Xe-F-Xe Bond Angle - Is it 104.5?

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The bond angle for Xe-F-Xe is approximately 109.5 degrees, which aligns with the ideal tetrahedral geometry for molecules with four electron domains. This angle is slightly reduced due to the presence of lone pairs on the central xenon atom, which introduces repulsive forces. Although 104.5 degrees is mentioned as a reasonable approximation, it does not accurately reflect the actual bond angle. The discussion highlights the complexity of bonding in xenon fluorides, particularly with two sigma bonds and lone pairs involved. Overall, the bond angle is influenced by the electron arrangement around the xenon atom.
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What is the bond angle for Xe- F- Xe ?..is it 104.5?

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Fluorine has 7 outer electrons, 2 of which are spent in bonding (sigma) to the 2 xenon atoms. That makes the number of electrons equal to 5 (2 pairs, 1 unpaired). So if FXe2 were to exist (and I am not saying it doesn't exist for I don't know myself) then it would have a funny geometry...2 sigma bonds, 2 lone pairs and 1 unpaired electron (counting basically).
 
: The bond angle for Xe-F-Xe is not exactly 104.5 degrees, but it is close. The actual bond angle is around 109.5 degrees, which is the ideal tetrahedral bond angle for molecules with four electron domains. The Xe-F-Xe bond angle is slightly smaller due to the lone pair of electrons on the central xenon atom, which creates a repulsive force and causes the bond angle to decrease. However, 104.5 degrees is a reasonable approximation for this bond angle.
 

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