Question

A light green coloured salt soluble in water gives black precipitate on passing ${{\text{H}}_{\text{2}}}{\text{S\;}}$ which dissolves readily in ${\text{HCl}}$. The metal ion present is:
A.$C{o^{2 + }}$
B.$F{e^{2 + }}$
C.$N{i^{2 + }}$
D.$A{g^ + }$

Answer 1

To answer this question, you should recall the concept of salt analysis. We know that, in qualitative analysis when ${{\text{H}}_{\text{2}}}{\text{S}}$ is passed through an aqueous solution of salt acidified with ${\text{HCl}}$, a black precipitate is obtained.

Complete Step by step solution:
We know that ${{\text{H}}_{\text{2}}}{\text{S}}$ is identified by its odour and its precipitation of coloured sulphides of various metal ions. Sulphides or hydrogen sulphide also are oxidized to elemental sulphur giving black precipitate and sulphate by oxidizing agents such as permanganate, nitric acid, sulfuric acid, ferrous, etc. The reaction can be represented as:
$3{H_2}S\left( {aq} \right){\text{ }} + {\text{ }}2{H^ + }\left( {aq} \right){\text{ }} + {\text{ }}2N{O_3}^ - \left( {aq} \right){\text{ }} \to {\text{ }}2NO\left( g \right){\text{ }} + {\text{ }}4{H_2}O{\text{ }} + {\text{ }}3S\left( s \right)$
On passing ${H_2}S$ the salt of metal ion converts to its sulphide. According to the question, the sulphide formed should be soluble in ${\text{HCl}}$.
Taking a look at the options, only $FeS\;$is soluble in ${\text{HCl}}$ and $F{e^{2 + }}$ salts are green.
Therefore, we can conclude that the correct answer to this question is option B.

Additional Information:
We know that majority of all the coloured compounds have transition element(s) ions in them. The colour of transition metal ions is due to the presence of unpaired electron in it. These electrons absorb radiations of one colour and undergo the transition from one level to another within the d-subshell. Due to this electronic transition, coloured light is emitted which is the complementary colour of the light absorbed. Down you'll find the list of absorbed colour & the complementary colour which is given out.

Note: We should know the colour of important cations and also remember the reactions of important anions: $C{O_3}^{2 - }{\text{ }},{\text{ }}C{H_3}CO{O^-},{\text{ }}{C_2}{O_4}^{2 - },P{O_4}^{3 - },{\text{ }}S{O_4}^{2 - }$as they are most commonly asked in competitive examinations.
And when we have to determine the colour of salt, we should know the complementary colour scheme:
1.Violet Yellow
2.Blue Orange
3.Green Red
4.Yellow Violet
5.Orange Blue
6.Red Green
Other important coloured salts are:

Colour of salt	Cation Present
Deep Green or purple	${\text{C}}{{\text{r}}^{{\text{3 + }}}}$
Whitish pink	${\text{M}}{{\text{n}}^{2 +}}$
Deep red	${\text{C}}{{\text{o}}^{2 + }}$
Green	${\text{F}}{{\text{e}}^{{\text{3}} + }}$
Brown or yellow	${\text{F}}{{\text{e}}^{2 +}}$
Dark blue	${\text{C}}{{\text{o}}^{2 + }}$
Green	${\text{N}}{{\text{i}}^{2 + }}$
Proper blue	${\text{C}}{{\text{u}}^{2 + }}$
Green or blue	${\text{C}}{{\text{u}}^{2 + }}$