+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)
0 in the elemental form, +2 or +4 in its compounds
O is 2- and there are two of them so Pb would have to be 4+
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)
0 in the elemental form, +2 or +4 in its 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.