Problem 11
Question
$$ \text { Which of the following molecules is (are) paramagnetic: } \mathrm{O}_{2}{\underline{\phantom{xx}}}^{2+} ; \mathrm{Be}_{2}{\underline{\phantom{xx}}}^{2+} ; \mathrm{F}_{2}{\underline{\phantom{xx}}}^{2+} \text { ? } $$
Step-by-Step Solution
Verified Answer
\( \mathrm{F}_{2}^{2+} \) is paramagnetic.
1Step 1: Understand Paramagnetism
Paramagnetism occurs when a substance has unpaired electrons in its molecular orbitals. Substances with paired electrons are typically diamagnetic.
2Step 2: Apply Molecular Orbital Theory
To determine the magnetic properties of a molecule, we need to apply molecular orbital (MO) theory and fill in the electron configuration for each molecule using their total number of electrons.
3Step 3: Analyze \( \mathrm{O}_{2}^{2+} \)
The molecule \( \mathrm{O}_{2} \) normally has 16 electrons. Removing 2 electrons to form \( \mathrm{O}_{2}^{2+} \) gives it 14 electrons. The MO configuration up to 14 electrons ends with: \( (\pi_{2p_x}^2)(\pi_{2p_y}^2) \). There are no unpaired electrons as the electrons fill up the orbitals symmetrically.
4Step 4: Analyze \( \mathrm{Be}_{2}^{2+} \)
The molecule \( \mathrm{Be}_{2} \) typically has 8 electrons. Removing 2 electrons, \( \mathrm{Be}_{2}^{2+} \) has 6 electrons. The MO configuration is: \( (\sigma_{1s}^2)(\sigma^*_{1s}^2)(\sigma_{2s}^2) \). All electrons are paired, hence it is diamagnetic.
5Step 5: Analyze \( \mathrm{F}_{2}^{2+} \)
The molecule \( \mathrm{F}_{2} \) normally has 18 electrons. Removing 2 electrons form \( \mathrm{F}_{2}^{2+} \) gives it 16 electrons. The MO configuration with 16 electrons is: \( (\sigma_{2s}^2)(\sigma^*_{2s}^2)(\sigma_{2p_z}^2)(\pi_{2p_x}^2)(\pi_{2p_y}^2)(\pi_{2p_x}^{*1})(\pi_{2p_y}^{*1}) \). The last two electrons are unpaired.
Key Concepts
Molecular Orbital TheoryElectron ConfigurationUnpaired Electrons
Molecular Orbital Theory
Molecular Orbital (MO) Theory provides a detailed understanding of how electrons are distributed in a molecule. This theory combines atomic orbitals from each atom in a molecule to form molecular orbitals that are spread over the entire molecule, not just localized between two atoms. This concept helps us better understand which molecules are paramagnetic or diamagnetic.
In MO theory, electrons fill the molecular orbitals starting from the lowest energy level to the highest, similar to electron filling in atomic orbitals. The molecular orbitals can be bonding or antibonding. Bonding orbitals are lower in energy and lead to stability by helping hold atoms together. Antibonding orbitals are higher in energy and can destabilize the molecule.
In MO theory, electrons fill the molecular orbitals starting from the lowest energy level to the highest, similar to electron filling in atomic orbitals. The molecular orbitals can be bonding or antibonding. Bonding orbitals are lower in energy and lead to stability by helping hold atoms together. Antibonding orbitals are higher in energy and can destabilize the molecule.
- Bonding Orbitals: Electrons here increase molecule stability.
- Antibonding Orbitals: Electrons here can decrease molecule stability.
Electron Configuration
Electron configuration in molecular orbital theory involves placing electrons in the molecular orbitals according to their energy levels. This step-by-step filling based on available electrons in a molecule's orbitals helps describe its bonding and magnetic properties.
For example:
For example:
- In the case of \( ext{O}_2^{2+}\), which has 14 electrons, a specific sequence of MO filling helps determine that all orbitals are completely filled and paired, making it diamagnetic.
- Comparatively, in the case of \( ext{F}_2^{2+}\) with 16 electrons, electrons are placed in a way that results in unpaired electrons remaining. This configuration suggests a paramagnetic nature due to these unpaired electrons.
Unpaired Electrons
Unpaired electrons play a critical role in determining the magnetic properties of molecules. When electrons in a molecule's molecular orbitals are unpaired, the molecule becomes paramagnetic. Paramagnetic substances are attracted to magnetic fields due to these unpaired electrons.
In the case of \( ext{F}_2^{2+}\), the occurrence of unpaired electrons in its MO configuration makes the molecule paramagnetic. Conversely, when all electrons in a molecule are paired, it is diamagnetic, like \( ext{O}_2^{2+}\) and \( ext{Be}_2^{2+}\), which are not influenced by magnetic fields.
In summary:
In the case of \( ext{F}_2^{2+}\), the occurrence of unpaired electrons in its MO configuration makes the molecule paramagnetic. Conversely, when all electrons in a molecule are paired, it is diamagnetic, like \( ext{O}_2^{2+}\) and \( ext{Be}_2^{2+}\), which are not influenced by magnetic fields.
In summary:
- Unpaired Electrons: Lead to paramagnetic behavior.
- Paired Electrons: Result in diamagnetic properties.
Other exercises in this chapter
Problem 4
Calculate the radius of a palladium atom, given that Pd has an FCC crystal structure, a density of \(12.0 \mathrm{~g} / \mathrm{cm}^{3}\), and an atomic weight
View solution Problem 5
Cite the indices of the direction that results from the intersection of each of the following pair of planes within a cubic crystal: (a) (110) and (111) planes;
View solution Problem 13
Which ions or atoms of the following pairs have the greatest radius: \(\mathrm{K} / \mathrm{K}^{+}\); \(\mathrm{O} / \mathrm{O}^{2-} ; \mathrm{H} / \mathrm{He}
View solution Problem 15
Show the centers of positive and negative charge in (i) \(\mathrm{CCl}_{4}\), (ii) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\), and (iii) \(\mathrm{CH}_{3
View solution