Q.Match the statements given in Column I with the oxidation states given in Column II.
Column I:

Concept: Lanthanide Contraction & Variable Oxidation States of Mn

Reasoning:

1. In $MnO_2$, oxygen has oxidation state $-2$. Let Mn’s state be $x$: $x + 2(-2) = 0 \Rightarrow x = +4$.
2. Mn has electronic configuration $[Ar]\,3d^5 4s^2$. The half-filled $3d^5$ subshell is exceptionally stable, making $+2$ (by losing both 4s electrons) the most stable oxidation state in aqueous solution.
3. In oxides, Mn shows multiple states; $+4$ (as in $MnO_2$) and $+7$ (as in $Mn_2O_7$) are common, but $+4$ is the most stable in solid oxides. …

Let’s unpack each match one by one. The lanthanide contraction is central here — it’s the steady decrease in ionic radii across the lanthanoid series due to poor shielding by 4f electrons. This makes the +3 state the most stable for all lanthanoids, because removing three electrons gives a compact, well-shielded ion. For manganese, things are different: it’s a transition metal with multiple accessible oxidation states, and stability depends on the environment (e.g., oxide vs. aqueous solution).

1. Oxidation state of Mn in $MnO_2$

   Oxygen almost always has an oxidation state of –2 (except in peroxides). In $MnO_2$, there are two oxygen atoms, so total contribution from oxygen is $2 \times (-2) = -4$. The molecule is neutral, so Mn must balance this: $x + (-4) = 0$, giving $x = +4$.

   Tip

   A quick check: $MnO_2$ is manganese(IV) oxide — the Roman numeral in the name directly gives the oxidation state.

   So (i) matches with (c) +4.

2. Most stable oxidation state of Mn

   Manganese has configurations from $3d^5 4s^2$ in the atom. The +2 state ($3d^5$) is exactly half-filled — a particularly stable arrangement due to exchange energy and symmetry. In aqueous solution, $Mn^{2+}$ is the most common and resists further oxidation or reduction. While +7 (in permanganate) is also stable in acidic conditions, +2 is the most stable overall across different environments.

   So (ii) matches with (a) +2.

3. Most stable oxidation state of Mn in oxides …

Method: Oxidation State Determination via Electronic Configuration & Stability Rules

Step 1: Recall the electronic configurations

• Mn (Z = 25): $[Ar] 3d^5 4s^2$
• Lanthanoids (general): $[Xe] 4f^n 6s^2$ (n = 1 to 14)

Step 2: Match each statement systematically

(i) Oxidation state of Mn in $MnO_2$

• Oxygen has oxidation state $-2$ (each).
• Let Mn oxidation state be $x$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$
• Answer: (c) +4

(ii) Most stable oxidation state of Mn

• Mn has half-filled $3d^5$ configuration → +2 (loss of two 4s electrons) gives stable $3d^5$.
• Also +7 is common but less stable in aqueous medium.
• Most stable in general chemistry: +2
• Answer: (a) +2

(iii) Most stable oxidation state of Mn in oxides

• In oxides, Mn shows +4 (as in $MnO_2$) and +7 (as in $Mn_2O_7$).
• Among oxides, $MnO_2$ (+4) is most stable at room temperature.
• Answer: (c) +4

(iv) Characteristic oxidation state of lanthanoids …

Here are the common mistakes students make on this matching question, along with how to avoid each.

Mistake 1: Confusing "most stable" with "most common" for Mn

• The Error: Students match (ii) "Most stable oxidation state of Mn is" with +7 (option e), because they remember that $KMnO_4$ is a common compound. They forget that "most stable" in aqueous/standard conditions refers to the state that is least reactive, not the one that is most famous.
• The Correct Logic: In aqueous solution, the +2 state ($Mn^{2+}$) is the most stable because it has a half-filled d-subshell ($d^5$). All other states (+3, +4, +6, +7) tend to get reduced to +2.
• How to Avoid: Always link "most stable" to electronic configuration (half-filled or fully-filled d/f orbitals) and the Latimer diagram trend. For Mn, +2 is the thermodynamic sink.

Mistake 2: Misreading "in oxides" for Mn

• The Error: Students match (iii) "Most stable oxidation state of Mn in oxides is" with +2 (option a), applying the same logic as above. They ignore the phrase "in oxides".
• The Correct Logic: In oxide compounds, the +4 state ($MnO_2$) is the most stable. $MnO_2$ is a common, stable mineral (pyrolusite), while $MnO$ (+2) is easily oxidized in air, and $Mn_2O_7$ (+7) is explosive.
• How to Avoid: Underline the key phrase "in oxides". Remember that oxide environment stabilizes higher oxidation states (e.g., $CrO_3$, $MnO_2$, $V_2O_5$). For Mn in oxides, +4 is the stable middle ground.

Mistake 3: Forgetting the "characteristic" state of lanthanoids

• The Error: Students match (iv) "Characteristic oxidation state of lanthanoids is" with +4 (option c) or +2 (option a), remembering exceptions like $Ce^{4+}$ or $Eu^{2+}$.
• The Correct Logic: The characteristic (most common across the entire series) oxidation state for lanthanoids is +3 (option b). While +2 and +4 occur for some elements (Eu, Yb for +2; Ce, Tb for +4), +3 is the norm for all 15 elements.
• How to Avoid: Distinguish between "characteristic" (general trend) and "possible" (exceptions). For lanthanoids, +3 is the rule; +2 and +4 are the exceptions due to half/full f-shell stability.

Mistake 4: Confusing the oxidation state of Mn in $MnO_2$

• The Error: Students calculate incorrectly (e.g., thinking O is -1 in peroxides) or match it with +7 because they confuse $MnO_2$ with $MnO_4^-$.
• The Correct Logic: In $MnO_2$, oxygen is -2 (not peroxide). Let $x$ be the oxidation state of Mn:

So (i) matches with (c) +4. …