Question:

The reaction of aqueous $KMnO_4$ with $ H_2O_2 $ in acidic conditions gives

Updated On: Jul 13, 2024
  • $Mn^{4+} $ and $ O_2 $
  • $Mn^{2+}$ and $O_2 $
  • $Mn^{2+}$ and $ O_3 $
  • $Mn^{4+}$ and $ MnO_2 $
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The Correct Option is B

Solution and Explanation

The reaction of aqueous $KMnO_4 \, with \, H_2O_2 \, in $ acidic medium is $3H_2SO_4+2KMnO_4+5H_2O_2 \rightarrow $
$\hspace5mm 5O_2+2MnSO_4+8H_2O+K_2SO_4 $
In the above reaction, $KMnO_4 \, oxidises \, H_2O_2 \, to \, O_2 $ and itself $[MnO_4^-]$ gets reduced to $Mn^{2+} $ ion as $MnSO_4. $ Hence, aqueous solution of $KMnO_4. $ with $H_2O_2 \, yields \, Mn^{2+} \, and \, O_2 $ in acidic conditions.
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Concepts Used:

Properties of D Block Elements

  • Multiple oxidation states- The oxidation states of d block elements show very few energy gaps; therefore, they exhibit many oxidation states. Also, the energy difference between s and d orbital is very less. Therefore both the electrons are involved in ionic and covalent bond formation, which ultimately leads to multiple oxidation states.
  • Formation of complex compounds- Ligands show a binding behavior and can form so many stable complexes with the help of transition metals. This property is mainly due to:
    • Availability of vacant d orbitals.
    • Comparatively small sizes of metals.
  • Hardness- Transition elements are tough and have high densities because of the presence of unpaired electrons.
  • Melting and boiling points- Melting and boiling points of transition are very high because of the presence of unpaired electrons and partially filled d orbitals. They form strong bonds and have high melting and boiling points.
  • Atomic radii- The atomic and ionic radius of the transition elements decreases as we move from Group 3 to group 6. However, it remains the same between group 7 and group 10, and from group 11 to group 12 increases.
  • Ionization enthalpy- The ionization enthalpies of the transition elements are generally on the greater side as compared to the S block elements