Question

The highest oxidation state of a transition metal is usually exhibited in its oxide and fluoride.

Answer 1

Hint: Oxidation state of an element is defined as degree of oxidation means ability to lose electrons is a chemical compound. Transition metals show a wide variety of oxidation states in their compounds.
Maximum number of oxidation state are shown by manganese’s i.e.,
$ + 2, + 3, + 4, + 5, + 6, + 7.$


Step by step answer: Oxygen and florin have the ability to oxidase metals to their highest oxidation state.
Oxygen and fluorine have highest electronegativity electronegativity is property by which atom attracts electron is the process in which atom loses electron and forms positive ions.
$M \to {M^{n + }} + n{e^ - }$
Metal metal ion
Since oxygen and fluorine have smaller atomic size and higher electronegativity.
Therefore, they can oxidize metals to their oxidation state.
So highest oxidation states one shown in the oxides and fluorides of transition metals.
Fluoride stabilize higher oxidation state due to higher lattice energy (example: $CO{F_3}$) and higher bond enthalpy in terms of higher covalent compound like $V{F_5}Cr{F_6}.$
And ability to stabilize higher oxidation states by oxygen is due to form multiple bonds with metals.
$\therefore $Oxides are superior to fluorides.
Since we know that metals having higher oxidation state have higher charge density and are unstable.
Fluorides and oxides are also highest negative charge stability.
When then combined a lot of energy is released and the compound is stabilized.

Additional Information: Transition elements exhibit a wide variety of oxidation state in their compound.
Number of oxidation states depend on the availability of unpaired electrons in d-orbitals. If fewer electrons are there they show less number of oxidation states.
Example:
$Sc$ shows only $ + 3$oxidation state
$Mn$ shows oxidation state from \[ + 2\] to \[ + 7.\;\]

Note: Transition metals show variable oxidation state in their compound because they have a very energy difference between $(n - 1)$ and ns orbitals.
So as the number of unpaired electrons increases the number of oxidation states also increases.