LiBrO3 this decomposes to Li^(+) & BrO3^(-)
The bromate anion has a charge of '-1'
Use the standard for oxygen at '-2'
Since there are 3 oxygens then the oxygen moiety is 3 X -2 = -6
Creating a little sum
Br + -6( oxygen moiety) = -1( anion charge)
Br - 6 = -1
Add '6 'to both sides
Br = (+)5 The oxidation state of bromine.
The symbol for a bromide ion is Br-.
If Br had an oxidation number of +7, the net charge on the ion would be +1, and not -1. Thus, the oxidation number for Br in BrO3- should be 5+.
Oxidation number of Br in BrO3 is 6. BrO3 doesn't exist. It should actually be BrO3- ion with +5 oxidation number for Br.
Each Br atom has an oxidation number of zero.
The Potassium (K) has an oxidation number of +1. The Bromine (Br) has an oxidation number of -1.
The symbol for a bromide ion is Br-.
If Br had an oxidation number of +7, the net charge on the ion would be +1, and not -1. Thus, the oxidation number for Br in BrO3- should be 5+.
Oxidation number of Br in BrO3 is 6. BrO3 doesn't exist. It should actually be BrO3- ion with +5 oxidation number for Br.
-2 for each O, +5 for Br
Each Br atom has an oxidation number of zero.
The Potassium (K) has an oxidation number of +1. The Bromine (Br) has an oxidation number of -1.
+1 for Na -1 for Br
Each Br atom has an oxidation number of zero.
dcn
The molecular formula should be CBr4. The oxidation numbers are -1 for each Br, +4 for C.
-1 in bromide (most common). It can exhibit oxidation numbers from -1 to +7 (in HBrO4)
-1 for Cl and +1 for Br