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Bond Angles Around a Tetravalent Atom Robert K Bohn, and Wesley D. Allen J. Phys. Chem. A, Just Accepted Manuscript • DOI: 10.1021/jp5073999 • Publication Date (Web): 07 Oct 2014 Downloaded from http://pubs.acs.org on October 8, 2014

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The Journal of Physical Chemistry

Bond Angles around a Tetravalent Atom Robert K. Bohn1* and Wesley D. Allen2* 1 2

Department of Chemistry, University of Connecticut, Storrs, CT 06269-3060 Center for Computational Quantum Chemistry & Department of Chemistry, University of Georgia, Athens, Georgia 30602

Abstract Relationships among the six bond angles about a central tetravalent atom depend on symmetry, ranging from the most symmetrical Td point group to the least symmetrical C1 point group having only the identity element. Exact relationships are derived here in two ways: (1) a purely algebraic treatment of the general mathematical conditions among the bond angles, followed by factorizations that arise from various symmetry constraints; and (2) a reverse approach based on geometric analysis, starting with the most symmetrical Td case and relaxing constraints stepwise to lower point groups. The mathematical formulas show systematically how the degrees of freedom among the bond angles increase from zero to a maximum of five as symmetry is relaxed from Td symmetry.

Key words: tetravalent bond angles, bond angles and symmetry, bond angle redundancies *Corresponding authors: [email protected], [email protected]

ACS Paragon Plus Environment


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I. Introduction There are six bond angles around a tetravalent central atom. Depending upon the symmetry of the system, all six can be different, and the number of independent angles ranges from zero to five. While some relationships among these angles have been previously reported in numerous sources, a systematic series of exact formulas for all point groups is needed. We present two approaches to this problem that yield the same results. First, a general algebraic approach treats the most general case of four bonds about a central atom with no symmetry constraints. The resulting general redundancy equation is then factorized into simpler forms by imposing symmetry constraints. Second, we derive the bond angle relationships using geometric arguments starting from the most symmetrical case, Td, and adding deformations stepwise to the least symmetric. Mastryukov and co-workers1 have investigated quantum chemical effects on the geometry surrounding specific tetravalent atoms. A co-author of that study is James E. Boggs, whom we are honoring here. In contrast, here we account for the constraints imposed solely by geometry independent of the chemical identity of the attached atoms. Another group2 states that the sum of the six bond angles around a tetravalent atom, 6 x 109.471° = 656.827°, is constant. However, this relation is only a first-order redundancy condition, one that appears repeatedly in the literature3. The equations reported here are both complete and exact. II. Methods Consider a central atom A bonded to n other atoms. Attach a Cartesian axis system to the molecule with the central atom at the origin, atom i along the z axis, and atom j in the xz plane. The unit vectors (nk) pointing from A to the peripheral atoms k = 1, 2, ..., n can be written in terms of polar (θ) and azimuthal (φ) angles as

n k = ( cosφk sinθ k ,sinφk sin θ k ,cosθ k )

(1)

where θi = 0, φi is arbitrary, and φj = 0. Thus, p = 2n–3 independent (θ, φ) angles fix the orientation of the peripheral atoms about the central atom. Because there are q = 12 n(n − 1) central angles between the bonds, there must be r = q − p = 12 (n − 3)(n − 2) redundancy conditions among the bond angles. In the case of a tetravalent central atom (n=4), p = 5, q = 6, and r = 1, so that only one redundancy condition need be derived. The conditions relating the central bond angles to the unit vectors are

cosθ ik = n i ⋅n k = cosθ k

(2)

cosθ jk = n j ⋅ n k = cosφk sin θ j sin θ k + cosθ j cosθ k

(3)

cosθ kl = n k ⋅ n l = ( cosφk cosφl + sin φk sinφl ) sin θ k sinθ l + cosθ k cosθ l

(4)

Equation (2) provides θik = θk. Consequently, Eq. (3) yields cosθ jk − cosθ ij cosθ ik cosφ k sinθ ik = sin θ ij


(5)

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Rearrange Eq. (4) to

( cosθ

kl

− cosθ ik cosθ il ) − cosφk cosφl sin θ ik sinθ il = sinφk sinφl sin θ ik sinθ il ,

(6)

then square both sides:

( cosθ

− cosθ ik cosθ il ) − 2cosφk cosφl sin θ ik sinθ il ( cosθ kl − cosθ ik cosθ il ) + cos 2 φk cos 2 φl sin 2 θ ik sin 2 θ il (7) = sin 2 φk sin 2 φl sin 2 θ ik sin 2 θ il 2

kl

Use the identity sin 2 φ = 1− cos 2 φ to arrive at

( cosθ

(

− cosθ ik cosθ il ) − 2cosφk cosφl sin θ ik sin θ il ( cosθ kl − cosθ ik cosθ il ) 2

kl

)

.

