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Question

Dichromate ion is treated with base, the oxidation number of Cr in the product formed in the following reaction is:

Cr2O7-2+2OH-2CrO4-2+H2O


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Solution

Step 1: The given reaction is

Cr2O7-2+2OH-2CrO4-2+H2O

Step 2: Determine the oxidation number of Cr in CrO4-2:

Formula to find the oxidation number of an atom in a compound is

Totalcharge=[NumberofChrominumatoms(OxidationnumberofChromium)]+[NumberofOxygenatoms(OxidationnumberofOxygen)]

Let 'x' be the oxidation number of Cr,

-2=1x+4-2x=-2+8x=+6x=+6

Oxidation number of Cr in CrO4-2 is +6.


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