Question

In which of the following compounds iron has lowest oxidation state

• A. $Fe(CO{)}_5$
• B. $F{e}_{2}O$
• C. [$K_{4}Fe(CN{)}_6$]
• D. $FeS{O}_4(N{H}_4{)}_{2}S{O}_4.6H_{2}O$

Answer 1

The correct option is A $Fe(CO{)}_5$

Analysing the options
Option (A)
$FeS{O}_4(N{H}_4{)}_{2}S{O}_4.6H_{2}O$
Let oxidation no. of Fe be x.
And charge on $S{O}_4^{2-}$ is -2.
charge on $N{H}_4^{+}$ is +1.
charge on $H_{2}O$ is 0.
Thus ,
$\Rightarrow x+(-2)+2\times (+1)+(-2)+6\times 0=0$
$\Rightarrow x-4+2=0$
$\therefore x=+2$

Option (B)
CO is a neutral molecule thus charge on CO is 0.
Let Oxidation state of Fe be x.
Thus, oxidation state of Fe in $Fe(CO{)}_5$ will be :
$\Rightarrow x+5\times 0=0$
$\therefore x=0$

Option (C)
$K_{4}Fe(CN{)}_6$
Let Oxidation state of Fe be x
Oxidation state of Fe in $K_{4}Fe(CN{)}_6$ will be:
$\Rightarrow 4\times (+1)+x+6\times (-1)=0$
$\Rightarrow 4+x-6=0$
$\therefore x=+2$

Option (D)
$F{e}_{2}O$
Let Oxidation state of Fe be x
Oxidation state of Fe in $F{e}_{2}O$ will be:
$\Rightarrow 2\times x+(-2)=0$
$\Rightarrow 2x-2=0$
$\therefore x=+1$
So, the correct answer is option (B).