Question

A violet compound of manganese (P) decomposes on heating to liberate oxygen and compounds (Q) and (R) of manganese is formed. Compound (R) reacts with $KOH$ in the presence of potassium nitrate to give compound (Q). On heating compound (R) with conc.${H_2}S{O_4}$ and $NaCl$, chlorine gas is liberated and a compound (S) of manganese along with other products is formed. Compounds P to S are:
A. P - $KMn{O_4}$, Q - ${K_2}Mn{O_4}$, R - $MnC{l_2}$, S - $Mn{O_2}$
B. P - ${K_2}Mn{O_4}$, Q - $Mn{O_2}$, R - $KMn{O_4}$, S - $MnC{l_2}$
C. P - $KMn{O_4}$, Q - ${K_2}Mn{O_4}$, R - $Mn{O_2}$, S - $MnC{l_2}$
D. P - ${K_2}Mn{O_4}$, Q - $KMn{O_4}$, R - $Mn{O_2}$, S - $MnC{l_2}$

Answer 1

The first line itself, which says that P is a violet compound of manganese gives us a hint that it may be potassium permanganate. As oxygen is liberated, we can understand that reduction takes place, and manganese will change from a higher oxidation state to a lower one. Also, R gets oxidised to form Q, indicating that manganese in R has a lower oxidation state than in Q. We can arrive at a confirmation by checking if oxidation states match for each reaction.

Complete step by step answer:
As we know from experiences in our laboratories and at home, a violet compound of manganese is commonly always potassium permanganate, which has the chemical formula $KMn{O_4}$ . Let us consider this as the compound P. Thermal decomposition of $KMn{O_4}$ is an important reaction of manganese in which the oxidation state of manganese changes from $+ 7$ to $+ 6$ and $+ 4$ . Thermal decomposition is the process by which a compound breaks up into two or more smaller, different compounds, due to the action of heat. The reaction proceeds as follows:
$2KMn{O_4}\xrightarrow{\Delta }{K_2}Mn{O_4} + Mn{O_2} + {O_2}$
We have to now identify which is Q and which is R. The next line says that R reacts with $KOH$ in the presence of potassium nitrate to give compound Q. This is the oxidation reaction of manganese dioxide ($Mn{O_2}$) which yields ${K_2}Mn{O_4}$ as the product, where oxidation state changes from $+ 4$ to $+ 6$. Hence, R is $Mn{O_2}$ and Q is ${K_2}Mn{O_4}$. The reaction is as follows:
$Mn{O_2}\; + \;KN{O_3}\; + \;2\;KOH\; \to \;{K_2}Mn{O_4}\; + \;KN{O_2}\; + \;{H_2}O$
And finally, it is given that R reacts with hydrochloric acid and sodium chloride. This reaction is used to produce manganese dichloride and chlorine gas. In this, sodium chloride acts as a reducing agent, changing the oxidation state of manganese from $+ 4$ to $+ 2$ . It proceeds as follows:
$Mn{O_2} + \;4\;NaCl + \;4\;{H_2}S{O_4}\;\xrightarrow{\Delta }\;MnC{l_2} + \;4\;NaHS{O_4} + \;C{l_2} + \;2\;{H_2}O$
Hence, the compounds from P to S are:
P - $KMn{O_4}$, Q - ${K_2}Mn{O_4}$, R - $Mn{O_2}$, S - $MnC{l_2}$
Therefore, the correct option is C.

Additional Information: Atoms of the same element are known to have differing numbers of neutrons in their nucleus. For instance, stable helium atoms exist which contain either one or two neutrons, but both the atoms have two protons. These different types of helium atoms have different masses (3 or 4 atomic mass units), and they are called isotopes. For any given isotope, the sum of the numbers of protons and neutrons in the nucleus is called the mass number.

Note:
The above reactions are all simultaneous oxidation-reduction reactions, otherwise known as redox reactions. Manganese is able to form a wide range of compounds as it can exist in many oxidation states, due to the presence of d-orbital. Note that the violet colour of $KMn{O_4}$ is not due to d-d transition, but due to charge transfer between the oxygen and manganese atom in $+ 7$ oxidation state.