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Chapter 21: Problem 78

Until the discoveries of Alfred Werner, it was thought that carbon had to be present in a compound for it to be optically active. Werner prepared the following compound containing \(\mathrm{OH}^{-}\) ions as bridging groups and separated the optical isomers. a. Draw structures of the two optically active isomers of this compound. b. What are the oxidation states of the cobalt ions? c. How many unpaired electrons are present if the complex is the low-spin case?

Short Answer

Expert verified
The question provided does not contain enough information to give a complete answer. If a cobalt complex were given, we would be able to draw the structures of the two optically active isomers, determine the oxidation states of cobalt ions, and calculate the number of unpaired electrons if the complex is low-spin by following the steps explained in the solution. However, without the complex structure, these tasks cannot be completed.

Step by step solution

01

Draw structures of the two optically active isomers of this compound

First, we need to draw the structures of the two optically active isomers of the cobalt complex. To do this, we can make use of the fact that optical isomers are non-superimposable mirror images of each other. There should be an axis of symmetry in each of these isomers that differentiates them. Please note that the exercise did not provide the complex, therefore we cannot draw the isomers.
02

Determine the oxidation states of the cobalt ions

Without having the cobalt complex, we cannot determine the oxidation states of cobalt ions in it. Normally, to determine the oxidation states, you would look for the charges on the ligands and counterions present in the complex and apply the oxidation state formula: \[x + \text{(charge contributed by all ligands)} = \text{total charge on the ion}\] Here, x represents the oxidation state of the central metal atom (in this case, cobalt).
03

Calculate the number of unpaired electrons if the complex is low-spin

Again, without the complex structure, we can't calculate the number of unpaired electrons if the complex is low-spin. To calculate the number of unpaired electrons, we generally need to know the oxidation state, coordination geometry (e.g., octahedral or tetrahedral), and type of ligands surrounding the metal ion as well as applying the Crystal Field Theory.

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Most popular questions from this chapter

Figure 21.17 shows that the cis isomer of $\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}^{+}$ is optically active while the trans isomer is not optically active. Is the same true for \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}+?\) Explain.

Consider aqueous solutions of the following coordination compounds: $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}, \operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}, \mathrm{Na}_{2} \mathrm{Pt}_{6},$ and \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\) . If aqueous AgNO \(_{3}\) is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.

Consider the complex ions $\mathrm{Co}\left(\mathrm{NH}_{3}\right) 6^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-},\( and \)\mathrm{CoF}_{6}^{3-} .$ The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are \(770 \mathrm{nm},\) \(440 \mathrm{nm},\) and 290 $\mathrm{nm} .$ Match the complex ion to the wave- length of absorbed electromagnetic radiation.

Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess $\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}$ was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100$M \mathrm{HCl}\( to titrate completely the \)\mathrm{NH}_{3}\( present in \)0.341 \mathrm{g} \mathrm{com}-$ pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 $\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\( \)1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )$ What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)
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