The Hydrogen fluoride/hydrofluoric acid reacts with many metals, usually forming hydrogen gas and a metal fluoride, for example: Mg + 2HF -> MgF2 + H2. It also reacts readily with most metal oxides.
Can anyone provide a list of metals and metal oxides which do not react with hydrogen fluoride/hydrofluoric acid? And explain why most metals and metal oxides react with Hydrogen fluoride/hydrofluoric acid while the others do not react?
The literature survey (Pls. confirm) leads to:
[1] Au and Pt are the only metals which are 100% resistant to pure HF. [2] Ni and its alloys as well as Ag metal are resistant to the marketed samples of HF which are 48-52% by weight; although the industrially prepared HF samples are70% by weight. [3] The literature survey did not name any metal oxide which is completely HF resistant.
M=O bonds, no doubt, very stable yet get replaced by the more stable M-F bonds as F being the most electronegative element and F(-) having the smallest size(excepting H) can attain the maximum coordination number to form very stable crystal structures which makes the following reaction to proceed in the forward direction: MO+2HF →MF2 +H2O; ∆H= highly – ve .
Dear Manohar Sehgal
Thank you for your answers
Would you please explain in theory why Au and Pt are 100% resistant to pure HF while the other metals are not 100% resistant to pure HF?
[I]The one word answer should be that both these metals are INERT. The word inertness means nonparticipation toward chemical reactions. Some common reasons for their inertness may be explained as: Both these metals possess very compact ccp crystal lattice, quite high values of IE (Au=890.0,Pt=870.0 k j/mol), bond enthalpies in the gaseous form[224.7(Au),357.3(Pt)kJ/mol],very low values of SRP[Au/Au(1)= - 1.83 V, Pt/Pt(II)= -1.88 V] All these factors will make them very reluctant to lose electron/s.
[II]The inertness of Pt is all the pronounced and the metal be termed as a noble metal as will be illustrated by its outermost electronic configuration. Platinum, being a congener of Ni should be a 6s2 5d8. However, as you go up the energy levels of electron states, the s and d get closer together in energy, so the half filled orbital rule breaks down, making 5d to fill completely to attain 6s0 5d10 state rather than a 6s2 5d8. It can also be looked as that the 5d sub shell electrons are shielded better by the completely filled 4f sub shell electrons to make 5d to be lower energy than the 6s orbital.
[III]. Relativistic effects, such as the contraction of s orbitals are a major factor in making gold less reactive than silver, and in lowering the oxidation potential of gold.
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