Draw the Lewis Structure for XeO4 and Decide if the Molecule is Polar or Nonpolar

To draw the Lewis structure for xenon tetraoxide (XeO₄), we start by calculating the total number of valence electrons. Xenon has 8 valence electrons, and each oxygen atom has 6 valence electrons. Since there are four oxygen atoms, that gives us:

• 8 (from Xe) + 4 × 6 (from O) = 32 valence electrons

Next, we arrange the atoms. Xenon will be the central atom, surrounded by the four oxygen atoms. We can start by placing a single bond between xenon and each oxygen:

• 4 single bonds account for 8 electrons (4 bonds × 2 electrons per bond).

This leaves us with:

• 32 – 8 = 24 valence electrons remaining.

Next, we will complete the octets of the oxygen atoms. Each oxygen will need 6 more electrons to complete its octet, which uses up:

• 4 × 6 = 24 electrons.

Now, each oxygen has its octet satisfied, and so does the xenon, which accommodates more than 8 electrons.

The final Lewis structure looks as follows:

• Xe with 4 double bonds to each of the 4 oxygens (O=).

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