In MoO4 Mo would have an ON of +8 and oxygen -2. You probably mean MoO3 where Mo has an ON of +6 or conceivably the MoO42- ion where Mo also has an ON of +6
The oxidation number of oxygen is typically -2, so for the whole MoO4^2- ion (which has a charge of -2), the total oxidation number must sum up to -2. This means that the oxidation number of molybdenum (Mo) in MoO4^2- would be +6.
The oxidation number of oxygen is typically -2. Since there are four oxygen atoms in MoO4-2 and the overall charge is -2, the oxidation number of molybdenum (Mo) is calculated to be +6 to balance the negative charge.
The oxidation number of molybdenum (Mo) in MoO4^2- is +6. This is because the overall charge of the compound is -2, and each oxygen atom has an oxidation number of -2. By setting up an equation (x + 4(-2) = -2), we find that x = +6.
The oxidation state of Mo in MoO4 is +6. This is because each oxygen atom has an oxidation state of -2, and since there are four oxygen atoms in the compound, the total oxidation state contributed by oxygen is -8. In order to balance the charge of the compound, the oxidation state of Mo must be +6.
If you mean MoO4 the oxidation number of Mo is 8 and the compound would be called Mo(VIII) oxide. This compound does not exist, it is theoretical- teachers and examiners often use such nonexistent compounds as teaching aids- why I don't know- i think it is misleading as there are plenty of real compounds about.
The oxidation number of oxygen is typically -2, so for the whole MoO4^2- ion (which has a charge of -2), the total oxidation number must sum up to -2. This means that the oxidation number of molybdenum (Mo) in MoO4^2- would be +6.
The oxidation number of oxygen is typically -2. Since there are four oxygen atoms in MoO4-2 and the overall charge is -2, the oxidation number of molybdenum (Mo) is calculated to be +6 to balance the negative charge.
The oxidation number of molybdenum (Mo) in MoO4^2- is +6. This is because the overall charge of the compound is -2, and each oxygen atom has an oxidation number of -2. By setting up an equation (x + 4(-2) = -2), we find that x = +6.
The oxidation state of Mo in MoO4 is +6. This is because each oxygen atom has an oxidation state of -2, and since there are four oxygen atoms in the compound, the total oxidation state contributed by oxygen is -8. In order to balance the charge of the compound, the oxidation state of Mo must be +6.
If you mean MoO4 the oxidation number of Mo is 8 and the compound would be called Mo(VIII) oxide. This compound does not exist, it is theoretical- teachers and examiners often use such nonexistent compounds as teaching aids- why I don't know- i think it is misleading as there are plenty of real compounds about.
-1. In simple inorganic compounds other than peroxides and superoxides, oxygen almost always has an oxidation number of -2, but since calcium cations always have an oxidation number of +2, in this compound, the two oxygen atoms must have only -1 charge each. This compound is calcium peroxide, one of the exceptions noted above.
Hydrogen's oxidation number is +1.Chlorin's oxidation number is +1.Oxygen's oxidation number is -2.
The oxidation number of acetate (CH3COO-) is -1. The carbon atom has an oxidation number of +3, each hydrogen atom has an oxidation number of +1, and the oxygen atoms have an oxidation number of -2.
The oxidation number of each hydrogen in H2CO2 is +1, while the oxidation number of each carbon in CO2 is +4. This is because hydrogen usually has an oxidation number of +1, and oxygen usually has an oxidation number of -2.
Silicon's oxidation number is +4.Oxygen's oxidation number is -2
The oxidation number of nitrosyl (NO) is +1. Nitrogen typically has an oxidation number of -3, and oxygen typically has an oxidation number of -2. In NO, nitrogen has a -3 oxidation number and oxygen has a -2 oxidation number, leading to an overall oxidation number of +1 for the nitrosyl ion.
Oxidation number of Nb is +4. Oxidation number of O is -2.