15 Molecular symmetry
|
|
- Lewis Casey
- 7 years ago
- Views:
Transcription
1 5 Molecular symmetry Solutions to exercises Discussion questions E5.(b) Symmetry operations Symmetry elements. Identity, E. The entire object 2. n-fold rotation 2. n-fold axis of symmetry, C n 3. Reflection 3. Mirror plane, σ 4. Inversion 4. Centre of symmetry, i 5. n-fold improper rotation 5. n-fold improper rotation axis, S n E5.2(b) E5.3(b) E5.4(b) E5.5(b) A molecule may be chiral, and therefore optically active, only if it does not posses an axis of improper rotation, S n. An improper rotation is a rotation followed by a reflection and this combination of operations always converts a right-handed object into a left-handed object and vice-versa; hence an S n axis guarantees that a molecule cannot exist in chiral forms. See Sections 5.4(a) and (b). The direct sum is the decomposition of the direct product. The procedure for the decomposition is the set of steps outlined in Section 5.5(a) on p. 47 and demonstrated in Illustration 5.. Numerical exercises CCl 4 has 4 C 3 axes (each C Cl axis), 3 C 2 axes (bisecting Cl C Cl angles), 3 S 4 axes (the same as the C 2 axes), and 6 dihedral mirror planes (each Cl C Cl plane). E5.6(b) Only molecules belonging to C s, C n, and C nv groups may be polar, so... (a) CH 3 Cl (C 3v ) may be polar along the C Cl bond; (b) HW 2 (CO) 0 (D 4h ) may not be polar (c) SnCl 4 (T d ) may not be polar E5.7(b) The factors of the integrand have the following characters under the operations of D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 σ h 3σ d 3σ v p x z p z Integrand E5.8(b) The integrand has the same set of characters as species E u, so it does not include A g ; therefore the integral vanishes We need to evaluate the character sets for the product A g E 2u q, where q = x, y,orz E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 σ h 3σ d 3σ v A g E 2u (x, y) Integrand
2 244 INSTRUCTOR S MANUAL To see whether the totally symmetric species A g is present, we form the sum over classes of the number of operations times the character of the integrand c(a g ) = (4) + 2( ) + 2() + ( 4) + 3(0) + 3(0) + (4) +2( ) + 2() + ( 4) + 3(0) + 3(0) = 0 E5.9(b) Since the species A g is absent, the transition is forbidden for x- ory-polarized light. A similar analysis leads to the conclusion that A g is absent from the product A g E 2u z; therefore the transition is forbidden. The classes of operations for D 2 are: E, C 2 (x), C 2 (y), and C 2 (z). How does the function xyz behave under each kind of operation? E leaves it unchanged. C 2 (x) leaves x unchanged and takes y to y and z to z, leaving the product xyz unchanged. C 2 (y) and C 2 (z) have similar effects, leaving one axis unchanged and taking the other two into their negatives. These observations are summarized as follows E C 2 (x) C 2 (y) C 2 (z) xyz A look at the character table shows that this set of characters belong to symmetry species A E5.0(b) A molecule cannot be chiral if it has an axis of improper rotation. The point group T d has S 4 axes and mirror planes (= S ), which preclude chirality. The T h group has, in addition, a centre of inversion (= S 2 ). E5.