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Redox Reactions question

2021 · 26 Feb · Shift 1 · Q18
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Redox Reactions question

2021 · 26 Feb · Shift 1 · Q18

JEE MainChemistryRedox ReactionsNumerical+4 / −1
Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 6

  1. The dichromate ion is \ceCr2O72−\ce{Cr2O7^{2-}}\ceCr2O72−.

  2. When dichromate ion is treated with base, it converts to chromate ion:

\ceCr2O72−+2OH−−>2CrO42−+H2O\ce{Cr2O7^{2-} + 2OH^- -> 2CrO4^{2-} + H2O}\ceCr2O72−+2OH−−>2CrO42−+H2O

  1. The product formed is chromate ion, \ceCrO42−\ce{CrO4^{2-}}\ceCrO42−.

  2. Let the oxidation number of chromium in \ceCrO42−\ce{CrO4^{2-}}\ceCrO42− be xxx.

x+4(−2)=−2x + 4(-2) = -2x+4(−2)=−2

x−8=−2x - 8 = -2x−8=−2

x=+6x = +6x=+6

  1. Therefore, the oxidation number of Cr in the product formed is:

6\boxed{6}6​

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