# How can I generally determine which bonds are longest/shortest in a molecule?

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.

• 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. Dec 13 '13 at 4:31
• 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. Dec 13 '13 at 4:35
• Is my explanation reasonable? Dec 13 '13 at 4:35
• I'm not sure about the "conjugation effect" part, do you mind clearing that up? Dec 13 '13 at 4:42
• 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? Dec 13 '13 at 4:46

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