For an atom with 30 protons, which is the element Zinc (Zn), there are 7 orbitals and a total of 14 electrons for which the magnetic quantum number ($m$) is equal to 0.
Understanding Electronic Configuration and Quantum Numbers
To determine the number of orbitals and electrons with $m=0$, we first need to establish the electronic configuration of an atom with 30 protons. This atom is Zinc (Zn), and its ground state electronic configuration is $1s^22s^22p^63s^23p^63d^{10}4s^2$. This configuration shows the distribution of all 30 electrons across various energy levels and subshells.
The magnetic quantum number ($m_l$) describes the orientation of an orbital in space. Its possible values depend on the azimuthal (or angular momentum) quantum number ($l$), which defines the shape of the orbital (s, p, d, f). The allowed values for $m_l$ range from $-l$ to $+l$, including 0.
Let's break down each subshell present in the electronic configuration of Zinc and identify how many orbitals within each subshell have an $m_l$ value of 0.
Subshells and Their $m_l$ Values
- s-subshell ($l=0$):
- For any s-orbital, $l=0$, which means the only possible $m_l$ value is 0.
- Therefore, every s-orbital (1s, 2s, 3s, 4s) contains one orbital with $m_l=0$.
- p-subshell ($l=1$):
- For p-orbitals, $l=1$, so the possible $m_l$ values are -1, 0, +1.
- This means one of the three p-orbitals in any p-subshell has an $m_l$ value of 0.
- d-subshell ($l=2$):
- For d-orbitals, $l=2$, so the possible $m_l$ values are -2, -1, 0, +1, +2.
- Consequently, one of the five d-orbitals in any d-subshell has an $m_l$ value of 0.
Calculating Total Orbitals with $m_l = 0$
Based on Zinc's electronic configuration ($1s^22s^22p^63s^23p^63d^{10}4s^2$), we can count the orbitals with $m_l=0$:
- 1s subshell: Contains 1 orbital with $m_l=0$.
- 2s subshell: Contains 1 orbital with $m_l=0$.
- 2p subshell: Contains 1 orbital with $m_l=0$.
- 3s subshell: Contains 1 orbital with $m_l=0$.
- 3p subshell: Contains 1 orbital with $m_l=0$.
- 3d subshell: Contains 1 orbital with $m_l=0$.
- 4s subshell: Contains 1 orbital with $m_l=0$.
Summing these up: $1 + 1 + 1 + 1 + 1 + 1 + 1 = \textbf{7 orbitals}$ with $m_l=0$.
This can be summarized in the table below:
| Subshell | Azimuthal Quantum Number ($l$) | Possible $m_l$ Values | Orbital with $m_l=0$? |
|---|---|---|---|
| 1s | 0 | 0 | Yes (1 orbital) |
| 2s | 0 | 0 | Yes (1 orbital) |
| 2p | 1 | -1, 0, +1 | Yes (1 orbital) |
| 3s | 0 | 0 | Yes (1 orbital) |
| 3p | 1 | -1, 0, +1 | Yes (1 orbital) |
| 3d | 2 | -2, -1, 0, +1, +2 | Yes (1 orbital) |
| 4s | 0 | 0 | Yes (1 orbital) |
| Total | 7 orbitals |
Calculating Total Electrons with $m_l = 0$
According to the Pauli Exclusion Principle, each atomic orbital can hold a maximum of two electrons, provided they have opposite spins. Since we have identified 7 orbitals with $m_l=0$, the total number of electrons occupying these orbitals will be:
7 orbitals $\times$ 2 electrons/orbital = 14 electrons.
This analysis confirms that for an atom with 30 protons (Zinc), there are indeed 7 orbitals and 14 electrons that have a magnetic quantum number ($m$) of 0.
For further reading on quantum numbers and their significance in atomic structure, you can refer to reputable chemistry resources such as LibreTexts Chemistry's section on Atomic Orbitals and Quantum Numbers.
