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Chapter 20: Problem 56

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 SCN \(^{-}\) lie in the spectrochemical series relative to \(\mathrm{CN}^{-} ?\)

Short Answer

Expert verified
In conclusion, SCN \(-\) lies below CN \(-\) in the spectrochemical series. This difference in their positions accounts for the difference in their paramagnetism and the unpaired electron count in the two given complex ions.

Step by step solution

01

Understand the Ligands in Complex Ions

In the given exercise, we have two complex ions: \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) and \(\mathrm{Fe}(\mathrm{SCN})_{6}^{3-}\). To understand the electronic configurations of these complex ions and their paramagnetism, we need to know about the ligands, which are the CN \(^{-}\) and SCN \(^{-}\) ions.
02

Recall the Spectrochemical Series

The spectrochemical series is an order of ligands based on the strength of their interaction with the central metal ion. Strong-field ligands result in a larger splitting of the d-orbitals, while weak-field ligands cause lesser splitting. A larger splitting leads to the pairing of electrons, reducing the number of unpaired electrons in a complex ion. Conversely, smaller splitting results in a higher number of unpaired electrons.
03

Observe the Paramagnetism of Complex Ions

From the given information, we have that \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is paramagnetic with one unpaired electron, and \(\mathrm{Fe}(\mathrm{SCN})_{6}^{3-}\) is also paramagnetic but with five unpaired electrons. This information tells us that the CN \(^{-}\) ligand causes a large splitting in the d-orbitals, leading to a higher degree of electron pairing, while the SCN \(^{-}\) ligand results in a smaller splitting, producing a higher number of unpaired electrons.
04

Compare CN \(^{-}\) and SCN \(^{-}\) in the Spectrochemical Series

Based on the number of unpaired electrons and their effect on the splitting of d-orbitals, it is clear that CN \(^{-}\) is a stronger field ligand than SCN \(^{-}\). Thus, SCN \(^{-}\) lies lower in the spectrochemical series relative to CN \(^{-}\). In conclusion, SCN \(^{-}\) lies below CN \(^{-}\) in the spectrochemical series. This difference in their positions accounts for the difference in their paramagnetism and the unpaired electron count in the two given complex ions.

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Most popular questions from this chapter

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

In which of the following is(are) the electron configuration(s) correct for the species indicated? a. Cu \([\mathrm{Ar}] 4 s^{2} 3 d^{9}\) b. \(\mathrm{Fe}^{3+} \quad[\mathrm{Ar}] 3 d^{5}\) c. Co \([\mathrm{Ar}] 4 s^{2} 3 d^{7}\) d. La \([\mathrm{Ar}] 6 s^{2} 4 f^{1}\) e. \(\mathrm{Pt}^{2+} \quad[\mathrm{Xe}] 4 f^{14} 5 d^{8}\)

Name the following coordination compounds. a. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\) b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) c. \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\) d. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

When an aqueous solution of KCN is added to a solution containing \(\mathrm{Ni}^{2+}\) ions, a precipitate forms, which redissolves on addition of more KCN solution. Write reactions describing what happens in this solution. [Hint: \(\mathrm{CN}^{-}\) is a Brönsted-Lowry base \(\left.\left(K_{\mathrm{b}} \approx 10^{-5}\right) \text {and a Lewis base. }\right]\)

Qualitatively draw the crystal field splitting for a trigonal bipyramidal complex ion. (Let the \(z\) axis be perpendicular to the trigonal plane.)
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