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A platinum-containing compound, known as Magnus's green salt, has the formula \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]\) (in which both platinum ions are \(\mathrm{Pt}^{2+}\) ). Name the cation and the anion.

Short Answer

Expert verified
The cation is "tetrammineplatinum(II)" and the anion is "tetrachloroplatinate(II)."

Step by step solution

01

Identify the Cation

The compound \[\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]\]consists of a cation and an anion. In this compound, the cation is surrounded by ammonia ligands (\(\mathrm{NH}_3\)), resulting in \[\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\]. The cation is named based on the central metal atom and its ligands. Since ammonia is a neutral ligand and the oxidation state of \(\mathrm{Pt}\) is \(+2\), it is simply called tetrammineplatinum(II). Therefore, the cation is named "tetrammineplatinum(II)."
02

Identify the Anion

The anion in the given compound is\[\left[\mathrm{PtCl}_{4}\right]^{2-}.\] The anion is also a complex ion where \(\mathrm{Pt}\) is the central metal atom surrounded by chloride ligands (\(\mathrm{Cl}^-\)). The complex anion is named based on the central metal atom, its ligands and its oxidation state. Since the oxidation state of \(\mathrm{Pt}\) here is \(+2\) and chloride is a negative ligand, the anion is named "tetrachloroplatinate(II)."

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Ions
Complex ions are fascinating structures in chemistry. They consist of a central metal ion, which can be a transition metal like platinum, surrounded by molecules or ions known as ligands. Ligands are capable of donating a pair of electrons to the metal, forming what is called a coordinate covalent bond. This interaction leads to the formation of a stable complex.

In the exercise, we examined a complex ion with platinum at its center. This ion is enclosed by ammonia ligands in the cation, and chloride ions in the anion. Complex ions are noted for their diverse geometries and colors, which arise due to the d-orbitals of the metal ion and ligand-field interactions.
  • Central metal ions often dictate the properties of the entire complex, including color and reactivity.
  • Different ligands can stabilize the metal in different oxidation states, affecting the overall charge of the complex ion.
Understanding complex ions is key to grasping larger concepts in coordination chemistry.
Naming Coordination Compounds
The naming of coordination compounds follows specific rules established by the International Union of Pure and Applied Chemistry (IUPAC). It's essential for ensuring clear communication among chemists. Let's break down how we named the example compound in this exercise.

For the cation \[\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\]1. **Start with Naming the Ligands**: Ammonia ligands are neutral and named 'ammine'.
2. **Specify the Number of Each Ligand**: There are four, so 'tetraammine'.
3. **Name the Metal and Its Oxidation State**: Platinum in oxidation state +2 is specified by '(II)'.
4. **Combine for Full Name**: "Tetrammineplatinum(II)".

For the anion,\[\left[\mathrm{PtCl}_{4}\right]^{2-}\]
  • Similar procedure is followed where chloride is the ligand called 'chloro'.
  • Four ligands again become 'tetrachloro'.
  • Include metal, platinum, and oxidation state, resulting in 'tetrachloroplatinate(II)'.
These clear naming conventions help chemists to accurately describe the composition and structure of coordination compounds globally.
Platinum Chemistry
Platinum is a significant element within chemistry, known for its applications and vibrant complexes. As a transition metal, platinum features prominently in coordination chemistry.

This metal is utilized widely due to its chemical properties:
  • **Catalysis**: Platinum is a core component in catalysts, especially in reduction reactions.
  • **Medical Use**: Compounds like cisplatin are crucial in cancer treatment.
  • **Versatile Bonds**: It can form stable complexes with various ligands.
In our example, platinum forms complex ions with ammonia and chloride. The interest in platinum's coordination compounds comes from their rich geometries, relatively inert nature to many reactions, and attractive electronic properties.

Additionally, studying platinum compounds like Magnus's green salt deepens our understanding of uses and innovations in industrial and research chemistry.
Oxidation States
Oxidation states indicate the degree of oxidation of an atom in a compound. They guide us in understanding electron distribution and charge balance within molecules.

For coordination compounds, knowing the oxidation states helps us name the structures accurately. It's especially crucial in complexes since the overall charge results from the metal's oxidation state along with the charges of surrounding ligands.
  • Stable oxide states for platinum include +2 and +4, among others.
  • In our complex, both Pt ions display an oxidation state of +2, essential for naming the complex ions correctly.
Recognizing oxidation states is also crucial for determining reactivity and potential applications within platinum complexes. In broader terms, understanding these states helps chemists manipulate compounds for desired reactions and purposes.

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Most popular questions from this chapter

In water, the titanium(III) ion, \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\) has a broad absorption band centered at about \(500 \mathrm{nm}\). What color light is absorbed by the ion?

Give the formula of the coordination complex formed from one \(\mathrm{Co}^{3+}\) ion, two ethylenediamine molecules, one water molecule, and one chloride ion. Is the complex neutral or charged? If charged, give the net charge on the ion.

As mentioned on page \(1047,\) transition metal organometallic compounds have found use as catalysts. One example is Wilkinson's catalyst, a rhodium compound \(\left[\mathrm{RhCl}\left(\mathrm{PR}_{3}\right)_{3}\right]\) used in the hydrogenation of alkenes. The steps involved in the catalytic process are outlined below. Indicate whether the rhodium compounds in each step have 18 - or 16 -valence electrons. (See Study Question \(34 .)\) Step \(1 .\) Addition of \(\mathrm{H}_{2}\) to the rhodium center of Wilkinson's catalyst. (For electron-counting purposes \(\mathrm{H}\) is considered a hydride ion, \(\left.\mathrm{H}^{-}, \text {a two-electron donor. }\right)\) Step \(2 .\) Loss of a PR \(_{3}\) ligand (a two-electron donor) to open a coordination site. (PR \(_{3}\) is a phosphine such as \(\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) triphenylphosphine.) Step \(3 .\) Addition of the alkene to the open site. Step 4. Rearrangement to add H to the double bond. (Here the \(-\mathrm{CH}_{2} \mathrm{CH}_{3}\) group is a two-electron donor and can be thought of as a \(\left[\mathrm{CH}_{2} \mathrm{CH}_{3}\right]^{-}\) anion for electron counting purposes.) Step \(5 .\) Loss of the alkane. Step \(6 .\) Regeneration of the catalyst. $$\text { Net reaction: } \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}$$

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: $$\begin{aligned} &\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\\\ &K_{f}=10^{8} \end{aligned}$$ $$\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq}) \quad \quad K_{\mathrm{f}}=10^{18}$$ The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{r} H^{\circ}\) for the NH seaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn},\) and \(\Delta_{r} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}\) -rxn. Is the difference in \(\Delta_{r} H^{\circ}\) sufficient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

One of the following nitrogen compounds or ions is not capable of serving as a ligand: \(\mathrm{NH}_{4}^{+}, \mathrm{NH}_{3}, \mathrm{NH}_{2}^{-} .\) Identify this species, and explain your answer.

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