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From Chemistry of the Elements by Greenwood and Earnshaw [1, p. 833]:

The mean bond energy for X-F
Figure 17.9 Mean bond energies of halogen fluorides.

This shows the mean bond energy for $\ce{X-F}$, where $\ce{X}$ is a halogen. I have several questions:

  1. Why is the $\ce{X-F}$ bond in $\ce{XF}$ stronger than that in $\ce{XF3}$ or $\ce{XF5}$? I think it is because the $\ce{XF}$ bond has bond order 1, but $\ce{XF3}$ and $\ce{XF4}$ both have 3c-4e bonds. However, I am not too sure about this explanation. It also doesn't really explain why $\ce{XF5}$ is weaker than $\ce{XF3}$.

  2. Why is the order of bond enthalpy generally $\ce{IF} > \ce{BrF} > \ce{ClF}$? I thought it would be the reverse, because smaller atoms would be able to have better atomic orbital overlap.

References

  1. Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.
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  • $\begingroup$ When appending an illustration from a third-party source, make sure it's readable and the source is properly cited (see my edit; also feel free to use it as an example for your future posts). $\endgroup$ – andselisk Jan 12 at 9:20
  • $\begingroup$ For me this graph is very logical and provides insight into actual properties of these compounds. For example it's generally more difficult to get higher and higher oxidation states of halogens. $\endgroup$ – Mithoron Jan 13 at 16:04

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