(b) The group multiplication table of group C 4v is E C + 4 C 2 σ v (x) σ v (y) σ d (xy) σ d ( xy) E E C + 4 C 2 σ v (x) σ v (y) σ d (xy) σ d ( xy) C + 4 C + 4 C 2 E σ d (xy) σ ( xy) σ v (y) σ v (x) E C 2 C + 4 σ d ( xy) σ (xy) σ v (x) σ v (y) C 2 C 2 C + 4 E σ v (y) σ v (x) σ d ( xy) σ d (xy) σ v (x) σ v (x) σ d ( xy) σ d (xy) σ v (y) E C 2 C + 4 σ v (y) σ v (y) σ d (xy) σ d ( xy) σ v (x) C 2 E C + 4 σ d (xy) σ d (xy) σ v (x) σ v (y) σ d ( xy) C + 4 E C 2 σ d ( xy) σ d ( xy) σ v (y) σ v (x) σ d (xy) C + 4 C 2 E E5.2(b) See ig. 5.. (a) Sharpened pencil: E,C,σ v ; therefore C v (b) Propellor: E,C 3, 3C 2 ; therefore D 3 (c) Square table: E,C 4, 4σ v ; therefore C 4v ; Rectangular table: E,C 2, 2σ v ; therefore C 2v (d) Person: E,σ v (approximately); therefore C s E5.3(b) We follow the flow chart in the text (ig. 5.4). The symmetry elements found in order as we proceed down the chart and the point groups are (a) Naphthalene: E, C 2,C 2,C 2, 3σ h,i; D 2h (b) Anthracene: E, C 2,C 2,C 2, 3σ h,i; D 2h
3 MOLECULAR SYMMETRY 245 (a) (b) (c) (d) igure 5. (c) Dichlorobenzenes: (i),2-dichlorobenzene: E, C 2,σ v,σ v ; C 2v (ii),3-dichlorobenzene: E, C 2,σ v,σ v ; C 2v E5.4(b) (a) (iii),4-dichlorobenzene: E, C 2,C 2,C 2, 3σ h,i; D 2h H (b) (e) I (c) O Xe O (f) T d (d) OC OC CO e CO OC CO e OC OC CO The following responses refer to the text flow chart (ig. 5.4) for assigning point groups. (a) H: linear, no i,so C v (b) I 7 : nonlinear, fewer than 2C n with n>2, C 5, 5C 2 perpendicular to C 5,σ h,so D 5h
4 246 INSTRUCTOR S MANUAL (c) XeO 2 2 : nonlinear, fewer than 2C n with n>2, C 2,noC 2 perpendicular to C 2,noσ h,2σ v, so C 2v (d) e 2 (CO) 9 : nonlinear, fewer than 2C n with n>2, C 3, 3C 2 perpendicular to C 3,σ h,so D 3h (e) cubane (C 8 H 8 ): nonlinear, more than 2C n with n>2, i, noc 5,so O h (f) tetrafluorocubane (23): nonlinear, more than 2C n with n>2, no i, so T d E5.5(b) (a) Only molecules belonging to C s, C n, and C nv groups may be polar. In Exercise 5.3b ortho-dichlorobenzene and meta-dichlorobenzene belong to C 2v and so may be polar; in Exercise 5.0b, H and XeO 2 2 belong to C nv groups, so they may be polar. (b) A molecule cannot be chiral if it has an axis of improper rotation including disguised or degenerate axes such as an inversion centre (S 2 ) or a mirror plane (S ). In Exercises 5.9b and 5.0b, all the molecules have mirror planes, so none can be chiral. E5.6(b) In order to have nonzero overlap with a combination of orbitals that spans E, an orbital on the central atom must itself have some E character, for only E can multiply E to give an overlap integral with a totally symmetric part. A glance at the character table shows that p x and p y orbitals available to a bonding N atom have the proper symmetry. If d orbitals are available (as in SO 3 ), all d orbitals except dz 2 could have nonzero overlap. E5.7(b) The product Ɣ f Ɣ(µ) Ɣ i must contain A (Example 5.7). Then, since Ɣ i = B, Ɣ(µ) = Ɣ(y) = B 2 (C 2v character table), we can draw up the following table of characters E C 2 σ v σ v B 2 B B B 2 = A 2 E5.8(b) Hence, the upper state is A 2, because A 2 A 2 = A. (a) Anthracene H H H H H D 2h H H H H H The components of µ span B 3u (x), B 2u (y), and B u (z). The totally symmetric ground state is A g. Since A g Ɣ = Ɣ in this group, the accessible upper terms are B 3u (x-polarized), B 2u (y-polarized), and B u (z-polarized). (b) Coronene, like benzene, belongs to the D 6h group. The integrand of the transition dipole moment must be or contain the A g symmetry species. That integrand for transitions from the ground state is A g qf, where q is x,y,orzand f is the symmetry species of the upper state. Since the ground state is already totally symmetric, the product qf must also have A g symmetry for the entire integrand to have A g symmetry. Since the different symmetry species are orthogonal, the only way qf can have A g symmetry is if q and f have the same symmetry. Such combinations include za 2u, xe u, and ye u. Therefore, we conclude that transitions are allowed to states with A 2u or E u symmetry.
