QJEEM25Ch7d89: JEE Main 2025 Chemistry - Class 11 JEEM25Ch7d89 Solution: Match List-I with List-II. .tg…

(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.

Answer: Option A