oxidation state of Fluorine is always -1.
In compounds, magnesium has a +2 oxidation state; in most but not quite all compounds, oxygen has a -2 oxidation state. In peroxides, oxygen has a -1 oxidation state. In oxygen fluorides, oxygen has positive oxidation states.
Reduction-Oxidation Reaction: If looking at MnF2(s) -> Mn(s) + F2(g), we can see that in the reactant, MnF2, the oxidation state of F is -1, and that of Mn is +2. However, in the products, both of them have oxidation states of 0 (because they are in their elemental form). Thus, F has been oxidized and Mn has been reduced. Mn is the oxidizing agent, F is the reducing agent.
Li = +1; Al = +3; F = -1 but I think your formula is wrong Li3AlF6
F-1 (flouride). Flourine has the strongest attraction for electrons of any element, so the oxidation state of -1 is the only one that flourine uses.
O = -2 oxidation state H = +1 oxidation state
Oxygen has an oxidation state of zero (0) in HOF. F has an oxidation state of -1 and H has +1.
In compounds fluorine, F, has an oxidation number of -1.
oxidation number of I is -1. oxidation number of F is +1.
It is solid
sikkim
The oxidation number of the ion F1- is -1.
Total electrons --> 7+42-1=48 Cl as a central atom One lone pair on Cl (on the side) 4 F's up and down and the other two on the opposite side of lone pair. F I :- Cl - F - F I F This is the best I can do with plain text.
no it was the 36th state
Himalayas
Cu is +2 and F is -1
-1 for F +6 for S
+1 for K -1 for F