Q.Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

The key idea is that the actinoid series ends where the $5f$ subshell is complete, and the last actinoid's filled $5f^{14}$ core makes +3 its most stable oxidation state.

Reasoning:

1. The actinoid series runs from Th ($Z=90$) to Lr ($Z=103$). The last element is Lawrencium (Lr).
2. Electronic configuration of Lr (as given in NCERT's own table of actinoid configurations): $[\text{Rn}]\,5f^{14}\,6d^{1}\,7s^{2}$. …

The actinoid series spans elements from atomic number 90 (Thorium) to 103 (Lawrencium). These are the elements where the $5f$ subshell is progressively filled. The last member of this series is the element with $Z = 103$, which is Lawrencium (Lr).

Why is Lawrencium the last? Because after filling the $5f$ subshell (which holds 14 electrons), the next element (Z=104, Rutherfordium) begins filling the $6d$ subshell and belongs to the transition series (group 4). So the actinoid series ends exactly when the $5f$ subshell is complete.

1. Electronic configuration of Lawrencium (Z = 103)

Based on the Aufbau principle, and matching NCERT's own printed table of actinoid configurations, the ground-state configuration is:

The $5f$ subshell is already completely full ($5f^{14}$) by the time we reach Lr; the last electron added goes into the $6d$ orbital.

2. Possible oxidation states of Lawrencium

Like all actinoids, Lawrencium shows a +3 oxidation state as the most stable and common one. This is because losing the $6d^1$ and the two $7s$ electrons gives a stable $5f^{14}$ core — a completely filled $f$-subshell, which is especially stable, exactly the same reasoning used for why +3 dominates across the whole actinoid series.

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Method: Periodic Trend Analysis + Electronic Configuration Rule

Method Name: Electronic Configuration & Oxidation State Prediction for f-Block Elements

Step 1 — Identify the last actinoid element

The actinoid series starts with Thorium (Z=90) and ends with Lawrencium (Z=103).

Last element: Lawrencium ($\text{Lr}$, atomic number 103)

Step 2 — Write the electronic configuration

For Lawrencium, NCERT's own table of actinoid configurations gives:

• Noble gas core: $[\text{Rn}]$ (Radon, Z=86)
• Electrons fill: 5f (already complete, $5f^{14}$), then 6d, then 7s

Step 3 — Comment on possible oxidation states

• The most stable oxidation state of Lawrencium is +3 (like most actinoids).
• Reason: Loss of the $6d^1$ electron and the two $7s$ electrons (3 electrons total) gives a stable $5f^{14}$ core (fully filled f-subshell). …

Here are the common mistakes students make on this question, along with clear strategies to avoid each.

1. Confusing the Last Element with the End of the Periodic Table

The Mistake:

Students write a later element — even Oganesson (Og, Z=118) — as the last actinoid, treating the series as if it ran on to the end of the periodic table.

Why it happens:

The actinoid series is a specific inner transition series in which the $5f$ subshell is being filled; it ends once the $5f$ subshell is complete.

How to Avoid:

• Memorise the range: Actinoids = Z = 90 (Th) to Z = 103 (Lr). The last element is Lawrencium (Lr).
• Remember the rule: The series ends when the $5f$ subshell is full ($5f^{14}$). For Lr, NCERT's own table gives the configuration as $[Rn] 5f^{14} 6d^1 7s^2$.

Correct Answer:

Lawrencium (Lr) is the last element in the actinoid series.

2. Writing the Wrong Electronic Configuration

The Mistake:

Writing the last electron into the wrong orbital, or writing the subshells out of order.

Why it happens:

The $5f$ subshell is already complete ($5f^{14}$) for Lr, so students sometimes keep adding to $5f$ instead of moving on to $6d$.

How to Avoid:

• Learn the exact configuration (per NCERT's own table): For Lr (Z=103):

• Write in order of increasing principal quantum number: $5f$ (complete), then $6d$, then $7s$.

Correct Answer:

Electronic configuration: $[Rn] 5f^{14} 6d^1 7s^2$

3. Forgetting the +3 Oxidation State is Most Common

The Mistake:

Listing only +2 or +4 oxidation states, or saying Lr shows only +1.

Why it happens:

Students confuse Lr with earlier actinoids like U or Pu which show multiple states. Lr is at the end of the series — its $5f$ subshell is full, making it stable.

How to Avoid:

• Remember the trend: Across the actinoid series, the +3 oxidation state becomes increasingly stable. For Lr, the $5f^{14}$ configuration is very stable, so losing the $6d^1$ and $7s^2$ electrons (total 3) gives a stable $[Rn] 5f^{14}$ core.
• Predict based on configuration: Lr has 3 outer electrons ($6d^1 7s^2$). The most stable oxidation state is +3 (removing all three).

Correct Answer:

The most common oxidation state is +3 — losing the $6d^1$ and $7s^2$ electrons leaves the stable $5f^{14}$ core.

4. Importing Nobelium's +2 Behaviour into Lawrencium

The Mistake:

Because nobelium (No, Z=102) has a famously stable +2 state, students assume the neighbouring Lr must also favour +2.

Why it happens:

The two elements sit side by side at the end of the series, and both involve a filled $5f^{14}$ shell.

How to Avoid:

• For No: losing just two electrons already leaves the filled $5f^{14}$ — hence its stable +2. …