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Chapter 21: Problem 13

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

Short Answer

Expert verified
The ligands capable of linkage isomerism are \(\mathrm{SCN}^{-}\) (through S or N), \(\mathrm{N}_{3}^{-}\) (through the terminal N), \(\mathrm{NO}_{2}^{-}\) (through N or O), and \(\mathrm{OCN}^{-}\) (through O or N).

Step by step solution

01

Ligand 1: \(\mathrm{SCN}^{-}\)

The ligand can connect to the metal through the sulfur (S) or nitrogen (N) atom.
02

Ligand 2: \(\mathrm{N}_{3}^{-}\)

The ligand can connect to the metal through the terminal nitrogen (N) atoms.
03

Ligand 3: \(\mathrm{NO}_{2}^{-}\)

The ligand can connect to the metal through the nitrogen (N) or one of the oxygen (O) atoms.
04

Ligand 4: \(\mathrm{NH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\)

This is an example of a bidentate ligand, with two nitrogen (N) atoms as potential binding sites but they will bind simultaneously to the metal center, so there is no linkage isomerism.
05

Ligand 5: \(\mathrm{OCN}^{-}\)

The ligand can connect to the metal through the oxygen (O) or nitrogen (N) atom.
06

Ligand 6: \(\mathrm{I}^{-}\)

This ligand has only one atom, iodine (I), which can bind to the metal center, so it cannot exhibit linkage isomerism. #Step 2: Determine which ligands can exhibit linkage isomerism#
07

Linkage isomerism-capable ligands

The ligands capable of linkage isomerism are: 1. \(\mathrm{SCN}^{-}\) (through S or N) 2. \(\mathrm{N}_{3}^{-}\) (through the terminal N) 3. \(\mathrm{NO}_{2}^{-}\) (through N or O) 4. \(\mathrm{OCN}^{-}\) (through O or N) Thus, the ligands \(\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{OCN}^{-}\) are capable of linkage isomerism.

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Most popular questions from this chapter

A coordination compound of cobalt(III) contains four ammonia molecules, one sulfate ion, and one chloride ion. Addition of aqueous \(\mathrm{BaCl}_{2}\) solution to an aqueous solution of the compound gives no precipitate. Addition of aqueous \(\mathrm{AgNO}_{3}\) to an aqueous solution of the compound produces a white precipitate. Propose a structure for this coordination compound.

The ferrate ion, \(\mathrm{FeO}_{4}{ }^{2-}\), is such a powerful oxidizing agent that in acidic solution, aqueous ammonia is reduced to elemental nitrogen along with the formation of the iron(III) ion. a. What is the oxidation state of iron in \(\mathrm{FeO}_{4}{ }^{2-}\), and what is the electron configuration of iron in this polyatomic ion? b. If \(25.0 \mathrm{~mL}\) of a \(0.243 \mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution is allowed to react with \(55.0 \mathrm{~mL}\) of \(1.45 M\) aqueous ammonia, what volume of nitrogen gas can form at \(25^{\circ} \mathrm{C}\) and \(1.50 \mathrm{~atm}\) ?

Draw all geometrical and linkage isomers of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\).

Would it be better to use octahedral \(\mathrm{Ni}^{2+}\) complexes or octahedral \(\mathrm{Cr}^{2+}\) complexes to determine whether a given ligand is a strong-field or weak-field ligand by measuring the number of unpaired electrons? How else could the relative ligand field strengths be determined?

The complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}\) is paramagnetic with one unpaired electron. The complex ion \(\mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-}\) has five unpaired electrons. Where does \(\mathrm{SCN}^{-}\) lie in the spectrochemical series relative to \(\mathrm{CN}^{-}\) ?
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