If you like us, please share us on social media.
The latest UCD Hyperlibrary newsletter is now complete, check it out.
MindTouch
Copyright (c) 20062014 MindTouch Inc.
http://mindtouch.com
This file and accompanying files are licensed under the MindTouch Master Subscription Agreement (MSA).
At any time, you shall not, directly or indirectly: (i) sublicense,
resell, rent, lease, distribute, market, commercialize or otherwise
transfer rights or usage to: (a) the Software, (b) any modified version
or derivative work of the Software created by you or for you, or (c)
MindTouch Open Source (which includes all nonsupported versions of
MindTouchdeveloped software), for any purpose including timesharing or
service bureau purposes; (ii) remove or alter any copyright, trademark
or proprietary notice in the Software; (iii) transfer, use or export the
Software in violation of any applicable laws or regulations of any
government or governmental agency; (iv) use or run on any of your
hardware, or have deployed for use, any production version of MindTouch
Open Source; (v) use any of the Support Services, Error corrections,
Updates or Upgrades, for the MindTouch Open Source software or for any
Server for which Support Services are not then purchased as provided
hereunder; or (vi) reverse engineer, decompile or modify any encrypted
or encoded portion of the Software.
A complete copy of the MSA is available at http://www.mindtouch.com/msa
This module will focus on the how Molecular Orbital theory applied to BH_{3}. First, we would go through the the structure, symmetry elements that BH_{3} has. Next, we would talk about the symmetry labels due to the orbitals of the B atoms in BH_{3}. Lastly, we would use all of the information about symmetry labels in BH_{3} to construct a BH_{3} diagram.
Molecular Orbital theory is used to show how bonds between atoms in a molecules are formed from the orbital perspective. This theory is very important in understanding whether a molecule is paramagnetic or diamagnetic since Valence Bond theory can not establish this. We usually see the MO diagram of diatomic such H _{2,} F_{2} or HBr. With this module we would learn how make MO diagram for polyatomic molecules and particularly BH3. The first step of making MO diagram for BH3 is to know the structure and symmetry elements BH3 has.
The molecular structure of BH_{3} is trigonal planar and it belongs to the point group D_{3h}. The symmetry elements are included:
Figure 1: Molecule structure of BH_{3} (Used with permission from Dean.H Johnston)
D_{3H}  E  2C_{3}  3C_{2}  σh  2S_{3}  σv 


