- A$5BiO_3^- + 22H^+ +Mn^{2+} \to 5Bi^{3+} + 7H_2O + MnO_4^-$
- ✓$5BiO_3^- + 14H^+ + 2Mn^{2+} \to 5Bi^{3+} + 7H_2O + 2MnO_4^-$
- C$2BiO_3^- + 4H^+ +Mn^{2+} \to 2Bi^{3+} + 2H_2O + MnO_4^-$
- D$6BiO_3^- + 12H^+ + 3Mn^{2+} \to 6Bi^{3+} + 6H_2O + 3MnO_4^-$
$5 \mathrm{BiO}_{3}^{-}+14 \mathrm{H}^{+}+2 \mathrm{Mn}^{2+} \rightarrow 5 \mathrm{Bi}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{MnO}_{4}^{-}$
In this equation, the number of atoms of each element on the left side are equal to the number of atoms on the right side.
The charges are also balanced on the two sides of equation.
Hence, option B is the correct answer.
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Assertion $(A)$ : The first ionization enthalpy of $3 d$ series elements is more than that of group $2$ metals
Reason $(R)$ : In $3d$ series of elements successive filling of d-orbitals takes place.
In the light of the above statements, choose the correct answer from the options given below :