State the number of valence electrons for an iron(II) ion in an ionic bond.
To determine the number of valence electrons for an iron(II) ion in an ionic bond, we need to first understand the electronic configuration of iron and how it forms ions.
Iron (Fe) is a transition metal and typically has two common oxidation states: +2 and +3. In this case, we are dealing with an iron(II) ion, which means it has a +2 charge.
To determine the number of valence electrons for an iron(II) ion, we can refer to the periodic table. Iron is located in group 8 (VIII) of the periodic table, specifically in the d-block. The group number indicates the number of valence electrons for elements in that group, excluding the transition metals. However, since iron is a transition metal, we need to consider the electron configuration.
The electron configuration for iron is [Ar] 3d6 4s2. Here, the 3d orbitals are responsible for the valence electrons. In the case of iron(II), it loses two electrons from the 4s orbital, leaving behind an electron configuration of [Ar] 3d6.
So, in an iron(II) ion, there are six valence electrons, specifically in the 3d orbitals.
In summary, the number of valence electrons for an iron(II) ion in an ionic bond is 6.