Let's solve the problem step-by-step.
To determine the second quantum number (azimuthal quantum number, [tex]\( l \)[/tex]) for a [tex]\( 1s^2 \)[/tex] electron in phosphorus (which has the electron configuration [tex]\( 1s^2 2s^2 2p^6 3s^2 3p^3 \)[/tex]), we need to understand the quantum numbers associated with electron orbitals.
1. Principal Quantum Number ([tex]\( n \)[/tex]): This indicates the main energy level or shell. For [tex]\( 1s^2 \)[/tex], [tex]\( n = 1 \)[/tex].
2. Azimuthal Quantum Number ([tex]\( l \)[/tex]): This indicates the shape of the orbital and can take on any integer value from 0 to [tex]\( n-1 \)[/tex].
- If [tex]\( l = 0 \)[/tex], the orbital is an [tex]\( s \)[/tex]-orbital.
- If [tex]\( l = 1 \)[/tex], the orbital is a [tex]\( p \)[/tex]-orbital.
- If [tex]\( l = 2 \)[/tex], the orbital is a [tex]\( d \)[/tex]-orbital.
- If [tex]\( l = 3 \)[/tex], the orbital is an [tex]\( f \)[/tex]-orbital.
For a [tex]\( 1s^2 \)[/tex] electron:
- The principal quantum number [tex]\( n \)[/tex] is 1.
- The second quantum number [tex]\( l \)[/tex] must be 0 because for [tex]\( s \)[/tex]-orbitals, [tex]\( l \)[/tex] is always 0.
Thus, the correct value of the second quantum number for a [tex]\( 1s^2 \)[/tex] electron is:
[tex]\[ l = 0 \][/tex]
So, the correct answer to the question is:
C. [tex]\( l = 0 \)[/tex]
