Chemistry · D And F Block Elements · NEET
To reach a high oxidation state, the metal must lose or share many electrons. Only a partner that pulls electrons very strongly and makes very strong bonds can do this. Oxygen and fluorine are the smallest and most electronegative elements, so they form the strongest bonds and give the highest lattice energy. Larger atoms like Cl, Br, I are less electronegative and their bonds are weaker, so they cannot hold the metal in a very high oxidation state. This is why the highest halides are TiX4, VF5 and CrF6 (mostly fluorides), and the highest oxides go up to Mn2O7.
Fluorine helps in two ways, depending on the compound. (1) In IONIC compounds it gives a very high LATTICE ENERGY because F- is small, so the crystal is very stable - example CoF3. (2) In COVALENT compounds it gives a high BOND ENTHALPY (very strong M-F bonds) - examples VF5 and CrF6. Both effects release a lot of energy, which pays for the energy needed to raise the metal to a high oxidation state. NCERT states this exactly.
Oxygen can form MULTIPLE bonds (M=O double bonds) with the metal because it has two lone pairs to donate into the metal's empty d-orbitals (p-d pi bonding). One oxygen can 'use up' two of the metal's valencies, so oxygen reaches higher oxidation states than fluorine even though fluorine is more electronegative. That is why the top state of Mn is Mn2O7 (an oxide, +7), while with fluorine Mn only manages MnO3F, an oxofluoride - not a simple fluoride.
Fluorine forms only single bonds, so to reach +7 with fluorine the metal would need 7 fluorines packed around a small Mn atom - there is not enough space (steric crowding). Oxygen forms double bonds, so fewer oxygen atoms are needed to reach +7. That is why Mn(+7) appears as Mn2O7 (oxide) and as the mixed oxofluoride MnO3F, but NOT as a simple fluoride like MnF7. For NEET, remember: oxygen's double bonds let it reach higher states than fluorine.
As the oxidation state of the metal INCREASES, the oxide becomes LESS basic and MORE acidic. Low oxidation state oxides like MnO or CrO are basic; high oxidation state oxides like Mn2O7 or CrO3 are acidic. Reason: a higher positive charge pulls the O-H electrons more, so it releases H+ (acidic) instead of OH-. NEET 2023 tested this exact idea (V2O3 to V2O4 to V2O5: basic character DECREASES).
Up to Mn, the maximum stable oxidation state equals the number of (4s + 3d) electrons available, so it equals the group number (Ti = +4, V = +5, Cr = +6, Mn = +7). After Mn, the d-electrons become too tightly held by the growing nuclear charge, so they are not easily lost. That is why beyond group 7 no oxide higher than Fe2O3 (+3) is common, and the elements settle into lower states like Fe(II,III), Co(II,III), Ni(II), Cu(I,II), Zn(II).
Which of the following statements are INCORRECT? A. All the transition metals except scandium form MO oxides which are ionic. B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc2O3 to Mn2O7. C. Basic character increases from V2O3 to V2O4 to V2O5. D. V2O4 dissolves in acids to give VO4^3- salts. E. CrO is basic but Cr2O3 is amphoteric.
The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to the change of
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Oxygen usually reaches a HIGHER maximum oxidation state because it can form double bonds (M=O), while fluorine forms only single bonds. So the very top state of Mn is Mn2O7 (oxide), while fluorine gives only VF5 and CrF6 lower down. Fluorine still wins where strong single bonds or high lattice energy matter.
An oxofluoride contains both O and F on the same metal. Mn cannot reach +7 with fluorine alone (no MnF7 - too crowded), but by using three double-bonded oxygens plus one fluorine it reaches +7 in MnO3F. It combines oxygen's double bonds with fluorine's electronegativity.
In an ionic compound like CoF3, the small F- ions pack tightly around the metal ion and release a very large lattice energy. This energy 'pays back' the large energy needed to pull extra electrons off the metal, so the high oxidation state becomes stable. NCERT names CoF3 as the example.
Only up to manganese (Ti +4, V +5, Cr +6, Mn +7), where it equals the number of 4s + 3d electrons. After Mn the d-electrons are held too tightly, so the maximum stable state drops and does not reach the group number.
Yes. NTA repeatedly asks about oxide/fluoride stability, the Sc2O3-to-Mn2O7 trend, and how basic character changes with oxidation state (asked in 2023). Knowing WHY oxygen and fluorine are special lets you eliminate wrong statement-type options quickly.