5 MOLECULAR SYMMETRY 247 E5.9(b) E 2C 3 3σ v A A 2 E 2 0 sin θ cos θ Linear combinations of sin θ and cos θ Product The product does not contain A,so yes the integral vanishes. Solutions to problems P5.3 Consider ig The effect of σ h on a point P is to generate σ h P, and the effect of C 2 on σ h P is to generate the point C 2 σ h P. The same point is generated from P by the inversion i,soc 2 σ h P = ip for all points P. Hence, C 2 σ h = i, and i must be a member of the group. igure 5.2 P5.6 Representation D(C 3 )D(C 2 ) = = = D(C 6 ) and from the character table is either A or A 2. Hence, either D(σ v ) = D(σ d ) = respectively. Representation 2 + or D(C 3 )D(C 2 ) = ( ) = = D(C 6 ) and from the character table is either B or B 2. Hence, either D(σ v ) = D(σ d ) = or D(σ v ) = D(σ d ) = respectively. P5.8 A quick rule for determining the character without first having to set up the matrix representation is to count each time a basis function is left unchanged by the operation, because only these functions give a nonzero entry on the diagonal of the matrix representative. In some cases there is a sign change, (... f...) (...f...); then occurs on the diagonal, and so count. The character of the identity is always equal to the dimension of the basis since each function contributes to the trace.
6 248 INSTRUCTOR S MANUAL E: all four orbitals are left unchanged; hence χ = 4 C 3 : One orbital is left unchanged; hence χ = C 2 : No orbitals are left unchanged; hence χ = 0 S 4 : No orbitals are left unchanged; hence χ = 0 σ d : Two orbitals are left unchanged; hence χ = 2 The character set 4,, 0, 0, 2 spans A + T 2. Inspection of the character table of the group T d shows that s spans A and that the three p orbitals on the C atom span T 2. Hence, the s and p orbitals of the C atom may form molecular orbitals with the four Hs orbitals. In T d, the d orbitals of the central atom span E + T 2 (character table, final column), and so only the T 2 set (d xy, d yz, d zx ) may contribute to molecular orbital formation with the H orbitals. P5.9 (a) In C 3v symmetry the Hs orbitals span the same irreducible representations as in NH 3, which is A + A + E. There is an additional A orbital because a fourth H atom lies on the C 3 axis. In C 3v, the d orbitals span A + E + E [see the final column of the C 3v character table]. Therefore, all five d orbitals may contribute to the bonding. (b) In C 2v symmetry the Hs orbitals span the same irreducible representations as in H 2 O, but one H 2 O fragment is rotated by 90 with respect to the other. Therefore, whereas in H 2 O the Hs orbitals span A + B 2 [H + H 2, H H 2 ], in the distorted CH 4 molecule they span A + B 2 + A + B [H + H 2, H H 2, H 3 + H 4, H 3 H 4 ]. In C 2v the d orbitals span 2A + B + B 2 + A 2 [C 2v character table]; therefore, all except A 2 (d xy ) may participate in bonding. P5.0 The most distinctive symmetry operation is the S 4 axis through the central atom and aromatic nitrogens on both ligands. That axis is also a C 2 axis. The group is S 4. P5.2 (a) Working through the flow diagram (ig. 5.4) in the text, we note that there are no C n axes with n>2 (for the C 3 axes present in a tetrahedron are not symmetry axes any longer), but it does have C 2 axes; in fact it has 2C 2 axes perpendicular to whichever C 2 we call principal; it has no σ h, but it has 2σ d. So the point group is D 2d. (b) Within this point group, the distortion belongs to the fully symmetric species A, for its motion is unchanged by the S 4 operation, either class of C 2,orσ d. (c) The resulting structure is a square bipyramid, but with one pyramid s apex farther from the base than the other s. Working through the flow diagram in ig. 5.4, we note that there is only one C n axis with n>2, namely a C 4 axis; it has no C 2 axes perpendicular to the C 4, and it has no σ h, but