Concept

If resonance energy is more that means the means resonance structures have varying energies and it should be less stable aka more energy ?

So more resonance energy = More Stability ?

Problem 8.16

 

Does the electron movement be of 1 electron and not a pair.

C needs 4 more electrons the 2 hooked arrow shows C is getting 1 extra electron.

Problem 8.17

1. Resonance structure needs to have same no. of unpaired electrons.

2. Why charges develop of II

3. Why is O in III +ve

4. Explain “does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable].”

Problem 8.18

In II why would O develop +ve charge

Explain the answer

Electrometric Effect

Book says transfer to Pie bond but pie bond atom always looses

Edited October 7Oct 7 by HbWhi5F

58 minutes ago, HbWhi5F said:

Problem 8.16

 

Does the electron movement be of 1 electron and not a pair.

An arrow represents the movement of a pair of electrons. A fish hook is used to represent the movement of a single electron.

1 hour ago, HbWhi5F said:

Concept

If resonance energy is more that means the means resonance structures have varying energies and it should be less stable aka more energy ?

So more resonance energy = More Stability ?

Problem 8.16

 

Does the electron movement be of 1 electron and not a pair.

Yes the concept of "resonance energy" indicates the amount by which the molecule is actually more stable than would be expected from any one of the individual canonical structures.

It is important to bear in mind though that "resonance" in this context is fictitious, i.e. nothing physically "resonates". Electron pairs do not flicker back and forth between canonical structures as shown by the arrows. What the concept conveys is that the real electronic structure of the molecule is a mixture of the canonical structures. It signifies that there are one or more molecular orbitals that extend across more of the molecule than simple 2-centre bonding implies, i.e. some of the the electrons are "delocalised" across several atoms. It is these more extended, lower energy orbitals that lead to greater stability than the simple canonical structures would predict. When the canonical structures are of equal energy the mixture will be 50:50. If one of them is significantly higher in energy than the other, it will contribute less to the mixture and the energy lowering effect will be less. There is a fuller explanation here: https://en.wikipedia.org/wiki/Resonance_(chemistry)

The arrows show movement of pairs of electrons, whether lone pairs or bonding pairs. As @KJW says, you get hooked arrows (with a half head on) to show movement of single electrons. Generally you will come across these in the context of free radical reactions, i.e. where unpaired electrons are involved.

17 hours ago, HbWhi5F said:

Problem 8.18

In II why would O develop +ve charge

Ask yourself how many electrons are in the valence shell of a neutral, free oxygen atom. And then count the electrons on the oxygen atom you are talking about: 3 lone pairs, plus a half share in the pair in the bond.

On 10/8/2025 at 2:52 PM, exchemist said:

Ask yourself how many electrons are in the valence shell of a neutral, free oxygen atom. And then count the electrons on the oxygen atom you are talking about: 3 lone pairs, plus a half share in the pair in the bond.

-ve O has 8 electrons so it's -2. +O has 8 electrons too and why it's negative.

1 hour ago, HbWhi5F said:

-ve O has 8 electrons so it's -2. +O has 8 electrons too and why it's negative.

OK, no, I asked you how many electrons there are in the valence shell of a neutral free oxygen atom. How many is that?

I now realise in my first reply I mistakenly spoke about the -ve O atom, not the +ve one you actually asked about, so apologies for confusing you. However the same principle applies. For working out electric charge you count 100% of the electrons in the lone pairs, as they are located 100% on the O atom, and you count 50% of the electrons in the bonds, as these are shared (almost) equally between the O atom and the atoms it is bonded to. Then you compare that with the number of valence electrons there are in a free neutral atom and see if you have the same number, or more, or fewer.

So the +ve O atom has one lone pair plus a half share of 3 bonding pairs right? How many is that in total?

Edited October 9Oct 9 by exchemist