Given: Limiting molar conductivity λ°H+ = 349.6 S cm2 mol–1Na+ = 50.1…

The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (I), (D) - (IV)

By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by the number of ions in one formula unit:

    \( \displaystyle \Lambda_m^{\circ}=\nu_{+}\lambda_{+}^{\circ}+\nu_{-}\lambda_{-}^{\circ} \)

• (A) NaCl — This matches (II) 126.4 S cm² mol^–1. One \( \ce{Na^{+}} \) and one \( \ce{Cl^{-}} \):

      \( \displaystyle \Lambda_m^{\circ}=50.1+76.3=126.4\ \mathrm{S\,cm^2\,mol^{-1}} \)

• (B) \( \ce{CaCl2} \) — This matches (III) 271.6 S cm² mol^–1. One \( \ce{Ca^{2+}} \) and two \( \ce{Cl^{-}} \):

      \( \displaystyle \Lambda_m^{\circ}=119+2(76.3)=271.6\ \mathrm{S\,cm^2\,mol^{-1}} \)

• (C) HCl — This matches (I) 425.9 S cm² mol^–1. One \( \ce{H^{+}} \) and one \( \ce{Cl^{-}} \); the very high value comes from the exceptionally mobile H⁺ ion:

      \( \displaystyle \Lambda_m^{\circ}=349.6+76.3=425.9\ \mathrm{S\,cm^2\,mol^{-1}} \)

• (D) \( \ce{MgCl2} \) — This matches (IV) 258.6 S cm² mol^–1. One \( \ce{Mg^{2+}} \) and two \( \ce{Cl^{-}} \):

      \( \displaystyle \Lambda_m^{\circ}=106+2(76.3)=258.6\ \mathrm{S\,cm^2\,mol^{-1}} \)

Shortcut: The pair (A)–(II) appears in Options (1) and (2), and (C)–(I) then leaves only Option (2).

NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – “… The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. … In general, if an electrolyte on dissociation gives ν₊ cations and ν_– anions then its limiting molar conductivity is given by:

    \( \displaystyle \Lambda_m^{\circ}=\nu_{+}\lambda_{+}^{\circ}+\nu_{-}\lambda_{-}^{\circ} \qquad (2.25) \)”