Question

$N{{a}_{2}}S+N{{a}_{2}}[Fe{{(CN)}_{5}}NO]\to X$(purple colour)
What is the formula of X?
A. $F{{e}_{4}}[Fe{{(CN)}_{6}}]$
B. $[Fe{{({{H}_{2}}O)}_{5}}NO]S{{O}_{4}}$
C. $[Fe{{({{H}_{2}}O)}_{2}}NO]S{{O}_{4}}$
D. $N{{a}_{4}}[Fe{{(CN)}_{5}}NOS]$

Answer 1

Hint: To find out the product of a given reactant, it is important to know the reaction mechanisms like addition, substitution etc. Since it is a direct inorganic reaction, it is advisable to remember the product or at least to remember the fact that it forms a thio-nitrosyl compound. The colour changes from red to purple as the reaction proceeds.

Step by step solution: Sodium nitroprusside reacts with the sulphide ion from the sodium sulphide present in the reactant and immediately gives a purple coloured thio nitroprusside anion.
 In the reaction the oxidation state of Fe changes from +2 to +4.
 The product formed is Sodium pentacyanonitrosylferrate(II) which is a basis for ${{S}^{2-}}$ test.
$N{{a}_{2}}[Fe{{(CN)}_{5}}NO]+{{S}^{2-}}\to {{[Fe{{(CN)}_{5}}NOS]}^{4-}}$
For the reactant, the charge of one Na ion =+1
Charge on ${{(CN)}^{-}}$=-1
Charge on ${{(NO)}^{+}}$=+1
$\therefore$2$\times$(+1) +5$\times$(-1) +1 +x=0
Or, x=+2
Therefore, charge on Fe in sodium nitroprusside is 2+
Similarly, for the product, the charge on Fe ion will be,
+2+x-5+1-2=0
Therefore x=+4
Therefore, the overall charge will be 4- on the product complex. Hence, option [D] is the correct answer.
Additional Information: The starting compound is an organic nitrosyl compound which generally exists in dihydrate form. It is a red colour sodium salt and readily dissolves in water and ethanol. Its chemical reactions are mainly associated with the NO ligand. Its ability to form the purple coloured product in presence of sulphide is of great organic use to detect the presence of sulphide ions in a solution.

Note: It is important to remember here that it is an additional reaction and the sulphide ion joins inside the co-ordination sphere forming a thi-nitrosyl anion.
In the option [B] and [C] there are water molecules present in the product which is clearly not present in the reactant. So It’s obvious that neither of them is the correct answer.