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My task is to arrange the bonds from longest to shortest. Although I have the answer (C,D,B,A), I'm having a very hard time understanding how to derive this answer. Any clues would be helpful, especially those that help me in other related problems.

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    $\begingroup$ From periodic table, the atomic radius $r_C>r_N>r_F$. Therefore C>B,D>A. From conjugation effect, we expect nitrogen in B and D are sp2 and sp3. The more s, the shorter. Therefore D>B. $\endgroup$
    – user26143
    Dec 13 '13 at 4:31
  • $\begingroup$ I have my final exam tomorrow and this was on a previous exam. So it's not homework, but I really need to understand the "why" behind the problem. $\endgroup$
    – Billy Joel
    Dec 13 '13 at 4:35
  • $\begingroup$ Is my explanation reasonable? $\endgroup$
    – user26143
    Dec 13 '13 at 4:35
  • $\begingroup$ I'm not sure about the "conjugation effect" part, do you mind clearing that up? $\endgroup$
    – Billy Joel
    Dec 13 '13 at 4:42
  • $\begingroup$ If the orbitals in nitrogen (the near benzene one) and benzene are coplanar, they will conjugate each other. If the nitrogen adopts sp2 hybridazation for s and two p, one p left. That p will be coplanar with the benzene ring. Is this explanation reasonable? $\endgroup$
    – user26143
    Dec 13 '13 at 4:46
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Because atomic radii of elements tend to decrease across a period (source), and bond length increases with atomic radius, the bond lengths are expected to decrease in the following order: C (C-C bond) > B, D (C-N) > A (C-F).

The difference between B and D is that in D, the nitrogen is $sp^3$-hybridized, while in B, it is $sp^2$-hybridized due to conjugation of its lone pair with the aromatic ring 2. This results in partial double bond character of the C(ring 2)-N bond, similar to aniline:

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More $s$ character leads to shorter bonds, and therefore the C-N bond B is shorter than D.

In summary, the bond lengths decrease in the following order: C > D > B > A.

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