How do you draw the Lewis diagrams for ClO4?

To draw the Lewis structure for the chlorate ion (ClO₄^–), you need to follow these steps:

1. Count the total valence electrons: Chlorine has 7 valence electrons, and each oxygen has 6. Since there are four oxygens, that’s 4 × 6 = 24. Therefore, the total is 7 + 24 = 31. But, because the ion has a -1 charge, you add one more electron, giving you 32 valence electrons.
2. Determine the central atom: Chlorine is the central atom, as it can expand its octet and can bond with more atoms than oxygen, which typically follows the octet rule.
3. Sketch the skeleton structure: Place the chlorine atom in the center and arrange the four oxygen atoms around it. Connect each oxygen to the chlorine with a single bond.
4. Add lone pairs: After forming single bonds, each oxygen should have three lone pairs to complete their octet. After this, you’ve used 8 electrons for the single bonds and 24 electrons for the six lone pairs (3 for each of the four O atoms), totaling 32 electrons used, which fits.
5. Check stability: The current structure has a central atom connected to four oxygens, with one being a double bond to avoid an incomplete octet on chlorine. You can convert one lone pair from any of the oxygen atoms into a double bond with chlorine to maintain octet stability. Thus, ClO₄^– has one Cl=O double bond and three Cl-O single bonds.

Final Structure: You end up with a structure showing a chlorine atom with a double bond to one oxygen and single bonds to the other three, with the overall charge showing that it is a negatively charged ion. The resulting structure is stable and complies with the octet rule.