The hybridization of water is sp3.

Explanation:
Water (H2O) consists of two hydrogen atoms bonded to one oxygen atom. In order to determine the hybridization of water, we need to consider the electron configuration of oxygen and the bonding in the molecule.

1. Electron configuration of oxygen:
The atomic number of oxygen is 8, indicating that it has 8 electrons. The electronic configuration of oxygen is 1s2 2s2 2p4. This means that oxygen has two electrons in its 1s orbital, two electrons in its 2s orbital, and four electrons in its 2p orbitals.

2. Bonding in water:
In water, oxygen is covalently bonded to two hydrogen atoms. Each hydrogen atom shares one electron with oxygen, resulting in two sigma bonds. Oxygen also has two lone pairs of electrons.

3. Determining the hybridization:
To determine the hybridization, we need to count the total number of sigma bonds and lone pairs around the oxygen atom. In water, there are two sigma bonds and two lone pairs.

The total number of sigma bonds and lone pairs determines the hybridization. The hybridization can be determined using the following formula:
Hybridization = 1/2 * (V + M - C + A)

- V = number of valence electrons of the central atom (oxygen in this case) = 6
- M = number of monovalent atoms bonded to the central atom (hydrogen in this case) = 2
- C = charge on the cation, if any (none in this case)
- A = charge on the anion, if any (none in this case)

Plugging in the values:
Hybridization = 1/2 * (6 + 2 - 0 + 0) = 1/2 * 8 = 4

The result is 4, which corresponds to sp3 hybridization. Therefore, the hybridization of water is sp3.