it has 4σ v. So the point group is C 4v. (d) Within this point group, the distortion belongs to the fully symmetric species A. The translation of atoms along the given axis is unchanged by any symmetry operation for the motion is contained within each of the group s symmetry elements. P5.4 (a) xyz changes sign under the inversion operation (one of the symmetry elements of a cube); hence it does not span A g and its integral must be zero (b) xyz spans A in T d [Problem 5.3] and so its integral need not be zero (c) xyz xyz under z z (the σ h operation in D 6h ), and so its integral must be zero
7 MOLECULAR SYMMETRY 249 P5.6 We shall adapt the simpler subgroup C 6v of the full D 6h point group. The six π-orbitals span A + B + E + E 2, and are a = (π + π 2 + π 3 + π 4 + π 5 + π 6 ) 6 b = (π π 2 + π 3 π 4 + π 5 π 6 ) 6 (2π π 2 π 3 + 2π 4 π 5 π 6 ) e 2 = 2 2 (π 2 π 3 + π 5 π 6 ) (2π + π 2 π 3 2π 4 π 5 + π 6 ) e = 2 2 (π 2 + π 3 π 5 π 6 ) The hamiltonian transforms as A ; therefore all integrals of the form ψ Hψ dτ vanish unless ψ and ψ belong to the same symmetry species. It follows that the secular determinant factorizes into four determinants A : H a a = 6 (π + +π 6 )H (π + +π 6 ) dτ = α + 2β B : H b b = 6 (π π 2 + )H (π π 2 + ) dτ = α 2β E : H e (a)e (a) = α β, H e (b)e (b) = α β, H e (a)e (b) = 0 Hence α β ε 0 = 0 solves to ε = α β(twice) 0 α β ε E 2 : H e2 (a)e 2 (a) = α + β, H e2 (b)e 2 (b) = α + β, H e2 (a)e 2 (b) = 0 Hence α + β ε 0 = 0 solves to ε = α + β(twice) 0 α + β ε P5.7 Consider phenanthrene with carbon atoms as labeled in the figure below d c e b f a g a b g f e c d (a) The 2p orbitals involved in the π system are the basis we are interested in. To find the irreproducible representations spanned by this basis, consider how each basis is transformed under the symmetry operations of the C 2v group. To find the character of an operation in this basis, sum the coefficients of the basis terms that are unchanged by the operation. a a b b c c d d e e f f g g χ E a a b b c c d d e e f f g g 4 C 2 a a b b c c d d e e f f g g 0 σ v a a b b c c d d e e f f g g 0 σ v a a b b c c d d e e f f g g 4
8 250 INSTRUCTOR S MANUAL To find the irreproducible representations that these orbitals span, multiply the characters in the representation of the orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 5.5(a)). The table below illustrates the procedure, beginning at left with the C 2v character table. E C 2 σ v σ v product E C 2 σ v σ v sum/h A A B B The orbitals span 7A 2 + B 2. To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 5.5(c). Refer to the table above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species irreproducible representation, sum the terms in the column, and divide by the order of the group. or example, the characters of species A are,,,, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of A symmetry. (No surprise here: the orbitals span only A 2 and B.) An A 2 SALC is obtained by multiplying the characters,,, by the first column: 4 (a a a + a) = 2 (a a ). The A 2 combination from the second column is the same. There are seven distinct A 2 combinations in all: /2(a a ), /2(b b ),...,/2(g g ). The B combination from the first column is: 4 (a + a + a + a) = 2 (a + a ). The B combination from the second column is the same. There are seven distinct B combinations in all: 2 (a + a ), 2 (b + b ),..., 2 (g + g ). There are no B 2 combinations, as the columns sum to zero. (b) The structure is labeled to match the row and column numbers shown in the determinant. The Hückel secular determinant of phenanthrene is: a b c d e f g g f e d c b a a α E β β b β α E β β c 0 β α E β d 0 0 β α E β e β α E β f β α E β g 0 β β α E β g β α E β β 0 f β α E β e β α E β d β α E β 0 0 c β α E β 0 b β β α E β a β β α E This determinant has the same eigenvalues as as in exercise 4.6(b)b.
