Which of the following is not an actinoid? (i) Curium (Z = 96) (ii) Californium (Z = 98) (iii) Uranium (Z = 92

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Q. Which of the following is not an actinoid?

(i) Curium (Z = 96)

(ii) Californium (Z = 98)

(iii) Uranium (Z = 92)

(iv) Terbium (Z = 65)

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Concept understanding — Periodic Table Blocks

The Intuition: Why "Blocks" at All?

Imagine you're building a house of cards. Each card has a specific shape and a specific place where it fits. The periodic table is like that house — but instead of cards, we have elements, and instead of shapes, we have electron configurations.

The periodic table is arranged in rows (periods) and columns (groups). But if you look closely, you'll notice that the table isn't a perfect rectangle. There's a detached island of elements (the f-block) floating below, and the main body has a strange "staircase" shape. That shape isn't random — it's dictated by which orbital the last electron enters.

That's the core idea: A block is a set of elements whose last electron enters the same type of orbital (s, p, d, or f).

The Precise Statement

Periodic Table Blocks are regions of the periodic table where elements share the same valence subshell — the subshell being filled as you move across that block.

There are four blocks, named after the four types of atomic orbitals:

Block	Orbital being filled	Location in the table	Number of groups
s-block	$ns$	Leftmost 2 columns (Groups 1 & 2)	2
p-block	$np$	Rightmost 6 columns (Groups 13–18)	6
d-block	$(n-1)d$	Middle 10 columns (Groups 3–12)	10
f-block	$(n-2)f$	Two rows below the main table (Lanthanides & Actinides)	14

Note

The "n" in the orbital notation refers to the principal quantum number (the period number). Notice how for d and f blocks, the orbital being filled has a lower n than the period you're in. That's because of the Aufbau principle — orbitals fill in order of increasing energy, and 4s fills before 3d, etc.

How to Read the Blocks

s-block (Groups 1 & 2)

Last electron enters an $s$ orbital.
Examples: Hydrogen ( $1s^1$ ), Lithium ( $2s^1$ ), Beryllium ( $2s^2$ ).
These are highly reactive metals (except H and He). They lose their $s$ electron(s) easily.

p-block (Groups 13–18)

Last electron enters a $p$ orbital.
Examples: Carbon ( $2p^2$ ), Oxygen ( $2p^4$ ), Chlorine ( $3p^5$ ).
This block contains metals, non-metals, and metalloids — the most chemically diverse block.

d-block (Groups 3–12)

Last electron enters a $d$ orbital — specifically, the $(n-1)d$ subshell.
Examples: Iron ( $3d^6$ ), Copper ( $3d^{10}$ ), Zinc ( $3d^{10}$ ).
These are transition metals. They often have variable oxidation states and form coloured compounds.

f-block (Lanthanides & Actinides)

Last electron enters an $f$ orbital — specifically, the $(n-2)f$ subshell.
Examples: Cerium ( $4f^1$ ), Uranium ( $5f^3$ ).
These are inner transition metals. They are placed below to keep the table from being absurdly wide.

Watch out

A common mistake: thinking that the block tells you the group number. It doesn't. The block tells you the orbital type, not the group. For example, both Carbon (Group 14) and Oxygen (Group 16) are in the p-block, but they're in different groups.

Why This Matters

Knowing the block of an element tells you three things instantly:

Which orbital is being filled — the heart of its electron configuration.
General chemical behaviour — s-block elements are electropositive, p-block are diverse, d-block are transition metals, f-block are inner transition metals.
Position in the table — you can locate any element just by knowing its block and period.

Tip

To find an element's block from its electron configuration: look at the last subshell that has electrons. If it ends in $s$ , it's s-block; if $p$ , p-block; if $d$ , d-block; if $f$ , f-block. For example, $[Ar] 4s^2 3d^6$ ends in $3d^6$ → d-block.

The Big Picture

The periodic table is not just a list — it's a map of how electrons fill orbitals. The blocks are the continents on that map. s-block on the left, p-block on the right, d-block in the middle, and f-block as the two islands below.

Once you see the blocks, the periodic table stops being a random grid and becomes a logical, predictable structure. Every element's position tells you its electron configuration, and every configuration tells you its block.

The s, p, d, and f block classification of the periodic table is a foundational topic in the NCERT Class 11 Chemistry chapter on Classification of Elements and Periodicity, and "periodic table blocks explained with examples" is a commonly searched revision topic for CBSE boards and JEE Main/NEET. Quickly identifying an element's block from its electron configuration is also a frequently tested skill in "periodic table important questions" for competitive chemistry.

