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Name the following coordination compounds. a. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\) b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) c. \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\) d. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

Short Answer

Expert verified
The short version answer for the given question is: a. Penta-aqua-bromo-chromium(III) tribromide b. Trisodium hexa-cyano-cobaltate(III) c. Di-ammine-di-nitrito-iron(III) chloride d. Tetra-ammine-di-iodido-platinum(IV) tetraiodidoplatinate(II)

Step by step solution

01

Naming Complex a: \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\)

First, list the ligands alphabetically, which are water, \(\mathrm{H}_{2}\mathrm{O}\) (called aqua) and bromide, \(\mathrm{Br}\). Since there are five water molecules, we use the prefix 'penta.' The complex formula is then written as penta-aqua-bromo-chromium(III). The overall name of this compound is penta-aqua-bromo-chromium(III) tribromide.
02

Naming Complex b: \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\)

For this compound, the ligand is cyanide which is named as cyano. We have six cyano ligands, so we use the prefix 'hexa' and name the complex ion as hexa-cyano-cobalt(III). Since there are three sodium ions, the full name of the compound is trisodium hexa-cyano-cobaltate(III).
03

Naming Complex c: \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\)

In this complex, we have the ligands en (ethylenediamine) and nitrito (NO鈧). We have two of each ligand, so we use the prefix 'di' for both. Since Fe is iron, and the complex ion has a negative charge, we add the suffix '-ate' to the metallic element name. The complex ion becomes di-ammine-di-nitrito-iron(III). Finally, add the chloride counterion, and the compound's name is di-ammine-di-nitrito-iron(III) chloride.
04

Naming Complex d: \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

This compound has two complex ions. For the first one, we have ammine (NH鈧) and iodido (I) as ligands with four ammines and two iodidos present, giving us tetra-ammine-di-iodido-platinum(IV). The second complex ion contains four iodido ligands, giving us tetraiodidoplatinum(II). The overall name of this compound is tetra-ammine-di-iodido-platinum(IV) tetraiodidoplatinate(II).

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Key Concepts

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

Naming Coordination Compounds
When it comes to naming coordination compounds, the process involves a specific set of rules to maintain consistency and clarity. Coordination compounds are combinations of central metal atoms or ions surrounded by nonmetal atoms or groups acting as ligands. The nomenclature for these compounds often follows a structured method. First, the ligands are named in alphabetical order, irrespective of their charge. Prefixes such as 'mono-', 'di-', 'tri-', 'tetra-' are used to indicate the number of each type of ligand present. After naming the ligands, we name the central metal atom. If the complex ion is an anion, the suffix '-ate' is often added to the metal. Additionally, the oxidation state of the metal in the complex is indicated in parentheses using Roman numerals immediately following the metal name. For example, in the complex \([\text{Cr(Cl}_3)\text{(H}_2\text{O)}_3]\), the name becomes "chlorotri-aquo-chromium(III)." This structured naming convention ensures that each element in the compound is accurately represented.
Ligands in Coordination Chemistry
In coordination chemistry, ligands play a crucial role. They are ions or molecules that bind to the central metal atom in a coordination complex. Each ligand possesses at least one donor atom, which donates a pair of electrons to the metal to form a coordinate bond. Ligands can be classified based on several factors:
  • Charge: Neutral (e.g., water, ammonia) or charged (e.g., chloride, cyanide).
  • Denticity: Monodentate, if they donate one pair of electrons from one donor atom (e.g., Cl鈦), bidentate like ethylenediamine, or polydenate if they can attach to the metal in more than two places.
  • Nature: Simple ligands like chloride, complex ligands like ethylenediamine, and chelating ligands which "hug" the central atom to form ring-like structures.
The ability of ligands to interact with metals gives rise to the extensive chemistry and diversity seen in coordination compounds.
Complex Ions Nomenclature
The nomenclature of complex ions is a well-defined part of coordination chemistry. Complex ions consist of a central metallic atom or ion closely bonded to surrounding ions or molecules known as ligands. In naming these complex ions, it's essential to follow a systematic approach.
1. Name the ligands first, in alphabetical order with appropriate prefixes based on their number (e.g., di-, tri-, tetra-).
2. The central metal is named next. For complex cations, the metal is named in its usual form. But for complex anions, the '-ate' suffix is added to the metal's root name.
3. Lastly, signify the oxidation number of the metal using Roman numerals within parentheses directly after the metal's name.
For example, in "hexa-amminecobalt(III) chloride," 'ammine' refers to the ammonia ligands, 'cobalt(III)' indicates cobalt in a +3 oxidation state, and the entire complex ion is treated as a cation.
Oxidation States in Coordination Compounds
Determining the oxidation state of a central metal within coordination compounds is crucial for correctly naming them. The oxidation state helps in ascertaining the amount of charge, positive or negative, the metal possesses in a given complex. To find the oxidation state:
  • Assign the known charges to the ligands and counter ions. Neutral ligands like ammonia (NH鈧) carry no charge, while anions like chloride are minus one (-1).
  • Set up an equation based on the total charge balance of the compound, which equals zero for neutral compounds or equals the charge on complex ions. Solve this equation to find the oxidation state of the metal.
For instance, in the compound \([\text{Co(NH}_3\text{)}_6]^{3+}\), each ammonia ligand is neutral, so their contribution is zero, leading to a total charge emanating solely from the cobalt ion, which is +3. Therefore, the oxidation state of cobalt here is +3. Recognizing and understanding these oxidation states is instrumental in both the naming and understanding of the chemical properties and reactivities of coordination compounds.

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Most popular questions from this chapter

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.

Iron is present in the earth's crust in many types of minerals. The iron oxide minerals are hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) and magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right) .\) What is the oxidation state of iron in each mineral? The iron ions in magnetite are a mixture of \(\mathrm{Fe}^{2+}\) and Fe \(^{3+}\) ions. What is the ratio of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) ions in magnetite? The formula for magnetite is often written as \(\mathrm{FeO} \cdot \mathrm{Fe}_{2} \mathrm{O}_{3} .\) Does this make sense? Explain.

Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin \((\mathrm{Hb})\) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\begin{aligned} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{kJ} \end{aligned}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$

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.

Which of the following crystal field diagram(s) is(are) correct for the complex given? a. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) (tetrahedral) b. \(\operatorname{Mn}(\mathrm{CN})_{6}^{3-}\) (strong field) c. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) (square planar, diamagnetic)

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