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Is it possible for a low-spin octahedral Fe(II) complex to be paramagnetic? Explain.

Short Answer

Expert verified
A low-spin octahedral Fe(II) complex has an electron configuration of \(t_{2g}^6 e_g^0\), with all 6 d-electrons in the lower-energy \(t_{2g}\) orbitals. Since there are no unpaired electrons in the complex, it exhibits diamagnetic behavior, not paramagnetic behavior. Therefore, it is not possible for a low-spin octahedral Fe(II) complex to be paramagnetic.

Step by step solution

01

Identify the electron configuration of the Fe(II) ion

Start by identifying the atomic number of Fe (iron), which is 26. The electron configuration of a neutral iron atom is \([Ar] 3d^6 4s^2\). When an iron atom loses 2 electrons to form the Fe(II) complex, its electron configuration becomes \([Ar] 3d^6\), which has 6 electrons in the d orbitals. 2.
02

Determine the electronic configuration in a low-spin octahedral complex

In an octahedral ligand field, the d orbitals split into two sets: \(t_{2g}\) (with \(d_{xy}\), \(d_{yz}\), and \(d_{zx}\) orbitals) and \(e_g\) (with \(d_{z^2}\) and \(d_{x^2-y^2}\) orbitals). The energy gap between these two sets of orbitals (螖) is relatively large in a low-spin complex. In a low-spin complex, electrons fill the lower energy \(t_{2g}\) orbitals first before occupying the higher energy \(e_g\) orbitals. This is due to strong ligand repulsion in high-spin complexes, which causes electrons to partially fill the \(t_{2g}\) orbitals before moving to \(e_g\) orbitals. For Fe(II), the electron configuration for a low-spin complex is \(t_{2g}^6 e_g^0\), with all 6 d-electrons in the lower-energy \(t_{2g}\) orbitals. 3.
03

Determine the number of unpaired electrons

In a low-spin complex, electrons fill up the three \(t_{2g}\) orbitals in pairs, with opposite spins. In the case of the Fe(II) complex, all 6 d-electrons are paired in the \(t_{2g}\) orbitals, resulting in no unpaired electrons. 4.
04

Analyze the magnetic properties of the Fe(II) complex

The magnetic properties of a complex depend on the number of unpaired electrons it possesses. Since there are no unpaired electrons in the low-spin octahedral Fe(II) complex, it exhibits no magnetic moment and is considered diamagnetic. 5.
05

Conclude whether the complex is paramagnetic or not

A low-spin octahedral Fe(II) complex has no unpaired electrons and exhibits diamagnetic behavior rather than paramagnetic behavior. Therefore, it is not possible for a low-spin octahedral Fe(II) complex to be paramagnetic.

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Key Concepts

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

Low-spin complexes
In coordination chemistry, understanding the concept of low-spin complexes is crucial. These are typically formed when the energy gap between the two main types of d orbitals in an octahedral complex鈥攏amely, the \( t_{2g} \) and \( e_{g} \) orbitals鈥攊s large. The size of this gap, known as the crystal field splitting energy (螖), influences whether the complex will be low-spin or high-spin.

- **Low-spin complexes** occur when the ligands create a strong crystal field, meaning the splitting energy 螖 is large.
- Electrons will pair up in the lower-energy \( t_{2g} \) orbitals before any are placed in higher \( e_{g} \) orbitals due to the significant energy requirement to pair electrons compared to occupying a higher orbital.
- This phenomenon leads to fewer unpaired electrons compared to high-spin counterparts.

Low-spin complexes are more common with strong-field ligands, such as CN鈦 and CO, which exert strong repulsion forces, creating a significant energy gap between the orbitals.

These complexes can have quite different properties from high-spin complexes, including lower magnetism due to the absence of unpaired electrons.
Electron configuration
Electron configuration refers to the distribution of electrons among different orbitals of an atom or ion. For the Fe(II) ion, formed by losing two electrons from iron (atomic number 26), the electron configuration directly affects the behavior of complexes it forms.

