Lewis Dot Structure Of So42 ⟶ [DIRECT]
$6 + 24 + 2 = \mathbf32$ valence electrons .
Therefore, the real ion is a . There are multiple equivalent resonance structures. In each structure, the two negative charges and the two double bonds are distributed among different pairs of oxygen atoms. The actual ion has: lewis dot structure of so42
Connect each oxygen to the sulfur using a single bond (2 electrons each). 4 bonds × 2 electrons = 8 electrons used. Remaining: 32 - 8 = 24 electrons. 4. Distribute Remaining Electrons $6 + 24 + 2 = \mathbf32$ valence electrons
At this stage, every oxygen has a full octet (2 bonding electrons + 6 lone electrons). However, we have a problem: It needs 4 more to complete its octet. In each structure, the two negative charges and
(with one double bond to fulfill resonance structures)
To reduce the formal charge on sulfur, we can form double bonds. Sulfur is in period 3 and can expand its octet (use empty 3d orbitals) to accommodate more than 8 electrons.
However, to fulfill the octet rule properly for sulfur and to reflect the actual structure more accurately:
