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There can be many resonating forms for a compound. It is to be noted that they are not in any equilibrium to each other. Each and every structures will contribute to the ultimate hybrid, according to their stability. The most important point is that resonating hybrid is not a mixture of the resonating structures, but rather an intermediate of them.

This is what my book writes. But if it is intermediate, it must contain at least some properties of each and every structure. Now, one structure has no charge, while other structure has one $+ve$ charge & one $-ve$ charge . In the final structure, it still contains both the charges. So, how can it be an intermediate as it got no contribution from the first?? Confused. Please help.

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  • $\begingroup$ What molecule are you referring to? $\endgroup$ – RBW Dec 1 '14 at 20:12
  • $\begingroup$ Brief answer about charge - if the molecule has no net formal charge, then it will always have no net formal charge. Every resonance structure must always have no net formal charge. However, individual atoms may have partial formal charges, and the net partial charge on an atom will be the weighted average over the formal and partial charges on that atom in each resonance contributor. $\endgroup$ – Ben Norris Dec 1 '14 at 20:14
  • $\begingroup$ @Marko: I asked the quo thinking about phenol. Phenol has 5 resonating str. 3 of them contain $+$ & $-$ charges;the other being neutral. In the hybrid also, charges remain present. So, if the hybrid is intermediate, what does it take from the two neutral strs.? $\endgroup$ – user5764 Dec 2 '14 at 1:53
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You can understand the relationship between the resonating forms for a compound and the ultimate hybrid by comparing it with the relationship between the different projections of a vector on the three coordinate axes and the vector itself. None of the projection can lonely represent the vector. But all of them can define the vector precisely in the space. Each projection contributes in defining the modulus of the vector according to its own value.

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  • $\begingroup$ That's a great analogy. Take an example: phenol. It has 5 resonating structures. 3 of them are charged; the other 2 being neutral. Now, like the original vector, the hybrid must be contributed by the 5 strs. But the hybrid is also charged. So, how can it be an intermediate, since it has no contribution from the 2 neutral strs?? $\endgroup$ – user5764 Dec 2 '14 at 2:00
  • $\begingroup$ And neutral str. is more stable than dipolar str. So, why don't the more stable neutral structures contribute to the hybrid of phenol?? $\endgroup$ – user5764 Dec 2 '14 at 2:27
  • $\begingroup$ First of all the the neutral stru. is not more stable than dipolar str. (phenol is a slightly polar molecule). Secondly, you have three charged resonating structures counter two uncharged. So, it's normal that the charged str. weighed more in the hybrid of phenol. You should also remember, that I've given you an analogy to explain the mesmerism. So, it has its own limits. It cannot explain everything. $\endgroup$ – Yomen Atassi Dec 2 '14 at 7:47
  • $\begingroup$ Sir, firstly, your analogy was great & has no flaw. Actually both the neutral ones contribute as the hybrid contains only partial charge,not whole charge. Now, regarding neutral being more stable, my book... $\endgroup$ – user5764 Dec 2 '14 at 8:24
  • $\begingroup$ ...cites an example: there are two resonanting strs of 1,3 di pentene . First,neutral and second the polar. Then the book says, the neutral one is more stable. $\endgroup$ – user5764 Dec 2 '14 at 8:27

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