-1
S = +4 oxidation state O = -2 oxidation state
The oxidation number is + for C and -2 for O.
KO2 is rare and is called potassium superoxide. It consists of K+ ion and O^2- ion. In this case, the oxidation number of O would be 1-. If, however, you meant to ask about K2O (potassium oxide), then the oxidation number of O would be 2-
C = +2 oxidation state O = -2 oxidation state
Yes, the reaction 2Al₂O₃ → 4Al + 3O₂ is a redox reaction. In this process, aluminum (Al) is reduced from its +3 oxidation state in Al₂O₃ to elemental aluminum (0 oxidation state), while oxygen (O) is oxidized from its -2 oxidation state in Al₂O₃ to molecular oxygen (0 oxidation state in O₂). The transfer of electrons between the aluminum and oxygen species characterizes it as a redox reaction.
The oxidation state of K in K2O2 is +1. In this compound, each K atom has a +1 oxidation state and each O atom has a -1 oxidation state.
S = +4 oxidation state O = -2 oxidation state
The oxidation number for H is +1, and the oxidation number for O is -1.
In N2O5, the oxidation number of N (nitrogen) is +5 and the oxidation number of O (oxygen) is -2.
The oxidation number of O in any elemental form (e.g. O2, O3) is zero. Typically in chemical compounds the oxidation number of O is -2 (oxide ion)
In ClO4-, the oxidation number of Cl is +7, and the oxidation number of each O is -2.
0. CO2 is electrically neutral. C is +4 and O is 2(-2)
In KO2, O has an oxidation number of -1, K has an oxidation number of +1. In CO2, O has an oxidation number of -2, C has an oxidation number of +4. In K2CO3, O has an oxidation number of -2, C has an oxidation number of +4, and K has an oxidation number of +1. In the given reaction, the oxidation numbers for each atom remain the same as in their individual compounds.
The oxidation number of oxygen (O) in most compounds, including sulfate (SO4), is typically -2. In Na2SO4, sodium (Na) has an oxidation number of +1. To calculate the oxidation number of oxygen in the compound, you would use the equation: 2(oxidation number of Na) + oxidation number of S + 4(oxidation number of O) = 0. By plugging in the known values, you can solve for the oxidation number of oxygen in Na2SO4.
Rb has an oxidation number of +1, Cl has an oxidation number of -1, and O has an oxidation number of -2 based on common rules. Therefore, the oxidation number for RbClO would be +1 for Rb, -1 for Cl, and -2 for O.
The overall oxidation number is -2 so the average oxidation number of O is -1.
The central carbon atom has an oxidation number of -2 (O is -2, H is +1)