JEE MAIN - Chemistry (2025 - 22nd January Morning Shift - No. 7)

Match List-I with List-II.

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

Choose the correct answer from the options given below :

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

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

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

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

Explanation

(A) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$

All four ions ($\mathrm{Al}^{3+},\mathrm{Mg}^{2+},\mathrm{Na}^{+},\mathrm{F}^{-}$) are isoelectronic (each has 10 electrons). For isoelectronic species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest ion here is $\mathrm{Al}^{3+}$ (Z=13) and the largest is $\mathrm{F}^{-}$ (Z=9). Thus, this ordering is one of increasing ionic radius.

$ \boxed{ (A) \;\longrightarrow\; \text{(IV) Ionic radii} } $

(B) $\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}$

Check first ionization enthalpies ($\mathrm{IE}_1$):

B: $\approx 801$\,kJ/mol

C: $\approx 1086$\,kJ/mol

O: $\approx 1314$\,kJ/mol

N: $\approx 1402$\,kJ/mol

Hence, the order of increasing $\mathrm{IE}_1$ is

$ B < C < O < N. $

This matches the given sequence exactly.

$ \boxed{ (B) \;\longrightarrow\; \text{(I) Ionisation enthalpy} } $

(C) $\mathrm{B}<\mathrm{Al}<\mathrm{Mg}<\mathrm{K}$

Consider metallic character (the tendency to lose electrons easily, show metallic properties). Across a period (left to right), metallic character decreases; down a group, it increases.

B (metalloid) has the least metallic character here.

Al (group 13 metal) is more metallic than B.

Mg (group 2 metal) is typically more metallic than Al.

K (group 1 metal) is the most metallic among these.

Thus, $\mathrm{B} < \mathrm{Al} < \mathrm{Mg} < \mathrm{K}$ is an order of increasing metallic character.

$ \boxed{ (C) \;\longrightarrow\; \text{(II) Metallic character} } $

(D) $\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$

Check electronegativities:

Si: $\approx 1.90$

P: $\approx 2.19$

S: $\approx 2.58$

Cl: $\approx 3.16$

They increase in the order

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

which matches the given sequence for increasing electronegativity.

$ \boxed{ (D) \;\longrightarrow\; \text{(III) Electronegativity} } $

Final Matching

$ (A) \to (IV),\quad (B) \to (I),\quad (C) \to (II),\quad (D) \to (III). $

Looking at the choices given:

Option A: $(A)-(IV), (B)-(I), (C)-(II), (D)-(III)$

This is exactly what we found.