Question

The spin-only magnetic moment of B₂ molecule is_____ μ_𝐵 .
(round off to two decimal places)

Answer 1

Correct Answer: 2.82 - 2.83

Spin-Only Magnetic Moment of B₂ Molecule 

To find the spin-only magnetic moment of the \( \text{B}_2 \) molecule, we begin by understanding its molecular orbital (MO) electron configuration. Boron has an atomic number of 5, and therefore \( \text{B}_2 \) will have 10 electrons.

Molecular Orbital Theory tells us that the electrons will fill the molecular orbitals as follows:

• \( \sigma_{1s}^2 \), \( \sigma^*_{1s}^2 \), \( \sigma_{2s}^2 \), \( \sigma^*_{2s}^2 \), and \( \pi_{2p}^2 \) (since both degenerate \( \pi \) orbitals will be filled first before the \( \sigma_{2p} \)).

The \( \pi_{2p} \) orbitals are of primary interest, possessing two electrons with parallel spins due to Hund's rule.

Formula for Spin-Only Magnetic Moment:

The spin-only magnetic moment \( \mu \) is given by:

\[ \mu = \sqrt{n(n+2)} \, \mu_B \]

Where \( n \) is the number of unpaired electrons.

Step 1: Number of Unpaired Electrons

Here, the number of unpaired electrons \( n = 2 \) (since the \( \pi_{2p} \) orbitals each contain one unpaired electron).

Step 2: Calculate the Magnetic Moment

Substitute \( n = 2 \) into the formula:

\[ \mu = \sqrt{2(2+2)} \, \mu_B = \sqrt{8} \, \mu_B = 2.828 \, \mu_B \]

Step 3: Final Answer

Upon rounding, \( \mu \approx 2.83 \, \mu_B \).

Verification:

The calculated value of \( 2.83 \, \mu_B \) falls within the given range of \( 2.82 \, \mu_B \) to \( 2.83 \, \mu_B \), confirming the correctness of our solution.