Sure! Let's determine the group to which an element with the electron configuration [tex]\( \text{1s}^2 \ \text{2s}^2 \ \text{2p}^6 \ \text{3s}^2 \ \text{3p}^6 \ \text{3d}^1 \ \text{4s}^2 \)[/tex] belongs.
1. Identify the Period: The electron configuration shows electrons filling up to the [tex]\( \text{4s} \)[/tex] orbital. Thus, the element is in the 4th period.
2. Analyze the Outermost Orbitals:
- The configuration [tex]\( \text{4s}^2 \)[/tex] indicates there are 2 electrons in the [tex]\( \text{4s} \)[/tex] orbital.
- Additionally, there is 1 electron in the [tex]\( \text{3d} \)[/tex] orbital ( [tex]\( \text{3d}^1 \)[/tex] ).
3. Determine the Group:
- According to the periodic table, the group number can be determined primarily by the electrons in the outermost shell (valence electrons).
- The [tex]\( \text{4s}^2 \)[/tex] configuration means we have 2 valence electrons in the 4s subshell.
- The [tex]\( \text{3d} \)[/tex] electron also influences the group, as it adds to the count of valence electrons affecting transition metals.
Given the structure and location of the electron configuration, the element falls under the d-block of the periodic table, specifically the 4th period.
- The first column of the d-block starts from Scandium ([tex]\( \text{Sc} \)[/tex]), where [tex]\( \text{3d}^1 \text{4s}^2 = \text{3} \)[/tex] electrons.
Therefore, the element with the electron configuration [tex]\(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^1 \ 4s^2 \)[/tex] belongs to Group 3 of the periodic table.
