Why is the seperation of lanthanoids difficult ?

Explanation

The reason are: (i) `I_(2)` is the least reactive of all the halogens because C-I bond formed is much weaker than C-Cl and C-Br bonds.
(ii) The reaction is reversible because HI produced during the reaction, being a reducing agent reduces
`C_(6)H_(6)+I_(2)hArr C_(6)H_(5)I+HI`
Iodobenzene thus formed back to benzene. therefore, to puch the reaction in the forward direction, iodination is usually carried out in presence of an oxidising agent such as nitric acid or iodic acid `(HIO_(3))` which oxidises HI to `I_(2)`. Iodination can also be carried out in presence of mercuric oxide which eliminates HI as insoluble mercuric iodide.
`2HI+2HNO_(3)to2NO_(2)+2H_(2)O+I_(2),5HI+HIO_(3)to3I_(2)+3H_(2)O`
`HgO+2HItoHgI_(2)+H_(2)O`