- **Neutral iron** has the electron configuration \[\text{[Ar]} 3d^6 4s^2\].
- When iron becomes Fe(II), it loses two 4s electrons, resulting in \[\text{[Ar]} 3d^6\].

In the specific context of low-spin octahedral complexes, understanding the positions of these electrons within the d orbitals is essential. Due to strong ligand interactions, the electrons in a low-spin complex occupy the lowest energy orbitals \( t_{2g} \) first, leading to a configuration like \( t_{2g}^6 e_g^0 \) for Fe(II).

This configuration fits all six d-electrons into the \( t_{2g} \) orbitals, matching up to create filled pairs. The absence of unpaired electrons in \[t_{2g}\] enables such complexes to exhibit unique magnetic properties.
Magnetic properties
Magnetic properties in coordination compounds are dictated by the presence or absence of unpaired electrons. Understanding these properties helps differentiate between paramagnetic and diamagnetic substances.

- **Diamagnetic substances** have all electrons paired with opposite spins, effectively canceling out any magnetic moment. As a result, they do not respond to magnetic fields.
- **Paramagnetic substances**, on the other hand, have one or more unpaired electrons, resulting in a net magnetic moment. This allows them to be attracted to external magnetic fields.

In the case of a low-spin octahedral Fe(II) complex, all six d-electrons are paired within the \( t_{2g} \) orbitals. The complete pairing of electrons ensures that the complex is diamagnetic.

Therefore, despite generally expecting iron-based compounds to exhibit magnetism, the arrangement in a low-spin Fe(II) complex negates this, leading to no overall unpaired electrons. The magnetic properties are crucial in scientific applications where magnetic interactions matter, demonstrating why precise electron arrangements in complexes play such an important role.

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Most popular questions from this chapter

Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}(o\) -phen \()\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \mathrm{Br}\) (c) \(\left[\mathrm{Cr}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\) (d) \(\left[\mathrm{Zn}(\mathrm{en})_{2}\right]\left(\mathrm{ClO}_{4}\right)_{2}\)

Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains \(\mathrm{Zn}^{2+}\) in its active site, is responsible for rapidly interconverting dissolved \(\mathrm{CO}_{2}\) and bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\). The zinc in carbonic anhydrase is tetrahedrally coordinated by three neutral nitrogen- containing groups and a water molecule. The coordinated water molecule has a \(\mathrm{p} K_{a}\) of \(7.5,\) which is crucial for the enzyme's activity. (a) Draw the active site geometry for the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase, just writing \({ }^{4} \mathrm{~N}^{n}\) for the three neutral nitrogen ligands from the protein. (b) Compare the \(\mathrm{p} K_{a}\) of carbonic anhydrase's active site with that of pure water; which species is more acidic? (c) When the coordinated water to the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase is deprotonated, what ligands are bound to the \(\mathrm{Zn}(\mathrm{II})\) center? Assume the three nitrogen ligands are unaffected. (d) The \(\mathrm{p} K_{a}\) of \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is \(10 .\) Suggest an explanation for the difference between this \(\mathrm{p} K_{a}\) and that of carbonic anhydrase. (e) Would you expect carbonic anhydrase to have a deep color, like hemoglobin and other metal-ion containing proteins do? Explain.

For each of the following metals, write the electronic configuration of the atom and its \(3+\) ion: (a) \(\mathrm{Ru},(\mathbf{b}) \mathrm{Mo},(\mathbf{c}) \mathrm{Co} .\) Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(3+\) ion, assuming a weak-field complex. How many unpaired electrons are there in each case?

A classmate says, "A strong-field ligand means that the ligand binds strongly to the metal ion." Is your classmate correct? Explain.

Sketch the structure of the complex in each of the following compounds and give the full compound name: (a) cis- \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{Ru}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{5}\right]\) (c) trans- \(\mathrm{NH}_{4}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (d) cis- \(\left[\mathrm{Ru}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\)

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