Limiting Molar Conductivities of ions :
$\lambda^0_{H^+} : 349.8 \text{ Sem}^2 \text{mol}^{-1}$
$\lambda^0_{Na^+} : 50.11 \text{ Sem}^2 \text{mol}^{-1}$
$\lambda^0_{K^+} : 73.52 \text{ Sem}^2 \text{mol}^{-1}$
$\lambda^0_{Ca^{2+}} : 119 \text{ Sem}^2 \text{mol}^{-1}$
$\lambda^0_{Mg^{2+}} : 106.12 \text{ Sem}^2 \text{mol}^{-1}$
Therefore correct order of limiting molar conductivity of cations will be -
$H^+>Ca^{2+}>Mg^{2+}>K^+>Na^+$
The question asks for the correct order of limiting molar conductivity for cations in water at 298 K. Limiting molar conductivity (\(\Lambda_m^0\)) is the conductivity of an ion when it is at infinite dilution, and no further dissociation occurs.
Among the given options, the correct order is determined by understanding the mobility of ions in water. The mobility and consequently the limiting molar conductivity of ions mainly depend on the size of the hydrated ions and the charge on the ions. Higher mobility results in higher conductivity.
Thus, the order based on the limiting molar conductivity is:
\(H^+ > Ca^{2+} > Mg^{2+} > K^+\)
Therefore, the correct answer is: \( H^+ > Ca^{2+} > Mg^{2+} > K^+ \)

Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]