+4 for lead
lead(II) chromate
The chemical formula for lead IV chlorite is Pb(ClO2)4. Lead IV chlorite is a compound containing lead with a +4 oxidation state and chlorite ions (ClO2-).
The ionic chemical formula of lead(IV) oxide is: (Pb)4+ + 2 O-.
The chemical name of SnCr2O7 is tin(IV) dichromate.
The oxidation number for Mn in H2MnO3 is +3. In this compound, oxygen is typically assigned an oxidation number of -2, and hydrogen is +1. By considering the overall charge of the compound and assigning hydrogen and oxygen their usual oxidation states, the oxidation number of Mn can be calculated as +3.
lead(II) chromate
lead (IV) ion is Pb4+ ion. Note that Pb4+ is never found as an ion- the (IV) is an oxidation number or oxidation state.
Plumbic = Lead(IV) Pb(Cr2O7)2
You probably mean lead chromate since you have mentioned an oxidation state of 2+ for the lead ion and the chromate ion has an oxidation state of 2-. Lead chromate is yellow. Please see the link.
Yes, Pb4O3 contains lead(II) ions, not lead(IV) ions. Lead typically forms a +2 oxidation state in its compounds.
The chemical formula for lead IV chlorite is Pb(ClO2)4. Lead IV chlorite is a compound containing lead with a +4 oxidation state and chlorite ions (ClO2-).
The chemical formula of lead (IV) is PbF4.
The chemical formula of lead(IV) chromate is PbCrO4.
The ionic chemical formula of lead(IV) oxide is: (Pb)4+ + 2 O-.
The chemical name of SnCr2O7 is tin(IV) dichromate.
The oxidation number for Mn in H2MnO3 is +3. In this compound, oxygen is typically assigned an oxidation number of -2, and hydrogen is +1. By considering the overall charge of the compound and assigning hydrogen and oxygen their usual oxidation states, the oxidation number of Mn can be calculated as +3.
Lead has two oxidation states (+4 and +2). As a result, there are two compounds composed of lead and sulfate: lead(IV) sulfate, which is Pb(SO4)2, and lead(II) sulfate, which is PbSO4.