"The other halogens are not as electronegative and so other hydrogen halides cannot form hydrogen bonds between molecules. Only London Forces are formed. - Therefore more energy is required to break the intermolecular forces in HF than the other hydrogen halides and so it has a higher boiling point."
not a hack link, just stating where i got your answer from! -
https://www.mytutor.co.uk/answers/17558/A-Level/Chemistry/Explain-the-unusually-high-boiling-point-of-HF/
Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂<span> (s) is x. For whole compound, the charge is zero.
Sum of oxidation numbers in all elements = Charge of the compound.
Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
x + 4*0 + 2*(-1) = 0
x + 0 - 2 = 0
x = +2
Hence the oxidation state of Fe in product </span>Fe(CO)₄I₂ (s) is +2.
Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
y = 0
Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.
