Question

${\text{Mn}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{\text{NaCl}}\,\,\mathop {\mathop \to \limits^{{{\text{H}}^ + }} }\limits_{{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}}$ choose incorrect statement for above reaction.
A. Mn goes from $+ 4$ to $+ 2$
B. ${\text{C}}{{\text{l}}^ - }$ is oxidized
C. ${\text{C}}{{\text{l}}_{\text{2}}}$ yellow gas is released
D. ${\text{SO}}_4^{2 - }$ reduces to ${\text{S}}{{\text{O}}_{\text{2}}}$

Answer 1

To determine the answer we have to write the equation of reaction of manganese oxide with sodium chloride and sulphuric acid. After writing the equation, we check the oxidation state of manganese and chlorine and sulphur to determine whether the statements are correct or not.

Complete solution: The equation of reaction of manganese oxide with sodium chloride and sulphuric acid is as follows:
${\text{Mn}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{\text{NaCl}}\,\,\mathop {\mathop \to \limits^{{{\text{H}}^ + }} }\limits_{{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}} {\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{\text{NaHS}}{{\text{O}}_{\text{4}}}\, + \,{\text{C}}{{\text{l}}_2}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
Manganese oxide with sodium chloride and sulphuric acid to give manganese sulphate, sodium hydrogen sulphate, chlorine gas and water.
The oxidation state of Mn in ${\text{Mn}}{{\text{O}}_{\text{2}}}$is as follows:
${\text{x}}\,\,{\text{ + }}\,\,\left( { - {\text{2}} \times {\text{2}}} \right)\, = \,0$
${\text{x}}\,\, = \, + 4$
So, the oxidation state of Mn in ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is $+ 4$.
The oxidation state of Mn in ${\text{MnS}}{{\text{O}}_{\text{4}}}$ is as follows:
${\text{x}}\,\,{\text{ + }}\,\,\left( { - {\text{2}}} \right)\, = \,0$
${\text{x}}\,\, = \, + 2$
So, the oxidation state of Mn in ${\text{MnS}}{{\text{O}}_{\text{4}}}$is $+ 2$.
So, ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is converting into ${\text{MnS}}{{\text{O}}_{\text{4}}}$ during the reaction and oxidation state of Mn is changing from$+ 4$ to $+ 2$ so, statement A is true.
The oxidation state of Cl in ${\text{NaCl}}\,$is $- 1$ and in ${\text{C}}{{\text{l}}_2}$is zero. Chlorine is converting from ${\text{C}}{{\text{l}}^ - }\,$to ${\text{C}}{{\text{l}}_2}$ during the reaction and oxidation state of chlorine is changing from$- 1$ to $0$. It means chlorine is losing electrons, so the oxidation of chlorine is taking place, so statement B is true.
Chlorine has valence electronic configuration ${\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{5}}}$so, the d-orbital of chlorine is vacant. Thus, chlorine absorbs energy and electrons get excited to the higher energy level. When electrons come back to lower energy levels it emits radiation. So, we observe yellow colour. So, statement C is true.
Sulphur is present in ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ as ${\text{SO}}_4^{2 - }$ and in ${\text{NaHS}}{{\text{O}}_{\text{4}}}$as ${\text{SO}}_4^{2 - }$ so, oxidation state of sulphur is the same in both reactant and product. So, statement D is false.
So, the statement D, ${\text{SO}}_4^{2 - }$ reduces to ${\text{S}}{{\text{O}}_{\text{2}}}$, is incorrect.

Thus, the correct options is (D).

Note: The charge of an ion is known as its oxidation state. The species which lose electrons get oxidized and are known as reducing agents. The species which accept electrons get reduced and known as oxidising agents. During oxidation, the oxidation state increases. During reduction oxidation state decreases. Manganese is a transition metal, so its oxidation state is variable. The oxidation state of oxygen in oxide is $- 2$.