We work on ultrathin nanostructured devices, so our interested thickness is below 50 nm.
Cold, dilute (5-10%) hydrofluoric acid active dissolves the titanium, but it does not react with gold and hardly reacts with the aluminum.
Here I think two different mechanisms are operating if I check the Pauling’s electro negativity values reported in the literature. Ti has 1.54 compared to gold 2.54, that makes it large enough difference in electronegative to form Titanium tetra fluoride: Ti + 4 HF= TiF4 + 2H2. , Aluminum (1.61) exposed to hydrofluoric acid may have already very thin layer of Al3O2 oxide, which may prevent the reaction between Al and HF solvent. Gold is novel metal and it has outer electronic configuration as (d10s2)
Gold metal should react with hydrofluoric, F2, or bromine, Br2, to form the trihalides such as AuF3 gold (III) fluorite, AuBr3, respectively. On the other hand, gold metal reacts with iodine, I2, to form the monohalide gold (I) iodine.
Gold metal dissolves in aqua regia, a mixture of hydrochloric acid, HCl, and concentrated nitric acid, HNO3, in a 3:1 ratio. Therefore 5-10% solution to be too dilute for the reaction at the room temperature.
For selective etching of Ti while preserving Al and Au, you may want to consider: Br2 + ethyl acetate (hot); I2 + MeOH (hot); and NH4OH + H2O2 (1:2). About last option ─ also a suitable etchant for TiO2 ─ you may check: J.-P. Rioult, R. Fabien (Assignee: U.S. Philips Corporation); “Method for selective etching of titaniumdioxide relative to aluminum”, U.S. Patent 4,322,264; Mar. 30, 1982.
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