Q41 P - Chemistry: Molecular Nature Of Matter And Change

Question

When a basic solution of $\mathit{C}\mathit{r}{\mathbf{\left(}\mathit{O}\mathit{H}\mathbf{\right)}}_{\mathbf{4}}^{\mathbf{-}}$ ion is slowly acidified, solid $\mathit{C}\mathit{r}{\mathbf{\left(}\mathbf{O}\mathbf{H}\mathbf{\right)}}_{\mathbf{5}}$ precipitates and then redissolves in excess acid. If  width="69" height="29" style="max-width: none; vertical-align: -11px;" $\mathit{C}\mathit{r}{\mathbf{\left(}\mathit{O}\mathit{H}\mathbf{\right)}}_{\mathbf{4}}^{\mathbf{-}}$ exists as $\mathit{C}\mathit{r}{{}^{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{)}}_{\mathbf{2}}{\left(OH\right)}_{\mathbf{4}}^{\mathbf{-}}$, write equations that represent these two reactions.

Step-by-Step Solution

Answer

The solution of given statement is 

$\begin{array}{rcl} Cr{\left(OH\right)}_{4\left(aq\right)}^{-}+H_{\left(aq\right)}^{+}& \to & Cr{\left(OH\right)}_{3\left(s\right)}+H_2{O}_{\left(l\right)}\\ Cr{\left(H_{2}O\right)}_2{\left(OH\right)}_{4\left(aq\right)}^{-}+H_{\left(aq\right)}^{+}& \to & Cr{\left(OH\right)}_{3\left(s\right)}+3H_2{O}_{\left(l\right)}\end{array}$

Step 1: Definition of Coordination Compound

A coordination complex consists of a central atom or ion, which is typically metallic and is named the coordination centre, and a surrounding array of bound molecules or ions, that are successively called ligands or complexing agents.

Step 2: Solve the given statements

Basic $Cr{\left(OH\right)}_4^{-}$ turns into solid $Cr{\left(OH\right)}_3$ when slowly acidified. One of the OH - groups of the complex ion reacts with H⁺ in an acid-base neutralization reaction to form water:

$\begin{array}{rcl}Cr{\left(OH\right)}_{4\left(aq\right)}^{-}+H_{\left(aq\right)}^{+}& \to & Cr{\left(OH\right)}_{3\left(s\right)}+H_2{O}_{\left(l\right)}\end{array}$

The same happens when $Cr{\left(OH\right)}_4{\left(H_{2}O\right)}_2^{-}$ is given. The complex still turns into solid $Cr{\left(OH\right)}_3$ when slowly acidified; aside from releasing one of the OH groups it also releases two moles of water into the solution:

$\begin{array}{rcl}Cr{\left(H_{2}O\right)}_2{\left(OH\right)}_{4\left(aq\right)}^{-}+H_{\left(aq\right)}^{+}& \to & Cr{\left(OH\right)}_{3\left(s\right)}+3H_2{O}_{\left(l\right)}\end{array}$