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Chapter 24: Problem 25

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) \(\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{2}\left[\mathrm{TiCl}_{6}\right]\) (c) \(\mathrm{MoOCl}_{4}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{Br}_{2}\)

Short Answer

Expert verified
The names of the coordination complexes are: (a) tetrachlorobis(ammine)rhodium(III) chloride (b) potassium hexachlorotitanate(IV) (c) molybdenum(VI) tetrachlorooxide (d) dibromido-tetraaqua(oxalato)platinum(IV)

Step by step solution

01

Name compound (a)

Compound (a) is \(\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\). The metal is Rhodium (Rh), which means that we should use the name "Rhodium" in the compound's name. The ligands are \(\mathrm{NH}_{3}\), named "ammine" and \(\mathrm{Cl}\), named "chloro". There are 4 ammine ligands and 2 chloro ligands. The complex carries one negative charge, so the counterion is simply \(\mathrm{Cl}\), named "chloride". The name of the complex in alphabetical order is "tetrachlorobis(ammine)rhodium(III) chloride".
02

Name compound (b)

Compound (b) is \(\mathrm{K}_{2}\left[\mathrm{TiCl}_{6}\right]\). The metal in the complex is Titanium (Ti). The ligand attached is 6 chloro ligands (\(\mathrm{Cl}\)). The complex ion has a charge of -2. The counterions are 2 potassium ions (\(\mathrm{K}\)) naming "potassium". The name of the compound in alphabetical order is "potassium hexachlorotitanate(IV)".
03

Name compound (c)

Compound (c) is \(\mathrm{MoOCl}_{4}\). It consists of a Molybdenum (Mo) metal with oxygen and chloro ligands. There is no counterion since it's a neutral complex, so there is no need to indicate any cation. We have 1 oxygen atom and 4 chloro ligands. The name of the compound in alphabetical order is "molybdenum(VI) tetrachlorooxide".
04

Name compound (d)

Compound (d) is \(\left[\mathrm{Pt}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{Br}_{2}\). The metal is Platinum (Pt). The ligands are \(\mathrm{H}_{2}\mathrm{O}\), named "aqua" and \(\mathrm{C}_{2}\mathrm{O}_{4}\), named "oxalato". There are 4 aqua ligands and 1 oxalato ligand. The complex carries two positive charges, so the counterion is \(\mathrm{Br}_{2}\), each named "bromide". The name of the complex in alphabetical order is "dibromido-tetraaqua(oxalato)platinum(IV)".

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Most popular questions from this chapter

Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains \(\mathrm{Zn}^{2+}\), is responsible for rapidly interconverting dissolved \(\mathrm{CO}_{2}\) and bicarbonate ion, \(\mathrm{HCO}_{3}^{-} .\) The zinc in carbonic anhydrase is coordinated by three nitrogen-containing groups and a water molecule. The enzyme's action depends on the fact that the coordinated water molecule is more acidic than the bulk solvent molecules. Explain this fact in terms of Lewis acid-base theory (Section 16.11).

A palladium complex formed from a solution containing bromide ion and pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (a good electronpair donor), is found on elemental analysis to contain \(37.6 \%\) bromine, \(28.3 \%\) carbon, \(6.60 \%\) nitrogen, and \(2.37 \%\) hydrogen by mass. The compound is slightly soluble in several organic solvents; its solutions in water or alcohol do not conduct electricity. It is found experimentally to have a zero dipole moment. Write the chemical formula, and indicate its probable structure.

(a) What is the difference between Werner's concepts of primary valence and secondary valence? What terms do we now use for these concepts? (b) Why can the \(\mathrm{NH}_{3}\) molecule serve as a ligand but the \(\mathrm{BH}_{3}\) molecule cannot?

Sketch the structure of the complex in each of the following compounds: (a) \(c i s-\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{Ru}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{5}\right]\) (c) \(\operatorname{trans}-\mathrm{NH}_{4}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (d) \(c i s-\left[\operatorname{Ru}(e n)_{2} C l_{2}\right]\)

Suppose that a transition-metal ion was in a lattice in which it was in contact with just two nearby anions, located on opposite sides of the metal. Diagram the splitting of the metal \(d\) orbitals that would result from such a crystal field. Assuming a strong field, how many unpaired electrons would you expect for a metal ion with six \(d\) electrons? (Hint: Consider the linear axis to be the z-axis).
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