Oxidation Numbers

The oxidation number of N in N2 is 0. Each nitrogen atom shares equally in the bonding electrons, leading to a neutral charge on each nitrogen atom.

In the compound BrO3-, bromine is more electronegative than oxygen and has an oxidation number of +5 to satisfy the overall charge of the anion which is -1. Each oxygen atom has an oxidation number of -2, so the total oxidation numbers in BrO3- sum up to -1.

The oxidation number of Na in NaNO3 is +1, since Na typically has a +1 oxidation state in compounds. The oxidation number of N in NO3 is +5, since oxygen is usually assigned a -2 oxidation state and there are three oxygen atoms bonded to nitrogen in NO3.

The oxidation number of V in V2O7-4 is +5, since the overall charge of the ion 4- is balanced by two vanadium atoms with an oxidation state of +5 each. The oxidation number of O is -2, as it typically has a -2 charge in compounds.

The oxidation number of ZnS (zinc sulfide) is 0. Zinc (Zn) typically has an oxidation number of +2, while sulfur (S) typically has an oxidation number of -2. In a compound like ZnS where there is no charge indicated for the compound, the total oxidation number must add up to 0.

The oxidation number of hydrogen in a compound other than a hydride is typically +1. This is because hydrogen usually forms covalent bonds where it shares an electron with another nonmetal, resulting in a slight positive charge on the hydrogen atom.

One example of a compound with an oxidation number of +5 is phosphorus pentachloride (PCl5). In this compound, phosphorus has an oxidation number of +5 due to the electronegativity difference with chlorine.

The oxidation number of an atom is a measure of the amount of electrons it has gained, lost, or shared in a chemical reaction. It helps in determining the extent of an element's electron loss or gain, which provides insight into its reactivity and bonding patterns in chemical compounds. The oxidation number is represented as a positive or negative integer.

In CrBr (chromium monobromide) Cr would have an oxidation number of +1. This compound is not known perhaps you meant CrBr3, where chromium has an oxidation number of +3

The oxidation number of sulfur in Al2S3 is -2. This is because aluminum has an oxidation number of +3, and the overall compound is neutral, so the total oxidation number contribution from sulfur must be -6 to balance the charge.

In FeSO4, the oxidation number of Fe is +2, and the oxidation number of oxygen is -2. To find the oxidation number of S, we can set up an equation: 2(-2) + x + 4(-2) = 0 (overall charge of the compound is zero). Solving this equation gives the oxidation number of S as +6.

Reduction is addition of electrons.

Iron is [Ar]4s23d6 and can lose electrons (or be oxidized) very easily.

Reduction or addition of electrons does not make it stable. The next inert gas is Kr with 36 electrons. It would thus have to gain 10 electrons (be reduced) to get to this structure. This is energetically very unfavorable. All metals as a general rule prefer to lose electrons than gain them.

The oxidation number of CO3 in NiCO3 is -2. In this compound, each oxygen atom has an oxidation number of -2, and since there are three oxygen atoms in CO3, the overall charge for the carbonate ion is -2. Therefore, the oxidation number of CO3 is -2 in NiCO3.

under one method - P has oxidation number of -3, this is where H is assigned +1 except in metal hydrides. Using the electronegativity method and using Pauling EN H is slightly more electronegative than P so on that method P would be +3. It all depends where you are taught.

Ru has oxidation number +8, O has oxidation number of -2

The oxidation number of each phosphorus atom in P2O5 is +5. This is because oxygen is assigned an oxidation number of -2 in most compounds, and the overall charge of P2O5 is 0. Since there are two phosphorus atoms, each must have an oxidation number of +5 to balance the -10 oxidation contributed by the five oxygen atoms.

I assume you mean the oxidation number of Mn in the permanganate ion , MnO4-

The sum of the oxidation numbers is the charge on a polyatomic ion so Mn has an oxidation number of +7 as each O is assigned -2.

Oxidation numbers can help predict the way an element will bond by indicating the charge an element is likely to obtain when forming a compound. Elements tend to bond in a way that results in achieving a more stable oxidation state, such as by gaining or losing electrons to achieve a full valence shell. By knowing the possible oxidation states of an element, one can anticipate how it will bond with other elements to achieve a balanced charge in a compound.

In diamond, the oxidation number of a carbon atom is 0 because it is bonded to four other carbon atoms, resulting in a balanced charge. In graphite, the oxidation number of a carbon atom is also 0 for the same reason.

Graphite is pure carbon.So it has 0 oxidation state.

The oxidation number of oxygen in OF2 is +2. In this compound, each fluorine atom has an oxidation number of -1, so the total oxidation numbers must add up to zero. Since there are two fluorine atoms at -1 each, the oxygen atom must have an oxidation number of +2 to balance out the charges.

The oxidation number of phosphorus in phosphorus pentoxide (P4O10) is +5. This is because each oxygen atom has an oxidation number of -2, leading to a total charge of -10 for the oxygen atoms in the compound. The sum of the oxidation numbers in a neutral compound like P4O10 is zero, so the oxidation number of phosphorus must be +5 to balance the overall charge.

In HIO4 (periodic acid), the oxidation number of hydrogen (H) is +1, of iodine (I) is +7, and of oxygen (O) is -2. Therefore, the overall oxidation number of HIO4 is zero since the sum of the oxidation numbers in a neutral molecule must equal zero.

The oxidation number of chromium in potassium dichromate is +6. Each oxygen has an oxidation number of -2, and since the compound is neutral, the oxidation number of potassium is +1. This means the two chromium atoms in potassium dichromate each have an oxidation number of +6 to balance the charges.

In oxidation, an element loses electrons, typically becoming more positively charged. This process often involves oxygen combining with other elements to form oxides.