Question

$Na _{2} S + Na _{2}[ Fe ( CNNO ]$ $\rightarrow$ $Na _{4}[ Fe ( CNNOS ]$
Find sum of oxidation number of $Fe$ in reactant (complex) and product (complex) are.

Answer 1

To find oxidation state of iron in reactant :
$2\times 1+x+5\times \left(\right.-1\left.\right)+1=0x=+2$ (NO has uni positive charge)
To find oxidation state of iron in product:
$4\times 1+x+5\times \left(\right.-1\left.\right)+\left(\right.-1\left.\right)=0x=+2$ (NOS has uni negative charge)
Sum of oxidation state of iron is $4$