What would be the Van’t Hoff factor for a molecule such as I2?

It's not trivial, but it SHOULD just be pretty much #1#... In actuality, it is typically around #1.02#, but you can use #1#.

Iodine undergoes the following hydrolysis in water at #25^@ "C"# in #"0.2 M NaClO"_4# and a #"pH" = 5.15 - 7.62# phosphate buffer:

#3("I"_2(aq) + "H"_2"O"(l) rightleftharpoons "HOI"(aq) + "I"^(-)(aq) + "H"^(+)(aq))#
#ul(3"HOI"(aq) rightleftharpoons "IO"_3^(-)(aq) + 2"I"^(-)(aq) + 3"H"^(+)(aq))#
#3"I"_2(aq) + 3"H"_2"O"(l) rightleftharpoons "IO"_3^(-)(aq) + 5"I"^(-)(aq) + 6"H"^(+)(aq)#

The reaction has #log K = -47.61#, or #K = 10^(-47.61) = 2.45 xx 10^(-48)#. Suppose that #["I"_2] = 1.8 xx 10^(-4) "M"#. Then we solve the equilibrium:

#2.45 xx 10^(-48) = (["IO"_3^(-)]["I"^(-)]^5["H"^(+)]^6)/(["I"_2]^3)#

#= (x(5x)^5(6x)^6)/(1.8 xx 10^(-4) "M" - 3x)^3#

#= (1.458 xx 10^18 cdot x^12)/(1.8 xx 10^(-4) "M" - 3x)^3#

Since #K# is obviously very small, #3x# #"<<"# #1.8 xx 10^(-4)# so that:

#2.45 xx 10^(-48) = (1.458 xx 10^18 cdot x^12)/(1.8 xx 10^(-4) "M")^3#

Therefore,

#x = (((2.45 xx 10^(-48))(1.8 xx 10^(-4) "M")^3)/(1.458 xx 10^18))^(1//12) = 3.83 xx 10^(-7) "M"#

That means (ignoring the autoionization of water):

#["IO"_3^(-)] = 3.83 xx 10^(-7) "M"#
#["I"^(-)] = 1.91 xx 10^(-6) "M"#
#["H"^(+)] = 2.30 xx 10^(-6) "M"#
#["I"_2] = 1.79 xx 10^(-4) "M"#

The percent ionization of #"I"_2# under these assumptions is

#(3cdot3.83 xx 10^(-7) "M")/(1.80 xx 10^(-4) "M") xx 100% = 0.639%#

Starting with #3# iodine particles, the maximum number of particles obtainable at equilibrium is #12#.

If the percent dissociation was #100%#, then the van't Hoff factor would have been #4.000# and the concentration at equilibrium would have been #7.20 xx 10^(-4) "M"# in total.

In actuality, the total concentration of all species in solution for this equilibrium is

#["IO"_3^(-)] + ["I"^(-)] + ["H"^(+)] + ["I"_2] = 1.84 xx 10^(-4) "M"#,

so the van't Hoff factor is about #color(blue)(1.02)#:

#i = (1.84 xx 10^(-4) "M IO"_3^(-) + "I"^(-) + "H"^(+) + "I"_2)/(1.80 xx 10^(-4) "M I"_2) ~~ 1.02#