Q.General electronic configuration of actinoids is $(n-2)f^{1-14}(n-1)d^{0-2}ns^2$. Which of the following actinoids have one electron in 6d orbital?

Step 1 — Which element is being asked for?

Every period ends with the $p$-block (group 18). So the last $p$-block element of the 6th period is the noble gas of period 6, i.e. radon, $Z = 86$.

Step 2 — Build its configuration by the Aufbau order.

After xenon ($[Xe]$, $Z=54$) the filling order is

so

The outermost/valence-shell region is therefore $4f^{14}5d^{10}6s^2 6p^6$.

Step 3 — Eliminate the distractors.

• (B) $7s^2 7p^6$ — principal quantum number $7$ ⇒ this is the 7th period (oganesson), not the 6th.
• (C) $5f^{14}6d^{10}7s^2 7p^6$ — again $n=7$; this is Og ($Z=118$), the last $p$-block element of period 7. …

Step 1 — Classify by the orbital being filled.

The periodic table is blocked by the sub-shell that receives the last electron:

• s-block $\to$ groups 1–2,
• p-block $\to$ groups 13–18,
• d-block $\to$ transition elements (filling $(n-1)d$),
• f-block $\to$ inner-transition elements (filling $(n-2)f$).

Step 2 — Identify the 4f series.

The lanthanoids — cerium ($Z=58$) through lutetium ($Z=71$) — follow lanthanum and have the general configuration

Here the electrons are progressively added to the 4f orbital. So this is the answer.

Step 3 — Reject the others.

• (A) Actinoids (Th, $Z=90$ → Lr, $Z=103$): $[\text{Rn}]\,5f^{\,1-14}\,6d^{\,0-1}\,7s^{2}$ — they fill 5f, not 4f. …

The lanthanides (elements 57–71, from lanthanum to lutetium) are characterised by the gradual filling of the 4f subshell. Their chemistry is dominated by a single, recurring pattern: the +3 oxidation state. Why is this so universal?

The key lies in the electronic configuration. A typical lanthanide atom has the outer configuration $[\text{Xe}]\,4f^{n}\,6s^{2}$ (with minor exceptions for lanthanum, gadolinium, and lutetium). The 6s electrons are the highest in energy and are always lost first when forming ions. Removing both 6s electrons gives a $+2$ ion. However, the 4f electrons are also relatively loosely held — they are not as deeply buried as inner core electrons. For most lanthanides, losing one additional 4f electron to reach the $+3$ state is energetically favourable enough to make this the dominant oxidation state in aqueous solution and in most solid compounds.

The $+3$ state is so stable that it is the only oxidation state common to all lanthanides. Some lanthanides do show $+2$ or $+4$ states, but these are exceptions tied to special stability of half-filled or empty 4f subshells (e.g., $\text{Eu}^{2+}$ has $4f^{7}$, $\text{Ce}^{4+}$ has $4f^{0}$). No lanthanide shows a $+5$ state under normal conditions.

Let’s check each option:

1. Option (A): $+2$ — This is not common for all. Only europium ($\text{Eu}^{2+}$), ytterbium ($\text{Yb}^{2+}$), and occasionally samarium ($\text{Sm}^{2+}$) show stable $+2$ states. Most lanthanides do not form $+2$ ions readily. …

Step 1 — Read the configuration carefully.

The configuration $[\mathrm{Ar}]3d^5$ belongs to the ion $X^{3+}$, not to the neutral atom. This is the trap in the question.

Step 2 — Electrons in the ion.

$[\mathrm{Ar}]$ = argon core = $18$ electrons.

Step 3 — Electrons in the neutral atom.

A $+3$ charge means three electrons were removed, so add them back:

Step 4 — Identify and cross-check.

$Z = 26$ is iron. Its ground-state configuration is

$\mathrm{Fe}: [\mathrm{Ar}]3d^{6}4s^{2}$ …