In the most stable Lewis structure for sulfuric acid, the formal charges are zero on every atom.
The is the charge that an atom appears to have when we count the electrons according to certain arbitrary rules:
The most stable Lewis structure for sulfuric acid is
The easy way to work out the formal charge is to cut each bond in half. Then each atom “owns” all the electrons on its side of the dividing line.
The H atom in sulfuric acid “owns” one of the electrons in the H-O bond.
An isolated H atom has one electron. So the H atom in sulfuric acid has a formal charge of zero.
The O atom next to H “owns” the four lone pair electrons, plus one electron from the O-H bond, plus one atom from the O-S bond. This makes a total of six electrons.
An isolated O atom has six electrons. So this O atom in sulfuric acid has a formal charge of zero.
The double-bonded O atom “owns” the four lone pair electrons, plus two electrons from the S=O double bond. This makes a total of six electrons.
So this O atom in sulfuric acid has a formal charge of zero.
The S atom “owns” half of the S=O electrons and half of the S-O electrons. This makes a total of six electrons.
An isolated S atom has six electrons. So the S atom in sulfuric acid has a formal charge of zero.
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