Question

Following order is observed in oxidizing power of certain ions: $VO_{2}^{+}The reason for this increasing order of oxidizing power is:
(A) increasing stability of the lower species to which they are reduced
(B) increasing stability of the higher species to which they are oxidized
(C) increasing stability of the higher species to which they are reduced
(D) increasing stability of the lower species to which they are oxidized

Answer 1

Hint: Take a look at d-block elements in the periodic table and write the electronic configurations of vanadium, chromium and manganese. Calculate the oxidation states of them in the given oxide forms. Recollect what is oxidizing power to obtain the answer.


Complete step by step solution:
- Oxidizing power of a species indicates the ability of that species to oxidize another moiety by itself getting reduced. Basically, an oxidizing agent has oxidizing power.
- In the question, we have three ions consisting of transition elements, Manganese, Vanadium and Chromium. Let’s write their electronic configurations.
\[{}^{23}V=\left[ Ar \right]3{{d}^{3}}4{{s}^{2}}\]
\[{}^{24}Cr=\left[ Ar \right]3{{d}^{4}}4{{s}^{2}}=\left[ Ar \right]3{{d}^{5}}4{{s}^{1}}\]
\[{}^{25}Mn=\left[ Ar \right]3{{d}^{5}}4{{s}^{2}}\]
- Now, let’s calculate the oxidation states of $VO_{2}^{+}$, $C{{r}_{2}}O_{7}^{2-}$, $MnO_{4}^{2-}$.
- Let’s consider the transition metal in their respective oxides as ‘x’
- Therefore, for $VO_{2}^{+}$,
\[\begin{align}
  & x+\left( 2\times -2 \right)=+1 \\
 & x=+5 \\
\end{align}\]
- For $C{{r}_{2}}O_{7}^{2-}$,
\[\begin{align}
  & 2x+\left( 7\times -2 \right)=-2 \\
 & x=+6 \\
\end{align}\]
-For $MnO_{4}^{2-}$,
\[\begin{align}
  & x+\left( 4\times -2 \right)=-2 \\
 & x=+6 \\
\end{align}\]
- From the electronic configuration, we can see that vanadium has +2, +3, +4 and +5 oxidation states. Out of them, the most stable oxidation states of vanadium are +2 and +5.
- Chromium has +1, +2, +3, +4, +5 and +6 oxidation states. Out of them, +1 is the most stable oxidation state because then 3d orbitals are half-filled. +6 oxidation state is also stable.
- Manganese has +2, +3, +4, +5, +6 and +7 oxidation states. Out of them, +2 and +7 are the most stable oxidation states. +2 oxidation state is very stable because 3d orbitals are half-filled.
- In the given species, $VO_{2}^{+}$, $C{{r}_{2}}O_{7}^{2-}$ and $MnO_{4}^{2-}$, vanadium has +5 oxidation state and is stable so, it will have the least oxidizing power because it can gain electrons and get reduced but 3d orbitals won’t be half-filled.
- Manganese has +6 oxidation state which is relatively not stable so it has a high tendency to accept four electrons to attain the most stable +2 oxidation state.
- Chromium has +6 oxidation state and it has half-filled 3d orbitals but it will still have some tendency to accept electrons and attain lower oxidation state.
- Therefore, order observed in oxidizing power of certain ions is: $VO_{2}^{+}

- Therefore, the correct option is option (A).


Note: Remember the most stable oxidation states of vanadium are +2 and +5, of chromium are +1, +2 and +6 and of manganese are +2 and +7. For such problems, always determine oxidation states of the given species and compare with stable oxidation states of those elements.