Vocab Dictionary

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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