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PostPosted: Sat, 3 Dec 2011 16:15:27 UTC 
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BH3 <-- Boron trihydride
If you are to draw a Lewis structure, you would see that just a single bond is needed b/ween each B-H bond. Is Boron coordinate covalent bonding in this situation?

Also, its electron configuration before bonding is 1s^2 2s^2 2p^1
After bonding with those 3 Hydrogens, is its valence configuration going to be:

1s^2 2s^0 2p^0 OR

Does it hybridize and we get:

[e-, ][e-, ][e-, ] <--- sp^2 [e-, ] <--- p

What I am not understanding is how does Boron's valence configuration change before bonding versus after bonding?


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PostPosted: Sat, 3 Dec 2011 17:03:51 UTC 
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A-R-Q wrote:
BH3 <-- Boron trihydride
If you are to draw a Lewis structure, you would see that just a single bond is needed b/ween each B-H bond. Is Boron coordinate covalent bonding in this situation?

Also, its electron configuration before bonding is 1s^2 2s^2 2p^1
After bonding with those 3 Hydrogens, is its valence configuration going to be:

1s^2 2s^0 2p^0 OR

Does it hybridize and we get:

[e-, ][e-, ][e-, ] <--- sp^2 [e-, ] <--- p

What I am not understanding is how does Boron's valence configuration change before bonding versus after bonding?


Indeed B-H bond is covalent (the difference in electronegativity is small), the boron is sp^2-hybridised (resulting in BH3 having a trigonal planar structure).

In practice, BH3 readily dimerises to form diborane B2H6. The dimer make up of the difficient electrons by forcing an H from each BH3 to act as a "bridge" between the two borons so the boron become effectively like sp^3 hybridised (technically we have a three-centre two-electron bond involving these H).

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 5 Dec 2011 01:37:19 UTC 
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outermeasure wrote:
A-R-Q wrote:
BH3 <-- Boron trihydride
If you are to draw a Lewis structure, you would see that just a single bond is needed b/ween each B-H bond. Is Boron coordinate covalent bonding in this situation?

Also, its electron configuration before bonding is 1s^2 2s^2 2p^1
After bonding with those 3 Hydrogens, is its valence configuration going to be:

1s^2 2s^0 2p^0 OR

Does it hybridize and we get:

[e-, ][e-, ][e-, ] <--- sp^2 [e-, ] <--- p

What I am not understanding is how does Boron's valence configuration change before bonding versus after bonding?


Indeed B-H bond is covalent (the difference in electronegativity is small), the boron is sp^2-hybridised (resulting in BH3 having a trigonal planar structure).

In practice, BH3 readily dimerises to form diborane B2H6. The dimer make up of the difficient electrons by forcing an H from each BH3 to act as a "bridge" between the two borons so the boron become effectively like sp^3 hybridised (technically we have a three-centre two-electron bond involving these H).


Ok so if it is sp^2 hybridizing, it will get this (Assume that "e-" is an electorn, and their seperated by commas around a square bracket that designates that that is an orbital)

sp^2 orbitals [e-,e-] [e-,e-] [e-,e-]
unhybridized p-orbital [ ]

Bold designates shared electrons from hydrogen
So Boron is stable like this, despite having its unhybridized p-orbital completely empty?


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PostPosted: Mon, 5 Dec 2011 05:56:17 UTC 
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A-R-Q wrote:
outermeasure wrote:
A-R-Q wrote:
BH3 <-- Boron trihydride
If you are to draw a Lewis structure, you would see that just a single bond is needed b/ween each B-H bond. Is Boron coordinate covalent bonding in this situation?

Also, its electron configuration before bonding is 1s^2 2s^2 2p^1
After bonding with those 3 Hydrogens, is its valence configuration going to be:

1s^2 2s^0 2p^0 OR

Does it hybridize and we get:

[e-, ][e-, ][e-, ] <--- sp^2 [e-, ] <--- p

What I am not understanding is how does Boron's valence configuration change before bonding versus after bonding?


Indeed B-H bond is covalent (the difference in electronegativity is small), the boron is sp^2-hybridised (resulting in BH3 having a trigonal planar structure).

In practice, BH3 readily dimerises to form diborane B2H6. The dimer make up of the difficient electrons by forcing an H from each BH3 to act as a "bridge" between the two borons so the boron become effectively like sp^3 hybridised (technically we have a three-centre two-electron bond involving these H).


Ok so if it is sp^2 hybridizing, it will get this (Assume that "e-" is an electorn, and their seperated by commas around a square bracket that designates that that is an orbital)

sp^2 orbitals [e-,e-] [e-,e-] [e-,e-]
unhybridized p-orbital [ ]

Bold designates shared electrons from hydrogen
So Boron is stable like this, despite having its unhybridized p-orbital completely empty?


No, it is not stable this way, that's why it dimerises readily.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Tue, 6 Dec 2011 02:21:24 UTC 
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outermeasure wrote:
A-R-Q wrote:
outermeasure wrote:
A-R-Q wrote:
BH3 <-- Boron trihydride
If you are to draw a Lewis structure, you would see that just a single bond is needed b/ween each B-H bond. Is Boron coordinate covalent bonding in this situation?

Also, its electron configuration before bonding is 1s^2 2s^2 2p^1
After bonding with those 3 Hydrogens, is its valence configuration going to be:

1s^2 2s^0 2p^0 OR

Does it hybridize and we get:

[e-, ][e-, ][e-, ] <--- sp^2 [e-, ] <--- p

What I am not understanding is how does Boron's valence configuration change before bonding versus after bonding?


Indeed B-H bond is covalent (the difference in electronegativity is small), the boron is sp^2-hybridised (resulting in BH3 having a trigonal planar structure).

In practice, BH3 readily dimerises to form diborane B2H6. The dimer make up of the difficient electrons by forcing an H from each BH3 to act as a "bridge" between the two borons so the boron become effectively like sp^3 hybridised (technically we have a three-centre two-electron bond involving these H).


Ok so if it is sp^2 hybridizing, it will get this (Assume that "e-" is an electorn, and their seperated by commas around a square bracket that designates that that is an orbital)

sp^2 orbitals [e-,e-] [e-,e-] [e-,e-]
unhybridized p-orbital [ ]

Bold designates shared electrons from hydrogen
So Boron is stable like this, despite having its unhybridized p-orbital completely empty?


No, it is not stable this way, that's why it dimerises readily.


Ok so for my purposes (1st Year General Chemistry), in which we have not yet heard of this dimer or bridge phenomena yet, can I consider that Boron is "stable" in this way? Otherwise, I don't see how we can assume that BH3 can be formed.


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PostPosted: Tue, 6 Dec 2011 06:06:12 UTC 
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A-R-Q wrote:
Ok so for my purposes (1st Year General Chemistry), in which we have not yet heard of this dimer or bridge phenomena yet, can I consider that Boron is "stable" in this way? Otherwise, I don't see how we can assume that BH3 can be formed.


Well, it is stable but it isn't stable, if that is what you want to hear.

It is stable because it is what happen when you have high enough temperature. But it is not stable because the boron will want to seek an extra pair of electrons from *somewhere*, the fact that BH3 exists in the gaseous state is just that there is enough energy to break free whenever the boron clings on to something.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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