Problem 134
Question
Hydrogen atoms have one electron. Does this mean that hydrogen gas is paramagnetic? Why or why not?
Step-by-Step Solution
Verified Answer
Explain your answer.
Answer: Hydrogen gas (H2) is not paramagnetic. This is because the H2 molecule consists of two paired electrons in its molecular orbital, with no unpaired electrons leading to any net magnetic moment. As a result, H2 is classified as a diamagnetic substance, meaning it is weakly repelled by an external magnetic field.
1Step 1: Determine the H2 molecule configuration
First, let's consider the hydrogen atom's electron configuration. Hydrogen has only one electron on the first energy level (1s orbital). When two hydrogen atoms come together to form a hydrogen gas molecule (H2), their electrons will pair up in the 1s molecular orbital.
2Step 2: Check for unpaired electrons in H2
In the molecular orbital diagram for H2, we can see that both electrons are paired. The two electrons from the individual hydrogen atoms combine to form a stable bonding molecular orbital. There are no unpaired electrons in the H2 molecule.
3Step 3: Determine if hydrogen gas is paramagnetic or not
Since there are no unpaired electrons in the hydrogen gas molecule (H2), it does not have any net magnetic moment. Therefore, hydrogen gas is not paramagnetic.
4Step 4: Conclusion
Hydrogen gas (H2) is not paramagnetic, as it does not have unpaired electrons in its molecular form. It is a diamagnetic substance, which implies that it is weakly repelled by an external magnetic field.
Key Concepts
Hydrogen MoleculeElectron PairingMolecular Orbital Theory
Hydrogen Molecule
The hydrogen molecule, often expressed as H extsubscript{2}, is a simple yet fascinating example in chemistry. Hydrogen atoms, each containing one electron, come together in a special way to form this molecule. The process begins when two hydrogen atoms share their single electrons. This sharing creates a bond that holds the atoms together as a molecule.
- Each hydrogen atom starts with one electron in its 1s orbital.
- When forming H extsubscript{2}, these electrons pair up.
- This pairing occurs in the molecular orbital formed between the two hydrogen nuclei.
Electron Pairing
Electron pairing is a crucial concept when discussing molecules like hydrogen. When two electrons occupy the same space, or orbital, they are said to be "paired." In the context of the H extsubscript{2} molecule, both electrons from the hydrogen atoms share the 1s orbital in the molecular structure.
- Paired electrons have opposite spins, which means they can coexist in the same orbital.
- The opposite spins result in a stable configuration, reducing the energy of the system.
- This stable arrangement of paired electrons means there are no unpaired electrons in H extsubscript{2}.
Molecular Orbital Theory
Molecular Orbital Theory provides a framework to understand how molecules like H extsubscript{2} form. It explains the combining of atomic orbitals to create molecular orbitals. This helps predict the magnetic and chemical properties of molecules. In the case of H extsubscript{2}:
- Each hydrogen atom contributes its 1s orbital to form a bonding and an antibonding molecular orbital.
- The bonding molecular orbital is filled first with the two electrons, resulting in a lower energy state.
- The antibonding molecular orbital remains empty in H extsubscript{2} due to the lack of additional electrons.
Other exercises in this chapter
Problem 132
To model the bonding in \(\mathrm{SF}_{6}\) gas, some chemists assume the existence of \(\mathrm{SF}_{4}^{2+}\) cations surrounded by two \(\mathrm{F}^{-}\) ion
View solution Problem 133
Which of the unstable nitrogen oxides \(\mathrm{N}_{2} \mathrm{O}_{2}, \mathrm{N}_{2} \mathrm{O}_{5},\) and \(\mathrm{N}_{2} \mathrm{O}_{3}\) are polar molecule
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Draw the molecular orbital diagram of the valence shell of a \(\mathrm{F}_{2}+\) ion, and use it to determine the bond order in the ion.
View solution Problem 136
Use molecular orbital diagrams to determine the bond order of the peroxide \(\left(\mathrm{O}_{2}^{2-}\right)\) and superoxide \(\left(\mathrm{O}_{2}^{-}\right)
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