If an iron (III) complex with the formula has no electron in its orbital, then the value of x + y is

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Solution

Q.
If an iron (III) complex with the formula ${[Fe {({NH}_{3})}_{x} {(CN)}_{y}]}^{-}$ has no electron in its $e_{g}$ orbital, then the value of x + y is [2024]

1 4

2 3

3 5

4 6

Ans.
(4)

Iron in +3 oxidation state forms complexes with C.N. = 6.

So $x + y = 6 .$

${[{Fe(NH}_{3})_{x} {(CN)}_{y}]}^{-}$

Ammonia is a neutral ligand and cyanide is a unit negatively charged ligand.

Since sum of oxidation number = charge

$O.N of Fe + x \times 0 + y \times (- 1) = - 1$

$3 + (- y) = - 1$

$y = 4$

Thus $x = 2$

The complex is ${[{Fe(NH}_{3})_{2} {(CN)}_{4}]}^{-}$

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