Question

Which one of the elements with following outer electronic configurations may exhibit the largest number of oxidation states?

(1) `3d^5 4s^1`
(2) `3d^5 4s^2`
(3) `3d^2 4s^2`
(4) `3d^3 4s^2`

Answer 1

For transition metal elements, a general rule is => The max number of oxidation states = no. unpaired d-orbital electrons + the 2s electrons.

Explanation:

As noted, the maximum number of oxidation states a given transition metal is (generally) the number of unpaired d-orbitals plus the 2 s-orbital electrons. This then finds the transition elements with the higher numbers of oxidation states in the middle of the transition series.

For the question presented & written in order of Aufbau Diagram:
`4s^(1)3d^5` => `4suarr3duarruarruarruarruarr` => 6 oxidation states
`4s^(2)3d^5` => `4suarrdarr3duarruarruarruarruarr` => 7 oxidation states
`4s^(2)3d^2` => `4suarrdarr3duarruarr` => 4 oxidation states
`4s^(2)3d^3` => `4suarrdarr3duarruarruarr` => 5 oxidation states

Example:
Iron has 4 unpaired electrons and 2 paired electrons. To find one of its oxidation states, we can use the formula:

Oxidation State of Fe = 4 + 2 = +6

Indeed, +6 is one of the oxidation states of iron, but it is very rare. Other possible oxidation states for iron includes: +5, +4, +3, and +2.

Here are some oxidation states of elements in the 1st Transition Series
![https://chem.libretexts.org/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/1b_Properties_of_Transition_Metals/Electron_Configuration_of_Transition_Metals/Oxidation_States_of_Transition_Metalsenter image source here]()