Step by step
- Known values: O = −2 (oxygen is −2 in most compounds); O (peroxide) = −1 (oxygen is −1 in a peroxide O–O link).
- Let the oxidation number of Cr be x. The sum over all atoms equals the overall charge, 0 for a neutral compound.
- 1(−2) + 4(−1) + x = 0
- −6 + x = 0, so x = +6
- x = +6
CrO₅ has two peroxide groups. Counting all O as −2 would give +10, more than chromium's six outer electrons allow.
Check any other formula with the oxidation number calculator.
Common oxidation numbers
| Element in compound | Oxidation number |
|---|---|
| Mn in KMnO4 | +7 |
| Cr in K2Cr2O7 | +6 |
| Cr in Cr2O72− | +6 |
| Cr in CrO42− | +6 |
| Mn in MnO42− | +6 |
| Mn in MnO2 | +4 |
| S in H2SO4 | +6 |
| S in SO2 | +4 |
| S in SO3 | +6 |
| S in H2S | −2 |
| S in Na2S2O3 | +2 |
| S in S4O62− | +5/2 |
| S in H2SO5 | +6 |
| N in HNO3 | +5 |
| N in NH3 | −3 |
| N in NH4+ | −3 |
| N in NO3− | +5 |
| N in NO2 | +4 |
| N in HNO2 | +3 |
| N in N2O | +1 |
| N in N2O5 | +5 |
| N in N2H4 | −2 |
| N in NH2OH | −1 |
| N in HN3 | −1/3 |
| P in H3PO4 | +5 |
| P in H3PO3 | +3 |
| P in H3PO2 | +1 |
| Cl in HClO4 | +7 |
| Cl in HClO | +1 |
| Cl in KClO3 | +5 |
| Cl in CaOCl2 | 0 |
| Fe in Fe3O4 | +8/3 |
| Fe in K4[Fe(CN)6] | +2 |
| Fe in K3[Fe(CN)6] | +3 |
| Cr in CrO5 | +6 |
| O in H2O2 | −1 |
| O in OF2 | +2 |
| O in KO2 | −1/2 |
| O in Na2O2 | −1 |
| H in NaH | −1 |
| Pb in Pb3O4 | +8/3 |
| C in CO2 | +4 |
| C in CO | +2 |
| C in CH4 | −4 |
| C in C6H12O6 | 0 |
| C in HCOOH | +2 |
| Ni in Ni(CO)4 | 0 |
| Cu in [Cu(NH3)4]2+ | +2 |
| Co in [Co(NH3)6]Cl3 | +3 |
| Xe in XeF6 | +6 |
| Os in OsO4 | +8 |
| Al in LiAlH4 | +3 |
Questions students ask
What is the oxidation number of Cr in CrO₅?
+6. CrO₅ has two peroxide groups. Counting all O as −2 would give +10, more than chromium's six outer electrons allow.
How do you find the oxidation number of chromium in CrO₅?
Known values: O = −2 (oxygen is −2 in most compounds); O (peroxide) = −1 (oxygen is −1 in a peroxide O–O link). Let the oxidation number of Cr be x. The sum over all atoms equals the overall charge, 0 for a neutral compound. 1(−2) + 4(−1) + x = 0 −6 + x = 0, so x = +6 x = +6