Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar ?

First, recall the definition: the first ionization enthalpy $$I_1$$ is the energy required to remove the outermost electron from a neutral gaseous atom.

We make use of two empirical rules which come straight from periodic-table trends.

Rule 1 - Down any group, $$I_1$$ decreases.
Inside one group each lower element possesses an extra shell, its outer electrons are farther from the nucleus, the shielding is larger, the nuclear attraction felt by the valence electron is weaker, so less energy is needed to remove that electron.

Rule 2 - Along a period from left to right, $$I_1$$ increases.
While moving across a period the principal quantum number does not change, but the nuclear charge increases steadily, pulling the electron cloud closer. The atomic radius shrinks, shielding is almost constant, and therefore the attractive force on the outermost electron grows, demanding more energy for its removal.

Let us apply Rule 1 to the two alkaline-earth metals present:

Both calcium and barium lie in group 2. Calcium occupies period 4 whereas barium lies in period 6, i.e. farther down the group. Following Rule 1 we write

$$I_1(\text{Ba}) \lt I_1(\text{Ca}).$$

Now apply Rule 1 again to the chalcogens (group 16): sulfur is in period 3, selenium in period 4. Hence

$$I_1(\text{Se}) \lt I_1(\text{S}).$$

Up to this point we possess two separate mini-orders:

$$\text{Ba} \lt \text{Ca} \quad\text{and}\quad \text{Se} \lt \text{S}.$$

Next we have to compare group 2 elements with group 16 elements that stand in the same or neighbouring periods. For this the left-to-right trend (Rule 2) is decisive. Inside one period any element that stands to the right has a higher first ionization enthalpy than an element standing to the left.

Inside period 4 we meet the sequence

$$\text{Ca (group 2)} \; \longrightarrow \; \cdots \; \longrightarrow \; \text{Se (group 16)},$$

so Rule 2 gives

$$I_1(\text{Ca}) \lt I_1(\text{Se}).$$

Inside period 3 the sequence is

$$\text{(Na Mg Al Si P) S (Cl) Ar},$$

and again, moving to the right raises $$I_1$$. Therefore we can combine

$$I_1(\text{Ca}) \lt I_1(\text{Se}) \lt I_1(\text{S}).$$

Finally, argon is the noble gas at the extreme right of period 3. Noble gases have completely filled shells. Their electrons are held most tightly, so they display the highest first ionization enthalpies in their respective periods. Consequently

$$I_1(\text{S}) \lt I_1(\text{Ar}).$$

Collecting every individual comparison we have written, we merge them into a single continuous inequality:

$$I_1(\text{Ba}) \lt I_1(\text{Ca}) \lt I_1(\text{Se}) \lt I_1(\text{S}) \lt I_1(\text{Ar}).$$

That sequence matches exactly the qualitative numerical data (in kJ mol−1): $$502 (\text{Ba}) \lt 590 (\text{Ca}) \lt 941 (\text{Se}) \lt 1000 (\text{S}) \lt 1521 (\text{Ar}).$$

Reading from left to right we obtain the required order of increasing first ionization enthalpy:

$$\text{Ba} \lt \text{Ca} \lt \text{Se} \lt \text{S} \lt \text{Ar}.$$

This ordering appears in option A.

Hence, the correct answer is Option A.