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 :

<h3>(A) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$</h3> <p>All four ions ($\mathrm{Al}^{3+},\mathrm{Mg}^{2+},\mathrm{Na}^{+},\mathrm{F}^{-}$) are <strong>isoelectronic</strong> (each has 10 electrons). For isoelectronic species, ionic radii <strong>decrease</strong> as the positive nuclear charge <strong>increases</strong>, 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 <strong>increasing ionic radius</strong>.</p> <p>$ \boxed{ (A) \;\longrightarrow\; \text{(IV) Ionic radii} } $</p> <hr /> <h3>(B) $\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}$</h3> <p>Check <strong>first ionization enthalpies</strong> ($\mathrm{IE}_1$):</p> <p><p>B: $\approx 801$\,kJ/mol </p></p> <p><p>C: $\approx 1086$\,kJ/mol </p></p> <p><p>O: $\approx 1314$\,kJ/mol </p></p> <p><p>N: $\approx 1402$\,kJ/mol </p></p> <p>Hence, the order of increasing $\mathrm{IE}_1$ is </p> <p>$ B < C < O < N. $ </p> <p>This matches the given sequence exactly.</p> <p>$ \boxed{ (B) \;\longrightarrow\; \text{(I) Ionisation enthalpy} } $</p> <hr /> <h3>(C) $\mathrm{B}<\mathrm{Al}<\mathrm{Mg}<\mathrm{K}$</h3> <p>Consider <strong>metallic character</strong> (the tendency to lose electrons easily, show metallic properties). Across a period (left to right), metallic character <strong>decreases</strong>; down a group, it <strong>increases</strong>.</p> <p><p><strong>B</strong> (metalloid) has the least metallic character here. </p></p> <p><p><strong>Al</strong> (group 13 metal) is more metallic than B. </p></p> <p><p><strong>Mg</strong> (group 2 metal) is typically more metallic than Al. </p></p> <p><p><strong>K</strong> (group 1 metal) is the most metallic among these.</p></p> <p>Thus, $\mathrm{B} < \mathrm{Al} < \mathrm{Mg} < \mathrm{K}$ is an order of <strong>increasing metallic character</strong>.</p> <p>$ \boxed{ (C) \;\longrightarrow\; \text{(II) Metallic character} } $</p> <hr /> <h3>(D) $\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$</h3> <p>Check <strong>electronegativities</strong>:</p> <p><p>Si: $\approx 1.90$ </p></p> <p><p>P: $\approx 2.19$ </p></p> <p><p>S: $\approx 2.58$ </p></p> <p><p>Cl: $\approx 3.16$</p></p> <p>They increase in the order </p> <p>$ \mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}, $</p> <p>which matches the given sequence for <strong>increasing electronegativity</strong>.</p> <p>$ \boxed{ (D) \;\longrightarrow\; \text{(III) Electronegativity} } $</p> <hr /> <h2>Final Matching</h2> <p>$ (A) \to (IV),\quad (B) \to (I),\quad (C) \to (II),\quad (D) \to (III). $</p> <p>Looking at the choices given:</p> <p><strong>Option A</strong>: $(A)-(IV), (B)-(I), (C)-(II), (D)-(III)$</p> <p>This is exactly what we found.</p> <hr /> <h3><strong>Answer: Option A</strong></h3>