Question

$F{e^{2 + }}$ ion can be distinguished by $F{e^{3 + }}$ ion by
(A) $N{H_4}SCN$
(B) $AgN{O_3}$
(C) $BaC{l_2}$
(D) None of these

Answer 1

Hint: In the given question the chemical molecule that needs to be differentiated should produce a different colour when combined with reagent. To do so we take salts of $F{e^{2 + }}$ and $F{e^{3 + }}$ and make them react with different reagents and note down the colour of the product so obtained.

Complete Step by Step Solution:
First of all it is important to be clear that $F{e^{2 + }}$ is known as ferrous ion and $F{e^{3 + }}$ is known as ferric ion.

We know that, $N{H_4}SCN$ (known as ammonium thiocyanate) dissociates into $N{H^{4 + \,}}$ and $SC{N^ - }$ .

When $F{e^{3 + }}$ is made to react with the reagent $N{H_4}SCN$ we get $Fe{\left( {SCN} \right)_3}$ as one of the product. It is deep red in colour and called iron thiocyanate.
Similarly, when $F{e^{2 + }}$ reacts with $N{H_4}SCN$ , we see that there is no change in the colouration of the product.

Therefore, we notice that there is a difference in colour of the products obtained when ferrous and ferric ions react with the reagent individually. This is how we distinguish between ferrous and ferric ions.

Although there is not just one reagent that could show such a result. $NaOH$ (an alkali) and $N{H_4}OH$ also show similar results and can be used in place of $N{H_4}SCN$ .
For example, if we take salts of ferrous ion, say, $FeS{O_4}$ and make it react with the reagent $N{H_4}OH$ then $Fe{\left( {OH} \right)_2}$ is obtained as one of the products. It is green in colour and also insoluble in the solvent.

Similarly, if we take a ferric salt, say, $FeC{l_3}$ and make it react with the reagent $N{H_4}OH$ then $Fe{\left( {OH} \right)_3}$ (ferric hydroxide) is obtained as one of the two products formed. It is reddish brown in colour.

So we can say that if green precipitate is formed then it has ferrous ions and if red precipitate is formed then it has ferric ions. Thus, we see that again the two ions give different colour products upon reaction with the reagent and hence get distinguished.
Hence, option A. is the answer.

Note: One must note that there can be more than two reagents that help in distinguishing any two elements but it is not necessary that they show the same colouration for the same ion when reacted with different reagents. There are many tests for distinguishing ions whose products and colouration might be different but the ultimate result is obtained.