Consider the van der Waal's constants, a and b, for the following gases.

Which gas is expected to have the highest critical temperature?

We recall the van der Waals relation between the critical temperature $$T_c$$ and the constants $$a$$ and $$b$$:

$$T_c=\frac{8a}{27Rb}$$

Here $$R$$ is the universal gas constant. Since $$R$$ is the same for all gases being compared, the factor $$\dfrac{8}{27R}$$ is common and need not be calculated explicitly. Thus, for the purpose of comparison, the critical temperature is directly proportional to the ratio $$\dfrac{a}{b}$$.

So, to find which gas has the highest $$T_c$$, we need to compute $$\dfrac{a}{b}$$ for each gas. The given data are:

Argon (Ar): $$a = 1.3\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 3.2\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Neon (Ne): $$a = 0.2\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 1.7\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Krypton (Kr): $$a = 5.1\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 1.0\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Xenon (Xe): $$a = 4.1\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 5.0\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Now we evaluate the ratios one by one.

For Ar: $$\frac{a}{b}=\frac{1.3}{3.2\times10^{-2}}=\frac{1.3}{0.032}=40.625$$

For Ne: $$\frac{a}{b}=\frac{0.2}{1.7\times10^{-2}}=\frac{0.2}{0.017}=11.7647$$

For Kr: $$\frac{a}{b}=\frac{5.1}{1.0\times10^{-2}}=\frac{5.1}{0.01}=510$$

For Xe: $$\frac{a}{b}=\frac{4.1}{5.0\times10^{-2}}=\frac{4.1}{0.05}=82$$

We observe that:

$$\frac{a}{b}\;(\text{Ne}) < \frac{a}{b}\;(\text{Ar}) < \frac{a}{b}\;(\text{Xe}) < \frac{a}{b}\;(\text{Kr})$$

Since $$T_c \propto \dfrac{a}{b}$$, the gas with the largest $$\dfrac{a}{b}$$ value will possess the highest critical temperature. The highest ratio is clearly for Krypton (Kr) with a value of 510.

Hence, the correct answer is Option A.