What Is The Lewis Structure Of XeO4?

The Lewis structure of XeO4 can be derived by first determining the total number of valence electrons in the molecule. Xe (Xenon) has 8 valence electrons, and each O (oxygen) has 6 valence electrons, for a total of:

8 + 4(6) = 32 valence electrons

To construct the Lewis structure, we start by placing the Xenon atom in the center and connecting it to each Oxygen atom with a single bond. This uses up 4 electrons, leaving 28. Next, we place the remaining electrons around the atoms in pairs (as lone pairs or as double bonds), until all electrons are used up and each atom has a full octet.

The final arrangement for XeO4 is:

O

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O--Xe--O

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O

Each Oxygen atom is bonded to the Xenon atom via a single bond, and there are two lone pairs of electrons on each Oxygen atom to complete the octet. The Xenon atom has no lone pairs, but is surrounded by four Oxygen atoms.