Knowing this that the bond length is 1.39 Å and the dipole moment is 1.60 D? With fully negative charge on fluorine, the dipole moment = the charge of an electron x distance in cm between the atoms, or (4.8 x 10-10 esu) (1.39 x 10-8 cm) = 6.97 10- 18 esu cm = 6.97 D. Since the dipole moment is 1.60 D, we can calculate the partial negative charge on the fluorine atom: 1.60 / 6.97 = 0.23 This should be the answer. But how do you get that answer?
Answer
Hi Joana,
I get a slightly different answer, but of the same order of magnitude.
First we convert all units to the SI system:
A dipole moment is by definition charge X distance. So in the SI system the unit of dipole moment is C(oulomb) xm(eter). The charge of 1 electron (then we assume a monovalent negative fluorine atom) is 1.6022×10^(-19) C. The distance is 1.39 Angstroms or 1.39×10^(-10) m.
The dipole moment is then 2.227×10^(-29) Cm
We convert this to D(ebeye) : 1 D = 3.34×10^(-30) Cm
So our dipole moment is 6.67 D
The measured dipole moment is 1.60 D so the partial charge is 1.60/6.67 = 0.24 (that’s about right).
There may be some rounding errors here and there.
Answered by
Dr Etienne Jooken
Chemistry
Old Market 13 3000 Leuven
https://www.kuleuven.be/
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