Lewis structure of H₂S
8 valence electrons. Two S-H single bonds, with two lone pairs on S.
Lewis structure
8 valence electronsFormal charges: none
Why this shape
- S has 6 valence electrons. It uses two in two single bonds to H, so four are left as two lone pairs.
- Steric number = two sigma bonds + two lone pairs = 4, so S is sp³ hybridised.
- Four electron pairs point to the corners of a tetrahedron (tetrahedral electron geometry). With two lone pairs in place of atoms, the atoms make a bent shape.
- S is larger and less electronegative than O, so the bond pairs sit farther from S and repel less. The angle closes to about 92 degrees, much less than 104.5 in water.
- The bond dipoles do not cancel in the bent shape, so H₂S is polar.
How to draw it
- Count the valence electrons: S 6 + 2 × H 1 = 8 valence electrons, 4 pairs.
- Put S in the middle and join it to the two 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.
- Result: two S-H single bonds, with two lone pairs on S.
- Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Lewis structure of H2S: quick answers
How many valence electrons are in H2S?
S 6 + 2 × H 1 = 8 valence electrons, 4 pairs.
What are the formal charges in the Lewis structure of H2S?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of H2S?
2 in all, 2 of them on the central S.