Actinoids are the fourteen elements from thorium ( $Z = 90$ ) to lawrencium ( $Z = 103$ ) in which the $5f$ subshell is progressively filled. Terbium ( $Z = 65$ ) lies in the lanthanoid series, not the actinoid series. The answer is (D).

The actinoid series is defined by the progressive filling of the $5f$ orbitals, just as the lanthanoid series is characterized by the filling of $4f$ orbitals. Understanding where these series begin and end in the periodic table immediately tells us which elements belong to each family.

The actinoids span from thorium ( $Z = 90$ ) to lawrencium ( $Z = 103$ ), occupying the bottom row of the f-block. These elements follow actinium ( $Z = 89$ ) and are characterized by electrons entering the $5f$ subshell (though there are some irregularities in electron configurations due to the close energy levels of $5f$ , $6d$ , and $7s$ orbitals).

Let me examine each option:

Curium ( $Z = 96$ ): This element falls squarely in the middle of the actinoid series. With 96 protons, it lies between americium ( $Z = 95$ ) and berkelium ( $Z = 97$ ), well within the $Z = 90$ to $103$ range. Curium is definitely an actinoid.
Californium ( $Z = 98$ ): Similarly, californium sits comfortably within the actinoid series. Named after California (where it was first synthesized), it continues the $5f$ filling pattern. This is an actinoid.
Uranium ( $Z = 92$ ): The most famous actinoid, uranium is the heaviest naturally occurring element in significant quantities. At $Z = 92$ , it's the third member of the actinoid series (after thorium and protactinium). Uranium is certainly an actinoid.
Terbium ( $Z = 65$ ): Here's the outlier. With only 65 protons, terbium falls far short of the actinoid range. In fact, $Z = 65$ places it in the lanthanoid series (also called the rare earth elements), which runs from cerium ( $Z = 58$ ) to lutetium ( $Z = 71$ ). Terbium belongs to the $4f$ block, not the $5f$ block.

Note

The lanthanoids occupy the first row of the f-block (elements $58–71$ ), while the actinoids occupy the second row (elements $90–103$ ). The two series are often placed below the main periodic table to keep the table compact.

✓Final answer

The correct option is (D) — Terbium is a lanthanoid, not an actinoid.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2025Set eng-2025-04284 marksMCQ
$^{2}p^{1-6}$
›Reveal solutionSolution

Halogens are Group 17 elements, not Group 16 — statement (C) is the incorrect one.

Reviewing the statements: p-block spans Groups 13–18 (A, correct); outer configuration $ns^2p^{1-6}$ (B, correct); Ga and Br are liquids near room temperature (D, correct); Group 18 with $ns^2p^6$ are inert (E, correct). Statement (C) wrongly places halogens in Group 16, whereas the halogens (F, Cl, Br, I, At) belong to Group 17. Hence (C) is incorrect.

✓Final answer
$The correct option is (C) .$
KEAM 2025Set eng-2025-04294 marksMCQ
$(n - 1)f^{1-14}(n - 1)d^{0-1}ns^2$
›Reveal solutionSolution

For f-block (inner transition) elements the differentiating electron enters the $(n-2)f$ subshell, giving the general outer configuration $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$ . Statement (E) wrongly writes $(n-1)f$ , so it is incorrect.

The f-block elements have their last electrons filling the antepenultimate $f$ -subshell, $(n-2)f$ . For actinoids this is $5f^{1-14}6d^{0-1}7s^2$ , i.e. the general form $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$ . Statement (E) mislabels it as $(n-1)f^{1-14}(n-1)d^{0-1}ns^2$ , which is wrong. The remaining statements are correct: f-block by definition fills f-orbitals (A); atomic numbers 58–71 (lanthanoids) and 90–103 (actinoids) (B); actinoids are radioactive (C); and there are 28 f-block elements, 14 in each series (D).

✓Final answer
$The correct option is (E) .$
KEAM 2025Set pha-2025-0424F4 marksMCQ
$Q. Main group elements are (A) s-block elements only (B) p-block elements only (C) both s-and p-block elements (D) d-block elements only (E) 4f and 5f-block elements$
›Reveal solutionSolution
The main group / representative elements are the s-block and p-block elements together (Groups 1, 2 and 13–18).

Elements are classified by the sub-shell being filled:

s-block — Groups 1 and 2.

p-block — Groups 13 to 18.

d-block — transition metals (Groups 3–12).

f-block — lanthanoids and actinoids.

The term main group (or representative) elements refers to Groups 1, 2 and 13–18, i.e. both the s- and p-block elements. The d- and f-blocks are the transition and inner-transition elements, not main group.

✓Final answer
$The correct option is (C) .$

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Q.Which of the following is not an actinoid?

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