(8)

= 1− cos 2 φk − cos 2 φl sin 2 θ ik sin 2 θ il

Employing Eq. (5) in Eq. (8) provides  cos θ − cosθ cosθ   cosθ − cosθ cosθ  2 ( cosθ kl − cosθ ik cosθ il ) − 2  jk sinθ ij ik   jl sinθ ij il  ( cosθ kl − cosθ ik cosθ il )    ij ij 2

(9)

2

 cosθ jk − cosθ ij cosθ ik   cosθ jl − cosθ ij cosθ il  2 2 = sin 2 θ ik sin 2 θ il −   sin θ il −   sin θ ik sinθ ij sinθ ij    

Multiply Eq. (9) by sin 2 θ ij to find

(cosθ

(

)( − ( cosθ

) − cosθ cosθ ) sin θ

− cosθ ik cosθ il ) sin 2 θ ij − 2 cosθ jk − cosθ ij cosθ ik cosθ jl − cos θ ij cosθ il ( cos θ kl − cosθ ik cos θ il ) 2

kl

(

)

2

= sin θ ik sin θ il sin θ ij − cosθ jk − cosθ ij cosθ ik sin θ il 2

2

2

2

2

jl

ij

(10)

2

il

ik

Employ sin 2 θ = 1− cos2 θ to obtain a form that only involves cosine functions:

(cosθ

(

(

) ( )(1− cos θ )(1− cos θ ) − ( cosθ

)( ) − cos θ cosθ ) (1− cos θ ) − ( cosθ − cos θ cosθ ) (1− cos θ )

− cosθ ik cosθ il ) 1− cos 2 θ ij − 2 cos θ jk − cos θ ij cos θ ik cosθ jl − cos θ ij cos θ il ( cos θ kl − cosθ ik cos θ il ) 2

kl

= 1− cos θ ik 2

2

2

2

il

ij

jk

ij

2

2

ik

il

jl

ij

il

(11)

2

ik

After algebraic simplification, we ascertain the exact redundancy condition 1 = cos 2 θ ij + cos 2 θ ik + cos 2 θ il + cos 2 θ jk + cos 2 θ jl + cos 2 θ kl − cos 2 θ ij cos 2 θ kl − cos 2 θ ik cos 2 θ jl − cos 2 θ il cos 2 θ jk

(12)

− 2cosθ ij cosθ ik cosθ jk − 2cosθ ij cosθ il cosθ jl − 2cos θ ik cosθ il cosθ kl − 2cosθ jk cosθ jl cosθ kl + 2cosθ ik cosθ il cosθ jk cosθ jl + 2cosθ ij cosθ il cosθ jk cosθ kl + 2cosθ ij cosθ ik cosθ jl cosθ kl

The derivation of Eq. (12) would appear to require that atoms i, j, and A are not colinear

(

) (

)

( sin θ ij ≠ 0 ). However, this colinear case requires cosθ ij ,cosθ jk ,cosθ jl = ± 1,cosθ ik ,cosθ il , and Eq. (12) is always satisfied if this condition holds. Eq. (12) is an exact redundancy condition that holds without assumptions for all possible sets of four peripheral atoms, regardless of their orientation. Eq. (12) can be represented as

(

)

R xij , xik , xil , x jk , x jl , xkl = 0 where R is the determinant


(13)


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(

)

R xij , xik , xil , x jk , x jl , xkl =

1

xij

xik

xil

xij

1

x jk

x jl

xik

x jk

1

xkl

xil

x jl

xkl

1

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= 1− xij2 − xik2 − xil2 − x 2jk − x 2jl − xkl2 + xij2 xkl2 + xik2 x 2jl + xil2 x 2jk +2xij xik x jk + 2xij xil x jl

