oxidation
The Pb K-egde Xanes data reveals that Pb is in a mixed valence state of Pb4+ and Pb2+. However in literature Pb is claimed to be in Pb2+ state. The Pb 6s2 electrons hybridize with the O 2p electrons to form strong covalent bonding which results in the relative displacement of Pb cage with respect to the O-octahedron. This results in increase ferroelectric properties of PbTiO3.However the question is that in the covalent state of the Pb2+ will it appear as Pb4+ state in the Pb K-edge? The reason argued here is that the Pb will lose the 6s2 electrons to form the bond and hence appear to be Pb4+. Hence the argument placed by this pool of thought is from the EXAFS data what appears to be Pb4+ is actually the covalently bonded Pb2+ while what appears to be Pb2+ is actually the ionic type Pb2+.What is the oxidation state of Pb and Ti in PbTiO3 ?
The oxidation state 4+ is not stable in PbCl4; the reaction is:Pb4+ + 2 e---------Pb2+
Yes. They react to form lead (II) oxide.
6.3 x 10-6
Lead (IV) ion
Cr(s) | Cr3+(aq) Pb2+(aq) | Pb(s)
Pb2+
Formula: Pb2+
Pb2+
It is very slightly soluble in water.In a saturated solution:[Pb2+] = 1.2x10-2 mol/L[Br-] = 2.4x10-2 mol/Lbecause [Pb2+]*[Br-]2 = Ks = 6.3*10-6 and [Br-] = 2*[Pb2+]
in Pb3O4 the oxygen has an oxidation number of -2, the average oxidation number is a fraction (+8/3) so this must be a mixed oxidation sate compound. Teh two common Ox numbers of Pb are +2 and +4. the only way to make +8 overall is to have one Pb4+ and two Pb2+ This compound is a lovely red colour and is often called "red lead". Full marks to teacher - this is a tricky question!
It is the log mean value of the inert B in counterdiffusion usually used to find a flux. P = pA1 + pB1 = pA2 + pB2 thus, pB1 = P - pA1, and pB2 = P - pA2, so: pBM = [pB2 - pB1]/ln(pB2/pB1) = [pA1 - pA2]/ln[(P-pA2)/(P-pA1)] happy engineering :)