Answer:
There are [tex]12[/tex] electrons with [tex]l = 1[/tex] in each [tex]\rm _{21}Sc[/tex] atom in its ground state.
Explanation:
The azimuthal quantum number (with symbol [tex]l[/tex]) of an electron in an atom specifies the type of the orbital that contains this electron.
For example:
- [tex]l = 0[/tex] corresponds to an [tex]s[/tex] orbital.
- [tex]l = 1[/tex] corresponds to a [tex]p[/tex] orbital.
- [tex]l = 2[/tex] corresponds to a [tex]d[/tex] orbital.
The question is asking for the number of electrons with [tex]l = 1[/tex] in each [tex]\rm _{21}Sc[/tex] atom. Because [tex]l = 1\![/tex] corresponds to [tex]p[/tex] orbitals, this question is equivalently asking for the number of electrons in [tex]p\![/tex] orbitals in one such atom.
Find the electron configuration of a ground-state [tex]\rm _{21}Sc[/tex] atom:
[tex]1s^{2}\; 2s^{2}\; 2p^{6}\; 3\, s^{2}\; 3\, p^{6}\; 3d^{1}\; 4s^{2}[/tex].
In other words, there are [tex]12[/tex] electrons in [tex]p[/tex] orbitals in total in a ground-state [tex]\rm _{21}Sc[/tex] atom in total; [tex]l = 1[/tex] for each of these electrons because of the
Therefore, a ground-state [tex]\rm _{21}Sc[/tex] atom would include [tex]12[/tex] electrons with [tex]l = 1[/tex].