(14)

+ 2xik xil xkl + 2x jk x jl xkl − 2xij xik x jl xkl − 2xij xil x jk xkl − 2xik xil x jk x jl and

(x ,x ,x ,x ij

ik

il

jk

) (

, x jl , xkl = cosθ ij ,cosθ ik ,cosθ il ,cosθ jk ,cosθ jl ,cosθ kl

)

(15)

If all of the xij are known except one, Eq. (14) merely constitutes a quadratic equation for the unknown variable. For example, to obtain xij from the other quantities, one solves the equation

Ax xij2 + Bx xij + Cx = 0

(16)

Ax = xkl2 − 1

(17)

where

(

)

(18)

Cx = 1− xik2 − xil2 − x 2jk − x 2jl − xkl2 + xik2 x 2jl + xil2 x 2jk + 2xik xil xkl + 2x jk x jl xkl − 2xik xil x jk x jl

(19)

Bx = 2 xik x jk + xil x jl − xik x jl xkl − xil x jk xkl

In previous work, Cihla and Pliva4 derived a general expression for bond-angle redundancy conditions of a single central atom of valency n. Applying Eq. (5) therein to the tetravalent case, the Cihla and Pliva condition is

cT D −1c = 1 , which involves the vector

(

c = cosθ il ,cosθ jl ,cosθ kl

(20)

)

(21)

and the matrix

 1 cos θ ij cos θ ik    1 cos θ jk  . (22) D =  cosθ ij  cosθ  cosθ jk 1 ik   By algebraically inverting D, Eq. (20) is equivalent to the explicit form of Eq. (12). One merit of Eq. (12) is that it holds regardless of whether the matrix D is singular or not. Lopez-Gonzalez and co-workers5 used linear dependence considerations of vectors in three-dimensional space to elegantly derive a determinant representation equivalent to Eq. (14) for bond-angle redundancy conditions. In particular, for any set of peripheral atoms (a,b,c,d)


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whose unit vectors can be written as linear combinations e m = Cmi e i + Cmj e j + Cmk e k + Cml e l for all m in the set, then

where

xai

xaj

xak

xal

xbi

xbj

xbk

xbl

xci

xcj

xck

xcl

xdi

xdj

xdk

xdl

(x

mi

=C

1

xij

xik

xil

xij

1

x jk

x jl

xik

x jk

1

xkl

xil

x jl

xkl

1

=0

) (

(23)

)

, xmj , xmk , xml = cosθ mi ,cosθ mj ,cosθ mk ,cosθ ml .

(24)

Equation (23) provides general quartic redundancy conditions for as many as eight sets of atoms at a time. The Lopez-Gonzalez paper imposed (a=i, b=j, c=k) and used the resulting five-atom equations to derive second-order redundancy conditions for bond-angle displacements of trigonal bipyramidal and octahedral molecules. III. Results and Discussion Special Cases of Local Symmetry for a Tetravalent Atom For all of the local symmetries discussed below, the general algebraic results and those derived from Euclidean geometry are in complete agreement. In each case a symmetry condition is inserted into Eq. (14), allowing R to be factorized. A relevant root to Eq. (13) is then chosen, producing the desired mathematical relationship. An alternative form derived from the Euclidean geometry approach is then listed. (1) Td symmetry; e.g., CH4 Under Td symmetry, all six angles are equal and have the value 1 109.4712206…° This result has been derived in the literature many times. A 2002 paper by ten Hoor5 notes that the bond angle for a tetrahedral central atom has been derived in 27 previous papers in J. Chem. Ed. alone.

α

2

3 4

Symmetry conditions: α = θ12 = θ 34 = θ13 = θ 23 = θ14 = θ 24 , a = cos α

(

) (

)(

Factorization: R a,a,a,a,a,a = 3a +1 1− a

)

3

Relevant root of R: a = cos α = − 13

( )

Equivalent alternative form: sin 12 α =

2 3

, α = 109.4712206…°

(2) C3v symmetry; e.g., CH3X


Projection


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(

)

Symmetry conditions: α = θ12 = θ13 = θ 23 , β = θ14 = θ 24 = θ 34 , (a,b) = cos α ,cos β

) (

(

)(

)

