Question

Four different octahedral chromium coordination compounds exist that all have the same oxidation state for chromium and have$H_{2}O$ and as $C{l}_2$the ligands and counterions. When 1 mole of each of the four compounds is dissolved in water, how many moles of silver chloride will precipitate upon addition of excess $AgN{O}_3$?

Short answer

Four different octahedral chromium coordinate compounds having $H_{2}O$and $C{l}_2$are as follows:

$$\begin{gathered} a.\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3 \\ b.\left[Cr{\left(H_{2}O\right)}_{5}Cl\right]C{l}_2 \\ c.\left[Cr{\left(H_{2}O\right)}_{4}C{l}_2\right]Cl \\ d.\left[Cr{\left(H_{2}O\right)}_3\right]C{l}_3 \end{gathered}$$

Step-by-step solution

Reaction

When one of the first compounds (a) is dissolved in water, 3 moles of silver chloride will precipitate out in excess of $AgN{O}_3$.

Reaction is as follows:

$\left[Cr(H_{2}O{)}_6\right]C{l}_3\left(aq\right)+3AgN{O}_3\left(aq\right)\to [Cr(H_{2}O{)}_6{]}^{+3}\left(aq\right)+3N{O}_{3^{-}}\left(aq\right)+3AgCl\left(s\right)$

Mole of silver cloride

Same in on addition of second compound (b) 2 moles of silver chloride will precipitate out.

In compound (c) 1 m0 mole of silver chloride will precipitate out and in compound (d) 0 mole of silver chloride will precipitate out.