A Draw A Lewis Structure For Xeo4 In Which All Atoms Obey The Octet Rule

To draw a Lewis structure for XeO4, we need to follow these steps:

1. Determine the total number of valence electrons:

Xenon (Xe) has 8 valence electrons and each oxygen atom (O) has 6 valence electrons, giving us a total of

8 + 4(6) = 32 valence electrons.

2. Choose the central atom:

Since Xenon is the least electronegative element among the atoms in XeO4, it will be the central atom.

3. Form single bonds between the central atom and each surrounding atom:

Each oxygen atom will form a single bond with the central xenon atom. This uses up 4 electrons (1 from each oxygen atom), leaving us with 28 valence electrons.

4. Place any remaining electrons on the outer atoms:

Each oxygen atom needs two more electrons to complete its octet. We can achieve this by placing two lone pairs of electrons on each oxygen atom. This uses up 16 electrons (4 pairs of electrons), leaving us with 12 valence electrons.

5. Place any remaining electrons on the central atom:

The central xenon atom still needs to complete its octet, so we place two lone pairs of electrons on it. This uses up the remaining 12 valence electrons.

The final Lewis structure for XeO4 is:

O Xe

Each oxygen atom has 2 lone pairs of electrons and one shared pair of electrons with xenon, while xenon has 2 lone pairs of electrons and four shared pairs of electrons with oxygen. All atoms in this Lewis structure obey the octet rule, meaning they have eight electrons in their outermost shell.