The electronic configuration with the highest ionization enthalpy is:

We need to find which electronic configuration has the highest ionization enthalpy among the given options.

Let us first identify each element:

Option A: $$[Ar]\, 3d^{10}\, 4s^2\, 4p^3$$ — This is Arsenic (As, Z = 33)

Option B: $$[Ne]\, 3s^2\, 3p^1$$ — This is Aluminium (Al, Z = 13)

Option C: $$[Ne]\, 3s^2\, 3p^3$$ — This is Phosphorus (P, Z = 15)

Option D: $$[Ne]\, 3s^2\, 3p^2$$ — This is Silicon (Si, Z = 14)

Ionization enthalpy depends on several factors:

1. Nuclear charge: Higher nuclear charge means stronger attraction, so higher ionization enthalpy.

2. Atomic size: Smaller atom means electrons are closer to nucleus, so higher ionization enthalpy.

3. Shielding effect: More inner shells mean more shielding, so lower ionization enthalpy.

4. Stability of electronic configuration: Half-filled and fully-filled subshells have extra stability, making it harder to remove an electron.

All options B, C, and D are in the third period ($$[Ne]$$ core), while option A (Arsenic) is in the fourth period with additional $$3d^{10}$$ shielding. Being in a higher period with more shielding, Arsenic has a lower ionization enthalpy than the third-period elements with half-filled p orbitals.

Among the third period elements (B, C, D):

Aluminium (B): $$3s^2 3p^1$$ — The lone $$3p$$ electron is easy to remove due to the higher energy of the $$p$$ subshell compared to $$s$$. Ionization enthalpy is relatively low.

Silicon (D): $$3s^2 3p^2$$ — Has higher nuclear charge than Al, so higher ionization enthalpy than Al.

Phosphorus (C): $$3s^2 3p^3$$ — Has a half-filled $$3p$$ subshell, which gives extra stability. The nuclear charge is also higher than Si. This makes its ionization enthalpy the highest among the given options.

The known ionization enthalpies confirm this:

Al = 577 kJ/mol, Si = 786 kJ/mol, P = 1012 kJ/mol, As = 947 kJ/mol

Phosphorus has the highest ionization enthalpy due to the extra stability of its half-filled $$3p^3$$ configuration.

Therefore, the answer is $$[Ne]\, 3s^2\, 3p^3$$, which is Option C.