Count the valence electrons in the sulfate ion, SO₄²⁻.
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Count the valence electrons in the sulfate ion, . Draw a structure with four S-O single bonds and find the formal charge of sulfur. Then explain why a structure with two S=O double bonds is often preferred.
Need a reminder? The key ideaShow
Lewis structure: count the valence electrons, adding one per negative charge and taking one per positive charge. Join the outer atoms to the central atom and complete their octets. Put the rest on the centre, and make multiple bonds if the centre is short. Formal charge: . Exceptions to the octet rule: incomplete octet, odd electrons, expanded octet.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Count. electrons.
Step 2. Single bonds. Four bonds use 8 electrons. The other 24 go on the oxygens as 3 lone pairs each.
Step 3. Formal charges. Sulfur: . Each oxygen: . The sum, , is the ion's charge.
Step 4. Double bonds. With two S=O bonds, sulfur has . Two oxygens become 0 and two stay at . The formal charges are smaller, and sulfur can expand its octet to 12.
Answer: 32 electrons; sulfur is with single bonds, and 0 with two double bonds.
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