Explanation:
The ionization energy is the energy required to remove an electron from an atom or ion. It is influenced by several factors such as the nuclear charge, the distance between the electron and the nucleus, and the electron configuration.

Comparing the Electronic Configurations:
a) Ne [3s23p2]
b) Ar [3d104s24p3]
c) Ne [3s23p1]
d) Ne [3s23p3]

Among these configurations, the electronic configuration with the highest ionization energy will have the highest effective nuclear charge (Zeff) and the smallest atomic radius.

Effective Nuclear Charge (Zeff):
The effective nuclear charge is the net positive charge experienced by an electron in an atom. It can be calculated using the formula Zeff = Z - S, where Z is the atomic number and S is the screening constant.

Atomic Radius:
The atomic radius is the distance between the nucleus and the outermost electron shell in an atom. It decreases as we move across a period from left to right in the periodic table.

Comparing the Options:
a) Ne [3s23p2]
- Zeff = 10 - 2 = 8
- Atomic Radius: Ne is the second element in the 2nd period, so its atomic radius is smaller than the elements in the 3rd period.

b) Ar [3d104s24p3]
- Zeff = 18 - 10 = 8
- Atomic Radius: Ar is the third element in the 3rd period, so its atomic radius is larger than Ne.

c) Ne [3s23p1]
- Zeff = 10 - 1 = 9
- Atomic Radius: Ne is the second element in the 2nd period, so its atomic radius is smaller than the elements in the 3rd period.

d) Ne [3s23p3]
- Zeff = 10 - 3 = 7
- Atomic Radius: Ne is the second element in the 2nd period, so its atomic radius is smaller than the elements in the 3rd period.

Conclusion:
Among the given options, the electronic configuration with the highest ionization energy is option d) Ne [3s23p3]. This is because it has the highest effective nuclear charge (Zeff = 7) and the smallest atomic radius among the options.

D