What is the oxidation number of iodine in #I_2#?

1 Answer
Nov 15, 2015

Answer:

The oxidation number of #I# in the iodine molecule is #0#.

Explanation:

By definition, oxidation number is the charge left on the central atom when all its bonds are (conceptually) cleaved with the charge going to the most electronegative atom. So let's cleave the #I-I# bond:

#I-I rarr 2I*# . We get two iodine atoms, two neutral iodine radicals, #2xxI*#. Since there is no charge, the oxidation number is a big fat #0#. On the other hand, when we do the same operation with the interhalogen #I-Cl#, the charge goes to the more electronegative atom, #Cl#:

#I-Cl rarr I^+ + Cl^-#. To give formal oxidation states of #I^+# of iodine in the molecule, and #I^-# for chlorine (note these are Roman numerals). The same oxidation states would pertain for #Br-Cl#, as chlorine is more electronegative than bromine. What are the oxidation numbers of fluorine and oxygen in #OF_2#? Please report back the answers here.