Q.Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

The key idea is that the stability of multiple oxidation states in transition metals arises from the availability of unpaired $d$-electrons for bonding, combined with the relative energies of the $3d$ and $4s$ orbitals.

Reasoning:

1. Manganese ($\text{Mn}$, atomic number 25) has the electronic configuration $[\text{Ar}]\,3d^5 4s^2$. The $3d$ subshell is half-filled, which gives it special stability.
2. Because the $3d$ and $4s$ orbitals are very close in energy, Mn can lose electrons from both shells across a wide range — from all $4s$ electrons up to all $3d$ electrons.
3. This allows Mn to exhibit oxidation states from $+2$ up to $+7$, covering six different states ($+2, +3, +4, +5, +6, +7$). The $+2$ state (half-filled $d^5$) and $+7$ state ($d^0$) are particularly stable. …

The question asks which 3d transition metal shows the largest number of oxidation states and the reason behind it. This is a classic concept from the chapter on d-block elements. The key lies in the electronic configuration and the stability of half-filled and fully-filled subshells.

Why this approach works

Transition metals show variable oxidation states because the $(n-1)d$ and $ns$ electrons have similar energies. As you move across the 3d series (Sc to Zn), the number of available electrons for bonding changes. The metal that can lose the maximum number of electrons from its valence shell (4s + 3d) without destabilising the nucleus will show the most oxidation states. But it's not just about counting electrons — the stability of the resulting electronic configuration matters enormously.

1. List the 3d series elements and their ground state configurations

   Sc: $3d^1 4s^2$

   Ti: $3d^2 4s^2$

   V: $3d^3 4s^2$

   Cr: $3d^5 4s^1$ (exception due to half-filled stability)

   Mn: $3d^5 4s^2$

   Fe: $3d^6 4s^2$

   Co: $3d^7 4s^2$

   Ni: $3d^8 4s^2$

   Cu: $3d^{10} 4s^1$ (exception due to fully-filled stability)

   Zn: $3d^{10} 4s^2$

2. Identify the metal with the maximum number of unpaired + paired valence electrons

   The total number of electrons in the 4s and 3d orbitals that can be lost (in principle) is:

   • Sc: 3 electrons
   • Ti: 4 electrons
   • V: 5 electrons
   • Cr: 6 electrons (but one 4s electron is already lost in the exceptional configuration)
   • Mn: 7 electrons (5 from 3d, 2 from 4s)
   • Fe: 8 electrons (but losing all 8 is not observed because the +8 state would be too highly charged)
   • Co: 9 electrons (similarly impractical)
   • Ni: 10 electrons (impractical)
   • Cu: 11 electrons (impractical)
   • Zn: 12 electrons (impractical, and Zn only shows +2)

   So Mn has the largest number of valence electrons that can actually be removed in stable steps.

3. Check the observed oxidation states for Mn

   Manganese shows oxidation states from +2 to +7, and even +1 and -1 in some special compounds. The commonly known states are:

   +2 (MnCl$_2$), +3 (Mn$_2$O$_3$), +4 (MnO$_2$), +5 (Na$_3$MnO$_4$), +6 (K$_2$MnO$_4$), +7 (KMnO$_4$).

   No other 3d element shows such a wide range. For comparison:

   • V shows +2 to +5 (4 states)
   • Cr shows +2 to +6 (5 states, but +5 and +6 are less common)
   • Fe shows +2 to +6 (5 states, but +6 is rare)
   • Mn shows 6 common states (+2 to +7), which is the maximum.

