The ionization energy will be higher when the electron is removed from .........if other factors being equal.

Options

A. `s`-orbital
B. `p`-orbital
C. `d-`orbital
D. `f`-orbital

Explanation

To determine which orbital will have the highest ionization energy when an electron is removed, we need to consider the following factors: the energy of the orbitals and their penetration power.

### Step-by-Step Solution:

1. **Understanding Ionization Energy**: Ionization energy is the energy required to remove an electron from an atom in the gas phase. The higher the ionization energy, the more difficult it is to remove an electron.

2. **Energy of Orbitals**: The energy levels of the orbitals can be ranked as follows:
- S < P < D < F
This means that the S orbital has the lowest energy, while the F orbital has the highest energy.

3. **Penetration Power**: Penetration power refers to how close an electron can get to the nucleus. The closer an electron is to the nucleus, the stronger the nuclear attraction it experiences. The order of penetration power is:
- S > P > D > F
This indicates that S orbitals have the highest penetration power, followed by P, D, and F.

4. **Attraction to the Nucleus**: Since the S orbital has the highest penetration power, electrons in the S orbital are held more tightly by the nucleus. This means that it requires more energy to remove an electron from an S orbital compared to electrons in P, D, or F orbitals.

5. **Conclusion**: Therefore, if all other factors are equal, the ionization energy will be highest when the electron is removed from the S orbital.

### Final Answer:
The ionization energy will be higher when the electron is removed from the S orbital.