9 MOLECULAR SYMMETRY 25 (c) The ground state of the molecule has A symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has A character. If a transition is to be allowed, the transition dipole must be non-zero, which in turn can only happen if the representation of the product f µ i includes the totally symmetric species A. Consider first transitions to another A wavefunction, in which case we need the product A µa.nowa A = A, and the only character that returns A when multiplied by A is A itself. The z component of the dipole operator belongs to species A,soz-polarized A A transitions are allowed. (Note: transitions from the A ground state to an A excited state are transitions from an orbital occupied in the ground state to an excited-state orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (A 2 or B ) to the other; in that case, the excited-state wavefunction will have symmetry of A B = B 2 from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A µb 2 = µb 2, and the only species that yields A when multiplied by B 2 is B 2 itself. Now the y component of the dipole operator belongs to species B 2, so these transitions are also allowed (y-polarized). P5.2 (a) ollowing the flow chart in ig. 5.4, not that the molecule is not linear (at least not in the mathematical sense); there is only one C n axis (a C 2 ), and there is a σ h. The point group, then, is C 2h. b d f h j k i g e c a a c e g i k j h f d b (b) The 2p z orbitals are transformed under the symmetry operations of the C 2h group as follows. a a b b c c... j j k k χ E a a b b c c... j j k k 22 C 2 a a b b c c... j j k k 0 i a a b b c c... j j k k 0 σ h a a b b c c... j j k k 22 To find the irreproducible representations that these orbitals span, we multiply the characters of orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 5.5(a)). The table below illustrates the procedure, beginning at left with the C 2h character table. E C 2 i σ h product E C 2 i σ h sum/h A g A u B g B u The orbitals span A u + B g. To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 5.5(c). Refer to the above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species irreproducible representation, sum the terms in the column, and divide by the order of the group. or example, the characters of species A u are,,,, so the
10 252 INSTRUCTOR S MANUAL columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of A g symmetry. (No surprise: the orbitals span only A u and B g ). An A u SALC is obtained by multiplying the characters,,, bythe first column: 4 (a + a + a + a) = 2 (a + a ). The A u combination from the second column is the same. There are distinct A u combinations in all: /2(a + a ), /2(b + b ),.../2(k + k ). The B g combination from the first column is: 4 (a a a + a) = 2 (a a ). The B g combination from the second column is the same. There are distinct B g combinations in all: /2(a a ), /2(b b ),.../2(k k ). There are no B u combinations, as the columns sum to zero. (c) The structure is labeled to match the row and column numbers shown in the determinant. The Hückel secular determinant is: a b c... i j k k j i... c b a a α E β b β α E β c 0 β α E i α E β j β α E β k β α E β k β α E β j β α E β i β α E c α E β 0 b β α E β a β α E The energies of the filled orbitals are α β, α β, α β, α β, α β, α +.365β, α β, α β, α β, α β, and α β. The π energy is β. (d) The ground state of the molecule has A g symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has A g character. If a transition is to be allowed, the transition dipole must be nonzero, which in turn can only happen if the representation of the product f µ i includes the totally symmetric species A g. Consider first transitions to another A g wavefunction, in which case we need the product A g µa g.nowa g A g = A g, and the only character that returns A g when multiplied by A g is A g itself. No component of the dipole operator belongs to species A g,sonoa g A g transitions are allowed. (Note: such transitions are transitions from an orbital occupied in the ground state to an excited-state orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (A u or B g ) to the other; in that case, the excited-state wavefunction will have symmetry of A u B g = B u from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A g µb u = µb u, and the only species that yields A g when multiplied by B u is B u itself. The x and y components of the dipole operator belongs to species B u, so these transitions are allowed.
