Find the oxcidation number of elements in (NH4)2Cr2O7
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The total oxidation state of an element is always zero. The oxidation state of an ion is equal to its charge. The ionic charge of nitride ion is -3 so its oxidation no. Is -3. Similarly the oxidation no. H+ ion is +1. Hence the oxidation no. Of nh4+ is +1. From above u can conclude that the oxidation no. Of oxygen is -2.
barbhuyan:
Now let the oxidation no. Of Cr be x. So (1)2+2x + (-2)7=0. :. X = +6. So the oxidation no. Of Cr is +6.
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