Na 2 S + Na 2[ Fe ( CNNO ] arrow Na 4[ Fe ( CNNOS ] Find sum of oxidation number of Fe in reactant (complex) and product (complex) are.

Question

Na 2 S + Na 2[ Fe ( CNNO ] arrow Na 4[ Fe ( CNNOS ] Find sum of oxidation number of Fe in reactant (complex) and product (complex) are. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

To find oxidation state of iron in reactant :

2 \times 1 + x + 5 \times (- 1) + 1 = 0 x = + 2

(NO has uni positive charge)
To find oxidation state of iron in product:

4 \times 1 + x + 5 \times (- 1) + (- 1) = 0 x = + 2

(NOS has uni negative charge)
Sum of oxidation state of iron is

4

Q. Na2​S+Na2​[Fe(CNNO] → Na4​[Fe(CNNOS] Find sum of oxidation number of Fe in reactant (complex) and product (complex) are.

To find oxidation state of iron in reactant : 2×1+x+5×(−1)+1=0x=+2 (NO has uni positive charge) To find oxidation state of iron in product: 4×1+x+5×(−1)+(−1)=0x=+2 (NOS has uni negative charge) Sum of oxidation state of iron is 4

Q. $N a_{2} S + N a_{2} [F e (CNNO]$ $\to$ $N a_{4} [F e (CNNOS]$
Find sum of oxidation number of $F e$ in reactant (complex) and product (complex) are.

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