1. Identify the elements from the electronic configurations

• A: $[\mathrm{Ne}]\,3s^2 3p^1$ corresponds to Aluminium (Al)
• B: $[\mathrm{Ne}]\,3s^2 3p^2$ corresponds to Silicon (Si)
• C: $[\mathrm{Ne}]\,3s^2 3p^3$ corresponds to Phosphorus (P)
• D: $[\mathrm{Ar}]\,3d^{10}4s^2 4p^3$ corresponds to Arsenic (As)

2. Recall the trend of ionization enthalpy

Ionization enthalpy generally:

• increases across a period from left to right,
• decreases down a group due to increase in atomic size and shielding.

Also, a half-filled subshell has extra stability.

3. Compare the given configurations

Option A: $[\mathrm{Ne}]\,3s^2 3p^1$ (Al)

• Relatively lower ionization enthalpy.
• First electron removed from a $3p$ electron.

Option B: $[\mathrm{Ne}]\,3s^2 3p^2$ (Si)

• Higher than Al because nuclear charge increases across the period.

Option C: $[\mathrm{Ne}]\,3s^2 3p^3$ (P)

• This has a half-filled $3p^3$ configuration, which is especially stable.
• Therefore, removing one electron requires extra energy.
• So its ionization enthalpy is higher than Si and Al.

Option D: $[\mathrm{Ar}]\,3d^{10}4s^2 4p^3$ (As)

• Also has a half-filled $p^3$ configuration.
• But As lies below P in the same group.
• Ionization enthalpy decreases down the group because of larger size and greater shielding.
• Therefore, As has lower ionization enthalpy than P.

4. Conclusion

Among all the given configurations, the one with the highest ionization enthalpy is

$[\mathrm{Ne}]\,3s^2 3p^3$

So the correct option is C.

5. Comparison with stored correct answer

• Derived answer: C
• Stored correct answer: C

They match.