Q. Which one is the most likely structure of CrCl₃ .6H₂O, if 1/3 of total chlorine of the compound is precipitated by adding AgNO₃ to its aqueous solution?

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A

$\text{CrCl}_{3} · 6 \text{H}_{2} \text{O}$

B

$\left[\text{Cr} \left(\left(\text{H}\right)_{2} \text{O}\right)_{3} \left(\text{Cl}\right)_{3}\right] · 3 \left(\text{H}\right)_{2} \text{O}$

C

$\left[\text{Cr} \left(\text{Cl}\right)_{2} \left(\left(\text{H}\right)_{2} \text{O}\right)_{4}\right] \text{Cl} \cdot \left(2 \text{H}\right)_{2} \text{O}$

D

$\left[\text{Cr} \text{Cl} \left(\left(\text{H}\right)_{2} \text{O}\right)_{5}\right] \left(\text{Cl}\right)_{2} · \left(\text{H}\right)_{2} \text{O}$

Solution:

$\text{CrCl}_{3} \cdot 6 \text{H}_{2} \text{O}$ has $3 - \text{Cl}$ atoms but $\frac{1}{3}$ $\text{Cl}$ atoms precipitated so one $1 \, \text{Cl }$ atom get precipitated.
According to Werner's theory, only those ions are precipitated which are attached to the metal atoms with ionic bonds and are present outside the co-ordination sphere.
$\left[\text{Cr} \left(\left(\text{H}\right)_{2} \text{O}\right)_{4} \left(\text{Cl}\right)_{2}\right] \text{Cl} \, \left(\right. \text{aq} \left.\right) \rightleftharpoons \text{ } \left(\left[\text{Cr} \left(\left(\text{H}\right)_{2} \text{O}\right)_{4} \left(\text{Cl}\right)_{2}\right]\right)^{+} \left(\text{aq}\right) + \left(\text{Cl}\right)^{-} \left(\text{aq}\right)$
$ \, \, \text{A}\left(\text{g}\right)^{+}+\text{C}\left(\text{l}\right)^{-} \, \rightarrow \text{A}\text{g}\text{C}\text{l} \, \downarrow\left(\text{w} \text{h} \text{i} \text{t} \text{e}\right)$

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