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Q4.25(ii):
Give examples and suggest reasons for the following features of the transition metal chemistry: (ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

Solution :

Theoretical Foundation: Stability of Oxidation States

The ability of a transition metal to exhibit its highest oxidation state is fundamentally governed by the electronegativity and the size of the ligand (anion) bonded to the metal center. According to the principles of chemical bonding and coordination chemistry, high oxidation states are stabilized by elements with high electronegativity and small ionic radii [Per Pauling's Electronegativity Scale and Hard-Soft Acid-Base (HSAB) Theory].

Step 1: Analyzing the Role of Fluorine and Oxygen

Fluorine ($F$) and Oxygen ($O$) are the two most electronegative elements in the periodic table. Their impact on the oxidation states of transition metals is dual-fold:

  • High Electronegativity: These atoms possess the ability to pull electron density effectively away from the metal center, reducing the electron-electron repulsion between the metal's $d$-electrons and the ligand electrons, thereby stabilizing the highly positive oxidation state of the cation.
  • Small Atomic/Ionic Size: Because these atoms are small, they exert high charge density and can accommodate a higher coordination number around the central metal atom without significant steric hindrance.

Step 2: Illustrative Examples

To support this, we observe the maximum oxidation states in the first transition series:

Metal Highest Oxidation State Example Oxide Example Fluoride
Manganese ($Mn$) $+7$ $Mn_2O_7$ $MnF_4$ (limited stability)
Chromium ($Cr$) $+6$ $CrO_3$ $CrF_6$
Osmium ($Os$) $+8$ $OsO_4$ $OsF_6$

Step 3: Geometric Representation of Metal-Ligand Stabilization

The following SVG illustrates the high coordination environment required to stabilize the high oxidation state ($M^{n+}$) of a transition metal by highly electronegative ligands ($L = O, F$).

M L L L L High Electronegativity (O, F) withdraws e- density, stabilizing M

Step 4: Chemical Argumentation

The ability to attain high oxidation states is restricted by the availability of orbitals and the lattice energy of the resulting compound. Oxygen is particularly effective at stabilizing high oxidation states via the formation of multiple bonds (specifically $p\pi-d\pi$ bonding), as seen in $Mn_2O_7$ ($Mn$ in $+7$) and $OsO_4$ ($Os$ in $+8$). Because fluorine has the highest electronegativity, it can force the metal to relinquish more electrons into the bonding regime, satisfying the octet or valence shell requirements of the high-valent metal ion.

Final Solution: Transition metals exhibit their highest oxidation states in combination with oxygen and fluorine because these elements possess high electronegativity and small ionic sizes, which facilitate the stabilization of high positive charges through effective electron withdrawal and high-coordination geometric arrangements.


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