Question

To a metal nitrate, when KI solution is added, a black precipitate is produced at first; on adding excess of KI, orange solution is produced. Identify the metal ion.

Answer 1

Consider an arbitrary element, say M ;Write down the possible reaction that can occur when the reagents are added to identify the ion. Color of the compound depends on the metal ion as well as the anion.

Complete answer:
Let us assume an arbitrary element M and proceed with the reactions that follow.
$MN{{O}_{3}}+KI\to MI+KN{{O}_{3}}$
The black precipitate is $MI$ as $KN{{O}_{3}}$ is colorless and soluble.
The Metal iodide is then subjected to excess of $KI$, changing the color to orange. Since there is no displacement reaction happening, the only possibility is the formation of a coordination compound.
The color of

-Hg{{I}_{2}}is\text{ }red \\\ -Bi{{I}_{3}}is\text{ }black \\\ -Cu{{I}_{2}}is\text{ }brown \\\ -Pb{{I}_{2}}is~bright\text{ }yellow \\\ \end{array}$$ From the above analysis, there is only one metal ion which will respond to the above reaction and that ion is $$B{{i}^{3+}}$$. We will now rewrite the reactions with the metal ion as $$B{{i}^{3+}}$$, $Bi{{(N{{O}_{3}})}_{3}}+3KI{{\to }_{{}}}Bi{{I}_{3}}+3KN{{O}_{3}}$ When excess KI is added to $Bi{{I}_{3}}$, it leads to the formation of a coordination compound. $Bi{{I}_{3}}+K{{I}_{(excess)}}{{\to }_{{}}}{{K}_{2}}[Bi{{I}_{5}}]$ **Therefore, the correct answer is option (B).** **Note:** At times students tend to consider only the metal ion while trying to find the color. This is not correct as $$Pb{{I}_{2}}$$ is yellow in color and $$PbS$$ is black in color although the oxidation of $$Pb$$ remains the same. Therefore, consider the both the radicals when trying to find the color to avoid making mistakes.