Why is the ability of oxygen greater than fluorine to stabilise higher oxidation states of transition metals?

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

Transition metals can form multiple bonds, including $\pi$-bonding, with oxygen. This strong $\pi$-bond overlap stabilizes higher oxidation states via oxo ligands. Despite fluorine's high electronegativity, it only forms single $\sigma$-bonds and does not engage in $\pi$-bonding with metals. Consequently, oxo ions, such as CrO$_4^{2-}$, are more effective at stabilizing Cr(+6) than their fluoride counterparts.

Why is the ability of oxygen greater than fluorine to stabilise higher oxidation states of transition metals?

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Solution and Explanation

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