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Periodic Table and Periodicity question

2025 · 22 Jan · Shift 1 · Q6
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Periodic Table and Periodicity question

2025 · 22 Jan · Shift 1 · Q6

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1

Match List-I with List-II.

List - I List - II
(A) Al3+<Mg2+<Na+<F−\mathrm{Al}^{3+}\lt \mathrm{Mg}^{2+}\lt \mathrm{Na}^{+}\lt \mathrm{F}^{-}Al3+<Mg2+<Na+<F− (I) Ionisation Enthalpy
(B) B<C<O<N\mathrm{B}\lt \mathrm{C}\lt \mathrm{O}\lt \mathrm{N}B<C<O<N (II) Metallic character
(C) B<Al<Mg<K\mathrm{B}\lt \mathrm{Al}\lt \mathrm{Mg}\lt \mathrm{K}B<Al<Mg<K (III) Electronegativity
(D) Si<P<S<Cl\mathrm{Si}\lt \mathrm{P}\lt \mathrm{S}\lt \mathrm{Cl}Si<P<S<Cl (IV) Ionic radii

Choose the correct answer from the options given below :

  1. A
    (A)-(IV), (B)-(I), (C)- (II), (D)-(III)
  2. B
    (A)-(III), (B)-(IV), (C)- (II), (D)-(I)
  3. C
    (A)-(II), (B)-(III), (C)- (IV), (D)-(I)
  4. D
    (A)-(IV), (B)-(I), (C)- (III), (D)-(II)
View written solutionFree

Correct answer: A

  1. Identify each trend in List-I

We match each sequence with the periodic property it represents.

  1. (A) Al3+<Mg2+<Na+<F−\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-}Al3+<Mg2+<Na+<F−

These are all isoelectronic species with 10 electrons.

For isoelectronic ions, radius decreases as nuclear charge increases. Thus,

Al3+<Mg2+<Na+<F−\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-}Al3+<Mg2+<Na+<F−

is the order of ionic radii.

So,

(A)→(IV)(A) \to (IV)(A)→(IV)
  1. (B) B<C<O<N\mathrm{B} < \mathrm{C} < \mathrm{O} < \mathrm{N}B<C<O<N

Check common periodic trends:

  • Ionisation enthalpy generally increases across a period.
  • In period 2, the order is roughly:
B<C<O<N\mathrm{B} < \mathrm{C} < \mathrm{O} < \mathrm{N}B<C<O<N

with nitrogen having especially high value due to half-filled 2p32p^32p3 stability.

So this corresponds to ionisation enthalpy.

Thus,

(B)→(I)(B) \to (I)(B)→(I)
  1. (C) B<Al<Mg<K\mathrm{B} < \mathrm{Al} < \mathrm{Mg} < \mathrm{K}B<Al<Mg<K

This is clearly increasing metallic character:

  • Metallic character increases down a group.
  • Metallic character decreases across a period from left to right.

Hence,

B<Al<Mg<K\mathrm{B} < \mathrm{Al} < \mathrm{Mg} < \mathrm{K}B<Al<Mg<K

represents increasing metallic character.

So,

(C)→(II)(C) \to (II)(C)→(II)
  1. (D) Si<P<S<Cl\mathrm{Si} < \mathrm{P} < \mathrm{S} < \mathrm{Cl}Si<P<S<Cl

Across period 3, electronegativity increases from left to right. Hence,

Si<P<S<Cl\mathrm{Si} < \mathrm{P} < \mathrm{S} < \mathrm{Cl}Si<P<S<Cl

is the order of electronegativity.

Therefore,

(D)→(III)(D) \to (III)(D)→(III)
  1. Final matching

So the correct matching is:

(A)−(IV),(B)−(I),(C)−(II),(D)−(III)(A)-(IV),\quad (B)-(I),\quad (C)-(II),\quad (D)-(III)(A)−(IV),(B)−(I),(C)−(II),(D)−(III)

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So, the answer agrees with the stored answer.

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