Factorization: R a,a,b,a,b,b = 2a + 1− 3b2 a − 1

2

Relevant root of R: 3b2 = 2a + 1 ⇒ 3cos 2 β = 2cos α +1 Alternative forms: sin ( 12 α ) =

3 2

sin β or cos α = -(3/2)sin2β + 1

Example: α = 112°, β = 106.8058…° The C3v relationship was derived by Gordy and Cook.6 This result and some others reported here have also been determined by Johnson.7 To demonstrate the derivation of the alternative form, we turn to Pythagoras. From his theorem applied to an isosceles triangle with unit sides and the angle between them α, the hypotenuse, called 1-2, is given by (1-2)2 = 1 + 1 - 2cos α = 2(1 – cos α) From the projection of the 3-fold group onto the yz plane, we derive (1-2)2 = sin2β + sin2β – 2sin2β (1 - cos 120) = 2 sin2β (1 - (-.5)) = 3 sin2β . Then, 2(1 – cos α) = 3 sin2β or cos α = -(3/2)sin2β + 1. (3) D2d symmetry; e.g., spiropentane, C5H8

Projection

(

Symmetry conditions: α = θ12 = θ 34 , β = θ13 = θ14 = θ 23 = θ 24 , (a,b) = cos α ,cos β

(

) (

)(

)(

)

Factorization: R a,b,b,b,b,a = a +1+ 2b a +1− 2b a −1 Relevant root of R: −2b = a + 1 ⇒ Alternative form: cos2

−2cos β = cos α +1

( α ) = − cos β 1 2

Example: α = 60°, β = 138.59…°


2

)

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(4) C2v symmetry; e.g., CH2X2

Projection

Symmetry conditions: α = θ12 , β = θ13 = θ14 = θ 23 = θ 24 , γ = θ 34

(

) ( )( = ( a + 1) ( c + 1)

)(

)(

)

Factorization: R a,b,b,b,b,c = a − 1 c − 1  a + 1 c + 1 − 4b2  Relevant root of R: 4b2

⇒

cos β = − cos ( 12 α ) cos ( 12 γ )

Example: α = 107°, γ = 112°, β = 109.43…° (5) D2 symmetry

Projection

( ) Factorization: R ( a,b,c,c,b,a ) = ( a + b + c +1) ( a − b − c + 1) ( −a + b − c +1) ( −a − b + c + 1) Symmetry conditions: α = θ12 = θ 34 , β = θ13 = θ 24 , γ = θ14 = θ 23 , (a,b,c) = cos α ,cos β ,cos γ

Relevant root of R:

a + b + c = cos α + cos β + cos γ = −1

Alternative form: cos β = cos2(α/2) – sin2(α/2) sin θ cos γ = cos2(α/2) + sin2(α/2) sin θ Example: α = 60°, θ = 45°, β = 124.975...° and γ = 157.938..° Projection

(6) C2 symmetry

Symmetry conditions: α = θ12 , β = θ13 = θ 24 , β ′ = θ14 = θ 23 , γ = θ 34 ,

(a,b,c,d) = ( cos α ,cos γ ,cos β ,cos β ′ )

(

) (

)(

) (

) (

)(

) (

)

2 2    Factorization: R a,c,d,d,c,b =  a + 1 b + 1 − c + d   a − 1 b − 1 − c − d    



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Relevant root of R:

( a +1) ( b +1) = ( c + d )

2

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⇒ cosβ + cosβ ′ = –2cos(α/2) cos(γ/2)

Alternative form: {–cosβ = cos(α/2) cos(γ/2) + sin(α/2)sinθ , –cosβ′ = cos(α/2) cos(γ/2) – sin(α/2)sinθ} Example: α = 107°, γ = 112°, θ = 45°, β = 143.50…° , and β′ = 97.97…° (7) Cs symmetry; Projection

Symmetry conditions: α = θ12 , β = θ13 = θ 23 , β ′ = θ14 = θ 24 , γ = θ 34 ,

(a,b,c,d) = ( cos α ,cos γ ,cos β ,cos β ′ )

(

) ( ) ( ) ( ) ( a + 1)(1− b ) = c + d − 2bcd ⇒

2 2 2 Factorization: R a,c,d,c,d,b = 1− a  a + 1 1− b − 2c − 2d + 4bcd 

Relevant root of R:

