Arrange the following oxidation states in increasing order
(A) characteristic oxidation state of lanthanoids
(B) most stable oxidation state of Mn in its oxides
(C) most stable oxidation state of Mn
(D) oxidation state of Mn in \( MnO_2 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C), (A), (D), (B)
Each item is first turned into a number.
(A) The characteristic oxidation state of the lanthanoids is +3. All the lanthanoids form \( \ce{Ln^3+} \) ions; +2 and +4 occur only where they give an empty, half-filled or filled 4f subshell.
(B) In its oxides Mn reaches its highest oxidation state, +7, in \( \ce{Mn2O7} \). Oxygen stabilises high oxidation states better than fluorine, because it can form multiple bonds to the metal; the highest fluoride of manganese is only \( \ce{MnF4} \).
(C) Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Mn^2+} \): \( \mathrm{3d^{5}} \), a half-filled 3d subshell. So the most stable oxidation state of Mn is +2.
(D) In \( \ce{MnO2} \): x + 2(−2) = 0, so x = +4.
In increasing order:
\( \displaystyle \text{(C)}\ {+2} \lt \text{(A)}\ {+3} \lt \text{(D)}\ {+4} \lt \text{(B)}\ {+7} \)
Shortcut: +2 for Mn is the lowest value, so the order starts with (C): Options (3) and (4). The +7 of \( \ce{Mn2O7} \) is the highest, so it ends with (B): Option (3).