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The acctylacctonate ion, (acac) . has no unpaircd clectrons. It forms compounds with Fe \(^{21}\) and with \(\mathrm{Fc}^{3+}\) ions whosc formulas are \(\mathrm{Fc}(\mathrm{acac})_{2}\) and Fe(acac) \(_{3}\), respectively. Explain which compound has the grcater attraction to a magnctic ficld.

Short Answer

Expert verified
Fe(acac) \(_3\) has greater attraction to a magnetic field due to more unpaired electrons.

Step by step solution

01

Understand the Electron Configuration

The magnetic properties of a compound depend on the presence of unpaired electrons. Fe atoms can exist in multiple oxidation states, notably as Fe\(^{2+}\) and Fe\(^{3+}\). In Fe\(^{2+}\), the electron configuration is [Ar] 3d\(^6\), which can have 4 unpaired electrons. In Fe\(^{3+}\), the electrons are [Ar] 3d\(^5\), which can also have 5 unpaired electrons, depending on spin pairing.
02

Define Paramagnetism and Diamagnetism

Substances with unpaired electrons are paramagnetic and attracted to magnetic fields. Those with all paired electrons are diamagnetic and repelled by magnetic fields. Thus, the more unpaired electrons a compound has, the stronger its paramagnetic properties.
03

Analyze Fe(acac)_2

The Fe in Fe(acac)\(_{2}\) is in the +2 oxidation state. Assuming high spin, Fe\(^{2+}\) typically has 4 unpaired electrons (3d\(^6\) configuration). This would make it strongly paramagnetic.
04

Analyze Fe(acac)_3

The Fe in Fe(acac)\(_{3}\) is in the +3 oxidation state. Assuming high spin, Fe\(^{3+}\) typically has 5 unpaired electrons (3d\(^5\) configuration). This means it should be even more paramagnetic due to the greater number of unpaired electrons.
05

Compare the Magnetic Properties

Fe(acac)\(_{3}\) with Fe\(^{3+}\) would have more unpaired electrons compared to Fe(acac)\(_{2}\) with Fe\(^{2+}\). Therefore, Fe(acac)\(_{3}\) should be attracted more strongly to a magnetic field.

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Key Concepts

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

Paramagnetism
Paramagnetism refers to the magnetic properties of materials that have unpaired electrons in their atomic or molecular structures. When a magnetic field is applied to such materials, the unpaired electrons align themselves along the magnetic field lines. This alignment results in a magnetic moment that increases the material's overall attraction to the magnetic field. Paramagnetic substances are generally characterized by their positive magnetic susceptibility, which is a measure of how much they are magnetized when exposed to a magnetic field.

In our context, **iron (Fe) compounds** with unpaired electrons exhibit paramagnetism. This is because the presence of unpaired electrons in these compounds allows them to interact with magnetic fields. It's important to note that the degree of paramagnetism in a material is directly related to the number of these unpaired electrons. As more unpaired electrons exist, the paramagnetic effect of a material becomes stronger.
Electron Configuration
The electron configuration of an atom or ion describes the distribution of its electrons across different orbitals. Understanding electron configurations is critical in predicting the magnetic properties of a substance. Electrons in an atom fill orbitals in a specific order, usually starting from those with lower energy levels to those with higher energy levels.

For **iron ions**, electron configurations can vary based on their oxidation states. For instance:
  • Fe2+: Has an electron configuration of [Ar] 3d6. This means there are six electrons in the 3d orbital.
  • Fe3+: Has an electron configuration of [Ar] 3d5. Here, there are five electrons in the 3d orbital.
The different configurations result in different magnetic properties. In both cases, the electron configuration determines the number of unpaired electrons available, which directly affects the paramagnetism of iron compounds.
Unpaired Electrons
Unpaired electrons are crucial in understanding the magnetic behavior of compounds. An unpaired electron is one that does not have a partner with opposite spin in an orbital. The presence of these electrons is the basis for paramagnetism. Unlike paired electrons, which produces no net magnetic moment due to opposite spins canceling each other out, unpaired electrons contribute directly to a net magnetic moment.

**Analysis of Fe Compounds:**
  • Fe(acac)2: Fe2+ typically has 4 unpaired electrons. In this case, the presence of the unpaired electrons results in paramagnetism.
  • Fe(acac)3: Fe3+ tends to have 5 unpaired electrons. This higher number of unpaired electrons makes Fe(acac)3 more strongly paramagnetic compared to Fe(acac)2.
By comparing these, you can determine that Fe(acac)3, with more unpaired electrons, will have a greater attraction to a magnetic field, demonstrating stronger paramagnetic properties.

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Most popular questions from this chapter

Arrange these elements in order of increasing atomic size: \(\mathrm{Al}, \mathrm{B}, \mathrm{C}, \mathrm{K}, \mathrm{Na}\). (Try arranging these without looking at Figure 5.25 and then check yourself by looking up the necessary atomic radii.)

Which ionic compound has the lowest melting point? Explain your choice. \(\begin{array}{lll}\mathrm{LiCl} & \mathrm{NaBr} & \mathrm{KCl}\end{array}\)

In a hypothetical universe, atoms have two \(s\) orbitals, four \(p\) orbitals, six \(d\) orbitals, and eight \(f\) orbitals. (a) Draw a diagram of the periodic table for this universe, a table that retains the general positions of each block relative to the others. (b) Write the electron configuration for the first \(f\) -block element in this universe.

The lanthanides, or rare earths, are only "medium rare," because all can be purchased for a reasonable pricc. Give clectron configurations for atoms of thesc lanthanides. (a) Europium (Eu), the most expensive of the rare earth elements; \(1 \mathrm{~g}\) ean be purchased for ahout $$\$ 1.40$$. (b) Ytterbium (Yb). Less expensive than Eu, Yb costs only about $$\$ 0.35$$ per gram. It was named for the village of Yuerby in Sweden, where a mineral source of the element was found.

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