The structure of the triiodide ion places a negative formal charge on the central iodine atom.
No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).
That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.
The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.
The structure of the triiodide ion places a negative formal charge on the central iodine atom.
No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).
That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.
The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.
is an electrically neutral entity, which would exclude the $\ce{I3-}$, even in Atkins definition. I made the same mistake though. For the rest of your statements I am happy to give you a thumbs up.More
@DavePhD All I am trying to say is that looking at the dipole moment for clarification is not enough. And I am saying that that classification does not add anything to the discussion anyway. Its too simple. Also, we might not want to have this discussion in the comment section. Maybe chat, but I am on the road tonight....More
@Martin if you define "polar" as having any multipole, you have to consider H-H, N-N, and every molecule as polar. "It is true, that for H2, N2 etc., wave mechanics, too, gives at least quadrupoles." Fritz London More
No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).
That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.
The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.
No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).
That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.
The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.
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