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What is the oxidation state of an individual bromine atom in NaBrO3?

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What is the oxidation state of an individual bromine atom in NaBrO3?

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The way to figure these out is to assign the oxidation state (valence) to everything that you know and see what’s left over: Na is always going to be +1 O is usually -2 unless something stronger is around Br is usually -1 unless something stronger is around You get these numbers from the periodic table (based on which column they’re in). If you add up the oxidation states, you have to get the charge of the compound. In this case it’s neutral or 0. So everything has to add up to 0. If they are just as I said above, then you have: Na = +1 O3 = 3 x -2 = -6 Br = -1 These add up to -6, so you know that something is not right Looking up bromine in wikipedia, it says that it can have oxidation states of 5, 4, 3, 1, -1. The best choice for Br is 5, since you have -6 (O3) +1 (Na) = -5, then the numbers are: +1 (Na), -2 (O) [-2 x 3 = -6], +5 (Br) These add up to zero. Ta da.

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