Lewis structure of H₂O
8 valence electrons. Two O-H single bonds, with two lone pairs on O.
Lewis structure
8 valence electronsFormal charges: none
Why this shape
- O 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 O 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.
- Repulsion falls in the order lone pair to lone pair > lone pair to bond pair > bond pair to bond pair. The lone pairs push the bond pairs closer, so the angle is 104.5°, less than the ideal 109.5°.
- The bond dipoles do not cancel in the bent shape, so H₂O is polar.
How to draw it
- Count the valence electrons: O 6 + 2 × H 1 = 8 valence electrons, 4 pairs.
- Put O 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 O-H single bonds, with two lone pairs on O.
- Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Lewis structure of H2O: quick answers
How many valence electrons are in H2O?
O 6 + 2 × H 1 = 8 valence electrons, 4 pairs.
What are the formal charges in the Lewis structure of H2O?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of H2O?
2 in all, 2 of them on the central O.