Lewis structure of H₂SO₄
32 valence electrons. Two S=O double bonds and two S-OH single bonds, with no lone pair on S, so S has 12 electrons around it (an expanded octet).
Lewis structure
32 valence electronsi
NCERT uses H2SO4 as an example of an expanded octet: S has 12 electrons around it.
Formal charges: none
Why this shape
- S has 6 valence electrons. It uses six in two double bonds to O and two single bonds to O, so none are left over.
- Steric number = four sigma bonds + no lone pairs = 4, so S is sp³ hybridised. A double or triple bond counts once: only its sigma bond uses a hybrid orbital.
- Four electron pairs point to the corners of a tetrahedron (tetrahedral electron geometry). With no lone pairs the atoms take the same shape: tetrahedral.
- There is no lone pair on S, but two bonds are S=O and two are S-OH, so the shape is a slightly distorted tetrahedron.
- The bond dipoles do not cancel in the tetrahedral shape, so H₂SO₄ is polar.
How to draw it
- Count the valence electrons: S 6 + 4 × O 6 + 2 × H 1 = 32 valence electrons, 16 pairs.
- Put S in the middle and join it to the four outer atoms with single bonds.
- Give each outer atom lone pairs until it has an octet (H needs only two electrons), then put any electrons left over on the central atom.
- Where an atom is still short of an octet, turn a lone pair of its neighbour into a shared pair: that makes a double or triple bond.
- Result: two S=O double bonds and two S-OH single bonds, with no lone pair on S, so S has 12 electrons around it (an expanded octet).
- Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Lewis structure of H2SO4: quick answers
How many valence electrons are in H2SO4?
S 6 + 4 × O 6 + 2 × H 1 = 32 valence electrons, 16 pairs.
What are the formal charges in the Lewis structure of H2SO4?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of H2SO4?
8 in all, none on the central S.