For each of the following diatomic molecules, use your electron configurations from question 2 to determine if they are paramagnetic or diamagnetic
1: H_2
2: Li_2
3: B_2
4: C_2
5: N_2
6: O_2
7: F_2

Question

BlackTom AI · Accepted Answer

We’re evaluating each diatomic molecule’s magnetic behavior (paramagnetic vs diamagnetic) based on their electron configurations in molecular orbitals.
Option 1: Diamagnetic
- H2: The total of 2+2 = 2 electrons occupy bonding orbitals with all electrons paired, so no unpaired electrons remain; this yields diamagnetism.
- Li2: With 4 valence electrons filling the MO set, all electrons are paired in the occupied orbitals, leaving no unpaired spins; thus, diamagnetic.
- N2: In the common MO ordering for N2 and lighter diatomics, all electrons are paired (two in sigma2s, two in sigma*2s, four in pi2p as paired pairs, and two in sigma2p as a pair), so there are zero unpaired electrons; this makes N2 diamagnetic.
- F2: All electrons are paired in the occupied MOs (including the pi2p and sigma2p levels for F2), producing no unpaired spins; therefore, F2 is diamagnetic.
- Summary note: In these four molecules, every occupied molecular orbital is paired, so the overall spin count is zero and they do not experience attraction to a magnetic field from unpaired electrons.

Option 2: Paramagnetic
- B2: In the MO picture for B2 (and similar light diatomics), there are unpaired electrons in the degenerate pi2p orbitals, leading to a net magnetic moment; this makes B2 paramagnetic.
- C2: Depending on the exact MO ordering used (and the number of electrons in the degenerate pi2p set), C2 also ends up with unpaired electrons in the pi2p set in many standard treatments, yielding paramagnetism.
- O2: A classic result is that O2 has two unpaired electrons in the pi*2p antibonding set, giving a clear paramagnetic behavior.
- Summary note: Each of these molecules hosts at least one unpaired electron in its molecular orbital configuration under typical MO schemes, which accounts for their paramagnetism.

Cross-checking the common teaching: the diamagnetic group (H2, Li2, N2, F2) lacks unpaired electrons in their MOs, while the paramagnetic group (B2, C2, O2) contains unpaired electrons in the pi or pi* sets, explaining the observed magnetic properties.

If you want, I can map each molecule to its specific MO occupancy to illustrate exactly where the unpaired electrons reside for the paramagnetic cases and show the paired arrangements for the diamagnetic cases, with orbital diagrams described in words.

1A W25 VidQuiz: MO Theory: Mixing P-orbitals

For each of the following diatomic molecules, use your electron configurations from question 2 to determine if they are paramagnetic or diamagnetic 1: H_2 2: Li_2 3: B_2 4: C_2 5: N_2 6: O_2 7: F_2

View Explanation

Log in for full answers

Similar Questions

Using the molecular orbital theory, predict how many unpaired electrons are in the B22– ion.

Using molecular orbital theory, predict the bond order for the O2+ ion.

Which of the following is the correct molecular electronic configuration for the diagram below?

Assuming an MO diagram with orbital energies like that of O2 what is the bond order and number of unpaired electrons for the molecule NO2- ?

Using the molecular orbital theory, predict how many unpaired electrons are in the O2+ ion.

Molecular Orbital Theory: Bonds are formed from interactions of atomic orbitals to form new molecular orbitals that describe the electronic structure of a molecule.

In a consumer society, many adults channel creativity into buying things

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler