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Chapter 21: Problem 35

Name the following coordination compounds. a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\) d. \(\mathrm{K}_{4}\left[\mathrm{PtCl}_{6}\right]\) b. \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}\) e. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) c. \(\mathrm{K}_{2}\left[\mathrm{PtCl}_{4}\right]\) f. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)

Short Answer

Expert verified
The names of the coordination compounds are: a. hexaamminecobalt(III) chloride d. potassium hexachloroplatinate(IV) b. hexaaquacobalt(III) iodide e. pentaamminechloridocobalt(III) chloride c. potassium tetrachloroplatinate(II) f. triamminetrititrocobalt(III)

Step by step solution

01

(a. Identifying elements and naming \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\))

In this coordination compound, the central metal atom is cobalt (Co) and the ligand is ammonia (NH\(_{3}\)) with a coordination number of 6. The compound has two chlorine (Cl) counterions. The name of this compound is: hexaamminecobalt(III) chloride. Notice that roman numeral III is used to represent the +3 oxidation state of cobalt.
02

(d. Identifying elements and naming \(\mathrm{K}_{4}\left[\mathrm{PtCl}_{6}\right]\))

In this coordination compound, the central metal atom is platinum (Pt) and the ligand is chloride (Cl) with a coordination number of 6. The compound has four potassium (K) counterions. The name of this compound is: potassium hexachloroplatinate(IV). Notice that roman numeral IV is used to represent the +4 oxidation state of platinum.
03

(b. Identifying elements and naming \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}\))

In this coordination compound, the central metal atom is cobalt (Co) and the ligand is water (H\(_{2}\)O) with a coordination number of 6. The compound has three iodide (I) counterions. The name of this compound is: hexaaquacobalt(III) iodide. Notice that roman numeral III is used to represent the +3 oxidation state of cobalt.
04

(e. Identifying elements and naming \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\))

In this coordination compound, the central metal atom is cobalt (Co). There are two types of ligands: ammonia (NH\(_{3}\)) with a coordination number of 5 and chloride (Cl) with a coordination number of 1. The compound has two chlorine (Cl) counterions. The name of this compound is: pentaamminechloridocobalt(III) chloride. Notice that roman numeral III is used to represent the +3 oxidation state of cobalt.
05

(c. Identifying elements and naming \(\mathrm{K}_{2}\left[\mathrm{PtCl}_{4}\right]\))

In this coordination compound, the central metal atom is platinum (Pt) and the ligand is chloride (Cl) with a coordination number of 4. The compound has two potassium (K) counterions. The name of this compound is: potassium tetrachloroplatinate(II). Notice that roman numeral II is used to represent the +2 oxidation state of platinum.
06

(f. Identifying elements and naming \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\))

In this coordination compound, the central metal atom is cobalt (Co). There are two types of ligands: ammonia (NH\(_{3}\)) and nitrite (NO\(_{2}\)) both with a coordination number of 3. The compound has no counterions. The name of this compound is: triamminetrititrocobalt(III). Notice that roman numeral III is used to represent the +3 oxidation state of cobalt.

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Most popular questions from this chapter

The complex ion Ru(phen) \({ }^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in \(\mathrm{Ru}(\mathrm{phen})_{3}^{2+} ?\) b. Ru(phen) \(_{3}^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+}\) ). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) 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 $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

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 \(\mathrm{A}\) ), and the following data are collected: i. When \(0.105 \mathrm{~g}\) of compound \(\mathrm{A}\) 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 \(0.100 \mathrm{M} \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{~g}\) compound \(\mathrm{A}\). 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 \(0.601\) g compound \(A\) was dissolved in \(10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) \(\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be sixcoordinate, with \(\mathrm{NH}_{3}\) and possibly \(\mathrm{I}^{-}\) as ligands. The \(\mathrm{I}^{-}\) ions will be the counterions if needed.)

The compound cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: $$\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)$$ Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as \(\mathrm{MoS}_{2}\), which is then converted to \(\mathrm{MoO}_{3}\). The \(\mathrm{MoO}_{3}\) can be used directly in the production of stainless steel for high-speed tools (which accounts for about \(85 \%\) of the molybdenum used). Molybdenum can be purified by dissolving \(\mathrm{MoO}_{3}\) in aqueous ammonia and crystallizing ammonium molybdate. Depending on conditions, either \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}\) or \(\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is obtained. a. Give names for \(\mathrm{MoS}_{2}\) and \(\mathrm{MoO}_{3}\). b. What is the oxidation state of Mo in each of the compounds mentioned above?
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