What are the oxidation numbers of the atoms in the Cr₂O₇²⁻ ion?

Question

Quizplus · Accepted Answer

To determine oxidation numbers, we first need to identify the elements present and their known oxidation states. From the formula, we can see that there are four elements present: Cr, S, U, and O. The oxidation states of S and P can be assumed to be -2 and +5, respectively.

For Cr, we can use the rule that the sum of oxidation numbers in a neutral compound is 0. We also know that the oxidation state of oxygen is usually -2, so there are a total of (-2 x 17) = -34 electrons from the O atoms. Summing up the oxidation states of the other elements should give us 34 to balance the equation.

Let x be the oxidation state of Cr, and y be the oxidation state of U. Thus, we have:
x + (-2 x 17) + y + (-2) + (5 x 12) - 2 = 0
x + y - 4 = 0
x + y = 4

From here, we can only make assumptions about the oxidation state of U, as it is not known. However, we do know that the oxidation state of Cr must be positive, as it is combining with negatively charged S and O atoms.

Option A can be eliminated, as this would mean the oxidation state of Cr is negative, which is not possible. Option B can also be eliminated, as there is no reason to believe that the oxidation state of O is -2 in this case. Option D can be eliminated, as a Cr oxidation state of +7 is not commonly observed.

Therefore, the best choice is C, where the oxidation state of Cr is +6 and the oxidation state of O is -2 to balance the charge.

What Are the Oxidation Numbers of the Atoms in the Cr2O72−

What Are the Oxidation Numbers of the Atoms in the Cr₂O₇²⁻