Dick,
A lot of answers are in Inorganic Chemistry Books.
I completely agree with you, I think things are improved faster when you
understand henk.thijs@eurocontrol.be (or feel) why you change processing steps to get what you want.
Why do you want to get K2PtCl6 instead of using K2PtCl4 ? I would think you
loose some sensitivity with K2PtCl6 , am I right ? The ferrous oxalate you
form while exposing as a given number of electrons to reduce the platinum
ions. That number is proportional to the amount of light that you gave to
the layer. The difference between K2PtCl4 and K2PtCl6 is that they contain
respectively Pt2+ and Pt4+. To reduce Pt2+ to metallic black platinum, you
need 2 electrons from the ferrous oxalate. To reduce Pt4+ you need 4
electrons. So, for a given exposure, you will get twice the black reduced
platinum with K2PtCl4 vs K2PtCl6.
Does it match to your experiences ?
Philippe
_____________________________________________________________________________
Philippe MONNOYER
Ph.D. Student
Laboratoire de RMN
Facultes Universitaires Notre-Dame de la Paix
rue de Bruxelles, 61
B-5000 Namur
Belgium
Phone: +32 81 724601
Fax: +32 81 724530
Mail: Philippe.Monnoyer@fundp.ac.be
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