Question Video: Determining the Angle Value between the Bonds in a Water Molecule | Nagwa Question Video: Determining the Angle Value between the Bonds in a Water Molecule | Nagwa

# Question Video: Determining the Angle Value between the Bonds in a Water Molecule Chemistry • Second Year of Secondary School

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What is the angle value between the two bonds of the water molecule?

02:52

### Video Transcript

What is the angle value between the two bonds of the water molecule? (A) 104.5 degrees, (B) 109.5 degrees, (C) 107 degrees, or (D) 90 degrees.

In order to determine the angle value between the two bonds of the water molecule, we can begin by drawing the Lewis structure for water. Using this Lewis structure, we can apply the valence shell electron pair repulsion model. This model can help us determine the three-dimensional shape of a molecule, which results from the minimization of repulsive electrostatic forces between electron pairs.

When applying the V-S-E-P-R model, we must use the Lewis structure to identify the number of bonded and lone electron pairs on the central atom. A bonded pair of electrons is shared between the central atom and an outer atom. The lone pairs on the central atom are pairs of valence electrons that are not shared in a covalent bond. So, we will determine the 3D shape, which will lead us to the angle value between the two bonds.

First, we can write the A-X-E notation for water. We can use the Lewis structure to determine the A-X-E notation. The A will represent the central atom, the X m term will represent the number of bonded pairs around the central atom, and the E n term will represent the number of lone pairs on our central atom. Oxygen is the central atom of a water molecule. Around this central oxygen, there are two bonded pairs, so we write X2. There are also 2 lone pairs on our central atom. We get the A-X-E notation for water to be AX2E2, which corresponds to a 3D bent shape.

The diagram given shows the molecule with a tetrahedron shape. There are four total repulsive electron domains around the central atom, and so we might expect a bond angle corresponding to a tetrahedron or 109.5 degrees. However, the lone pairs generate increased repulsion and cause the bonded pairs to have an angle smaller than 109.5 degrees. Because of the great repulsion from the lone pairs, the angle value corresponding to a bent molecule with the A-X-E notation of AX2E2 is 104.5 degrees.

Therefore, the angle value between the two bonds of the water molecule is answer choice (A), 104.5 degrees.

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