Q.Highest oxidation state of manganese in fluoride is +4 ($MnF_4$) but highest oxidation state in oxides is +7 ($Mn_2O_7$) because

The key idea is that oxygen, unlike fluorine, can form multiple bonds (double bonds) with manganese, which stabilises the metal in a very high oxidation state.

Reasoning:

1. In $Mn_2O_7$, each Mn is in the +7 state. This is achieved because oxygen can form $p\pi-d\pi$ double bonds with Mn, effectively sharing more electron density and stabilising the high positive charge.
2. Fluorine is highly electronegative but has no accessible d-orbitals and can only form a single sigma bond. It cannot engage in $\pi$-backbonding. …

Why this question matters

This is a classic exam trap that tests your understanding of how bonding capacity — not just electronegativity — determines the highest oxidation state a transition metal can achieve with a given ligand. Many students pick option (i) because "fluorine is more electronegative" sounds plausible, but that reasoning misses the real chemistry.

Let's unpack it properly.

1. What does "highest oxidation state" actually mean here?

Manganese in its ground state has the configuration $[Ar] 3d^5 4s^2$. To reach +7, it must lose all seven valence electrons — that's a huge electron deficiency. The only way to stabilise such a high positive charge is through strong covalent bonding with the surrounding atoms. The more bonds the ligand can form per atom, the more electron density it can donate to the metal, stabilising the high oxidation state.

2. Fluorine vs oxygen: the bonding difference

Fluorine has only one unpaired electron and can form only a single covalent bond (it has no d-orbitals to expand its octet). So in $MnF_4$, each fluorine contributes one bond — manganese gets four bonds total, which can at most support the +4 state.

Oxygen, on the other hand, has two unpaired electrons and can form a double bond (one sigma, one pi). In $Mn_2O_7$, each oxygen can share two electron pairs with manganese. The structure is:

    O
    ||
O = Mn - O - Mn = O
    ||         ||
    O          O

Each manganese is bonded to four oxygens — but because three of those oxygens are double-bonded, the total bond order per manganese is much higher than four. This allows manganese to reach +7.

3. Why not option (i) — electronegativity?

Fluorine is indeed more electronegative than oxygen (4.0 vs 3.5 on the Pauling scale). If electronegativity alone decided the highest oxidation state, fluorine should stabilise an even higher state than oxygen — but the opposite is observed. The reason is that a highly electronegative atom pulls electron density toward itself, which actually makes the metal even more electron-deficient. Without enough bonds to supply electron density, the high oxidation state becomes unstable.

4. Why not option (ii) — d-orbitals? …

Method: Electronic Configuration & Bonding Analysis

This method uses the electronic structure of the central atom and the bonding capacity of the ligand to explain oxidation state stability.

Steps:

1. Identify the central atom's available orbitals

   Manganese ($Mn$) has the configuration $[Ar]\,3d^5\,4s^2$. In higher oxidation states, it uses its $3d$ and $4s$ orbitals for bonding.

2. Analyze the ligand's bonding capability

   • Fluorine ($F$) has no $d$-orbitals (only $2s$ and $2p$). It can form only a single covalent bond (one electron pair shared).
   • Oxygen ($O$) has $2p$ orbitals and can form double bonds (one $\sigma$ and one $\pi$ bond) by using its lone pairs.

3. Relate bond type to oxidation state

   • To reach $Mn(VII)$, manganese must share 7 electrons.
     • With fluorine: only 7 single bonds are possible — but $MnF_7$ is sterically impossible (too many large atoms around small $Mn$). Hence, the maximum is $MnF_4$ ($+4$). …

Here are the common mistakes students make on this question about Lanthanide Contraction and Oxidation States of Manganese, along with how to avoid each.

Mistake 1: Choosing option (i) — “Fluorine is more electronegative than oxygen”

Why students pick it:

They think higher electronegativity means fluorine will pull electrons harder, so it should stabilize a higher oxidation state.

Why it’s wrong:

Fluorine is indeed more electronegative, but in this context, electronegativity alone does not explain the limit on oxidation state. If it did, fluorine would stabilize $+7$ even better than oxygen — but the opposite is true.

How to avoid:

Remember: Electronegativity helps explain bond polarity, not the maximum oxidation state accessible. The key is bond multiplicity — oxygen can form double bonds, fluorine cannot.

Mistake 2: Choosing option (ii) — “Fluorine does not possess d-orbitals”

Why students pick it:

They recall that fluorine lacks d-orbitals, and think this limits its ability to expand octet or stabilize high oxidation states.

Why it’s wrong:

The central atom (Mn) uses its own d-orbitals to achieve high oxidation states — not fluorine’s. Fluorine’s lack of d-orbitals is irrelevant here.

How to avoid:

Focus on the central atom’s orbitals, not the ligand’s. The limitation comes from bonding capacity, not orbital availability of the ligand.

Mistake 3: Choosing option (iii) — “Fluorine stabilises lower oxidation state”

Why students pick it:

They confuse this with the general trend that fluorine often stabilizes lower oxidation states in transition metals (e.g., $CoF_3$ vs $Co_2O_3$).

Why it’s wrong:

That trend applies to some metals, but here the question is about maximum possible oxidation state. Fluorine actually stabilizes $+4$ for Mn, which is not the highest possible — oxygen goes to $+7$.

How to avoid:

Distinguish between “stabilizes a lower state” and “limits the highest state.” The reason is structural: fluorine can only form single bonds, so it cannot provide enough electron withdrawal to reach $+7$.

Mistake 4: Not recognizing the correct answer (iv) — “Fluorine can form single bond only while oxygen forms double bond”

Why students miss it:

They don’t connect bond multiplicity to oxidation state stabilization. They think only electronegativity matters.

Why it’s correct: …