Solution :
Molar mass of 
is :
M = 6×12 + 6×1 g
M = 78 g
78 gram of contains 
molecules.
So, 89.5 gram of contains :
Now, from the formula we can see that one molecule of contains 2 hydrogen atom . So, number of hydrogen atom are :
Hence, this is the required solution.
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.
It is in period 3
It is in group 17
It is a chlorine atom because it has 17 electrons which means the atomic number is 17
