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simpleton

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orbitals and bonds formation queries
« on: August 13, 2009, 06:24:46 AM »

I've some questions regarding organic chemistry. Would really appreciate if you could clarify my doubts.

(1) If one of the p-orbitals is filled with only 1 electron will it still be shaped like a dumb-bell? Does the # of electrons in the p-orbital affect the shape of the orbital or that all p-orbitals are shaped as a dumb-bell regardless of the # of electrons in each orbital?

(2) Does bond formation usually occur when there are unpaired electrons?

(3) How do we know there is hybridisation? Do we infer that there is hybridisation based purely on the geometry structures of the compounds?

(4) For the case of NH3, it is a stable compound (having an octet) then how NH4+ can be formed? Does the H bond to the lone pair?

(5) I understand that H has a proton and an electron, so it means H is a neutral atom? But why H is always refer as H+ is it because it being in the Group I hence it is very prompt to losing its electron hence, it exists mostly in an ionic form?

I thank you very much in advance.
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Yggdrasil

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Re: orbitals and bonds formation queries
« Reply #1 on: August 13, 2009, 08:38:28 AM »

I've some questions regarding organic chemistry. Would really appreciate if you could clarify my doubts.

(1) If one of the p-orbitals is filled with only 1 electron will it still be shaped like a dumb-bell? Does the # of electrons in the p-orbital affect the shape of the orbital or that all p-orbitals are shaped as a dumb-bell regardless of the # of electrons in each orbital?

The shape of the p-orbital is independent of the number of electrons in the orbital.  A single electron can occupy either lobe of the p-orbital and can pass between the lobes despite having a zero probability of existing between the lobes.  This is part of the quantum mechanical weirdness that governs the behavior of electrons.

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(2) Does bond formation usually occur when there are unpaired electrons?

Yes.  Species with unpaired electrons (called radicals) are usually very reactive and often react to form new bonds.  However, the converse is not true.  Bond formation usually does not proceed through the bonding of unpaired electrons.  Many times, a pair of electrons in the highest occupied molecular orbital (HOMO) of one compound will overlap with the lowest unoccupied molecular orbital (LUMO) of another compound, and this HOMO-LUMO overlap forms new molecular orbitals and a chemical bond.

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(3) How do we know there is hybridisation? Do we infer that there is hybridisation based purely on the geometry structures of the compounds?

I'm not so familiar with the evidence for hybridization, so I don't know the answer.  The geometry of most compounds is certainly one argument, but I'm not sure if this is the only evidence.

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(4) For the case of NH3, it is a stable compound (having an octet) then how NH4+ can be formed? Does the H bond to the lone pair?

The nitrogen atom in NH4+ still has a stable octet.  As you said, an the unoccupied s orbital of the H+ ion will overlap with the lone pair on the NH3 molecule, forming NH4+.

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(5) I understand that H has a proton and an electron, so it means H is a neutral atom? But why H is always refer as H+ is it because it being in the Group I hence it is very prompt to losing its electron hence, it exists mostly in an ionic form?

As mentioned before, radicals are very unstable and reactive because the unpaired electrons on radicals would like to find an electron with which to pair.  Thus a H atom with one unpaired electron is very unstable and will quickly react to pair its electron.  One such reaction is giving up the electron to become an H+ ion.  Alternatively, it can gain an electron to become a hydride (H-) ion.  Both these ions are more stable than a H radical because they contain no unpaired electrons.
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MrTeo

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Re: orbitals and bonds formation queries
« Reply #2 on: August 13, 2009, 08:42:40 AM »

They're simply the area of the space in which the probability (given by the square value of $$\psi /$$, which is the wave function of the electron - I'm referring to Schrödinger's equation - ) is bigger than 90%. Physics consider it quite a picturesque fact that chemists call these areas "orbitals" as they've got no physical reality and moreover they're not even linked with the beahviour of orbits. Heisenberg's principle tells us that we can't really know where the electron is and how fast is he moving (nay: that the indeterminacy of the first measure is inversely proportional to the indeterminacy of the second one, so that if we increase precision in one the other is less accurate) so all we get is a probability of presence, it's more likely to find electrons there than anywhere else. Here is a visual aid to make you understand that there's no fixed area (no dumb-bell) but only a distribution:

1) So answering your first question the number of electrons doesn't affect the shape of obitals and we could also say that orbitals as we know "exist" even if there are no electrons on that level (every atom has a number of orbitals, obviously we don't bother to think what would the behaviour of our electron be if there was an electron on level 4f in H)

2) Yeah, because naturally unpaired electrons make atoms less stable, this means more potential energy and a bonds decreases it. As in nature a system always tends to the less energetic configuration it's easy to understand that a bond will soon be formed with another atom or molecule. Just think at the high reactivity of the radicals (an easy example of unpaired electrons).

3) You don't know that there is hybridisation as there's no real hybridization in a bond. It's a way we use to think at the bonds and understand them according to the VB theory. If you use the MO theory you won't find any sp3 but bonding and anti-bonding orbitals. So I think we could say that hybridisation accords quite well to experimental evidence especially when you have to easily plot a molecule's geometry but it's not based "purely" on that. Most of the molecules have resonance forms and the hybridised one is just one of the possible; while in other cases you can only use hybridisation to justify the bonding of an atom with other atoms (think at Xe compounds).

I'll leave the other three to someone other or I'll post again later... see if these answers are clear of if you nedd some other clarification

Edit: I see Yggdrasill it's a turbo-poster Well, I hope my information are useful anyway...
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simpleton

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Re: orbitals and bonds formation queries
« Reply #3 on: August 13, 2009, 03:23:20 PM »

i got it now. THANKS A LOT TO Yggdrasil and Mr Teo!

These information are great and enriching. Been in deep thoughts quite a few days and can't get my answers online finally it's all solved. THANKS ONCE AGAIN!
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simpleton

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Re: orbitals and bonds formation queries
« Reply #4 on: August 13, 2009, 03:25:33 PM »

oh ya, one last clarification. so it means NH4+ is formed when H+ (without any electron) form a bond with N where it has the lone pair bond right? so H+ purely just share the lone pair bond with N to make itself stable?
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Yggdrasil

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Re: orbitals and bonds formation queries
« Reply #5 on: August 13, 2009, 05:21:56 PM »

oh ya, one last clarification. so it means NH4+ is formed when H+ (without any electron) form a bond with N where it has the lone pair bond right? so H+ purely just share the lone pair bond with N to make itself stable?

Correct.  The bond between the H+ and NH3 consists of the two electrons from the nitrogen's lone pair.
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simpleton

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Re: orbitals and bonds formation queries
« Reply #6 on: August 13, 2009, 05:48:25 PM »

i see. THANKS A LOT
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