Ground state electron configuration that doesn’t quite follow the filling rules. The configuration is: [Rn]5f
6d
7s
. Neptunium does obey Hund’s rule. Use the table below to assign a set of four quantum numbers to each electron beyond the radon configuration: Electron # n l m
m
1 2 3 4 5 6 7
- [6 marks] The first transuranium element neptunium (Np, Z = 93) has a ground state electron configuration that doesn’t quite follow the filling rules. The configuration is: [Rn]5f
6d
7s
. Neptunium does obey Hund’s rule. Use the table below to assign a set of four quantum numbers to each electron beyond the radon configuration: Electron # n l m
m
1 2 3 4 5 6 7
The Correct Answer and Explanation is:
Correct Answer
The ground state electron configuration for Neptunium (Z = 93) is [Rn] 5f⁴ 6d¹ 7s². This means there are seven electrons beyond the stable Radon core (Z = 86). A valid set of quantum numbers for these seven valence electrons is assigned below, following Hund’s Rule and the Pauli Exclusion Principle.
| Electron # | n | l | m_l | m_s |
| 1 | 7 | 0 | 0 | +1/2 |
| 2 | 7 | 0 | 0 | -1/2 |
| 3 | 6 | 1 | -2 | +1/2 |
| 4 | 5 | 3 | -3 | +1/2 |
| 5 | 5 | 3 | -2 | +1/2 |
| 6 | 5 | 3 | -1 | +1/2 |
| 7 | 5 | 3 | 0 | +1/2 |
Note: The specific m_l values for the single 6d electron and the set of 5f electrons can vary. The set provided is one of several valid combinations, chosen by convention (e.g., filling from the lowest m_l value upwards).
Explanation
To assign a unique set of four quantum numbers (n, l, m_l, m_s) to each of Neptunium’s seven valence electrons, we must systematically apply the rules of quantum mechanics. The configuration [Rn] 5f⁴ 6d¹ 7s² dictates which orbitals these electrons occupy.
- The Principal Quantum Number (n): This number indicates the electron’s energy level or shell. For the 7s electrons, n=7. For the 6d electron, n=6. For the four 5f electrons, n=5.
- The Azimuthal Quantum Number (l): This number defines the shape of the orbital (subshell). It has values from 0 to n-1. For an s-orbital, l=0. For a d-orbital, l=2. For an f-orbital, l=3.
- The Magnetic Quantum Number (m_l): This number describes the orientation of the orbital in space and can have integer values from -l to +l.
- For the 7s electrons (l=0), m_l can only be 0.
- For the 6d electron (l=2), m_l can be -2, -1, 0, +1, or +2.
- For the 5f electrons (l=3), m_l can be -3, -2, -1, 0, +1, +2, or +3.
- The Spin Quantum Number (m_s): This number represents the intrinsic angular momentum of the electron, which can be either spin-up (+1/2) or spin-down (-1/2).
Assignment Process:
- 7s² electrons (Electrons 1 & 2): The two electrons in the 7s orbital have n=7, l=0, and m_l=0. According to the Pauli Exclusion Principle, no two electrons can have the same four quantum numbers. Therefore, they must have opposite spins: m_s = +1/2 and m_s = -1/2.
- 6d¹ electron (Electron 3): This single electron is in the 6d subshell (n=6, l=2). It occupies one of the five degenerate d-orbitals. We conventionally assign it an m_l value of -2 and a spin of +1/2.
- 5f⁴ electrons (Electrons 4-7): These four electrons are in the 5f subshell (n=5, l=3). The problem states that Neptunium obeys Hund’s Rule, which dictates that for a set of degenerate orbitals, electrons will first fill each orbital singly with parallel spins before any pairing occurs. This maximizes the total spin. Therefore, the four electrons are placed in four different f-orbitals (e.g., m_l = -3, -2, -1, and 0), and all are assigned the same spin (m_s = +1/2).
