Question

An \[{\text{X}}{{\text{O}}_2}\] is fused with an alkali metal hydroxide in presence of an oxidizing agent such as \[{\text{KN}}{{\text{O}}_3}\]; a dark green product is formed which disproportionate in acidic solution to afford a dark purple solution. X is:
A.Cr
B.V
C.Ti
D.Mn

Answer 1

Hint: The above reaction is a process of formation of an oxidizing agent. The metal belongs to group VIIB in the periodic table. In the compound, it is present in $ + 7$ oxidation state. The compound it forms is also used as a self-indicator in titration.

Complete step by step answer:
The above reaction is a process of formation of potassium permanganate.
When manganese oxide reacts with an alkali metal oxide that is potassium hydroxide, in the presence of the given oxidizing agent, potassium nitrate \[{\text{KN}}{{\text{O}}_3}\] forms potassium manganate which is green in colour. The reaction occurs as follow:
\[{\text{Mn}}{{\text{O}}_2} + 2{\text{KOH}} + {\text{KN}}{{\text{O}}_3} \to {{\text{K}}_2}{\text{Mn}}{{\text{O}}_4} + {\text{KN}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\]
\[{{\text{K}}_2}{\text{Mn}}{{\text{O}}_4}\] is potassium manganate which has green colour. In acidic solution it is disproportionate that it undergoes self oxidation and self reduction to form permanganate ions, which is deep purple in colour. The reaction occurs as follow:
\[{{\text{K}}_2}{\text{Mn}}{{\text{O}}_4} + 4{{\text{H}}^ + } \to {\text{KMn}}{{\text{O}}_4} + {\text{Mn}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}\]

Hence, the correct option is D that is Mn.

Additional information:
Cr is the chemical symbol for chromium, V is the chemical symbol for vanadium, Ti is the chemical symbol for titanium and Mn is the chemical symbol for manganese. They all are transition metals or d block elements. Mn has the atomic number 25 and belongs to 3 d series with group number 7.

Note:
 Potassium permanganate has the molecular formula \[{\text{KMn}}{{\text{O}}_4}\]. It is a very good oxidizing agent in acidic medium. The oxidation state of manganese is 7 in potassium permanganate and hence acts as oxidizing agent. Manganese has 7 electrons in its valence shell so it cannot lose more electrons than 7, so it can only take electrons and reduce itself and oxidize others.