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Chapter 21: Problem 79

Give formulas for the following. a. hexakis(pyridine)cobalt(III) chloride b. pentaammineiodochromium(III) iodide c. tris(ethylenediamine)nickel(II) bromide d. potassium tetracyanonickelate(II) e. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Short Answer

Expert verified
\([Co(\text{C}_5\text{H}_5\text{N})_6]Cl_3\), \([Cr(\text{NH}_3)_5 \text{I}]I_2\), \([Ni(\text{en})_3]Br_2\), \(K_2[Ni(\text{CN})_4]\), and \([Pt(\text{NH}_3)_4\text{Cl}_2][PtCl_4]\).

Step by step solution

01

Identify the ligand and the metal

In this complex, the ligand is pyridine and it is present six times, indicated by 'hexakis'. The metal is cobalt and its oxidation state is III. The complex is a cation thus named as it is and the counter anion, chloride, follows.
02

Write down the formula

Because there are six pyridine ligands, we denote it as \( \text{C}_5\text{H}_5\text{N} \) and since cobalt has an oxidation state of III, the corresponding Roman numeral is included in parentheses. The chloride ion is written as \(\text{Cl}^{-}\). Putting it all together: \([Co(\text{C}_5\text{H}_5\text{N})_6]Cl_3\). b. pentaammineiodochromium(III) iodide
03

Identify the ligand and the metal

In this complex, ammonia is the ligand, indicated by 'ammine', and it is present five times, indicated by 'penta'. The metal is chromium and its oxidation state is III. The complex is a cation, thus it is named first, and the counter anion, iodide (I-), follows.
04

Write down the formula

Because there are five ammonia ligands, write it as \( \text{NH}_3 \). As it's 'iodo' it explicitly tells we also have an iodine directly bound to the chromium, so \( \text{I} \) is also a part of the complex. After, the anion iodide is written as \(\text{I}^{-}\). Putting it all together: \([Cr(\text{NH}_3)_5 \text{I}]I_2\). c. tris(ethylenediamine)nickel(II) bromide
05

Identify the ligand and the metal

In this complex, the ligand is ethylenediamine and it is present three times, indicated by 'tris'. The metal is nickel and its oxidation state is II. The complex is cationic and the counter anion, bromide, follows.
06

Write down the formula

Because there are three ethylenediamine ligands, we denote it as \( \text{en} \). Putting it all together: \([Ni(\text{en})_3]Br_2\). d. potassium tetracyanonickelate(II)
07

Identify the ligand and the metal

In this complex, the ligand is cyanide and it is present four times, indicated by 'tetra'. The metal is nickel and its oxidation state is II. The complex is anionic and the counter cation, potassium, is named first.
08

Write down the formula

Because there are four cyanide ligands, we denote it as \( \text{CN} \). Putting it all together: \(K_2[Ni(\text{CN})_4]\). e. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)
09

Identify the ligands and the metals

In this complex, there are two complex ions. In the first one, ammonia is the ligand (indicated by 'ammine') presented four times (indicated by 'tetra') while chloride (indicated by 'dichloro') is present twice. The metal is platinum with oxidation state IV. In the second complex, the ligand is also chloride presented four times (indicated by 'tetrachloro'). The metal is again platinum but with oxidation state II.
10

Write down the formula

In the first complex, there are four ammonia ligands, denoted as \( \text{NH}_3 \), and two chloride ligands, denoted as \( \text{Cl} \). Thus, the first complex is \([Pt(\text{NH}_3)_4\text{Cl}_2]\). In the second complex, there are four chloride ligands, so this complex is \([PtCl_4]^{2-}\). Putting it all together: \([Pt(\text{NH}_3)_4\text{Cl}_2][PtCl_4]\).

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Most popular questions from this chapter

A blast furnace is used to reduce iron oxides to elemental iron. The reducing agent for this reduction process is carbon monoxide. a. Given the following data: \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) & \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) & & \Delta H^{\circ}=-23 \mathrm{~kJ} \\ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) & \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) & & \Delta H^{\circ}=-39 \mathrm{~kJ} \\ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) & \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g) & & \Delta H^{\circ}=18 \mathrm{~kJ} \end{aligned}\) determine \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ b. The \(\mathrm{CO}_{2}\) produced in a blast furnace during the reduction process actually can oxidize iron into \(\mathrm{FeO}\). To eliminate this reaction, excess coke is added to convert \(\mathrm{CO}_{2}\) into \(\mathrm{CO}\) by the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \longrightarrow 2 \mathrm{CO}(g) $$ Using data from Appendix 4 , determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is the conversion reaction of \(\mathrm{CO}_{2}\) into CO spontaneous at standard conditions?

The \(\mathrm{CrF}_{6}{ }^{4-}\) ion is known to have four unpaired electrons. Does the \(\mathrm{F}^{-}\) ligand produce a strong or weak field?

Acetylacetone (see Exercise 43, part a), abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac-, as shown below: Acetylacetone reacts with an ethanol solution containing a salt of europium to give a compound that is \(40.1 \% \mathrm{C}\) and \(4.71 \% \mathrm{H}\) by mass. Combustion of \(0.286 \mathrm{~g}\) of the compound gives \(0.112\) \(\mathrm{g} \mathrm{Eu}_{2} \mathrm{O}_{3}\). Assuming the compound contains only \(\mathrm{C}, \mathrm{H}, \mathrm{O}\), and Eu, determine the formula of the compound formed from the reaction of acetylacetone and the europium salt. (Assume that the compound contains one europium ion.)

Draw structures of each of the following. a. \(c i s\) -dichloroethylenediamineplatinum(II) b. trans-dichlorobis(ethylenediamine) cobalt(II) c. cis-tetraamminechloronitrocobalt(III) ion d. trans-tetraamminechloronitritocobalt(III) ion e. trans-diaquabis(ethylenediamine)copper(II) ion

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?
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