To determine the second quantum number (also known as the azimuthal quantum number, [tex]\( l \)[/tex]) for the 3p^1 electron in an aluminum atom, [tex]\( 1s^2 2s^2 2p^6 3s^2 3p^1 \)[/tex], we need to understand how the quantum numbers define the state of an electron in an atom.
1. Principal Quantum Number (n): This number describes the energy level of the electron. For the 3p^1 electron, the principal quantum number [tex]\( n \)[/tex] is 3.
2. Azimuthal Quantum Number (l): This number describes the shape of the orbital and can take any integer value from 0 to [tex]\( n-1 \)[/tex]. It determines the subshell where the electron resides:
- [tex]\( l = 0 \)[/tex] corresponds to the s orbital
- [tex]\( l = 1 \)[/tex] corresponds to the p orbital
- [tex]\( l = 2 \)[/tex] corresponds to the d orbital
- [tex]\( l = 3 \)[/tex] corresponds to the f orbital
Given that the electron in question is in a p orbital, we need the azimuthal quantum number corresponding to a p orbital.
From the information provided, since the given electron is a 3p^1 electron:
- The principal quantum number [tex]\( n \)[/tex] is 3.
- The azimuthal quantum number [tex]\( l \)[/tex] for a p orbital is 1.
Therefore, the correct answer is:
A. [tex]\( l = 1 \)[/tex].
