Oxidation Numbers

Elements in group 17 (halogens) have seven valence electrons and tend to gain one electron to achieve a full outer shell of eight electrons, resulting in an oxidation number of -1. This electron gain corresponds to achieving the stable electron configuration of a noble gas.

The order of increasing oxidation number of nitrogen is: -3 < -2 < -1 < 0 < +1 < +2 < +3 < +4 < +5.

In CrO3, the oxidation number of chromium (Cr) is +6 and the oxidation number of oxygen (O) is -2. This is because oxygen typically has an oxidation number of -2, and there are three oxygen atoms in CrO3 to give a total charge of -6. To balance the charge, the chromium atom must have an oxidation number of +6.

0. CO2 is electrically neutral. C is +4 and O is 2(-2)

The oxidation state of the potassium ion (K) in K2O is +1, as it typically loses one electron to achieve a stable electron configuration. The oxidation state of the oxygen ion (O) in K2O is -2, as it tends to gain two electrons to reach a stable electron configuration.

The overall oxidation number is -2 so the average oxidation number of O is -1.

The oxidation number of N in NO2 is +4. Each oxygen atom in NO2 has an oxidation state of -2, so the oxidation state of nitrogen can be calculated as follows: 2*(-2) + x = 0, where x is the oxidation state of N. Solving for x gives us x = +4.

The oxidation number for Na in NaOH is +1. This is because in NaOH, oxygen has an oxidation number of -2, and hydrogen has an oxidation number of +1. Therefore, the oxidation number of Na must be +1 to balance the charges and make the compound neutral.

It contains only two elements. Fe shows +3 number. Chlorine shows -1 as the oxidation number.

Zirconium typically exhibits an oxidation state of +4 in its compounds due to the loss of its four valence electrons. However, zirconium can also display lower oxidation states, such as +3, in certain compounds under specific conditions.

The chlorate ion is ClO3-. Some text books say that the oxidation number of the whole ion is equal to the charge, so in this cas it would be -1. However most teachers would sya that oxidation numbers only refer to atoms. So working out the oxidation numbers:-

oxygen is -2, (using the rule, or if you use the he electronegativity method- gives the same answer as O is more electronegatoive than Cl)

the sum of oxidation numbers of Cl and O are -1 (the charge on the ion) - I'll call the oxidation number of Cl OxCl

-1 =OxCl + (3* -2)= OxCl -6

therefore OxCl = +5

As it forms the I- ion froming an octet its most likely oxidation number is -1

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 chlorine in KCl is -1. This is because potassium (K) has an oxidation number of +1 and the overall compound is neutral, so the oxidation number of chlorine must be -1 to balance the charge.

The oxidation number of an element tells us the possible charge that the element possesses at that instant. So, in the case of Lithium, it generally has a +1 oxidation state, meaning that it exists generally as a positive ion with charge +1. In the case of Beryllium, which is a Group-2 element, it has a tendency of losing two electrons and remaining in a +2 ionic state. Hence, its oxidation number is +2. Like that, the general state of an element is determined taking into consideration their d-orbitals and stability.

For example the atomic number of oxygen is 8.

The oxidation number of Cr in MgCrO4 is +6. This is because oxygen is typically assigned an oxidation number of -2, and the overall charge of the compound is 0, so the oxidation number of magnesium (+2) and oxygen (-2) must be balanced by the oxidation number of Cr (+6).

In Na2SO4, the oxidation state of sodium (Na) is +1, the oxidation state of sulfur (S) is +6, and the oxidation state of oxygen (O) is -2. To calculate the oxidation state of the whole compound, you can use the rule that the sum of the oxidation states in a neutral compound is zero, so in this case it would be +1*2 + (-2)*4 = 0.

If the atom is neutral, the number of protons equal the number of electrons. The number of protons can be found by the mass number of the specific atom.

For Xe the mass number is 54, therefore there are 54 protons, and 54 electrons.

The oxidation number of the entire ion (P_2O_{8}^{2-}) is -2. Each oxygen atom has an oxidation number of -2, so in total, the 8 oxygen atoms contribute a total oxidation number of -16. This means the two phosphorus atoms must have an oxidation number that sums to +14 to balance out the charge of -2 on the ion.

Silver has plus one.Chlorine has minus one.

The oxidation number of hydrogen in KH is -1, because the only oxidation number for potassium in a simple inorganic compound is +1. A hydrogen atom can complete its outer electron shell by adding one electron acquire the electron configuration of helium, and this is what happens during the formation of alkali metal hydrides such as KH.

+1 for Na, +1 for H, -2 for each O, +4 for S

A non stoichiometric compound is one where an exact integer ratio of atoms is not achieved. An example is FeO- all laboratory specimens are "iron deficient" with formals of about Fe0.8 O this is a feature of non -stoichiometric compounds their composition is variable within narrow limits- some of the iron is Fe3+, the crystal lattice has "defects" that accomodate them. (Pure FeO can be made but it is pyrophoric (may spontaneously in flame)