Question

Two different compounds of $Cr\left(III\right)$ have the same empirical formula, $CrC{l}_3\cdot 6H_{2}O$ one $\left(A\right)$ is green and the other $\left(B\right)$ is violet. $\left(A\right)$ on reaction with excess $AgN{O}_3$ gives $1$ mole of $AgCl$ per mol of $\left(A\right)$ while $\left(B\right)$ on reaction with excess $AgN{O}_3$ gives $3$ moles $AgCl$ per mol of $\left(B\right)$. Comment on Conductance of $A$ and $B$.

Answer 1

$\underset{1mol}{\left(A\right)}+AgN{O}_3\to \underset{1mol}{AgCl}$
Thus, $\left(A\right)$ has only ionisable $Cl$ atom. Thus, $A$ is $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$.
$\underset{1mol}{\left(B\right)}+AgN{O}_3\to \underset{3mol}{AgCl}$
Thus, $\left(B\right)$ has three ionisable $Cl-$ atoms. Thus, $B$ is $\left[Cr\right(H_{2}O{)}_6]C{l}_3$- an anhydrous complex.
$A.\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$,
$B.\left[Cr\right(H_{2}O{)}_6]C{l}_3$
$A$ will change to $\left[Cr\right(H_{2}O{)}_4\cdot C{l}_2]Cl$ on heating being hydrated complex. $B$ has no effect of heating, all $H_{2}O$ molecules being ligands. $A$ will ionize as $\left[Cr\right(H_{2}O{)}_{4}C{l}_2{]}^{+}$ and $C{l}^{-}$. $B$ will ionize as $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$ and $3C{l}^{-}$. Hence, conductance of $B$ is greater than that of $A$.