Practice Problem 7

Assign the oxidation numbers of the atoms in the following compounds.

(a) Al₂O₃

(b) XeF₄

(c) K₂Cr₂O₇

Solution

(a) The sum of the oxidation numbers in Al₂O₃ must be zero because the compound is neutral. If we assume that oxygen is present in the -2 oxidation state, the oxidation state of the aluminum must be +3.

Al2O3:

2(+3) + 3(-2) = 0

(b) Because the oxidation number of the fluorine is -1, the xenon atom must be present in the +4 oxidation state.

XeF4:

(+4) + 4(-1) = 0

(c) Assigning oxidation numbers in this compound is simplified if you recognize that K₂Cr₂O₇ is an ionic compound that contains K⁺ and Cr₂O₇^2- ions. The oxidation state of the potassium is +1 in the K⁺ ion. Because the oxidation number of oxygen is usually -2, the oxidation state of the chromium in the Cr₂O₇^2- ion is +6.

Cr2O72-:

2(+6) + 7(-2) = -2

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