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secondary 4 | Chemistry
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Why is the answer B?
Date Posted: 2 years ago
Views: 210
Positive nuclei surrounded by a 'sea' of free mobile electrons.
Remember that the nuclei are fixed in their positions. Only the electrons move around in the orbitals.
To add on , as for graphite, all the carbon atoms are sp² hybridised. Each carbon atom is covalently bonded to 3 other carbon atoms adjacent to it. One electron in the unhybridised p-orbital of each C is delocalised across the layer each C is in. This is called π-conjugation.
We do not have positive ions in this case since graphite could be thought of as many sheets/layers of 'alternating benzene rings' (if you use Kekulé structures to represent them) with alternating double bonds
At any point of time, delocalisation is occurring. If an electron moves to the next carbon, it is being replaced by another. No carbcations can be isolated or formed.
As for metallic bonding or not,
I will leave a link here to a published research paper for further reading :
przyrbwn.icm.edu.pl/APP/PDF/112/a112z308.pdf
This is a chemistry question so descriptions like holes (which are more towards solid-state physics and) are not very apt to explain this.
Disclaimer :
Some people who don't know their concepts will jumble things up.
Also, they will only insist their conventions they learnt are right when other conventions/styles exist and are better to describe and explain certain concepts.
Being ignorant is okay. Being ignorant and brushing off comparable and alternative evidence as 'mistakes' is deplorable.
Furthermore, some people are hypocritical. Pretend to own up to their pitfalls but at other times hide their mistakes and subtly change their work, thinking that no one noticed.
Remember that the nuclei are fixed in their positions. Only the electrons move around in the orbitals.
To add on , as for graphite, all the carbon atoms are sp² hybridised. Each carbon atom is covalently bonded to 3 other carbon atoms adjacent to it. One electron in the unhybridised p-orbital of each C is delocalised across the layer each C is in. This is called π-conjugation.
We do not have positive ions in this case since graphite could be thought of as many sheets/layers of 'alternating benzene rings' (if you use Kekulé structures to represent them) with alternating double bonds
At any point of time, delocalisation is occurring. If an electron moves to the next carbon, it is being replaced by another. No carbcations can be isolated or formed.
As for metallic bonding or not,
I will leave a link here to a published research paper for further reading :
przyrbwn.icm.edu.pl/APP/PDF/112/a112z308.pdf
This is a chemistry question so descriptions like holes (which are more towards solid-state physics and) are not very apt to explain this.
Disclaimer :
Some people who don't know their concepts will jumble things up.
Also, they will only insist their conventions they learnt are right when other conventions/styles exist and are better to describe and explain certain concepts.
Being ignorant is okay. Being ignorant and brushing off comparable and alternative evidence as 'mistakes' is deplorable.
Furthermore, some people are hypocritical. Pretend to own up to their pitfalls but at other times hide their mistakes and subtly change their work, thinking that no one noticed.
Jonathan, updated and uploaded again. This version is correct. When mistake made, must admit... Sorry.
See 1 Answer
In my first solution, under option B, I had carelessly put "free electrons in a sea of positive ions.
Normally, because there are usually more free movable electrons than positive ions, so we like to describe the electrons, rather than the positive ions, as the 'sea'. So take note of that.
Normally, because there are usually more free movable electrons than positive ions, so we like to describe the electrons, rather than the positive ions, as the 'sea'. So take note of that.
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