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

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鈦 and SCN鈦

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

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

Since we have deduced that SCN鈦 has a weaker ligand field strength than CN鈦, we can deduce that SCN鈦 lies to the left of CN鈦 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鈦 lies to the left of CN鈦 in the spectrochemical series, as it has a weaker ligand field strength yielding more unpaired electrons in the complex ion than CN鈦.

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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鈥攁s 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鈥攁 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

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for the reaction that is used in production of gold: $$ 2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}{ }^{2-}(a q) $$ The relevant half-reactions are $$ \begin{aligned} \mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-1.26 \mathrm{~V} \end{aligned} $$

Consider the following data: \(\begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & \mathscr{E}^{\circ} &=1.82 \mathrm{~V} \\\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & K &=1.5 \times 10^{12} \\\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en}) 3^{3+} & K &=2.0 \times 10^{47} \end{aligned}\) where en \(=\) ethylenediamine. a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction $$ \mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} $$ b. Based on your answer to part a, which is the stronger oxidizing agent, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ? c. Use the crystal field model to rationalize the result in part b.

Give formulas for the following. a. hexakis(pyridine)cobalt(III) chloride b. pentaammineiodochromium(III) iodide c. tris(ethylenediamine)nickel(II) bromide d. potassium tetracyanonickelate(II) e. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as \(\mathrm{MoS}_{2}\), which is then converted to \(\mathrm{MoO}_{3}\). The \(\mathrm{MoO}_{3}\) can be used directly in the production of stainless steel for high-speed tools (which accounts for about \(85 \%\) of the molybdenum used). Molybdenum can be purified by dissolving \(\mathrm{MoO}_{3}\) in aqueous ammonia and crystallizing ammonium molybdate. Depending on conditions, either \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}\) or \(\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is obtained. a. Give names for \(\mathrm{MoS}_{2}\) and \(\mathrm{MoO}_{3}\). b. What is the oxidation state of Mo in each of the compounds mentioned above?

Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?

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