Discussion
In Frost diagram, NE° values for four different oxidation states of Thorium is found to be Th(II) \(\gt\) Th(I) \(\gt\) Th(III) \(\gt\) Th(IV), then :
Answer: C
The correct conclusion is that Th(II) can disproportionate to Th(I) and Th(III).
In a Frost diagram, nE° (proportional to the free energy of a species relative to the element) is plotted against oxidation state. The species with the lowest nE° is the most stable, and a species whose point lies above the line joining its two neighbours is unstable to disproportionation into them.
Th(II) has the highest nE° of all four, higher than both its neighbours Th(I) and Th(III), so it lies above the line joining them and will disproportionate. Th(II) is therefore the least stable state, not the most stable; Th(IV), with the lowest value, is the most stable, in line with thorium's real chemistry. Option B would need Th(III) to lie above the line joining Th(I) and Th(IV), which the given order does not establish, and the order alone does not fix the comproportionation in option D either.
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