Q23.142 CP - Chemistry: Molecular Nature Of Matter And Change

Question

You know the following about a coordination compound:

(1) The partial empirical formula is $\mathit{K}\mathit{M}\mathbf{\left(}\mathbf{C}\mathbf{r}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right)}\mathit{C}{\mathit{l}}_{\mathbf{2}}{\mathbf{\left(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{4}}$

(2) It has A (red) and  B (blue) crystal forms.

(3) When  $\mathbf{1}\mathbf{.0}\mathbf{ }\mathbf{\text{mol}}$of A or B reacts with $\mathbf{1}\mathbf{.0}\mathbf{ }\mathbf{\text{mol}}$ of $\mathit{A}{\mathbf{\text{gNOH}}}_{\mathbf{3}}$  $\mathbf{0}\mathbf{.50}\mathbf{ }\mathbf{\text{mol}}$of a red precipitate forms immediately.

(4) After the reaction in (3),  $\mathbf{1}\mathbf{.0}\mathbf{ }\mathbf{\text{mol}}$of  A reacts very slowly with $\mathbf{1}\mathbf{.0}\mathbf{ }\mathbf{\text{mol}}$ of silver oxalate $\mathbf{\left(}{\mathbf{A}}_{{\mathbf{g}}_{\mathbf{2}}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right)}$ to form $\mathbf{2}\mathbf{.0}\mathbf{ }\mathbf{\text{mol}}$ of a white precipitate. (Oxalate can displace other ligands.)

(5) After the reaction in $\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{,}\mathbf{1}\mathbf{.0}\mathbf{ }\mathbf{\text{mol}}$ of B does not react further with I.0 mol of ${\mathbf{\text{AgNO}}}_{\mathbf{5}}$

From this information, determine the following:

(a) The coordination number of M

(b) The group(s) bonded to M ionically and covalently

(c) The stereochemistry of the red and blue formulas

Step-by-Step Solution

Answer

 

(a) The coordination number is 6.

(b) ${\text{K}}^{+}$and ${\text{CrO}}_4^{2-}$are attached to the metal covalently

(c) The cis isomer is compound A and another is B.

Step 1: Given information

(a)The given information is that

$\text{K}\left[\text{M}\left({\text{CrO}}_4\right){\text{Cl}}_2{\left({\text{NH}}_3\right)}_4\right]$ reacts with ${\text{AgNO}}_3$ to form a precipitate

Step 2: Definition of formation of complex of given compound

The  $\text{K}\left[\text{M}\left({\text{CrO}}_4\right){\text{Cl}}_2{\left({\text{NH}}_3\right)}_4\right]$reacts with ${\mathbf{\text{AgNO}}}_{\mathbf{3}}$ to form a precipitate that is half the molar amount of the reactant given, the charge of the ion reacting with ${\mathbf{\text{Ag}}}^{\mathbf{+}}$must be $-2 \text{A}$  known red precipitate of  $\mathbf{\text{Ag}}$is${\text{AgCrOC}}_4$ . Chromate ions have a charge of  and are found in the complex, therefore it must be a counter ion in the complex and not connected to the complex itself. Potassium, a cation, is a metal that cannot bind to the metal, so it must also be a counter ion. Therefore, the complex can be: $\mathbf{\text{K}}\mathbf{\left[}{\mathbf{\text{MCl}}}_{\mathbf{2}}{\mathbf{\left(}{\mathbf{\text{NH}}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{\right]}{\mathbf{\text{CrO}}}_{\mathbf{4}}$

 

When compound A reacts with ${\mathbf{\text{AgC}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{4}}$ and forms a white precipitate, the precipitate formed can be AgCl, with the oxalate chelating the metal, therefore the chloride ions attached to the metal must be found on the same side in compound A for it to happen.

Step 3: Find the coordination number of M

(a)

The complex ion therefore is ${\left[{\text{MCl}}_2{\left({\text{NH}}_3\right)}_4\right]}^{+}$. The metal has 6 atoms attached to it, giving a coordination number of 6.

 

Step 4: Find the group bonded to M

 (b)

${\text{K}}^{+}$  and ${\text{CrO}}_4^{2-}$are attached to the metal covalently, while ${\text{Cl}}^{-}$ and ${\text{NH}}_3$atoms are attached to the metal via coordinate covalent bonds.

 

Step 5: Find the stereochemistry of compounds

(c) 

The cis isomer is compound so that the oxalate ion can attach to the metal, and displace the chloride ions to form the white precipitate.