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Chapter 21: Problem 88

For the process $$\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+}(a q)+\mathrm{NH}_{3}(a q)$$ what would be the expected ratio of cis to trans isomers in the product?

Short Answer

Expert verified
The expected ratio of cis to trans isomers in the product [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) is 0:1, as only the trans isomer is formed during the substitution process.

Step by step solution

01

Understand the initial complex structure

The initial complex in this process is [Co(NH\(_3\))\(_5\)Cl]\(^{2+}\), which means we have a central Co atom surrounded by 5 NH\(_3\) ligands and 1 Cl ligand. Considering the coordination geometry of Co, it has an octahedral structure. Now, let's understand the binding sites opened up by the substitution of one ammonia.
02

Analyze the substitution process

The given reaction represents the substitution of one NH\(_3\) molecule by a Cl\(^-\) ion. The substitution process occurs at an NH\(_3\) ligand that is trans to the initial Cl ligand. This means a Cl\(^-\) will replace an NH\(_3\) molecule which is opposite to the Cl\(^-\) attached to the central Co ion in the initial complex.
03

Analyze the product complex structure

After the substitution, the new complex becomes [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\). Here, we have a central Co ion surrounded by 4 NH\(_3\) ligands and 2 Cl ligands. In this structure, the two Cl ligands can be either adjacent (cis) or opposite (trans) to each other.
04

Determine the cis and trans isomers in the complex

Since the substitution takes place at an NH\(_3\) ligand that is trans to the initial Cl ligand in the [Co(NH\(_3\))\(_5\)Cl]\(^{2+}\), a trans isomer of [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) is formed. As there is no mention of any other reaction that would lead to the formation of cis isomers, the product will consist of only the trans isomer. Therefore, there will be no cis isomer formed.
05

Calculate the expected ratio of cis to trans isomers

Since there are no cis isomers formed in this process, the expected ratio of cis to trans isomers in the product is 0:1, where 0 represents the cis isomer and 1 represents the trans isomer. In conclusion, the expected ratio of cis to trans isomers in the product [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) is 0:1.

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Most popular questions from this chapter

Compounds of \(\mathrm{Sc}^{3+}\) are not colored, but those of \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) are. Why?

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosul-fate. The equation for the reaction is: $$\operatorname{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q) \longrightarrow \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q)$$ c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ Assume these copper complex ions have tetrahedral geometry.

Use standard reduction potentials to calculate $\mathscr{C}^{\circ}, \Delta G^{\circ},\( and \)K$ (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

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