Given below are two statements:
Statement I : Aniline is less basic than acetamide.
Statement II : In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton.
Choose the most appropriate option:

We begin by recalling the definition of basicity in organic molecules. A base is a species which can donate a lone pair of electrons to accept a proton ($$H^{+}$$). Hence, the more freely available the lone pair on nitrogen, the stronger is the basic character of the compound.

Now we consider the two compounds in the question:

(i) Acetamide : $$CH_{3}CONH_{2}$$

(ii) Aniline  : $$C_{6}H_{5}NH_{2}$$

For acetamide, the nitrogen lone pair interacts with the adjacent carbonyl group. We write the resonance structures first, stating the resonance principle: “If a lone pair on an atom is adjacent to a multiple bond, $$\pi$$-electron delocalisation can occur.” Using this rule we obtain

$$CH_{3}CONH_{2}\; \longleftrightarrow \; CH_{3}C^{(-)}=O^{(+)}NH_{2}$$

Because of this $$n \rightarrow \pi^{*}$$ conjugation, the lone pair is pulled toward the more electronegative oxygen. Thus it is very poorly available for protonation. Quantitatively, the conjugate acid $$(CH_{3}CONH_{3}^{+})$$ has a $$pK_{a}$$ around 0, showing extreme weakness of the base.

For aniline, the nitrogen lone pair is adjacent to an aromatic ring. Again applying the same resonance principle, we write the resonance forms:

$$C_{6}H_{5}NH_{2}\; \longleftrightarrow \; C_{6}H_{5}^{(-)} \!-\! NH_{2}^{(+)}$$

symbolically extended over the ring as

$$\begin{aligned} &\; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; N \!H_{2}\\ C_{6}H_{5}NH_{2}&\; \longleftrightarrow \; C_{6}H_{5}^{(-)}\!=\!NH_{2}^{(+)} \end{aligned}$$

Thus the nitrogen lone pair is delocalised into the benzene ring, reducing its electron density and consequently its tendency to accept a proton. Nevertheless, the withdrawal in aniline is less severe than in acetamide, because in acetamide the electron-withdrawing carbonyl oxygen is far more electronegative than any carbon in the aromatic ring.

Therefore the experimentally observed order of basicity is

$$\text{acetamide} \; < \; \text{aniline}$$

or, equivalently,

$$\text{aniline is more basic than acetamide.}$$

This conclusion directly contradicts Statement I, which claims “Aniline is less basic than acetamide.” Hence Statement I is false.

Statement II says, “In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton.” The resonance structures shown above confirm exactly this reasoning, so Statement II is true.

We have now established:

$$\text{Statement I : false}, \qquad \text{Statement II : true}$$

Among the given options, the description “Statement I is false but statement II is true” corresponds to Option B.

Hence, the correct answer is Option B.