Consider the following question: Why are there free electrons in a metal?

The answer suggests that conductors have free electrons because the atoms have a weak attraction on the electrons in the outer shell:

Compared to the s and p orbitals at a particular energy level, electrons in the d shell are in a relatively high energy state, and by that token they have a relatively "loose" connection with their parent atom; it doesn't take much additional energy for these electrons to be ejected from one atom and go zooming through the material,

Conversely, I'm assuming insulators have strong connections with the outer shell electrons. Since insulator atoms "want" electrons and conducting atoms "don't" why aren't all conductors charged since there always in contact with air (an insulator)?

If the insulator atoms want electrons and the conductor atoms don't, why doesn't the insulator pull the electrons away from the conductor? Why doesn't air pull the electrons away from conductors?

Please answer in layman's terms without using concepts that a high school freshman wouldn't understand.

  • $\begingroup$ Why are you assuming that insulators "want" electrons ? They already have their own .... $\endgroup$ – mannaia Feb 5 '14 at 7:42
  • 1
    $\begingroup$ Rather than insulators want electrons, a more accurate statement would be that Insulators are reluctant of giving away their electrons. $\endgroup$ – Satwik Pasani Feb 5 '14 at 8:51
  • $\begingroup$ @SatwikPasani How is that different? I thought the reason insulators are reluctant to give away electrons is because they are more stable when they have more electrons since they need electrons to complete their valence shell. And if they stabler with more electrons why wouldn't they want electrons? For example consider oxygen. It wants 2 electrons to complete its shell so its reluctant to give away electrons since that would make it more unstable but it should also want electrons to complete its shell (since adding 2 or less electrons brings it closer to completing its shell). $\endgroup$ – dfg Feb 5 '14 at 16:42
  • $\begingroup$ @mannaia To complete their valence shell $\endgroup$ – dfg Feb 5 '14 at 16:42

You have a few questions and a few misunderstandings in the way.

From the comments:

if they ["electronegative" atoms like Oxygen] [are] stabler with more electrons why wouldn't they want electrons?

They aren't. If they were, their ground state would be with those electrons attached to them. However, a bond between two or more atoms might be more stable than each of the atoms in their ground state. Oxygen sharing two extra electrons with another atom might be more stable than Oxygen alone. "$\ce{O^{-2}}$" has unbalanced charges, it isn't stable at all (unless bonded with another atom).

why aren't all conductors charged since there always in contact with air (an insulator)?

Because an uncharged insulator doesn't have any charge to charge an uncharged conductor.

However, as I understand it, your question is Why doesn't Oxygen steal the conductor's (free) electrons since it wants it to "fill it's outer shell", thus charging it?. You're confusing things charging each other and things reacting with each other.

Even if you had Oxygen to "steal the conductor's electrons", it would react with it, it'd form a bond, not just run away with the electrons. As mentioned above, it's not that $\ce{O^{-2}}$ is more stable than $\ce{O}$, it isn't. But a molecule of $\ce{O-Something}$ might be.

But you don't even have Oxygen in the first place. In our atmosphere, Oxygen is in the form of $\ce{O2(g)}$, which doesn't "want" two electrons to "complete it's outer shell". And air is mostly $\ce{N2(g)}$, actually.

However, some metals will react with $\ce{O2}$ at room temperature, "giving away" their electrons in the bond. Aluminum, for instance, will react to form Aluminum Oxide. I suggest you read about activation energy and chemical bond.

Why do free electrons exist?

You quoted the answer. If I'm right your question wasn't really about free electrons, but how weren't they attracted by "atoms that want to complete their outer shell".

  • $\begingroup$ Oh I see, that makes sense. Thanks a lot, appreciate the detail! $\endgroup$ – dfg Feb 9 '14 at 1:02

Refer to this link: http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles/Metallic_Bonding

So the outer d-orbitals (or later on in the periodic table f-orbitals) interact with each other to give new molecular orbitals, a bit like in covalent compounds like methane, except in this case there are loads and loads of them, and they are less directional and delocalised. This is in effect the picture you have given of "free" electrons, because the electrons are delocalised. This is what you have in metals (metallic bonding) and graphite, where there is a delocalised set of electrons which are easy to "move" by applying a voltage. That is a conductor.

In an insulator, like plastic - covalently bonded the electrons are either like the 1s (not involved in bonding and bound to the atoms) or are localised into molecular orbitals that are across only a few atoms (even the delocalised ring in benzene is localised to just 6 atoms). Here the electrons are in MO's and are not free to move due to voltage. These are made normally from s and p orbitals (or hybrids...), where the atom has a stronger attraction to them than in metals to the d-orbitals

This may be helpful/interesting: http://www.whfreeman.com/college/pdfs/raynercanhampdfs/RaynerCanhamch04.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.