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Chapter 21: Problem 52

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 \(\mathrm{CN}^{-}\) ?

Short Answer

Expert verified
In the spectrochemical series, the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}\) has fewer unpaired electrons (one) than the complex ion \(\mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-}\) (five), indicating that \(\mathrm{CN}^{-}\) is a stronger ligand than \(\mathrm{SCN}^{-}\). Thus, \(\mathrm{SCN}^{-}\) lies lower in the spectrochemical series relative to \(\mathrm{CN}^{-}\).

Step by step solution

01

Identify the metal ion and its oxidation state in both complexes

We are given the complex ions \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}\) and \(\mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-}\). The metal ion in both complexes is Iron (Fe). To find the oxidation state of the Fe ion in both complexes, we can set up the following equation: Fe + (oxidation state of ligand) * (number of ligands) = charge on complex For \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}\), the oxidation state of \(\mathrm{CN}^{-}\) is -1, so we have: Fe + (-1) * 6 = -3 => Fe = +3 For \(\mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-}\), the oxidation state of \(\mathrm{SCN}^{-}\) is -1, so we have: Fe + (-1) * 6 = -3 => Fe = +3 In both complexes, the oxidation state of Fe is +3.
02

Analyze the unpaired electrons in each complex

We are given that \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}\) has one unpaired electron and \(\mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-}\) has five unpaired electrons.
03

Determine the position of \(\mathrm{SCN}^{-}\) relative to \(\mathrm{CN}^{-}\) in the spectrochemical series

The unpaired electrons are determined by the energy difference between the paired and unpaired d-orbitals in a metal complex, which is affected by the ligand strength in the spectrochemical series. We are given that the complex ion with \(\mathrm{CN}^{-}\) has fewer unpaired electrons (one) than the complex ion with \(\mathrm{SCN}^{-}\) (five). This suggests that \(\mathrm{CN}^{-}\) is a stronger ligand than \(\mathrm{SCN}^{-}\) and causes a greater energy splitting between the d-orbitals. Therefore, \(\mathrm{SCN}^{-}\) lies lower in the spectrochemical series relative to \(\mathrm{CN}^{-}\), as it is a weaker ligand.

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Most popular questions from this chapter

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Zn}^{2+}\) b. \(\mathrm{Co}^{2+}\) (high and low spin) c. \(\mathrm{Ti}^{3+}\)

The complex ion \(\mathrm{PdCl}_{4}{ }^{2-}\) is diamagnetic. Propose a structure for \(\mathrm{PdCl}_{4}^{2-}\)

Name the following complex ions. a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}^{2+}\)

Name the following coordination compounds. a. \(\mathrm{Na}_{4}\left[\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) b. \(\mathrm{K}_{2}\left[\mathrm{CoCl}_{4}\right]\) c. \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}\) d. \(\left[\mathrm{Co}(\mathrm{en})_{2}(\mathrm{SCN}) \mathrm{Cl}\right] \mathrm{Cl}\)

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 \(1.50 \mathrm{~atm}\) ?
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