Step 1: List the successive ionization enthalpies.
The given values in kJ/mol are: $IE_1 = 1012$, $IE_2 = 1907$, $IE_3 = 2955$, $IE_4 = 4955$, $IE_5 = 6275$, $IE_6 = 21260$. Notice that while the values increase steadily from $IE_1$ to $IE_5$, there is a dramatic jump between $IE_5$ and $IE_6$.
Step 2: Interpret the sudden jump in ionization enthalpy.
A very large jump in successive ionization enthalpies indicates that we have removed all the valence electrons and the next electron must be pulled from a much more stable inner shell. Here, the big jump occurs after the 5th ionization, so the element has exactly 5 valence electrons.
Step 3: Identify the group from the number of valence electrons.
An element with 5 valence electrons belongs to Group 15 of the periodic table. The Group 15 elements are: N (Period 2), P (Period 3), As (Period 4), Sb (Period 5), Bi (Period 6).
Step 4: Match with the given options.
Among the options: Carbon (C) is in Group 14 (4 valence electrons), Phosphorus (P) is in Group 15 (5 valence electrons), Sulfur (S) is in Group 16 (6 valence electrons), Chlorine (Cl) is in Group 17 (7 valence electrons).
Step 5: Confirm by checking ionization data for phosphorus.
Phosphorus has the electronic configuration $[\text{Ne}] 3s^2 3p^3$. It has 5 valence electrons. The first five ionization enthalpies remove the valence electrons progressively, while the sixth requires breaking into the stable neon core, causing the massive jump in energy. This matches the data exactly.
Step 6: State the final answer.
The element $X$ is Phosphorus (P). \[ \boxed{X = \text{Phosphorus (P)}} \]