4. Why does Mn achieve this?

   The reason is twofold:

   • Electronic configuration: Mn has $3d^5 4s^2$. Losing the two 4s electrons gives Mn$^{2+}$ ($3d^5$), which is half-filled and extra stable. Further removal of 3d electrons (one by one) leads to Mn$^{3+}$ ($3d^4$), Mn$^{4+}$ ($3d^3$), Mn$^{5+}$ ($3d^2$), Mn$^{6+}$ ($3d^1$), and finally Mn$^{7+}$ ($3d^0$). Each step is possible because the nuclear charge (25) is high enough to hold the remaining electrons, and the half-filled $d^5$ configuration of Mn$^{2+}$ provides a stable intermediate.
   • Stabilisation by half-filled and fully-filled configurations: The +2 state (half-filled $d^5$) and the +7 state (noble gas configuration, $3d^0$) are particularly stable. The +4 state (MnO$_2$) is also very stable — +4 is a common state for many 3d metals, and MnO$_2$ is further stabilised by its high lattice energy. The key point is that Mn can access all these states without the energy penalty being too high. …

Method: Electronic Configuration Analysis for Oxidation State Stability

This method uses the Aufbau principle and Hund's rule to predict how many oxidation states a transition metal can exhibit based on its $3d$ and $4s$ electron configuration.

Steps

1. Write the ground-state electronic configuration

For the $3d$ series (Sc to Zn), the general configuration is:

2. Identify the element with maximum unpaired electrons in the $3d$ subshell

• The number of possible oxidation states is highest when many $3d$ and $4s$ electrons can be removed stepwise, with extra-stable configurations (half-filled $d^5$, empty $d^0$) available along the way to anchor the accessible states.
• Manganese (Mn) has configuration:

• It has 5 unpaired electrons in the $3d$ subshell — the maximum possible in the series.

3. Count the accessible oxidation states

Mn can lose electrons from both $4s$ and $3d$ orbitals to exhibit:

• This is 6 different oxidation states — the largest in the $3d$ series.

4. Explain why using stability reasoning

• The $3d^5$ configuration is half-filled, which is extra stable. …

Here are the most common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Picking the Wrong Element (e.g., Mn vs. Cr vs. Fe)

The Mistake: Students often guess Chromium (Cr) because they remember it has $+3$ and $+6$ states, or Iron (Fe) because of $+2$ and $+3$. They forget that Manganese (Mn) has the widest range.

Why it happens: Students memorize a few common states for each element but fail to systematically list all possible states for the entire series.

How to Avoid:

• Systematically list the possible oxidation states for the 3d series from Sc to Zn.
• Remember the key rule: Maximum oxidation state = number of 4s + 3d electrons (up to the group number).
• For Mn (Group 7): It has 2 (4s) + 5 (3d) = 7 valence electrons. It can lose all 7 to show $+7$ (in $\text{KMnO}_4$). It also shows $+2, +3, +4, +5, +6$.
• Result: Mn shows all states from +2 to +7 (total 6 states). Cr (max +6) and Fe (max +6, but unstable) show fewer.

Mistake 2: Forgetting the Lower States ($+2$ and $+3$)

The Mistake: Students only list the high oxidation states (like $+6$ or $+7$) and forget that the element also shows common lower states like $+2$ and $+3$.

Why it happens: The question asks for the "largest number," so students focus on the maximum possible value, not the count of distinct states.

How to Avoid:

• Count all distinct states, not just the highest.
• For Mn: $+2, +3, +4, +5, +6, +7$ → 6 states.
• For Cr: $+2, +3, +4, +5, +6$ → 5 states (Cr does not show $+7$).
• For Fe: $+2, +3, +4, +5, +6$ → 5 states (Fe(VI) is rare but known).

Mistake 3: Giving the Wrong Reason (Stability vs. Availability)

The Mistake: Students say "Mn has the largest number because it is the most stable" or "because it has the highest oxidation state." This is incorrect.

Why it happens: Confusing "number of states" with "stability of a particular state."

How to Avoid:

• The correct reason is electronic configuration.
• Mn has the $3d^5 4s^2$ configuration. This is a half-filled d-subshell.
• Key concept: A half-filled d-subshell ($d^5$) is extra stable. Therefore, Mn can lose electrons to reach this stable configuration from both higher and lower states.
• Result: Mn can easily gain or lose electrons to achieve $d^5$, allowing it to exist in many oxidation states. This is not about stability of the highest state, but about flexibility due to the half-filled stability.

Mistake 4: Ignoring the "Why" (Electronic Configuration)

The Mistake: Students give a vague reason like "because it has many electrons" or "because it is in the middle of the series."

Why it happens: Not connecting the property to the fundamental electronic structure. …