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The ionic resonance of the bonds are given as follows:

$$ \begin{array}{cr} \ce{A-B} & 24.3 \\ \ce{A-C} & 50.6 \\ \ce{A-D} & 102.3 \\ \ce{A-E} & 105.9 \end{array} $$

Which of the bonds is the most polar?

a) $\ce{A-B}$
b) $\ce{A-C}$
c) $\ce{A-D}$
d) $\ce{A-E}$

I thought that more the resonance energy, more is the electron delocalisation and less polar is the bond, but this is not the case with the answer given. I was getting a), but the answer is d).

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  • $\begingroup$ Ionic resonance means including contributions to the structure from species such as $A^+B^- $ or $A^-B^+$ in a molecule of type AB. In this case (d) would seem to be correct. $\endgroup$ – porphyrin May 2 at 18:03
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I'm not too familiar with resonance theory, but I'll try:

According to modern VBT, the general idea is that the state of the molecule is not described by a single state wave function, but rather a linear combination ("superposition" if you will) of multiple different state wave functions. In your particular case, the resonance is between a "dominant" covalent structure (involving a completely equal sharing of electron density) and a "minor contributor" ionic structure (with unequal electron density).

Since resonance energy is defined as the amount of energy needed to convert the true delocalized structure into that of the most stable contributing structure (which can be interpreted as the amount of stabilisation provided by the linear combination of the state wave functions), a higher resonance energy would mean a greater contribution (and hence weighing coefficient) from the ionic structure. This means that there is a greater ionic character to the bond (and it is more polar).

Hope this helped.

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