find oxidation number of N in (N2H5)2SO4
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Hey mate here is your answer
In this case, we know the oxoration no for H is +1. Then set this value equal to the overall net charge of the ion. Our equation now looks like this: 1(4) = 1, You use the multiplier of 4 to indicate that the ammonium ion has 4 hydrogen. Thus, the oxidation numberfor Nitrogen is -3.
In this case, we know the oxoration no for H is +1. Then set this value equal to the overall net charge of the ion. Our equation now looks like this: 1(4) = 1, You use the multiplier of 4 to indicate that the ammonium ion has 4 hydrogen. Thus, the oxidation numberfor Nitrogen is -3.
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