The energy required to remove an electron from the outermost shell of an isolate gaseous atom is known as `IE_(1)` of that atom. Similarly, the enrgy required for the removal of the electron from the unipositive ion, diapositive ion and tripositive ion are known as `IE_(2),IE_(3)` and `IE_(4)` respectively, and are called successive ionisation energies. The magnitude of the charge depends on the size of the orbital of electron. Electrons in smaller orbitals are on average close with each other and have more repulsion. Thus for `Be(2s^(2))`, the `IE_(1)` and `IE_(2)` are `9.3` and `18.2 eV "atom"^(1)`, whereas for `Ca(4s^(2))`, the values are `6.1` and `11.9 eV`.
The relationship between `IE_(4)` and `IE_(2)` of an element is

Explanation

To determine the relationship between `IE_(4)` and `IE_(2)` of an element, we need to understand the concept of successive ionization energies and how they relate to the atomic structure of the element.

### Step-by-Step Solution:

1. **Understanding Ionization Energies**:
- Ionization energy (IE) is the energy required to remove an electron from an atom or ion.
- `IE_(1)` is the energy required to remove the first electron from a neutral atom.
- `IE_(2)` is the energy required to remove the second electron from a unipositive ion (after the first electron has been removed).
- `IE_(3)` and `IE_(4)` follow similarly for dipositive and tripositive ions, respectively.

2. **Trends in Ionization Energies**:
- As electrons are removed from an atom or ion, the effective nuclear charge experienced by the remaining electrons increases. This is because there are fewer electrons to shield the positive charge of the nucleus.
- The increase in effective nuclear charge leads to a stronger attraction between the nucleus and the remaining electrons, making it more difficult to remove additional electrons.

3. **Comparing `IE_(2)` and `IE_(4)`**:
- Since `IE_(2)` refers to the removal of the second electron from a unipositive ion, and `IE_(4)` refers to the removal of the fourth electron from a tripositive ion, we can analyze the trends:
- After the removal of the first electron, the remaining electrons experience a higher effective nuclear charge.
- This trend continues as more electrons are removed, leading to an increase in ionization energy values.

4. **Conclusion**:
- Based on the trends and the explanation provided, we can conclude that `IE_(4)` will always be greater than `IE_(2)` for any element. This is due to the increased effective nuclear charge and decreased size of the ion as electrons are removed.

Thus, the relationship is:
- **`IE_(4) > IE_(2)`**

### Final Answer:
- The correct option is **Option A: `IE_(4)` is greater than `IE_(2)`**.