Oxidation Numbers

No as it doesn't as it doesn't form compounds.

The only possible oxidation number is zero for elemental helium

The oxidation number of carbon in hydrogen carbonate (HCO3-) is +4. This can be determined by considering the oxidation numbers of the other elements in the compound (hydrogen and oxygen) and applying the rule that the sum of oxidation numbers in a compound must equal zero.

The oxidation number of the sulfur atom in H2S is -2. Hydrogen is assigned an oxidation number of +1, and since the overall molecule has no charge (neutral), the oxidation number of sulfur must be -2 to balance the charges.

Roman Numeral

0 in the elemental form, +3 in its compounds

Yes. 0 in elemental form and +2 in its compounds.

Fluorine is the most electronegative element, so it tends to attract electrons strongly and form a single covalent bond with one electron from another element. This results in a stable electronic configuration, leading to an oxidation number of -1.

Nonmetals tend to gain electrons to achieve a stable electron configuration and become more like noble gases. This results in a negative oxidation number because they gain electrons rather than lose them.

The oxidation number of chromium in MgCrO4 is +6. Since the overall charge of the compound is 0, the sum of the oxidation numbers must equal 0. Therefore, the oxidation number of oxygen is -2.

The oxidation number of oxygen in H2O (water) is -2. Oxygen typically has an oxidation number of -2 in most compounds. In H2O, there are two hydrogen atoms each with an oxidation number of +1, resulting in a total charge of 0 as oxygen contributes -2.

O is 2- and there are two of them so Pb would have to be 4+

It is equal to zero. In pure elemental form.

The oxidation number of CF4 is +4 for carbon and -1 for each fluorine atom. This results in a total of zero, reflecting the neutral charge of the compound.

The oxidation number of sulfur in H2SO4 is +6. This is because hydrogen is typically assigned an oxidation number of +1 and oxygen -2, leading to the equation 2(1) + x + 4(-2) = 0. Solving for x gives x = +6 for sulfur.

The oxidation number for radon is typically 0, since it is a noble gas and does not readily form compounds where it gains or loses electrons.

The oxidation number for chromium can vary depending on the compound it is part of. In general, chromium can have oxidation states ranging from -2 to +6.

It is equal to the charge.It is minus one.

For Mg the oxidation no is +2 while it is -2 for oxygen.

Oxidation number is a concept that refers to atoms not molecules. The oxidation number of S in S8 is zero. The oxidation number of an atom in the pure element is always 0.

S8 is a form of sulfur, the most commonly encountered , there are many other forms (allotropes). The oxidation number of S in any of its allotrpes is zero.

The oxidation number of H+ is +1. This is because hydrogen typically has an oxidation number of +1 when it forms an ion by losing its single electron.

This is an EXTREMELY complex area, but there is a fairly basic rule, but there are MANY rule breakers. Let's look at it this way, an ion forms so that an element can achieve 8 electrons, they want to achieve this with the least amount of energy possible. (Keep in mind we are only working with s and p sublevel electrons, or the valence electrons, not the d's and f's). First you need to undestand this (this is a rough chart, as I am ignoring all of the rule breakers):

Group 1: 1 valence electron

Groups 2-12: 2 valence electrons

Group 13: 3 valence electrons

Group 14: 4 valence electrons

Group 15: 5 valence electrons

Group 16: 6

Group 17: 7

Group 18: 8 (noble gases)

Everything left of the metalloid line will LOSE valence electrons to achieve the state of the noble gas on the period before it, for example:

Calcium (Ca #20) has 2 valence electrons, so it will lose those two to achieve the electron configuration (E.C.) of Argon (Ar #18), and since it has two more positively charged protons than it does negatively charged electrons, it forms the ion Ca2+.

Everything to the right of the metalloid line will GAIN valence electrons to achieve the E.C. of the noble gas on its period, for example:

Selenium (Se #34) has 6 valence electrons, so instead of losing 6 electrons to achieve the state of Argon, like what Calcium did, it will gain two electrons to achieve the state of Krypton (Kr #36), and since its negatively charged electrons outnumber its positively charged protons it gets the oxidation number 2-, making it Se2-.

Just remember that, metals (to the left of the zigzag/metalloid line) lose electrons to form positive ions, and nonmetals (to the right of the zigzag/metalloid line) gain electrons to form negative ions. Hope this helped, and I hope you do not mind that I disregarded the rule breakers.

Molecules with transition metals like manganese, iron, or copper can have both positive and negative oxidation numbers. The oxidation number depends on the specific chemical environment and the oxidation state of the metal ion.

0 in the elemental form. +2 and +4 in its compounds

In the compound Co2O3, cobalt's oxidation number is +3. Each oxygen atom has an oxidation number of -2, and since the compound is neutral, the sum of the oxidation numbers must equal zero, leading to cobalt having an oxidation number of +3.

The oxidation number of the polyatomic ion sulfate (SO4) is -2. Each oxygen atom contributes -2, and since sulfate has a charge of -2, the single sulfur atom must have an oxidation number of +6 to balance the charges.