Use the `EAN` rule to predict the molecular formula for the simple carbonyls of (a) `Cr(Z = 24)`
(b) `Fe (Z =26)` and (c ) `Ni (Z =28)` (Assume that the oxidation state of the metals in the these carbonyls is zero) .

Explanation

Let molecular formulae for the simple carbony1 of `Cr` Fe and `Ni` are `Cr(CO)_(n) Ni(CO)_(n)` and `Ne(CO)_(n)` respectively
EAN of these metals in `Cr(CO)_(n)Ni(CO)_(n)` nad `Ne(CO)_(n)` respectively
`EAN` of these metals in `Cr(CO)_(n)Fe(CO)_(n)` and `Ni(CO)_(n)` respectively is equal to the atomic number of `Kr(Z =36)` next nearest noble gas)
In `Cr(CO)_(n) Z` of `Cr =24`
Oxidation state of `Cr =0` (given) No of electrons donated by n `CO` groups `=2n`
`EAN` of Cr in `Cr(CO)_(n) =24 - 0 + 2n = 36`
`2n = 36 -24`
`n = (36 - 24)/(2) = (12)/(2) =6`
Hence molecular formula of `Cr(CO)_(n) = Cr(CO)_(6)`
(b) In `Fe(CO)_(n)Z` of `Fe = 26`
Oxidation state of `Fe = 0`
No of electrons donated by n COgroups `=2n`
`EAN` of Fe in `Fe(CO)_(n) = 26 - 0 + 2n = 36`
`n = (36 -26)/(2) = 5`
Hence molecular formula of Fe `(CO)_(n) = Fe(CO)_(5)`
(c) In `Ni(CO)_(n) Z ` of `Ni = 28`
Oxidation state of `Ni = 0`
No of electrons donated by n `CO` groups `=2n`
`EAN` of Ni in `Ni (CO)_(n) = 28 - 0 + 2n = 36`
` n = (36 - 28)/(2) = 4`
Hence molecular formula of `Ni(CO)_(n)` =` Ni (CO)_(4)` .