The formula is CrBr2. This is derived by balancing the electrons from the atoms. Chromium II has a charge of +2 and Bromine has a charge of -1. Therefore in order for this molecule to be electrically balanced, there must be 2 bromine ions to offset the +2 charge of the chromium ion.
Cr(BrO3)6 I find no MSDS for this compound, so it is either not stable, or it does not form.
Three chromium bromides are known: CrBr2, CrBr3, CrBr4.
Chromium II bromide is CrBr2 .
Chromium III bromide is CrBr3 .
Formula: Cr(BrO3)3
Chromium III bromate is Cr(BrO3)3
CrBr3
Cr(BrO2)3
Cr3(BO3)2
NiBr2
the chemical formula for chromim(II) fluoride is CrF2
Chromium has four oxidation states: 2, 3, 4, and 6.Iodine has one, and it's -1.There will be a iodide for each oxidation state of chromium.CrI2 Chromium (II) iodideCrI3 Chromium (III) iodideCrI4 Chromium (IV) iodideCrI6 Chromium (VI) iodide
Chromium(II) Iodide
Cr(NO3)3 (aq) + Na3PO4 (aq) --> CrPO4 (s) + 3NaNO3 (aq)
NiBr2
the chemical formula for chromim(II) fluoride is CrF2
Cr2O3 is the chemical formula for chromium (III) oxide
That would be chromium(II) sulfate, but really, chromium mostly exists as Cr(III) or Cr(VI), which would change the formula.
Chromium has four oxidation states: 2, 3, 4, and 6.Iodine has one, and it's -1.There will be a iodide for each oxidation state of chromium.CrI2 Chromium (II) iodideCrI3 Chromium (III) iodideCrI4 Chromium (IV) iodideCrI6 Chromium (VI) iodide
CrO2this is the incorrect formula. For Chromium II oxide the formula is actually CrO without the 2. The reason for this is simple. the II after chromium indicates that it has a charge of +2, and the oxygen, we know is in group 16, has a charge of -2. The formula CrO2 is actually the formula for chromium IV oxide>
Chromium(II) oxide.
Chromium(II) Iodide
Chromium(III) Sulphate is Cr2(SO4)3 . Chromium(II) Sulphate (much rarer) is CrSO4
Cr(NO3)3 (aq) + Na3PO4 (aq) --> CrPO4 (s) + 3NaNO3 (aq)
CrNO3 ************2nd Opinion*********** To get the correct formula, you need to state the oxidation number of chromium in the compound, using a Roman numeral. It's likely to be chromium(III) nitrate, which is Cr(NO3)3
Chromium (II) Carbonate. Although I'm not sure Chromium (II) can form, there might be an overall charge on your formula there...