When arranging the species I₂, HI, HIO₄, and ICl by the oxidation number of iodine (I) in increasing order, we need to identify the oxidation number of iodine in each compound or molecule.

• In molecular iodine, I₂, iodine is in its elemental form, so its oxidation number is 0.
• In hydrogen iodide, HI, hydrogen usually has an oxidation number of +1 (except in metal hydrides, which is not the case here), and since the compound is neutral, iodine must have an oxidation number of -1 to balance the +1 from hydrogen.
• In iodine monochloride, ICl, chlorine usually has an oxidation state of -1 (except in compounds with oxygen or fluorine). Given that the compound is neutral, iodine must have an oxidation number of +1 to balance the -1 from chlorine.
• In periodic acid, HIO₄, oxygen has a typical oxidation state of -2 (except in peroxides, superoxides, or when bonded to fluorine). Considering there are 4 oxygen atoms, this gives a total oxidation state of -8 from oxygen. Hydrogen, as usual, is +1. To make the compound neutral, the oxidation state of iodine must be such that when added to the +1 from hydrogen and -8 from oxygen, the total is 0. Therefore,

  $+1 + (\text{Oxidation state of I}) - 8 = 0$

  Solving for the oxidation state of iodine:

  $ \text{Oxidation state of I} = 7$

So, the oxidation numbers of iodine in the given species are:

• I₂: 0
• HI: -1
• ICl: +1
• HIO₄: +7

Arranging them in increasing order of the oxidation number of iodine, we get:

HI (-1) < I₂ (0) < ICl (+1) < HIO₄ (+7)