Which electron configuration violates Hund’s rule

Which electron configuration violates Hund’s rule?
A) [He]2s2 2px2 2py2 2pz1
B) [He]2s2 2px2 2py1 2pz2
C) [He]2s2 2px1 2py2 2pz2
D) [He]2s2 2px2 2py2
E) [He]2s2 2px1 2py1 2pz2

The Correct Answer and Explanation is :

The correct answer is B) [He]2s² 2px² 2py¹ 2pz².

Explanation:

Hund’s Rule states that when electrons occupy degenerate orbitals (orbitals with the same energy level, such as the three p orbitals: 2px, 2py, and 2pz), electrons will fill them singly before pairing up. Additionally, they will have parallel spins, meaning electrons in these degenerate orbitals will first occupy them in a way that maximizes the number of unpaired electrons.

Let’s analyze each option:

1. Option A: [He]2s² 2px² 2py² 2pz¹

• Here, the electrons are distributed across the 2p orbitals: 2px and 2py each have 2 electrons, and 2pz has 1 electron.
• This configuration does not violate Hund’s Rule, as the electrons in the 2px and 2py orbitals are paired, and there is no conflict with unpaired electrons in the 2pz orbital.

1. Option B: [He]2s² 2px² 2py¹ 2pz²

• In this configuration, the 2px and 2pz orbitals are paired, but the 2py orbital has only one electron.
• This violates Hund’s Rule because according to Hund’s Rule, the electrons in the 2px, 2py, and 2pz orbitals should first occupy the orbitals singly before pairing up. In this case, the electron in 2py should have been placed into the 2pz orbital first to follow Hund’s Rule of maximizing unpaired electrons.

1. Option C: [He]2s² 2px¹ 2py² 2pz²

• This configuration is valid according to Hund’s Rule. Electrons in the 2px, 2py, and 2pz orbitals are all placed singly in their respective orbitals first, before pairing occurs.

1. Option D: [He]2s² 2px² 2py²

• This configuration does not involve the 2pz orbital, so it does not violate Hund’s Rule. It simply shows the 2px and 2py orbitals fully paired.

1. Option E: [He]2s² 2px¹ 2py¹ 2pz²

• Here, the electrons are distributed across the three p orbitals (2px, 2py, and 2pz), with each orbital having one electron. This adheres to Hund’s Rule because the electrons are occupying the orbitals singly and in parallel spins.

Thus, Option B violates Hund’s Rule because it pairs electrons in the 2px and 2pz orbitals before filling all orbitals singly.