# Why Mn atom show maximum valency of 4 with F atom?

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.

• Actually even MnF4 is unstable and easily decomposes to MnF3 and F2. – Mithoron Jan 26 '18 at 17:31
• For some reason $\ce{MnO4-}$ with $\ce{Mn}$ at valence $+7$ is stable but there is no $\ce{MnF7}$ or $\ce{Mn^{7+}}$. – Zhuoran He Jan 26 '18 at 18:25
• 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. – Mithoron Jan 27 '18 at 0:17
• @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."_ – Apurvium Jan 27 '18 at 17:32
• There's no problem with any orbitals, there can be even $\ce{MnF7^3-}$ anion. – Mithoron Jan 27 '18 at 19:30