To determine the second quantum number (also known as the azimuthal quantum number, [tex]\( l \)[/tex]) for a [tex]\( 1s^2 \)[/tex] electron in phosphorus with the electron configuration [tex]\( 1s^2 2s^2 2p^6 3s^2 3p^3 \)[/tex], let us follow these logical steps:
1. Identify the principal quantum number ([tex]\( n \)[/tex]):
- The principal quantum number [tex]\( n \)[/tex] represents the energy level or shell where the electron is located. For the [tex]\( 1s^2 \)[/tex] electron, the principal quantum number is [tex]\( n = 1 \)[/tex].
2. Identify the type of orbital from the electron configuration:
- The letter "s" in [tex]\( 1s^2 \)[/tex] specifies the type of orbital. In this notation:
- "s" orbitals have [tex]\( l = 0 \)[/tex]
- "p" orbitals have [tex]\( l = 1 \)[/tex]
- "d" orbitals have [tex]\( l = 2 \)[/tex]
- "f" orbitals have [tex]\( l = 3 \)[/tex]
3. Assign the azimuthal quantum number ([tex]\( l \)[/tex]):
- For an electron in an "s" orbital, [tex]\( l \)[/tex] is always 0.
Since the specific electrons we are considering are in a [tex]\( 1s^2 \)[/tex] orbital, corresponding to the [tex]\( n = 1 \)[/tex] energy level and the "s" orbital, the second quantum number [tex]\( l \)[/tex] must be:
[tex]\[ l = 0 \][/tex]
Thus, the correct answer is:
A. [tex]\( l = 0 \)[/tex]
