10
$\begingroup$

My first thought was there would be a dipole moment as chlorine and bromine are unequal in electronegativity. I have since looked up the electronegativity of both chlorine and bromine through the Pauling scale. The difference between the electronegativity is 0.20 according to the table. For reference, the difference in electronegativity between carbon and hydrogen is 0.25 on the same table.

Since this is the case, I am very confused about the dipole moment of this molecule. Since the carbon-hydrogen bond is so negligible, why should this difference be significant?

To add to my confusion I found a source stating Dipole moment = 0. And Wikipedia only states: Dipole moment = D.

However, my textbook questions lists it as having a dipole moment with the simple explanation that Br and Cl have differing electronegativities (question 1.61).

$\endgroup$
  • 1
    $\begingroup$ In my thought the dipole moment of 1-bromo-4-chlorobenzene is 0, there is not that much difference in the electronegativity of Chlorine(3.0) and bromine (2.8). $\endgroup$ – Khan Sep 23 '16 at 5:59
  • 4
    $\begingroup$ Well, the correct answer to the way you phrased your question is: of course it does have a dipole moment. Whether the dipole moment is significant enough for us to consider the molecule to be polar, is a different matter entirely. But that is not the question asked. The question is "does it have a dipole moment" and the answer is "yes". $\endgroup$ – orthocresol Sep 23 '16 at 6:08
  • $\begingroup$ Sorry for the confusion, I meant a permanent dipole moment. I will adjust the question to clarify this. $\endgroup$ – R. Mauban Sep 23 '16 at 6:21
  • 3
    $\begingroup$ There was no confusion at all; when you say "dipole moment", you do mean "permanent dipole moment", that's the default assumption. Yes, by the reasons listed above this molecule does have some dipole moment, but I would not be surprised if it is 0.00 when rounded to 2 decimal digits. $\endgroup$ – Ivan Neretin Sep 23 '16 at 7:42
8
$\begingroup$

As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places. Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$. \begin{array}{lr}\hline \text{Method} & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline \text{BP86/def2-SVP} & 0.07\\ \text{BP86/def2-TZVPP} & 0.11\\ \text{B3LYP/def2-TZVPP} & 0.09\\ \text{B3LYP+D3(BJ)/def2-TZVPP} & 0.09\\ \text{PBE0/def2-TZVPP} & 0.10\\ \text{M11/def2-TZVPP} & 0.10\\ \text{B2PLYP/def2-TZVPP} & 0.09\\ \text{MP2/def2-TZVPP} & 0.06\\ %\text{CCD//MP2/def2-TZVPP} & ----\\ \hline\end{array}

In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:

NBO charges CCSD(T)

We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.

 Normalization factor of the integral of electron density is    1.000001
 The atomic charges after normalization and atomic volumes:
      1 (C )    Charge:    0.071729     Volume:    68.316 Bohr^3
      2 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      3 (C )    Charge:    0.088508     Volume:    78.816 Bohr^3
      4 (C )    Charge:   -0.122884     Volume:    73.483 Bohr^3
      5 (C )    Charge:    0.088513     Volume:    78.816 Bohr^3
      6 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      7 (H )    Charge:    0.026701     Volume:    46.501 Bohr^3
      8 (H )    Charge:    0.025890     Volume:    46.475 Bohr^3
      9 (Br)    Charge:   -0.097472     Volume:   260.461 Bohr^3
     10 (H )    Charge:    0.025886     Volume:    46.475 Bohr^3
     11 (H )    Charge:    0.026702     Volume:    46.501 Bohr^3
     12 (Cl)    Charge:   -0.290521     Volume:   214.364 Bohr^3

However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine. This causes a larger charge difference than the polarisation of the bonding structure.

QTAIM MP2

Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.

$\endgroup$
0
$\begingroup$

There are two aspects to dipole moments: The first is the amount of charge transfer between the linked atoms - this is determined by the electronegativity difference. The second is the bond length. So, the greater difference in electronegativity between Cl and C vz Br and C is partially compensated by the longer Br-C bond vz Cl-C.

Many dipole moment are determined in solution and between different solvents they may differ by more than 0.2 units. I.e. 0 < μ < 0.2

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.