Molecular Symmetry 1
Molecular Symmetry 1 I. WHAT IS SYMMETRY AND WHY IT IS IMPORTANT? Some object are more symmetrical than others. A sphere is more symmetrical than a cube because it looks the same after rotation through
More informationGroup Theory and Molecular Symmetry
Group Theory and Molecular Symmetry Molecular Symmetry Symmetry Elements and perations Identity element E - Apply E to object and nothing happens. bject is unmoed. Rotation axis C n - Rotation of object
More informationSymmetry and group theory
Symmetry and group theory or How to Describe the Shape of a Molecule with two or three letters Natural symmetry in plants Symmetry in animals 1 Symmetry in the human body The platonic solids Symmetry in
More informationCHAPTER 5: MOLECULAR ORBITALS
Chapter 5 Molecular Orbitals 5 CHAPTER 5: MOLECULAR ORBITALS 5. There are three possible bonding interactions: p z d z p y d yz p x d xz 5. a. Li has a bond order of. (two electrons in a bonding orbital;
More informationGroup Theory and Chemistry
Group Theory and Chemistry Outline: Raman and infra-red spectroscopy Symmetry operations Point Groups and Schoenflies symbols Function space and matrix representation Reducible and irreducible representation
More informationC 3 axis (z) y- axis
Point Group Symmetry E It is assumed that the reader has previously learned, in undergraduate inorganic or physical chemistry classes, how symmetry arises in molecular shapes and structures and what symmetry
More informationMath 215 HW #6 Solutions
Math 5 HW #6 Solutions Problem 34 Show that x y is orthogonal to x + y if and only if x = y Proof First, suppose x y is orthogonal to x + y Then since x, y = y, x In other words, = x y, x + y = (x y) T
More informationVSEPR Model. The Valence-Shell Electron Pair Repulsion Model. Predicting Molecular Geometry
VSEPR Model The structure around a given atom is determined principally by minimizing electron pair repulsions. The Valence-Shell Electron Pair Repulsion Model The valence-shell electron pair repulsion
More informationMOLECULAR SYMMETRY, GROUP THEORY, & APPLICATIONS
1 MOLECULAR SYMMETRY, GROUP THEORY, & APPLICATIONS Lecturer: Claire Vallance (CRL office G9, phone 75179, e-mail claire.vallance@chem.ox.ac.uk) These are the lecture notes for the second year general chemistry
More informationAdding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors
1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number
More informationLecture 34: Symmetry Elements
Lecture 34: Symmetry Elements The material in this lecture covers the following in Atkins. 15 Molecular Symmetry The symmetry elements of objects 15.1 Operations and symmetry elements 15.2 Symmetry classification
More information1. Human beings have a natural perception and appreciation for symmetry.
I. SYMMETRY ELEMENTS AND OPERATIONS A. Introduction 1. Human beings have a natural perception and appreciation for symmetry. a. Most people tend to value symmetry in their visual perception of the world.
