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Calculate the oxidation number of the underlined element in the following ions. ul(N)H_(4)^(+) |
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Answer» Solution :`NH_(4)^(+)` : Let the oxidation number of N in `NH_(4)^(+)` bei X. The oxidation number of each hydrogen is +1. SINCE `NH_(4)^(+)` has a charge equal to +1, the sum of oxidation NUMBERS of all atoms in it must be equal to +1. Therefore, `(x)+(+1)xx4=+1` or `x=+1-4=-3` Hence, the oxidation number of N in `NH_(4)^(+)` ion is -3. |
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