1. Write the electronic configurations of the given second-period elements:

• $\mathrm{Li}: 1s^2 2s^1$
• $\mathrm{Be}: 1s^2 2s^2$
• $\mathrm{B}: 1s^2 2s^2 2p^1$
• $\mathrm{C}: 1s^2 2s^2 2p^2$
• $\mathrm{N}: 1s^2 2s^2 2p^3$
• $\mathrm{O}: 1s^2 2s^2 2p^4$
• $\mathrm{F}: 1s^2 2s^2 2p^5$

2. General trend of first ionization enthalpy across a period:

From left to right in a period, first ionization enthalpy generally increases because effective nuclear charge increases and atomic size decreases.

So the rough trend is: $\mathrm{Li} < \mathrm{Be} < \mathrm{B} < \mathrm{C} < \mathrm{N} < \mathrm{O} < \mathrm{F}$

But we must check the known exceptions.

3. Important exceptions:

(i) $\mathrm{Be}$ vs $\mathrm{B}$

• In $\mathrm{Be}$, the electron removed is from a filled $2s$ subshell: $2s^2$.
• In $\mathrm{B}$, the electron removed is from a higher-energy $2p$ subshell: $2p^1$.
• Since a $2p$ electron is easier to remove than a $2s$ electron, $\mathrm{B} < \mathrm{Be}$

(ii) $\mathrm{N}$ vs $\mathrm{O}$

• $\mathrm{N}$ has half-filled configuration: $2p^3$, which is especially stable.
• $\mathrm{O}$ has $2p^4$, so one orbital contains paired electrons, causing extra electron-electron repulsion.
• Therefore it is easier to remove an electron from $\mathrm{O}$ than from $\mathrm{N}$. So, $\mathrm{O} < \mathrm{N}$

4. Hence the correct increasing order of first ionization enthalpy is:

$\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$

5. Compare with options:

• A: $\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$ ❌ (wrong because $\mathrm{B}<\mathrm{Be}$)
• B: completely incorrect ❌
• C: $\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}<\mathrm{F}$ ❌ (wrong for both $\mathrm{Be}/\mathrm{B}$ and $\mathrm{N}/\mathrm{O}$)
• D: $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$ ✅

Therefore, the correct option is D.