1 2

2

2



2

cos2 ( 12 α ) sin 2 γ = cos 2 β + cos 2 β ′ − 2cos β cos β ′ cos γ Alternative forms: ( cos β − cos β ′ ) = 4sin 2 ( 12 γ )  cos 2 ( 12 α ) cos 2 ( 12 γ ) − cos β cos β ′  2

or {–cosβ = -cos(α/2) cos(γ/2 + θ) , –cosβ′ = -cos(α/2) cos(γ/2 – θ)} Example: α = 107°, γ = 112°, θ = 10°, β = 104.00...° , and β′ = 114.40...° (8) (8 C1 symmetry Projection

Notation: α = θ12, β1 = θ13, β2 = θ14; β3 = θ23, β4 = θ24, γ = θ34 Factorization: No factorization of Eq. (14) is generally possible. Alternative forms: {–cosβ1 = cos(α/2 – φ) cos(γ/2) + sin(α/2 − φ) sin(γ/2) sinθ , –cosβ2 = cos(α/2 + φ) cos(γ/2) − sin(α/2 + φ) sin(γ/2) sinθ , –cosβ3 = cos(α/2 – φ) cos(γ/2) − sin(α/2 − φ) sin(γ/2) sinθ, –cosβ4 = cos(α/2 + φ) cos(γ/2) + sin(α/2 + φ) sin(γ/2) sinθ }


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Example:

α = 107°, γ = 112°, θ = 45°, φ = 30°, β1 = 132.70...°, β2 = 60.22...°, β3 = 73.80...°, and β4 = 130.22...°

IV. Summary The geometric relationships and redundancies among bond angles about a tetravalent central atom have been derived via two different approaches for all relevant point groups. The general algebraic approach yields Eq. (12) as an exact result for any geometry. Factorization of this equation under various symmetry constraints provides a systematic series of bond angle redundancy conditions. The second approach derives the same or equivalent results by applying geometrical rules such as the Pythagorean Theorem. The equations and analysis reported in this paper will aid future studies of molecular structure and remove confusion over bond angle relationships and geometric degrees of freedom. Acknowledgments We honor the late Jim Boggs for his great service in establishing and organizing the biannual Austin Symposium on Molecular Structure for many years, 1966-2010, and Dieter Cremer and Elfi Kraka for continuing the symposium as the Austin Symposium on Molecular Structure and Dynamics @ Dallas (ASMSD@D). The 25th renewal of that meeting in March, 2014 brought the authors in contact and encouraged us to collaborate and write this paper. Each of us wishes to acknowledge the teaching and inspiration passed on many years ago in high school by our mathematics teachers, Miss Mildred Weseen at Analy Union High School, Sebastopol, CA for RKB and Mrs. Fran Allen at Dickson County High School, Dickson, TN for WDA. References 1. Mastryukov, V. S., Schaefer III, H. F., Boggs, J. E. Self-Consistent Changes of Geometrical Parameters: Experiment and Theory. Acc. Chem. Research, 1994, 27, 242249. 2. George, P., Glusker, J. P., Bock, C. W. Bond Angles Around Tetrahedrally Bonded Carbon, and Distortion of the Tetrahedron in CH3-X Structures. J. Molec. Struct. 1995, 338, 155-173. 3. ten Hoor, M. J. An Evergreen: The Tetrahedral Bond Angle. J. Chem. Ed. 2002, 79, 956. 4. Cihla, Z., Pliva, J. General Form of Conditions for Redundant Sets of Vibrational Coordinates. Collect. Czech. Chem. Commun. 1995, 26, 1903–1908. 5. Lopez-Gonzalez, J. J., Martinez-Sanchez, M., Fernandez-Gomez, M., Arenas, J. F. Further Considerations on the Calculation of Branching Redundancies. Collect. Czech. Chem. Commun. 1986 51, 1178–1186.



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6. Gordy, W., Cook, R. L., “Microwave Molecular Spectra”, Interscience, 1969 ed., p. 503 7. Russell Johnson (NIST), http://cccbdb.nist/gov/angles.asp


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For the Table of Contents of JPhysChem

CH2X2; C2v 2

α 1

β

cos β = -cos(α/2) cos(γ/2)

3

γ 4

If α = 107° and γ = 112°, then β = 109.43°.


Bond Angles around a Tetravalent Atom - The ... - ACS Publications

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