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.
elements have equal number of protons and neutrons
the condensed format is when the closest noble gas with the closest electron configuration is given, this closest noble gas atomic number should be lesser than the atoms atomic number
atomic number of Kr is 36
1 electron in 5s subshell and 7 electrons in 4d subshell.
there's a total of 36 + 1 + 7 = 44 electrons
atomic number of the atom is 44
element with atomic number 44 is Ruthenium - Ru
answer is Ru
C. A central body is the center of the universe.
