hoff factor $i$ is given by the formula: $i = 1 + (n-1)\alpha$. Given $i = 4$ and $\alpha = 0.75$: $4 = 1 + (x-1)(0.75)$ $3 = (x-1)(0.75)$ $x-1 = \frac{3}{0.75} = 4$ $x = 5$. The coordination number of the central metal ion $B$ is equal to the number of ligands $C$ attached to it,which is $x = 5$.

$A^{2+}$,$B^{2+}$ and $C^{-}$ form an ionic complex like $A_{x-2}[B(C)_x]_2$. If the complex is $75 \%$ dissociated in a solvent with $i=4$,the coordination number of $B$ is:

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