What is the second quantum number of the [tex]3p^1[/tex] electron in aluminum, [tex]1s^2 2s^2 2p^6 3s^2 3p^1[/tex]?

A. [tex]l = 2[/tex]
B. [tex]l = \sqrt{3}[/tex]
C. [tex]l = 3[/tex]
D. [tex]l = 0[/tex]



Answer :

To determine the second quantum number of a [tex]$3p^1$[/tex] electron in aluminum, we need to understand a few key concepts about electron configuration and quantum numbers.

1. Electron Configuration of Aluminum:
Aluminum has an atomic number of 13, which means it has 13 electrons. The electron configuration for aluminum is:
[tex]\[ 1s^2 2s^2 2p^6 3s^2 3p^1 \][/tex]
In this configuration, the electrons are filled in the increasing order of energy levels.

2. Quantum Numbers:
Each electron in an atom is described by four quantum numbers:
- The principal quantum number (n): This number describes the energy level and size of the orbital where the electron is located.
- The angular momentum quantum number (l): This number describes the shape of the orbital. It is also called the azimuthal quantum number.
- The magnetic quantum number (m_l): This number describes the orientation of the orbital in space.
- The spin quantum number (m_s): This number describes the electron's spin direction (either +1/2 or -1/2).

3. Second Quantum Number (Angular Momentum Quantum Number, [tex]\( l \)[/tex]):
The second quantum number [tex]\( l \)[/tex] can take integer values from 0 to [tex]\( n-1 \)[/tex], where [tex]\( n \)[/tex] is the principal quantum number.
The value of [tex]\( l \)[/tex] corresponds to the type of orbital:
- [tex]\( l = 0 \)[/tex] refers to an s-orbital
- [tex]\( l = 1 \)[/tex] refers to a p-orbital
- [tex]\( l = 2 \)[/tex] refers to a d-orbital
- [tex]\( l = 3 \)[/tex] refers to an f-orbital

4. Analyzing the Specific Electron:
For the [tex]$3p^1$[/tex] electron in aluminum:
- The principal quantum number [tex]\( n = 3 \)[/tex] because the electron is in the 3rd energy level.
- The angular momentum quantum number [tex]\( l \)[/tex] for a p-orbital is 1.

Therefore, the second quantum number [tex]\( l \)[/tex] for the [tex]$3p^1$[/tex] electron in aluminum is 1.

Hence, the correct answer is:
D. [tex]\( l = 1 \)[/tex]