Question

A black compound (A) in solid state is fused with $KOH$ and $KCl{O_3}$ . The aqueous extract of the fused mass is green colour solution (B). On passing $C{O_2}$ gas through it the pink colour of (C) is noticed along with some black insoluble mass of (A). The pink coloured solution is decolorized by $F{e^{2 + }}$ in acidic medium. What is the oxidation state of central metal (C)?

Answer 1

Hint : The given question requires the knowledge of d-block elements and their reaction and properties along with the concept of oxidation state which is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. Students are also required to keep in mind whether the medium given is acidic or basic.

Complete Step By Step Answer:
In the presence of $C{O_2}$ , the medium become acidic
\eqalign{ & 3MnO_4^{2 - } + 4{H^ + } \to Mn{O_2} + 2MnO_4^ - + 2{H_2}O \cr & MnO_4^ - + 5F{e^{2 + }} + 8{H^ + } \to M{n^{2 + }} + 5F{e^{3 + }} + 4{H_2}O \cr & {\text{ (C)}} \cr}
Therefore the reactant (c) is $MnO_4^ -$
Now in order to calculate the oxidation state of the central atom let its oxidation state be $x$ .
Hence we will form the following equation.
\eqalign{ & x + ( - 2)4 = - 1 \cr & x - 8 = - 1 \cr & x = - 1 + 8 \cr & x = 7 \cr}
The oxidation state of the central metal (C) is $+ 7$ .

Note :
$KMnO_4^ -$ is an inorganic compound also known as potassium permanganate which is also called Condy’s crystal or permanganate or potash. When the potassium permanganate crystals are dissolved in water the solution thus formed is purple in colour which is because the compound possesses an ionic bond between the potassium cation and the permanganate anion. It is an odourless, purple to magenta crystalline solid which is soluble in water, acetone, acetic acid, methanol and pyridine. Potassium permanganate occurs in the form of monoclinic prisms, almost opaque with a blue metallic lustre.