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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 conclusion, SCN⁻ lies to the left of CN⁻ in the spectrochemical series, as it has a weaker ligand field strength yielding more unpaired electrons in the complex ion than CN⁻.

Step by step solution

01

Understand paramagnetism in complex ions

Paramagnetism in complex ions happens when unpaired electrons are present. Greater the number of unpaired electrons, more paramagnetic the complex ion is. The unpaired electrons result from the splitting of d-orbitals in the metal ion due to the ligand field. The stronger the ligand field created by the ligand around the metal ion, the greater the splitting of d-orbitals, and fewer unpaired electrons will be left in the complex ion.
02

Compare ligand field strength of CN⁻ and SCN⁻

We are given that Fe(CN)₆³⁻ has one unpaired electron, which means the ligand field strength of CN⁻ is strong enough to create a splitting that leaves only one electron unpaired. On the other hand, Fe(SCN)₆³⁻ has five unpaired electrons; this implies that the ligand field strength of SCN⁻ is relatively weaker, allowing more unpaired electrons in the complex ion.
03

Determine the position of SCN⁻ relative to CN⁻ in the spectrochemical series

Since we have deduced that SCN⁻ has a weaker ligand field strength than CN⁻, we can deduce that SCN⁻ lies to the left of CN⁻ in the spectrochemical series. The spectrochemical series is an ordering of ligands based on their ligand field strength. Ligands on the left side of the series cause less splitting and have weaker field strength, while ligands on the right side cause greater splitting and have stronger field strength. In conclusion, SCN⁻ lies to the left of CN⁻ in the spectrochemical series, as it has a weaker ligand field strength yielding more unpaired electrons in the complex ion than CN⁻.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paramagnetism in Complex Ions
The phenomenon of paramagnetism in complex ions involves the magnetic response of ions due to the presence of unpaired electrons within their structure. Generally, each unpaired electron contributes to the magnetic moment of the ion, making the complex paramagnetic. An easy way to understand this is to picture each unpaired electron as a tiny magnet. When an external magnetic field is applied, these tiny magnets align in the direction of the field, causing the ion to be drawn towards it.

In complexes like \( \mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-} \) and \( \mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-} \) from our exercise, this property is used to determine the number of unpaired electrons and infer the strength of the ligands involved. The more unpaired electrons there are, the stronger the magnetic response and, by inference, the weaker the ligand field strength—as is the case with \( \mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-} \), which has five unpaired electrons.
Ligand Field Strength
Ligand field strength is a measure of a ligand's ability to split the d-orbitals of a transition metal ion. The concept of ligand field theory stemmed from the need to explain the colors and magnetism of metal complexes. To visualize how this works, think of each ligand as an athlete running towards a group of five doors representing the d-orbitals of the metal ion. Some athletes (strong field ligands like \( \mathrm{CN}^{-} \) ) have the strength to push the doors wide open, causing a large separation, while others (weak field ligands like \( \mathrm{SCN}^{-} \) ) can only open them slightly.

This 'door-opening' creates an energy gap between two sets of d-orbitals, known as 'd-orbital splitting'. In our exercise, \( \mathrm{CN}^{-} \) is a strong field ligand, significantly splitting the orbitals and thereby pairing electrons. Conversely, \( \mathrm{SCN}^{-} \) is a weak field ligand producing less splitting and resulting in a larger number of unpaired electrons.
D-Orbital Splitting
D-orbital splitting is a term describing the energy difference that arises between the d-orbitals of transition metal ions when they are surrounded by ligands—a framework called the crystal field or ligand field. In a nutshell, before ligands approach, the d-orbitals are all at the same energy level. But as ligands get closer, they interact electromagnetically with the d-orbitals causing them to split into two energy levels. For simplification, imagine a group of seats at a concert where everyone had equal view before, but then a screen is placed, obstructing the view for some of the seats.

This can result in a lower energy set of orbitals called 't2g' and a higher energy set called 'eg'. The difference in energy between the 't2g' and 'eg' orbitals decides how electrons are arranged and influences properties like color and magnetism in the complex. In our textbook exercise, the different splitting effects of \( \mathrm{CN}^{-} \) and \( \mathrm{SCN}^{-} \) dictate the number of unpaired electrons in the resulting complexes.
Unpaired Electrons in Transition Metal Complexes
The presence or absence of unpaired electrons in transition metal complexes plays a critical role in determining their chemical and physical properties, including color, magnetism, and reactivity. Unpaired electrons are those that do not have a paired spin in the same orbital. These 'lone rangers', so to speak, have profound implications. For instance, the existence of unpaired electrons typically renders a complex paramagnetic, as seen with the \( \mathrm{Fe}(\mathrm{SCN})_{6}{ }^{3-} \) complex which has five.

In contrast, the counterpart complex \( \mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-} \) with only one unpaired electron is less paramagnetic. The difference in their magnetisms tells a story about the electron's environment, specifically how ligands like \( \mathrm{CN}^{-} \) and \( \mathrm{SCN}^{-} \) influence d-orbital splitting and electron pairing. This concept is essential for students getting to grips with coordination chemistry and is a bedrock for understanding the behavior of these fascinating molecules.

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Most popular questions from this chapter

Compounds of \(\mathrm{Sc}^{3+}\) are not colored, but those of \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) are. Why?

One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

Write electron configurations for each of the following. a. \(\mathrm{Cr}, \mathrm{Cr}^{2+}, \mathrm{Cr}^{3+}\) b. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}\)

For the process $$ \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Co}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}_{2}^{+}+\mathrm{NH}_{3} $$ what would be the expected ratio of cis to trans isomers in the product?

Draw geometrical isomers of each of the following complex ions. a. \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}^{-}\) b. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}{ }^{2+}\) c. \(\mathrm{Ir}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\) d. \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}^{+}\)

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