Which has better dipole moment?
Since fluorine is much more electronegative my presumption was that $ceCH3F$ should have actually a bigger dipole moment.
On the other hand because the angle in between the dipole moment is less in 3-nitro phenol than in 4-nitrophenol, my presumption was the 3-nitro phenol has actually a greater dipole moment. But this contradicts the answer given.
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edited might 22 at 7:58
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The dipole is the vector sum of each atoms"s partial fee times its distance from the origin. Lacking this details individual bond dipole values have the right to be used
In the very first case it is that of $ceCH3$ vs. The halogen, and also then i m sorry halogen has actually the higher electronegativity. Spring up some bond worths it seems that a $ceC-F$ bond has the slightly larger dipole ($1.6$ vs $1.5$ Debye ) for this reason the difference is tiny and not easy to predict as the C-F bond is shorter than the C-Cl.
In the second molecule, geometry deserve to be supplied as the two groups are the very same on each molecule however positioned differently. The para molecule will have the smaller dipole vs. The meta if both groups act in the same direction, one of two people both to retract or add electron density with respect come the ring, as then castle act come cancel one an additional to part extent.
If they action in a various direction then the para molecule will have actually the bigger dipole vs the meta. Looking for dipole moments it seems that phenol has a dipole the $1.45$D pointing into the ring and also nitrophenol $3.95$D the end of the ring so plot in different directions and also so the para should have actually the bigger dipole.
(To add two vectors of size b and c at an edge $ heta$ the result c is found using the cosine ascendancy $c^2 = a^2 + b^2 -2abcos( heta)$ . If the dipoles point in the contrary direction make either a or b negative)