Question on bonding and formal charge

krackers

Hello! I was wondering why, when calculating a formal charge, you actually use that formula of (valence electrons in ground state) - lone electrons after bonding - 1/2(bonded electrons)

Lets take H2O.

H - O - H


The O has 6-4-2 = 0 formal charge.
However, the O originally had 6 electrons, and now it has 8, so shouldn't that mean it has a charge of -2 because it has 2 more electrons than before. Also, when that happens, doesn't that mean the number of electrons doesn't match with the number of protons, meaning the atom is now unstable? I would appreciate it if anyone could help to clear this doubt.

AGNuke

That is the "Formal" Charge, not an actual charge. Oxygen has 6 electrons, 2 of which them is "shared" with other atoms. They do not fully belongs to O-atom and the other 2 electrons also are shared, so it is not like that 6-proton 8-electron system.