A1'  1  1  1  1  1  1 
 x^{2}+y^{2}, z^{2} 
A2'  1  1  1  1  1  1  R_{z} 

E'  2  1  0  2  1  0  (x,y)  (x^{2}y^{2},xy) 
A1''  1  1  1  1  1  1 


A2''  1  1  1  1  1  1  z 

E''  2  1  0  2  1  0  (R_{x},R_{y})  (xz,yz) 
FIGURE2: Character table for the the point group D3h
B atom in BH3:
+sorbital: with the shape of the sphere, its function is x^{2}+y^{2}+z^{2}. Therefore, 2s orbital hasa_{1}' symmetry
+porbital: has 3 orbitals , p_{x}, p_{y}, p_{z}. Therefore, 2p_{z} orbital has a_{2}" symmetry
2p_{x} and 2p_{y} orbital are degenerate and have e' symmetry
3 Hydrogen atoms in BH3: (Ligand group orbitals)
a. Symmetry labels of LGOs:
With the symmetry operations of BH3 above, we can determine how many LGO unmoved by creating the following table:
D_{3h}  E  2C_{3}  3C_{2}  σ_{h}  2S_{3}  3σ_{v} 
LGO  3  0  1  3  0  1 
Next, with these values we can apply the following formula to identify the symmetry labels of the Ligan group orbitals
a= 1/h ∑[(N).Xr(R).Xi(R)]
h: the total number of coeficients of symmetry operation
N: the coeficient of the each symmetry operation
Xr(R): the character of the reducible representation corresponding to the R (values that just found in the LGO row
Xi(R): the character of the irreducible representation corresponding to the R (from the character table)
Calculation:
A1'= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(1)+(1)(3)(1)+(2)(0)(1)+(3)(1)(1)] = 1 A1'
A2'= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(1)+(1)(3)(1)+(2)(0)(1)+(3)(1)(1)]= 0 A2'
E'= 1/12 [(1)(3)(2)+(2)(0)(1)+(3)(1)(0)+(1)(3)(2)+(2)(0)(1)+(3)(1)(0)]= 1E'
A1''= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(1)+(1)(3)(1)+(2)(0)(1)+(3)(1)(1)]= 0 A1''
A2''= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(1)+(1)(3)(1)+(2)(0)(1)+(3)(1)(1)] = 0 A2''
E''= 1/12 [(1)(3)(2)+(2)(0)(1)+(3)(1)(0)+(1)(3)(2)+(2)(0)(1)+(3)(1)(0)] = 0 E ''
Thus, LGO's symmetry labels are a1' + e'
LGO's symmetry labels are a1' + e'. There are 3 LGOs that can be made out of these symmetry labels,one LGO is from a1' and two LGOs are from e' due to doubly degenerate. In order to determine the shape of each LGO, we would use the wavefunctions.
Three hydrogens in BH3 are assigned with Ψ1, Ψ2, Ψ3. Now lets look at how each Ψ is affected by the symmetry operations of the D3h and their results are completed in the following table:
D_{3h}  E  C_{3}  C^{2}_{3}  C_{2}  C_{2}’  C_{2}’’  σ_{h}  S_{3}  S^{2}_{3}  σ_{v}  σ_{v}’  σ_{v}’’ 
Ψ1  Ψ1  Ψ2  Ψ3  Ψ1  Ψ3  Ψ2  Ψ1  Ψ2  Ψ3  Ψ1  Ψ3  Ψ2 
a1’  1  1  1  1  1  1  1  1  1  1  1  1 
LGO1  Ψ1  Ψ2  Ψ3  Ψ1  Ψ3  Ψ2  Ψ1  Ψ2  Ψ3  Ψ1  Ψ3  Ψ2 
Ψ (a1') = 4Ψ1+4Ψ2+4Ψ3
= 4(Ψ1+Ψ2+Ψ3)
Ψ(a1')= 1/√3 (Ψ1+Ψ2+Ψ3)
The shape of the LGO1 is
D_{3h}  E  C_{3}  C^{2}_{3}  C_{2}  C_{2}’  C_{2}’’  σ_{h}  S_{3}  S^{2}_{3}  σ_{v}  σ_{v}’  σ_{v}’’ 
Ψ1  Ψ1  Ψ2  Ψ3  Ψ1  Ψ3  Ψ2  Ψ1  Ψ2  Ψ3  Ψ1  Ψ3  Ψ2 
e’  2  1  1  0  0  0  2  1  1  0  0  0 
LGO1  2Ψ1  Ψ2  Ψ3  0  0  0  2Ψ1  Ψ2  Ψ3  0  0  0 
Ψ(e') = 4 (Ψ1)2 (Ψ2)2 (Ψ3)
= 2[ 2(Ψ1)Ψ2Ψ3]
Ψ(e') = 1/ √6 (2 Ψ1Ψ2Ψ3)
The shape of the LGO2 is
Noticed that in the LGO2, we have 1 nodal plane which is the horizonal line between the positive charge and negative charge. Therefore, the LGO3 (doubly degenerate with e') would also 1 nodal plane and its wavefunction would be Ψ(e') = 1/√2 (Ψ2Ψ3). The shape of the LGO3 is
As we can see in this diagram, the energy level of 3 LGOs are higher than the 2s orbital and below the 2 p orbital dued to the electronegativy of both Boron and Hydrogen. Hydrogen has higher electronegativity than boron, therefore hydrogen would have lower energy level in the MO diagram.
In addition, B has 3 electrons in the valence electrons and 3 hydrogens have total 3 electrons. Therefore, the total number of electrons filled in orbitals are 6. With all of the informations above about symmetry labels of B atom and the 3 LGOs, we now construct the MO diagram of BH3. Noticed that, the bonding formation only happens to atoms that have the same symmetry labels. 2s orbital and LGO(1) would contribute 1 electron to give 2 spin pairs electrons at the a_{1}' energy level. 2p_{x} and 2p_{y} orbitals would bond to the LGO(2) and LGO(3), which give 2 spin pairs electrons at the e' energy level.
FIGURE 3: MO diagram for the formation of BH3
there is a detailed explaination that performed in this video
Analytical Chemistry
Biological Chemistry
Inorganic Chemistry
Organic Chemistry
Physical Chemistry
Theoretical Chemistry
Cal Poly Pomona
Diablo Valley College
Florida State U
Hope College
Howard University
Purdue
Sacramento City College
UC Davis
UC Irvine
Zumdahl 9^{ed}
Petrucci 10^{ed}
McQuarrie & Simon
Kotz et al.
Tro
Bruice 6^{ed}
An NSF funded Project
Unless otherwise noted, content in the UC Davis ChemWiki is licensed under a Creative Commons AttributionNoncommercialShare Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Questions and concerns can be directed toward Prof. Delmar Larsen (dlarsen@ucdavis.edu), Founder and Director. Terms of Use