What is the oxidation number of iodine in SO4 2-?

Answer 1

The oxidation number of a free element is zero. However, the oxidation number of elements (e.g. iodine) in compounds will not be zero. The actual oxidation number/state can be deduced if the chemical formula of the compound is given.

Answer 2

iodine has an oxidation number of zero in the elemental form (I2). In iodide salts containing I- it has an oxidation number of -1. There are a range of compounds with iodine in other oxidation states, for example the compounds with fluorine IF, IF3, IF5, IF7 where iodine has oxidation numbers of +1, +3, +5, +7

Answer 3

Iodine oxides are chemical compounds of oxygen and iodine. There exist multiple oxides as iodine has more than one oxidation state.

+2 in IO,

+4 in IO2,

+3 and +5 in I2O4,

+5 in I2O5,

+3 and +5 in I4O9

Answer 4

The oxidation number of iodine in SO4 2- is +7. In sulfate ion (SO4 2-), the overall charge is -2. Oxygen has an oxidation number of -2, so sulfur must have an oxidation number of +6 to balance the charge. Since there are four oxygen atoms with a total oxidation number of -8, the oxidation number of iodine must be +7 to balance the overall charge of -2.

Answer 5

0- as it is in elemental state

Anything in elemental state = 0 (oxidation number)

Hope that helped. Good Luck!

Answer 6

0 for the elemental form

It forms both cations and anions, depending on what it is bonding with. The most likely is the -1 anion.

Answer 7

0 oxidation state

Answer 8

There is no iodine in the sulfate ion!