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Which compound has larger $\ce{C-N}$ bond length, 4-aminobenzaldehyde (1) or aniline (2)?

Compounds

I think it's the second compound. $\ce{-CHO}$ is an electron withdrawing group. This means the electron density of the ring decreases. So the carbon attached to nitrogen becomes more electronegative. Hence its p-character decreases and the $\ce{C-N}$ bond length decreases.

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You think right, but for the wrong reason. The inductive effect is not that great and does not quite reach that far. It is the effect of conjugated $\pi$-system that really matters. See, we may draw an alternative resonance structure for I with minus on O, plus on N, and rearranged double bonds (quinone-like ring instead of aromatic). This means the C-N bond in I is kind of "partially double" (i.e., has a noticeable $\pi$ component), hence it is shorter.

Resonance structures

Indeed, there is some conjugation between the aromatic ring and nitrogen's lone pair in unsubstituted aniline as well. That's what makes aniline a weaker base than ammonia, and its aromatic ring more reactive towards electrophiles than benzene. It is just that the $\ce{-CH=O}$ group in para position makes that conjugation stronger yet, as we may see qualitatively by drawing the resonance structures, or (supposedly) quantitatively by means of quantum chemistry.

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