What are the lone pair angles in H2O according to VSEPR theory?

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SUMMARY

The angles between lone pairs in H2O, according to VSEPR theory, are approximately 104.45 degrees. While the bond angles in H2O are often cited as 109.5 degrees, this value applies to a tetrahedral arrangement, which does not accurately describe the bent geometry of the water molecule. The presence of lone pairs results in a repulsion that alters the angles, leading to the smaller angle of 104.45 degrees. This conclusion is supported by the understanding that lone pairs occupy more space than bonding pairs, affecting the overall molecular geometry.

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  • Basic understanding of VSEPR theory
  • Knowledge of molecular geometry
  • Familiarity with bond angles in tetrahedral structures
  • Concept of lone pair repulsion in molecular shapes
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Chemistry students, educators, and anyone interested in molecular geometry and VSEPR theory will benefit from this discussion.

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Homework Statement



What are the angles between lone pairs in H2O? I know that the bonds are separated by 109.5°. What about the lone pair angles? I googled everything. I found nothing. I cannot directly assume it is 109.5°... I have a basic knowledge of VSEPR.
 
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alingy1 said:
I know that the bonds are separated by 109.5°.
Really? I thought it was 104.45 degrees.
In VSEPR, both bonding and lone pairs should be distributed approximately along the corners of a tetrahedron. Lone pairs are a bit more space filling than bonding pairs, so that the angle between the lone pairs is somewhat larger.
 
I will double check my answers before posting.
 
Last edited:
Treedude said:
H2O is bent. 109.5

H2O is bent, but 109.5 is incorrect. Water molecule is not tetrahedral. DrDu already posted a correct value.
 

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