+1 in CuI or Cu2I2. +2 in CuI2.
+2. PbI2 could be named lead(II) iodide.
The chemical formula of uranium iodide is UI4. The valence of uranium is 4+.
Aluminum (III) Iodide the aluminum uses it's +3 oxidation state therefore it is named Aluminum (III) and since it is ionic the Iodine just uses Iodide
Reductant Iodide (I-) can be oxidised to other (more postive) oxidation numders by loosing electrons to the oxidant. Depending on the oxidant's strength this can change to the following numbers: 0 (zero) in I2 , and +1, +2, +3, +4 (in hypoiodite IO-, iodite IO2-, iodate IO3-, periodate IO4- respectively)
+1 in CuI or Cu2I2. +2 in CuI2.
+2. PbI2 could be named lead(II) iodide.
The oxidation state of iodine in the compound in the question is -1, as it is in any compound with a name that properly includes the word "iodide" instead of "iodine". "Iodide" is the name of the anion with formula I-1.
In Iodine, the element is in 0 oxidation state, but in Iodide it is in -1. Iodine = I2 , Iodide = I- Iodine can exist freely, but Iodide cannot.
The chemical formula of uranium iodide is UI4. The valence of uranium is 4+.
Sn4+ is the symbol for Tin(IV), that is, the element tin with a oxidation state of 4.
Cu ----> Cu 2+ + 2e-
By mild oxidation you'll get Iodine, I2
Aluminum (III) Iodide the aluminum uses it's +3 oxidation state therefore it is named Aluminum (III) and since it is ionic the Iodine just uses Iodide
Tin has two oxidation states (II and IV), and exhibits approximately equal stability in both its II and IV oxidation state. The chemical formula Tin (II) Iodide is SnI2. The chemical formula for Tin (IV) Iodide is SnI4.
Reductant Iodide (I-) can be oxidised to other (more postive) oxidation numders by loosing electrons to the oxidant. Depending on the oxidant's strength this can change to the following numbers: 0 (zero) in I2 , and +1, +2, +3, +4 (in hypoiodite IO-, iodite IO2-, iodate IO3-, periodate IO4- respectively)
Because it contains Iodine in its lower oxidation state of -1