Which of the following sequences contain atomic numbers of only representative elements? (Note: more than one

NCERT Exemplar · Q20

Q. Which of the following sequences contain atomic numbers of only representative elements? (Note: more than one of the given options may be correct.)

(i) 3, 33, 53, 87

(ii) 2, 10, 22, 36

(iii) 7, 17, 25, 37, 48

(iv) 9, 35, 51, 88

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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. …

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Representative elements are the s- and p-block elements (Groups 1, 2, 13–18), excluding d-block transition metals and f-block elements. Only (A) and (D) contain atomic numbers that are all representative elements.

Identify the block of each atomic number; a transition metal ( $d$ -block) or inner-transition element disqualifies the set.

(A) 3, 33, 53, 87: Li ( $Z=3$ , s), As ( $Z=33$ , p), I ( $Z=53$ , p), Fr ( $Z=87$ , s) — all s/p block → all representative ✓

(B) 2, 10, 22, 36: He ( $Z=2$ ), Ne ( $Z=10$ ), Ti ( $Z=22$ , d-block, transition metal), Kr ( $Z=36$ ) → disqualified by Ti ✗ …

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

KCET 2026Set D31 markMCQ

4^{th}

List-I (Element - Atomic number)	List-II (Position in periodic table)
a. Ra - 88	i. $4^{th}$ period, $13^{th}$ group
b. Ga - 31	ii. $6^{th}$ period, $6^{th}$ group
c. W - 74	iii. $5^{th}$ period, $10^{th}$ group
d. Pd - 46	iv. $7^{th}$ period, $2^{nd}$ group

›Reveal solution

Each element's period and group follow directly from the shell number and valence electron configuration reached by its atomic number.

Step 1 — Radium (Z = 88)

Radium's electron configuration ends at $7s^2$ , an alkaline earth metal. Its outermost shell is $n=7$ (7th period) and it belongs to Group 2 (two valence $s$ -electrons). This matches List-II item iv (7th period, 2nd group).

Step 2 — Gallium (Z = 31)

Gallium's configuration is $[Ar]3d^{10}4s^24p^1$ . The highest principal quantum number occupied is $n=4$ (4th period), and with the filled $3d^{10}$ plus $4s^24p^1$ it falls in Group 13. This matches item i (4th period, 13th group).

Step 3 — Tungsten (Z = 74) …

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KCET 2024Set B-21 markMCQ

Q. A metalloid is (A) Bi (B) Sb (C) P (D) Se

›Reveal solution

Metalloids sit along the metal/non-metal staircase of the periodic table; among the four options, only antimony (Sb) is a standard group-15 metalloid.

Step 1 — What a metalloid is.

A metalloid has properties intermediate between metals and non-metals — e.g. a lustrous but brittle solid, a semiconducting electrical conductivity that increases with temperature (unlike a true metal), and oxides that are amphoteric. The classic metalloids are B, Si, Ge, As, Sb, Te (and sometimes Po).

Step 2 — Locate each option.

Element	Group	Character
Bi (bismuth)	15, period 6	Metal — the heaviest member; typically metallic lustre, forms basic Bi $_2$ O $_3$
Sb (antimony)	15, period 5	Metalloid — brittle, lustrous, semiconducting; amphoteric Sb $_2$ O $_3$
P (phosphorus)	15, period 3	Non-metal — acidic P $_4$ O $_{10}$
Se (selenium)	16, period 4	Generally classed as a non-metal (photoconductive, but its oxides SeO $_2$ /SeO $_3$ are acidic)

Step 3 — Apply the group trend.

Down group 15 the metallic character increases: …

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COMEDK 2023Set 2023-E1 markMCQ
$(\mathrm{n}-1) \mathrm{d}^2 \mathrm{~ns}^2$
›Reveal solutionSolution

Titanium indeed belongs to group 4 of the periodic table.

Concept: for d-block elements, group number = (number of (n-1)d electrons) + (number of ns electrons).

Configuration (n-1)d^2 ns^2 with n = 4:

3d^2 4s^2 -> total 4 valence electrons -> Z = 22 -> Titanium …
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KCET 2020Set A-11 markMCQ
$Q. Silicon doped with gallium forms (A) p - type semiconductor (B) n - type semiconductor (C) both n and p type semiconductor (D) an intrinsic semiconductor$
›Reveal solutionSolution
Doping silicon with gallium creates a p‑type semiconductor because gallium has one fewer valence electron than silicon, leaving a hole that acts as the majority charge carrier.

The key idea is valence electron count. Silicon is a group‑14 element with four valence electrons. To form a crystalline semiconductor, each silicon atom bonds with four neighbours, sharing one electron per bond — this is the perfect four‑electron‑pair arrangement.

When you introduce an impurity (dopant) into the silicon lattice, what matters is whether the dopant atom has more or fewer valence electrons than silicon. If it has more, the extra electron becomes free to move — that gives an n‑type (negative) semiconductor. If it has fewer, one bond is left unsatisfied, creating a hole that can accept an electron — that gives a p‑type (positive) semiconductor.

Gallium is in group‑13 of the periodic table. It has three valence electrons. So when a gallium atom replaces a silicon atom in the lattice, it can only form three complete covalent bonds with its neighbours. The fourth bond site is empty — that missing electron is what we call a hole.

That hole behaves like a positive charge carrier. It can move through the crystal as neighbouring electrons jump into it, effectively shifting the hole around. Because holes are the majority carriers, the material is p‑type.

Watch out

A common mistake is to confuse the dopant’s group number with the type. Group‑13 elements (B, Al, Ga, In) always produce p‑type in silicon; group‑15 elements (P, As, Sb) produce n‑type. Don’t memorise “gallium = p‑type” blindly — understand why: three valence electrons → one hole per dopant atom.

Now let’s walk through the reasoning step by step.

Identify the host and dopant.

Silicon (Si) is in group 14 — it has 4 valence electrons. Gallium (Ga) is in group 13 — it has 3 valence electrons.

What happens when Ga replaces Si in the lattice?

In the pure silicon crystal, every Si atom is surrounded by four neighbours and forms four covalent bonds, each bond sharing one electron from Si and one from the neighbour. When a Ga atom sits in a Si site, it can only offer 3 electrons to form bonds. It will still try to bond with all four neighbours, but one bond will be missing an electron — that missing electron is a hole.

Hole as a charge carrier. …
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Q.Which of the following sequences contain atomic numbers of only representative elements? (Note: more than one of the given options may be correct.)

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