Xeo4 Valence Electrons

XeO4 is a compound that contains xenon as the central atom. Xenon belongs to group 18 of the periodic table, which means it has 8 valence electrons. Each oxygen atom contributes 6 valence electrons (since they belong to group 16), for a total of 24 valence electrons. Therefore, the total number of valence electrons in XeO4 is:

8 (from xenon) + 24 (from oxygen) = 32

To determine the Lewis structure of XeO4, we first need to arrange the atoms such that the xenon atom is in the center and the oxygen atoms are surrounding it. Then, we connect each oxygen atom to the central xenon atom with a single bond. This uses up 4 x 2 = 8 valence electrons.

Next, we add lone pairs of electrons on each oxygen atom until each oxygen atom has an octet (i.e., 8 valence electrons). This uses up another 4 x 6 = 24 valence electrons. We still have 32 - 8 - 24 = 0 valence electrons remaining, indicating that we have used all the available valence electrons.

The resulting Lewis structure of XeO4 consists of a central xenon atom surrounded by four oxygen atoms, each of which is connected to the central atom by a single bond. Each oxygen atom also has two lone pairs of electrons around it.