Identify the elements and their positions

• $\mathrm{B}$ = Boron, period 2, group 13
• $\mathrm{P}$ = Phosphorus, period 3, group 15
• $\mathrm{S}$ = Sulfur, period 3, group 16
• $\mathrm{F}$ = Fluorine, period 2, group 17

Recall the periodic trend for first ionization enthalpy

First ionization enthalpy generally:

• increases across a period from left to right,
• decreases down a group.

So we expect:

• $\mathrm{F}$ to have the highest ionization enthalpy among these.
• $\mathrm{B}$ to have the lowest among these, since it is relatively left in period 2 and has a smaller ionization enthalpy than $\mathrm{P}, \mathrm{S}, \mathrm{F}$.

Compare $\mathrm{P}$ and $\mathrm{S}$ carefully

This is the important exception.

Electronic configurations:

• $\mathrm{P}: [\mathrm{Ne}]\,3s^2 3p^3$
• $\mathrm{S}: [\mathrm{Ne}]\,3s^2 3p^4$

Phosphorus has a half-filled $3p^3$ configuration, which is extra stable. Sulfur has one paired electron in the $3p$ subshell, causing extra electron-electron repulsion.

Therefore, it is easier to remove an electron from sulfur than from phosphorus, so: $I_1(\mathrm{S}) < I_1(\mathrm{P})$

Construct the increasing order

Combining all trends: $\mathrm{B} < \mathrm{S} < \mathrm{P} < \mathrm{F}$

Match with the options

• A: $\mathrm{F} < \mathrm{S} < \mathrm{P} < \mathrm{B}$ ❌
• B: $\mathrm{P} < \mathrm{S} < \mathrm{B} < \mathrm{F}$ ❌
• C: $\mathrm{B} < \mathrm{P} < \mathrm{S} < \mathrm{F}$ ❌
• D: $\mathrm{B} < \mathrm{S} < \mathrm{P} < \mathrm{F}$ ✅

Final answer

The correct increasing order of first ionization enthalpy is: $\boxed{\mathrm{B} < \mathrm{S} < \mathrm{P} < \mathrm{F}}$ Hence, Option D is correct.