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Why is $\mu=0$ for p-dinitrobenzene ? How can the dipole moments of the two Nitro groups exactly cancel each other ? Shouldn't it have a non-zero, albeit feeble, dipole moment ?

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    $\begingroup$ They exactly cancel each other because they are exactly similar. $\endgroup$ – Ivan Neretin Feb 10 '16 at 7:21
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    $\begingroup$ Why would it be non-zero? The molecule is planar and the nitro groups are symmetric. $\endgroup$ – bon Feb 10 '16 at 10:09
  • $\begingroup$ @bon So both the N-O bonds in a nitro group are of exactly equal length. $\endgroup$ – Abhirikshma Feb 10 '16 at 13:02
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    $\begingroup$ @Abhirikshma Yes that is correct. $\endgroup$ – bon Feb 10 '16 at 13:03
  • $\begingroup$ The nitro groups most probably undergo free rotation and so on average are symmetrically disposed about the molecule, so no dipole. If there was some restricted rotational motion, due to say interaction with adjacent H atoms then the dipole could be very small but still approximate to zero. $\endgroup$ – porphyrin Jan 13 '17 at 9:44
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This paper (J. Sep. Sci. 2011, 34, 1489–1502) lists 1,4-Dinitrobenzene as having a very small dipole moment (0.01 Debye). No error is given on the dipole moments, but they do list compounds with exactly zero dipole moment (naphthalene for example).

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Let the two nitro groups have a different rotational excitation against the aryl core, and i'm sure you can construct a small dipole moment. Firstly, higher rotational states have different bond angles.

Or easier to imagine: Let one stand in plane with the aryl ring, and one perpendicular or rotating. Different interaction with the aromatic ring, clearly leads to slightly different electronic structure, ergo: a dipole moment.

The dipole moment should vanish at low temperatures, but i guess the stuff crystallises before.

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