Answer:
- The oxidation state of chromium in K₂Cr₂O₇ is 6⁺ (i.e. + 6).
Explanation:
You can calculate the oxidation number of most elements following some simple rules.
This is how you do it for chromium in K₂Cr₂O₇.
1) Rule: in a neutral compound the net oxidation number is zero (0).
Hence, sum of the oxidation numbers of K, Cr and O in K₂Cr₂O₇ is 0.
2) Rule: The most common oxidation number of oxygen in compounds, except in peroxides, is 2 ⁻ (negative 2).
3) Rule: the most common oxidation state of alkali metals is 1⁺ (positive 1)
4) Rule: multiply each oxidation state by the corresponding number of atoms in the compound (the subscripts)
↑ ↑ ↑
K Cr O
Hence, the oxidation number of chromium in this compound is 6⁺.
