Here's the question you clicked on:
Aninxahc
How do you find the oxidation number of Na2SeO3
Na would be 2(1) which is 2.... you get the 1 since Na is in the first group so the oxidation number will be 1 oxygen oxidation number is always -2 so you multiply it by 3(-2)= -6 2-6=4 so the oxidation number for Se is 4 cause thats the only way it will = 0 on the total
whose oxinumber do you mean?