Oxidation Numbers

Hydrogen is an element that can have an oxidation state of -1 when it is part of a hydride ion, such as in lithium hydride (LiH).

The oxidation number of PH4 cannot be determined because phosphorus (P) does not have a fixed oxidation number in this compound.

The oxidation number for oxygen in most compounds is -2. Since there are two oxygen atoms in F2O, this gives a total charge of -4. Therefore, the oxidation number for each fluorine atom in F2O is +2 to balance out the overall charge to zero.

The oxidation number of silicon (Si) in SiF4 is +4. This is because each fluorine atom in the compound has an oxidation number of -1, and since SiF4 is a neutral compound, the oxidation numbers must add up to zero.

The oxidation number of O in H2O2 is -1. Each hydrogen atom has an oxidation number of +1, and since the overall molecule has a neutral charge, the oxidation number of oxygen must be -1 for the equation to balance.

The oxidation number of ammonium ion (NH4+) is +1 and the oxidation number of cyanate ion (CNO-) is -1. Therefore, the overall oxidation number of NH4CNO is 0 because the sum of the oxidation numbers must equal zero for a neutral compound.

Assuming you mean NH3 since there is no N3H ....The charge on NH3 is zero; it doesn't have an oxidation number.

The oxidation number of N in NH3 is 3-

The oxidation number of each H in NH3 is 1+

The oxidation number for carbon in CHI3 compound is -2. In CHI3, iodine has an oxidation number of -1 and hydrogen has an oxidation number of +1, which allows carbon to have an oxidation number of -2 to balance the overall charge of the compound.

The oxidation number of Phosphorus (P) in H3P2O7 is +5. This can be determined by assigning hydrogen an oxidation number of +1 and oxygen an oxidation number of -2, then setting up an equation to solve for the unknown oxidation number of Phosphorus.

The sulfate ion is SO42 -. The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion. The sulfite ion is SO32-. The oxidation state of the sulfur is +4.

The oxidation number of aluminium in potash alum is +3. This is because in the compound potassium aluminum sulfate dodecahydrate (KAl(SO4)2·12H2O), aluminum is bonded to two sulfate ions (each with a charge of -2) and one potassium ion (with a charge of +1), resulting in a total oxidation state of +3 for aluminium.

The oxidation number of Na in Na2SO3 is +1, the oxidation number for S in SO3 is +4, and the oxidation number for O in SO3 is -2.

The oxidation number for hydrogen is usually +1, for nitrogen it is +5, and for oxygen it is typically -2. So in HNO, the oxidation numbers would be +1 for hydrogen, +5 for nitrogen, and -2 for oxygen.

Manganese (Mn) in Mn3O4 has an oxidation state of +3. Each oxygen atom has an oxidation state of -2, so the total negative charge from oxygen is -8. To balance the charge, the three manganese atoms must have a total positive charge of +8, making each manganese atom have an oxidation state of +3 in Mn3O4.

The oxidation number of silver (Ag) is +1 and the oxidation number of oxygen (O) is -2. To find the oxidation number of phosphorus (P) in Ag3PO4, we can set up an equation: 3(+1) + x + 4(-2) = 0. Solving for x, we find that the oxidation number of phosphorus in Ag3PO4 is +5.

The oxidation number for Cr in CrPO4 is +3. This is because the overall charge of the compound is neutral, and the oxidation numbers for oxygen (-2) and phosphorus (+5) are known. By calculation, we determine the oxidation number of Cr to be +3.

In K2MnO4, the oxidation state of oxygen is -2, and the overall charge of the compound is -1. Given that potassium has a +1 oxidation state, the oxidation state of manganese (Mn) in this compound is +7.

The oxidation number for oxygen (O) in KO2 is -1. Since the overall charge of the compound is 0, the oxidation number of potassium (K) is +1.

In the compound Al₂O₃, aluminum has an oxidation number of +3, and oxygen has an oxidation number of -2. This gives a total charge of zero for the compound, as it should be electrically neutral.

The oxidation number of Na in NaOCN is +1. In most compounds, the oxidation number of sodium is typically +1 due to its tendency to lose one electron.

The oxidation number of Cr in Cr2O5 is +6. This is because the oxidation number of oxygen is typically -2, and the total charge for the compound must be 0 since Cr2O5 is neutral. By setting up an equation with x representing the oxidation number of chromium, we get 2x + 5(-2) = 0, and solving for x gives +6 for Cr.

The oxidation number of Cs (Cesium) is +1, and the oxidation number of Br (Bromine) is -1. In CsBr, Cesium donates one electron to Bromine, resulting in Cs having an oxidation number of +1 and Br having an oxidation number of -1 in the compound.

KNO3 has no oxidation number.

However, the individual atoms do each have an oxidation number.

Using oxygen as the 'yardstick' at '-2' ; NB There are exceptions this rule.

There are 3 oxygens, so the oxygen component has an oxidation of 3 x -2 = -6.

So for it to be a neutrally charge molecule the potassium and the nitrogen must sum to +6.

Since potassium only loses one electron its oxidation number is '+1'. It follows that nitrogen must be in oxidation state '+5'

Summarising;-

Potassium(K) = +1

Nitrogen(N) = +5

Oxygen(O) = 3 x -2 = -6

+1+5-6 = 0

NB Nitrogen , like sulphur and phosphorus exhibit various oxidation states.

=

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 in oxyanions such as this is usually considered to be -2.