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Topic: VSEPR/Hybirdization/diagrams *delete me*!  (Read 14065 times)

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ROxy

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VSEPR/Hybirdization/diagrams *delete me*!
« on: November 22, 2004, 06:29:13 PM »
Sorry this is so long but I don't get any of it.

For these 2 molecules:
C2H4
NH2OH

1. Using VSEPR, how to u predict the shape?
2. Indicate hybridization of the central atom???
3. Sketch 3-D diagram an indicate bond angles
4. Demonstrate concept of hybirdization using the molecule H2CO3. Show electron configuration, hybird orbitals, hybird orbitals, a complete diagram, and sigma and pi bond (if any)


ssssss

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Re:VSEPR/Hybirdization/diagrams *delete me*!
« Reply #1 on: November 23, 2004, 01:32:50 AM »
Sorry this is so long but I don't get any of it.

For these 2 molecules:
C2H4
NH2OH

1. Using VSEPR, how to u predict the shape?
2. Indicate hybridization of the central atom???
3. Sketch 3-D diagram an indicate bond angles
4. Demonstrate concept of hybirdization using the molecule H2CO3. Show electron configuration, hybird orbitals, hybird orbitals, a complete diagram, and sigma and pi bond (if any)




All is given in your Chemistry book.

Offline Mitch

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Re:VSEPR/Hybirdization/diagrams *delete me*!
« Reply #2 on: November 23, 2004, 03:01:18 AM »
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

Offline Donaldson Tan

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Re:VSEPR/Hybirdization/diagrams *delete me*!
« Reply #3 on: November 23, 2004, 02:01:45 PM »
Refer to our link section. There is an online textbook that can teach you how to use VSEPR. I think it's Oxford Virtual Chemistry,
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Offline bluemonster

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Re: VSEPR/Hybirdization/diagrams *delete me*!
« Reply #5 on: July 19, 2006, 07:28:40 AM »
What the order of electronegative of Csp, sp2 and sp3 ?? Explain !?
hope to receive helps !
 ???
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Re: VSEPR/Hybirdization/diagrams *delete me*!
« Reply #6 on: July 19, 2006, 12:03:00 PM »
The more S-character the greater the electronegativity.

Offline bluemonster

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Re: VSEPR/Hybirdization/diagrams *delete me*!
« Reply #7 on: July 20, 2006, 01:04:00 AM »
Quote
The more S-character the greater the electronegativity
Why !?? I need the clear answer!
Since the more s-character, the less effect of the nuclear to the electrons. Remember the electrons in the s-orbital are atracted by the nuclear, because of their high symmetric sphere. So, they reduced the attracted force from the nuclear to the electrons, so decrease electronegative !?? But, this agument is wrong, so I need another !
hope to reply soon !  8) ???
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Re: VSEPR/Hybirdization/diagrams *delete me*!
« Reply #8 on: July 20, 2006, 06:36:30 AM »
S orbitals are more penetrating than p orbitals (of the same principal quantum number). This means that elctrons in an s orbital is not as shielded from the nucleus and are held more tightly. It follows that an electron in an orbital with higher s character experiences a greater effective nuclear charge, and is bound more tightly.
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Offline bluemonster

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Re: VSEPR/Hybirdization/diagrams *delete me*!
« Reply #9 on: July 22, 2006, 01:48:53 AM »
S orbitals are more penetrating than p orbitals (of the same principal quantum number). This means that elctrons in an s orbital is not as shielded from the nucleus and are held more tightly. It follows that an electron in an orbital with higher s character experiences a greater effective nuclear charge, and is bound more tightly.
But the electronnegative is a effect which the nuclear cause to electron bonding (or the electrons are outside nuclear). I just only know why the higher s character, the more electronagative!? I don't need the explaination about the electrons in s-characterictic hybridization! (as you say they are held more tightly).  ::) ;D 
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Re: VSEPR/Hybirdization/diagrams *delete me*!
« Reply #10 on: March 18, 2007, 06:25:08 AM »
Hi everybody !!!
It's a long time since I visited chemicalforums, and now, I can resolve my problem !!!
Really, in this case, I use MO modelling to resolve a problem of VB modelling, hix, may be not sensible !!!
Because, if we consider MO aspect, have never had combination between AOs in a atom (except combination between AOs in a atom with AO in other atom)! so, hybrided state do not exist in MO theory !
But, if we consider atoms should hybride to overlap with the other atoms in system (as exception above), I believe it's very easy to resolve problem !
Review the characteristic of combination AOs, the sp3 hybrided state is considered a combination between one  AO s with three  AO p, similarly for the case of sp2 and sp hybrided state !
Therefore, the energy of LUMO (lowest unoccupied MO) of three hybrided states are different, and different AOs of Carbon in ground state !
Hence, capacity of acceptor electrons outer atom will higher with LUMO energy is lower ! and electronegative of sp > sp2 > sp3 > C ground state !

Best regard !
Chemistry is a practical science, the theories can not make practices, they only use to explain the practices !
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