This compound is lead bromide - PbBr2.
Lead(II) bromide = PbBr2
Pb(BrO2)2 would be the correct formula.
The formula of lead (II) sulphide is PbS whilst the formula of lead (IV) sulphide is PbS2. The formula is: PbS and the number of Solubility product constants is 3x10^-25 if you're interested;)
Lead(II) sulfide
Pb10S10 The prefix deca in decasulfide means that there are 10 sulfide ions in the formula, each with a 2- charge. So the total negative charge is 10 x 2-, which equals 20-. To balance the total sulfide charge, you need enough lead II ions to equal 20+, so that means 10 lead II ions, which equals a charge of 20+. This is not really the correct way to name this compound. It should just simply be lead II sulfide, and the proper formula would be PbS, which, if you look at the ratio of lead to sulfide ions, which is 10:10, the formula for lead II decasulfide should be reduced to PbS.
The formula for the iodite anion is IO2-
The formula unit for lead (II) chloride is Pb(NO3)2. This formula shows that each formula unit contain one lead (II) ion. By the definition of molarity, a liter of a 0.250 molar solution of lead (II) nitrate therefore contains 0.25 gram formula units of lead (II) ions per liter of solution, so that 250 milliliters of such a solution will contain 250/1000 of this amount, or 0.0625 gram formula units. Assuming that chlorine is an ideal gas, 22.4 liters of it at STP contain one mole. Since gases are homogeneous, 14.6 liters of it will contain 14.6/22.4 or 0.652 moles. The formula of gaseous chlorine at STP is Cl2. Therefore, 0.652 moles of it contain twice this number of chlorine atoms, or 1.305 "moles" of such atoms. The formula unit for lead (II) chloride is PbCl2. Therefore, each gram formula unit of lead (II) requires two gram formula units of chloride ions. Comparison shows that there is a large excess of chlorine over that required to form lead (II) chloride from the 0.0625 gram formula units of lead (II) contained in the specified amount of solution. Lead (II) is therefore the limiting reagent in this combination, and 0.0625 gram formula units of PbCl2 can be produced. The gram atomic mass of lead is 207.2 and that of chlorine is 35.453. Therefore, the mass of lead (II) chloride that can be produced is 0.0625[207.2 + 2(35.453)] or 17.4 grams, to the justified number of significant digits.
Sodium bromite is NaBrO2
Pb2+ is the lead(II) ion formula
Lead(II) Bromide = PbBr2
Lead(II) Bromide = PbBr2
Lead (II) bisulfate has the formula Pb(HSO4)2
Mg(BrO2)2
Pb(ClO3)4
The formula for lead (II) nitrate is Pb(NO3)2.
The formula is MgI2 the compound would be ionic.
There are two lead phosphates: lead (II) metaphosphate with formula Pb(PO3)2 and lead (II) orthophosphate with formula Pb3(PO4)2
The formula for Lead Sulfide (or Lead Sulphide) is either: PbS for Lead(II) Sulphide, or PbS2 for Lead(IV) Sulphide.
copper (II) hypobromite