Question

Which transition element among the following shows the highest number of oxidation states?

• A. Scandium (Sc)
• B. Iron (Fe)
• C. Manganese (Mn)
• D. Zinc (Zn)

Hint:

Oxidation states of a transition metal come from removing both its $ns$ and $(n-1)d$ electrons. Compare the options by group number, which sets the highest possible oxidation state, and remember that a half-filled $d$-subshell resists pairing longer than others, allowing more intermediate states to form. A completely empty or completely filled $d$-subshell gives very little flexibility.

Answer 1

The Correct Option is C

Concept: Transition elements show variable oxidation states because both their $ns$ and $(n-1)d$ electrons can participate in bonding. This allows them to exhibit multiple oxidation numbers. Among transition elements, the number of oxidation states generally increases toward the middle of the transition series.

Step 1: Electronic configuration of manganese. Manganese ($Z = 25$) has the electronic configuration: \[ [Ar]\,3d^5\,4s^2 \] Because it has five $d$ electrons and two $s$ electrons, a total of seven electrons can potentially participate in bonding.

Step 2: Possible oxidation states. Manganese commonly shows the following oxidation states: \[ +2,\ +3,\ +4,\ +5,\ +6,\ +7 \] Examples include:
• $Mn^{2+}$ in $MnCl_2$
• $Mn^{4+}$ in $MnO_2$
• $Mn^{7+}$ in $KMnO_4$

Step 3: Compare with other options.
• Scandium → Mainly shows +3 oxidation state.
• Iron → Commonly +2 and +3.
• Zinc → Only +2 oxidation state (due to fully filled $3d^{10}$ configuration). Therefore, among the given elements, manganese shows the highest number of oxidation states.

Answer 2

Concept:

• The highest oxidation state a transition metal can show generally equals its group number, because all valence electrons (both $ns$ and $(n-1)d$) can potentially be removed. This rule works well up to Group 7, which is manganese.
• After manganese, increasing pairing energy in the partly filled $d$-subshell makes it harder to remove further electrons, so elements past Group 7 show fewer accessible oxidation states than their group number would suggest.
• A completely empty or completely filled $d$-subshell gives almost no flexibility, since extra oxidation states come specifically from a partially filled $d$-subshell.

Step 1: Assign each element its group number and check the $d$-subshell configuration.
Scandium is in Group 3, with configuration $3d^1 4s^2$.
Iron is in Group 8, with configuration $3d^6 4s^2$; the high pairing energy of this $d^6$ arrangement blocks most higher oxidation states.
Manganese is in Group 7, with configuration $3d^5 4s^2$; this half-filled arrangement keeps pairing energy low right up to the loss of all seven valence electrons.
Zinc has configuration $3d^{10} 4s^2$, a completely filled $d$-subshell.

Step 2: List the known oxidation states of each element.
Scandium shows mainly $+3$.
Iron commonly shows $+2$ and $+3$, with $+4$ and $+6$ appearing rarely.
Manganese shows $+2, +3, +4, +5, +6$ and $+7$.
Zinc shows only $+2$.

Step 3: Count the number of oxidation states for each element and compare.
Manganese shows six oxidation states, more than iron shows, and far more than scandium or zinc, each of which shows essentially one state.

Final Answer: Manganese (Mn)