Match List - I with List - II LIST-I (Atomic number) LIST-II (Block of periodic table) (A) $37$ I. $p$-block (B) $78$ II. $d$-block (C) $52$ III. $f$-block (D) $65$ IV. $s$-block Choose the correct answer from the options given below:

The orbital where the last electron enters defines the block in the periodic table where the elements belong to: Atomic number Block Electronic Configuration $37(K)$ $s$-block $[\mathrm{Ar}]4{s}^{1}$ $78(\mathrm{Pt})$ $d$-block $[\mathrm{Xe}]4{f}^{14}5{d}^{9}6{s}^{1}$ $52(\mathrm{Te})$ $p$-block $[\mathrm{Kr}]4{d}^{10}5{s}^{2}5{p}^{4}$ $65(\mathrm{Tb})$ $f$-block $[\mathrm{Xe}]4{f}^{9}6{s}^{2}$

JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqeasy

Match List - I with List - II

LIST-I (Atomic number)		LIST-II (Block of periodic table)
(A)	$37$	I.	$p$ -block
(B)	$78$	II.	$d$ -block
(C)	$52$	III.	$f$ -block
(D)	$65$	IV.	$s$ -block

Choose the correct answer from the options given below:

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

Options

A
A - II, B - IV, C - I, D - III
B
A - I, B - III, C - IV, D - II
C
A - IV, B - III, C - II, D - I
D
A - IV, B - II, C - I, D - III

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Solution

The orbital where the last electron enters defines the block in the periodic table where the elements belong to:

Atomic number	Block	Electronic Configuration
$37(K)$	$s$ -block	$[\mathrm{Ar}]4{s}^{1}$
$78(\mathrm{Pt})$	$d$ -block	$[\mathrm{Xe}]4{f}^{14}5{d}^{9}6{s}^{1}$
$52(\mathrm{Te})$	$p$ -block	$[\mathrm{Kr}]4{d}^{10}5{s}^{2}5{p}^{4}$
$65(\mathrm{Tb})$	$f$ -block	$[\mathrm{Xe}]4{f}^{9}6{s}^{2}$

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JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqeasy

Match List - I with List - II

LIST-I (Atomic number)		LIST-II (Block of periodic table)
(A)	$37$	I.	$p$ -block
(B)	$78$	II.	$d$ -block
(C)	$52$	III.	$f$ -block
(D)	$65$	IV.	$s$ -block

Choose the correct answer from the options given below:

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

Options

A
A - II, B - IV, C - I, D - III
B
A - I, B - III, C - IV, D - II
C
A - IV, B - III, C - II, D - I
D
A - IV, B - II, C - I, D - III

Did you get this right?

Solution

The orbital where the last electron enters defines the block in the periodic table where the elements belong to:

Atomic number	Block	Electronic Configuration
$37(K)$	$s$ -block	$[\mathrm{Ar}]4{s}^{1}$
$78(\mathrm{Pt})$	$d$ -block	$[\mathrm{Xe}]4{f}^{14}5{d}^{9}6{s}^{1}$
$52(\mathrm{Te})$	$p$ -block	$[\mathrm{Kr}]4{d}^{10}5{s}^{2}5{p}^{4}$
$65(\mathrm{Tb})$	$f$ -block	$[\mathrm{Xe}]4{f}^{9}6{s}^{2}$

JEE Main 2023 — Chemistry Inorganic Chemistry

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JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

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