To determine the valence electrons in an ionic compound like potassium bromide (KBr), we need to look at the electronic configurations of the individual atoms.
Potassium (K) is an alkali metal and is found in Group 1 of the periodic table. It has an atomic number of 19, which means it has 19 electrons. The electron configuration of potassium is [Ar] 4s^1, indicating that it has one valence electron in its outermost energy level (4s^1).
Bromine (Br) is a halogen and is found in Group 17 of the periodic table. It has an atomic number of 35, which means it has 35 electrons. The electron configuration of bromine is [Ar] 3d^10 4s^2 4p^5, indicating that it has seven valence electrons in its outermost energy level (4s^2 4p^5).
In the formation of potassium bromide (KBr), potassium donates its one valence electron to bromine. As a result, potassium forms a cation with a +1 charge (K+) since it loses one electron, and bromine forms an anion with a -1 charge (Br-) since it gains one electron.
So, in the ionic compound KBr, the valence electron configuration of potassium is 0 (since it donates its valence electron), and the valence electron configuration of bromine is 8 (since it gains one electron and achieves a stable electron configuration with eight valence electrons).
In summary, the valence electron configuration of KBr can be represented as K+ (0 valence electrons) and Br- (8 valence electrons).