Question

$FeC{{l}_{3}}$ solution gives blood red colour with $N{{H}_{4}}SCN$. If ${{H}_{2}}S$ gas is passed and filtered, then filtrate does not give red colour with $N{{H}_{4}}SCN$.
If true enter $1$, else enter $0$.

Answer 1

The hydrogen sulphide is a reducing agent as it helps in reduction of the atoms of other molecules by gaining electrons from other substances.
The metal complexes impart colour to the solution because of the presence of vacant orbitals in the central metal ions.

Complete step by step answer:
In order to understand the colour of the product, we need to understand the reaction which is taking place in the following reaction.
When $FeC{{l}_{3}}$ reacts with $N{{H}_{4}}SCN$, it forms a complex, where the central metal ion is iron, and two of the chlorine atoms are outside the coordination sphere while, iron, sulphur and cyanide are inside the coordination sphere. So, the reaction could be chemically expressed as,
$FeC{{l}_{3}}+N{{H}_{4}}SCN\to [Fe(SCN)]C{{l}_{2}}+N{{H}_{4}}Cl$
As we can see that the ammonium chloride is also formed as a side product of the reaction. This complex which is formed because of the ferric chloride has a red colour, which turns the colour of the solution red. The two chlorides which are present outside the coordination sphere, or the square brackets, are ionisable, meaning, if this complex is exposed to aqueous solution, the two chlorines will dissociate in water as chloride ions.
Now, in the ferric chloride solution if we pass hydrogen sulphide having the chemical formula ${{H}_{2}}S$, and the filter the solution, the ferric chloride which has the formula $FeC{{l}_{2}}$, will convert into ferrous chloride which does not form complex with $N{{H}_{4}}SCN$ unlike ferrous chloride. As a result of which the solution will not have any colour change after addition of $N{{H}_{4}}SCN$.

So the given statement would be true.

Note: The colour of the complex $[Fe(SCN)]C{{l}_{2}}$ is because of the presence of vacant orbitals in the central metal ion which is iron.
At first the electrons are excited in a higher state, and then while coming back to the ground state, it emits radiations which falls on the visible region, hence we can see the red colour of the complex $[Fe(SCN)]C{{l}_{2}}$.