The magnetic character of bulk matter is determined largely by electron spin magnetic moments, rather than by orbital dipole moments. (Nuclear contributions are negligible, as the proton's spin magnetic moment is about 658 times smaller than that of the electron.) If the atoms or molecules of a substance have unpaired electron spins, the associated magnetic moments give rise to paramagnetic behavior or to ferromagnetic behavior if the interactions between atoms or molecules are strong enough to align them in domains. If the atoms or molecules have no net unpaired spins, then magnetic perturbations of the electron orbits give rise to diamagnetic behavior. a) Molecular hydrogen gas \(\left(\mathrm{H}_{2}\right)\) is weakly diamagnetic. What does this imply about the spins of the two electrons in the hydrogen molecule? b) What would you expect the magnetic behavior of atomic hydrogen gas (H) to be?

Step by step solution

01

a) Implications of weak diamagnetism on electron spins in H₂

Molecular hydrogen gas (H₂) is weakly diamagnetic, which means the electrons in the molecule do not possess any net unpaired spins and the presence of an external magnetic field makes it repel slightly. In H₂, there are two electrons, and weak diamagnetism implies that these two electrons have paired (opposite) spins. When the spins are paired, their magnetic moments cancel each other out, leading to no net unpaired electron spin and resulting in diamagnetic behavior.

02

b) Predicting the magnetic behavior of atomic hydrogen gas (H)

Atomic hydrogen gas (H) has one electron in its outer shell. Since there is only one electron, it has an unpaired spin. An unpaired electron spin means that the magnetic moment is not canceled out, as there is no pairing with another electron. In such a scenario, the material exhibits paramagnetic behavior. Therefore, we can expect atomic hydrogen gas (H) to exhibit paramagnetic behavior due to the presence of an unpaired electron spin.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!