0 in the elemental form, +2 or +4 in its compounds
+2 for Pb
-2 for S
+2
+4 for Pb and -1 for each hydrogen
The oxidation number of lead (Pb) in the compound PbCl2 has to be what?
+2
Oxidation number of Pb (in PbF4) is +4, of F (in PbF4) it is -1. (Together the oxidation numbers add up to null in neutral compounds)
O is 2- and there are two of them so Pb would have to be 4+
+4 for Pb and -1 for each hydrogen
The oxidation number of lead (Pb) in the compound PbCl2 has to be what?
The oxidation number of lead (Pb) in the compound PbCl2 has to be what?
+2
Oxidation number of Pb (in PbF4) is +4, of F (in PbF4) it is -1. (Together the oxidation numbers add up to null in neutral compounds)
O is 2- and there are two of them so Pb would have to be 4+
No, Pb is not a transition metal and it has 2 oxidation states
Pb, lead, +2; S, sulfur, +6; O, oxygen, -2.
Oxygen is in this case -2. There are three oxygens in this problem, so the total is -6. Pb's oxidation number is the same as its ionic charge, which is +2. Everything must equal zero out if you add all the numbers together. +2(Pb)+(Sulfur's oxidation number)-6(Oxygen)= 0 -4+(4)=0 Pb=-2 S=+4 O=-2
It has to be Pb(NO3)2 with NaCl as Pb has a +II oxidation state and NO3 has -I oxidation state. The reaction is the following: Pb(NO3)2 +2NaCl ----> PbCl2 + 2NaNO3
Assuming the 2 oxidation state of lead. Pb + 2HNO3 --> Pb(NO3)2 + H2
lead (IV) ion is Pb4+ ion. Note that Pb4+ is never found as an ion- the (IV) is an oxidation number or oxidation state.