Chapter 23, Problem 1DE

(a)

Interpretation Introduction

To determine: The experiment to confirm the formation of $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ in the given reaction.

Answer to Problem 1DE

Solution: The reaction of the given product with ${\text{AgNO}}_3$ leads to the formation of one mole of $\text{AgCl}$ which confirms the formation of $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ .

Explanation of Solution

The reaction of the given coordination compound with the solution of ${\text{AgNO}}_3$ solution gives the following reaction:

$\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}+{\text{AgNO}}_3\left(aq\right)\to {\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]}^{+}+\text{AgCl}\left(s\right)+{\text{NO}}_3^{-}\left(aq\right)$

Whereas, if the compound formed is $\left[\text{Co}{\left(\text{en}\right)}_3\right]{\text{Cl}}_3$ then the addition of ${\text{AgNO}}_3$ solution leads to the reaction,

$\left[\text{Co}{\left(\text{en}\right)}_3\right]{\text{Cl}}_3+{\text{AgNO}}_3\left(aq\right)\to {\left[\text{Co}{\left(\text{en}\right)}_3\right]}^{3+}+3\text{AgCl}\left(s\right)+3{\text{NO}}_3^{-}\left(aq\right)$

This will confirm the formation of desired product.

Conclusion

The reaction of the given product with ${\text{AgNO}}_3$ leads to the formation of one mole of $\text{AgCl}$ which confirms the formation of $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ .

(b)

Interpretation Introduction

To determine: If the cobalt present in the given reaction is in $+3$ oxidation state or not; also the spin state of cobalt in the given complex.

Answer to Problem 1DE

Solution: The oxidation state of cobalt is $+3$ in the given reaction and the spin state of cobalt is zero.

Explanation of Solution

The oxidation state of cobalt in $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ is calculated as,
$\left(\begin{array}{l}\text{Oxidation}\text{state}\text{of}\text{Co}+2\times \text{Oxidation}\text{state}\text{of}\\ \text{ethylene}\text{diammine}+2\times \text{Oxidation}\text{state}\text{of}\text{chlorine}\end{array}\right)=\text{Total}\text{charge}\text{on}\text{the}\text{molecule}$

Substitute the values of oxidation state of ethylene diammine, chlorine and the total charge on the molecule in the above equation,

$\begin{array}{c}\text{Co}+2\times 0\left(\text{en}\right)+2\times -1\left(\text{Cl}\right)=+1\\ -2+\text{Co}=+1\\ \text{Co}=+3\end{array}$

Therefore, the oxidation state of cobalt in $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ is $+3$ .
The electronic configuration of ${\text{Co}}^{3+}$ is,

${\text{Co}}^{3+}=\left[\text{Ar}\right]3{\text{d}}^{6}4{\text{s}}^0$

The ethylene diammine present in the given complex is a strong field ligand which causes pairing of electrons in metal ions.

Figure 1

Since, there is no unpaired electron present in the given complex. Therefore, the spin state of the cobalt is zero in the given complex.

Conclusion

The oxidation state of cobalt is $+3$ in the given reaction and the spin state of cobalt is zero.

(c)

Interpretation Introduction

To determine: The existence of geometrical isomerism in the complex $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ ; also the way to distinguish them in the reaction product.

Answer to Problem 1DE

Solution: The geometrical isomerism exists in the given complex and the isomers are of different colors which confirm their presence in the mixture.

Explanation of Solution

The geometrical isomerism is the isomerism in which the structure and molecular formula of the molecule is same but the orientation of groups around the metal atoms changes.
The molecule $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ contains two groups of one kind and two of other kind. Therefore, there are two possible orientations present for this molecule.
The geometrical isomers of the $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ molecule is,

Figure 2

The isomers of $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ are different in color and can be easily distinguished in the reaction medium.

Conclusion

The geometrical isomerism exists in the given complex and the isomers are of different colors which confirm their presence in the mixture.

(d)

Interpretation Introduction

To determine: The way to know the isomer which is formed as a product in the given reaction out of the two isomers.

Answer to Problem 1DE

Solution: The color of trans isomer is green and of cis isomer is violet which gives the confirmation of the geometry of product.

Explanation of Solution

The color of cis isomer of $\left[{\text{CoCl}}_2{\left(\text{en}\right)}_2\right]\text{Cl}$ is green and the color of trans isomer is violet. Therefore, if the product of the given reaction contains only one isomer then this is confirmed by the color of the product. Since, the product formed in the given reaction is green. Therefore, the product formed is a cis isomer.

Conclusion

The color of trans isomer is green and of cis isomer is violet which gives the confirmation of the geometry of product.