Question:

Oxidation states of Cr in K2Cr2O7 and CrO5 are, respectively:

Updated On: Apr 15, 2025
  • +6, +5
  • +6, +10
  • +6, +6
  • None of these
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The Correct Option is C

Solution and Explanation

The correct answer is +6, +6.

Here's how we determine the oxidation states:

  • K2Cr2O7 (Potassium Dichromate):
    • The oxidation state of K is +1.
    • The oxidation state of O is -2.
    • Let the oxidation state of Cr be x.
    • The compound is neutral, so the sum of the oxidation states must be zero: 2(+1) + 2(x) + 7(-2) = 0 2 + 2x - 14 = 0 2x = 12 x = +6
    • Therefore, the oxidation state of Cr in K2Cr2O7 is +6.
  • CrO5 (Chromium Pentoxide):
    • CrO5 has a "butterfly" structure with two peroxide linkages (O-O) and one double-bonded oxygen. The structure can be written as CrO(O2)2.
    • Each peroxide linkage (O22-) has an oxidation state of -1 for each oxygen atom, hence -2 for the O2 group.
    • The oxidation state of the doubly bonded oxygen is -2.
    • Let the oxidation state of Cr be x.
    • The sum of oxidation states must be zero: x + (-2) + 2(-2) = 0 x - 2 - 4 = 0 x = +6
    • Therefore, the oxidation state of Cr in CrO5 is +6. (Note: Assigning a simple oxidation state of -2 to each of the five oxygens would incorrectly give chromium an oxidation state of +10. We need to know the real structure)

Therefore, the oxidation states of Cr in K2Cr2O7 and CrO5 are +6 and +6, respectively.

Correct Answer: +6, +6

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