What is the molecular geometry of “H”_2″O”? Draw its VSEPR structure.

Water has a bent molecular geometry.

Water’s is considered “bent”.

The structure has two bonds (one to each hydrogen from oxygen in the center) and two non- pairs on the oxygen.

The Lewis structure is similar but in drawings the VSEPR will appear three dimensional showing bond angles (in this case ##104.5^@##) and the Lewis will be flat (2-D). Check out the images at:

When drawing the structures, decide which atom belongs in the center and draw its electron dot structure, usually the single one but you can also determine this based on the electron dot of each type of atom. In the case of water, the oxygen gets the central position.

When a correct e-dot is drawn for oxygen, there should be two sides with single electrons (these will be bonding electrons since they are not “married” yet) and two sides with pairs of electrons (non-bonding since they are already “married”).

The hydrogen atoms with their single electrons will pair up with oxygen’s single electrons creating two single bonds.(Lewis structure)

An electrostatic repulsion exits between the bonded electrons and the non-bonding pairs (NBPs) on the oxygen. This repulsion pushes the bonds away from the NBPs creating the bond angle of ##104.5^@## (VSEPR structure).

The resulting water molecule is actually a special version of the tetrahedral shape. If the central atom has four bonds and no NBPs to repel bonds, the tetrahedral shape will result with larger bond angels of ##109.5^@##.


Imagine removing two atoms (leaving the electrons participating in the bonds) from a tetrahedral molecule. You now have two bonds and two NBPs (like water) with an electrostatic repulsion that reduces the bond angle to ##104.5^@##.

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