Why NCl3 is dipole-dipole in intermolecular force?

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3.0 is a rounded value, they are not exactly the same. The small difference is still more important than the weaker van-der-Waals forces.
This post suggests a notable difference between the electronegativity.
 
mfb said:
3.0 is a rounded value, they are not exactly the same. The small difference is still more important than the weaker van-der-Waals forces.
This post suggests a notable difference between the electronegativity.
But, according to https://en.wikipedia.org/wiki/Chemical_polarity ,
  • Polar bonds occur when the difference in electronegativity between the two atoms is between 0.4 and 1.7
In that post you suggested, the electronegativity difference is just 0.2 , still not enough to perform a polar bond..
 
There is no sharp line dividing polar and nonpolar. Some bonds are more polar than others.
In the same way, there is no sharp line dividing polar and ionic bonds.
 
mfb said:
There is no sharp line dividing polar and nonpolar. Some bonds are more polar than others.
In the same way, there is no sharp line dividing polar and ionic bonds.

If there is no sharp line dividing polar and non-polar, how to determine if such bond is polar?
By memorizing only??
 
I don't see no reason why NCl3, even with perfectly covalent bonds, should not have a permanent dipole moment (which we know from experimental measurements to be 0.6 D). It is a question of the asymmetry of the electronic charge distribution around the nuclei. The lone pair on N is especially important in this respect.
 
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terryds said:
If there is no sharp line dividing polar and non-polar, how to determine if such bond is polar?
By memorizing only??
Every bond between atoms of different types is a bit polar. You can set an arbitrary threshold for saying "this is sufficient to call it polar", which can be convenient sometimes, but you do not have to.