MCQ
The $EAN$ of $Fe$ in $K_4[Fe(CN)_6]$ is
- A$35$
- B$37$
- ✓$38$
- D$36$
Iron pentacarbonyl has $EAN$ number of $36= Z - X + Y =(26-0+2 x )$
[ $Z =$ atomic number, $X =$ oxidation state of metal, $Y =$ total electrons donated by ligand]
$\therefore x =5$
So, the formula will be $Fe ( CO )_5$.
Hence it exists in $Fe ( CO )_5$ form.
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Reason : $[Fe(CN)_6]^{3-}$ has $+3$ oxidation state while $[Fe(CN)_6]^{4-}$ has $+2$ oxidation state.