2
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Mn ($3d^5$, $4s^2$) can have maximum 7 unpaired electrons in excited state so it should have formed MnF7 molecule but it can form only MnF4 molecule.

Is it because: $d_{xy}, d_{yz}, d_{zx}$ orbitals are not axially oriented and therefore not suitable for head on sigma overlap but can form pi bonds as in Mn2O7 molecule.

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  • $\begingroup$ Actually even MnF4 is unstable and easily decomposes to MnF3 and F2. $\endgroup$
    – Mithoron
    Jan 26, 2018 at 17:31
  • $\begingroup$ For some reason $\ce{MnO4-}$ with $\ce{Mn}$ at valence $+7$ is stable but there is no $\ce{MnF7}$ or $\ce{Mn^{7+}}$. $\endgroup$
    – Zhuoran He
    Jan 26, 2018 at 18:25
  • $\begingroup$ Answer could be similar as in chemistry.stackexchange.com/questions/74528/… MnF4 is already about as strong oxidant as fluorine, MnF5 would be even stronger. $\endgroup$
    – Mithoron
    Jan 27, 2018 at 0:17
  • $\begingroup$ @Mithoron This is surprising that MnO2 even oxidize HF to F2. But my doubt is that MnF5 is even possible? considering _ "dxy, dyz and dzx orbitals are not axially oriented and therefore not suitable for head on sigma overlap but can form pi bonds as in Mn2O7 molecule."_ $\endgroup$
    – Apurvium
    Jan 27, 2018 at 17:32
  • $\begingroup$ There's no problem with any orbitals, there can be even $\ce{MnF7^3-}$ anion. $\endgroup$
    – Mithoron
    Jan 27, 2018 at 19:30

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