More informationMolecular Geometry and Chemical Bonding Theory
Chapter 10 Molecular Geometry and Chemical Bonding Theory Concept Check 10.1 An atom in a molecule is surrounded by four pairs of electrons, one lone pair and three bonding pairs. Describe how the four
More informationChapter 9. Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure
Chapter 9 Molecular Geometry & Bonding Theories I) Molecular Geometry (Shapes) Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure Molecular
More informationThe Unshifted Atom-A Simpler Method of Deriving Vibrational Modes of Molecular Symmetries
Est. 1984 ORIENTAL JOURNAL OF CHEMISTRY An International Open Free Access, Peer Reviewed Research Journal www.orientjchem.org ISSN: 0970-020 X CODEN: OJCHEG 2012, Vol. 28, No. (1): Pg. 189-202 The Unshifted
More informationC has 4 valence electrons, O has six electrons. The total number of electrons is 4 + 2(6) = 16.
129 Lewis Structures G. N. Lewis hypothesized that electron pair bonds between unlike elements in the second (and sometimes the third) row occurred in a way that electrons were shared such that each element
More informationACE PRACTICE TEST Chapter 8, Quiz 3
ACE PRACTICE TEST Chapter 8, Quiz 3 1. Using bond energies, calculate the heat in kj for the following reaction: CH 4 + 4 F 2 CF 4 + 4 HF. Use the following bond energies: CH = 414 kj/mol, F 2 = 155 kj/mol,
More informationMAT 200, Midterm Exam Solution. a. (5 points) Compute the determinant of the matrix A =
MAT 200, Midterm Exam Solution. (0 points total) a. (5 points) Compute the determinant of the matrix 2 2 0 A = 0 3 0 3 0 Answer: det A = 3. The most efficient way is to develop the determinant along the
More information1 Symmetries of regular polyhedra
1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an
More informationAn introduction to molecular symmetry
44 An introduction to molecular symmetry 4.2 igure 4.1 or answer 4.2: the principal axis of rotation, and the two mirror pianes in H^O. (a) E is the identity operator. It effectively identifies the molecular
More informationChapter 10 Molecular Geometry and Chemical Bonding Theory
Chem 1: Chapter 10 Page 1 Chapter 10 Molecular Geometry and Chemical Bonding Theory I) VSEPR Model Valence-Shell Electron-Pair Repulsion Model A) Model predicts Predicts electron arrangement and molecular
More informationSimilar matrices and Jordan form
Similar matrices and Jordan form We ve nearly covered the entire heart of linear algebra once we ve finished singular value decompositions we ll have seen all the most central topics. A T A is positive
More informationEXPERIMENT 17 : Lewis Dot Structure / VSEPR Theory
EXPERIMENT 17 : Lewis Dot Structure / VSEPR Theory Materials: Molecular Model Kit INTRODUCTION Although it has recently become possible to image molecules and even atoms using a high-resolution microscope,
More informationNonlinear Iterative Partial Least Squares Method
Numerical Methods for Determining Principal Component Analysis Abstract Factors Béchu, S., Richard-Plouet, M., Fernandez, V., Walton, J., and Fairley, N. (2016) Developments in numerical treatments for
More informationSHAPES OF MOLECULES (VSEPR MODEL)
1 SAPES MLEULES (VSEPR MDEL) Valence Shell Electron-Pair Repulsion model - Electron pairs surrounding atom spread out as to minimize repulsion. - Electron pairs can be bonding pairs (including multiple
More informationLaboratory 11: Molecular Compounds and Lewis Structures
Introduction Laboratory 11: Molecular Compounds and Lewis Structures Molecular compounds are formed by sharing electrons between non-metal atoms. A useful theory for understanding the formation of molecular
More informationName: Class: Date: 3) The bond angles marked a, b, and c in the molecule below are about,, and, respectively.
Name: Class: Date: Unit 9 Practice Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The basis of the VSEPR model of molecular bonding is. A) regions of
More information5. Which of the following is the correct Lewis structure for SOCl 2
Unit C Practice Problems Chapter 8 1. Draw the lewis structures for the following molecules: a. BeF 2 b. SO 3 c. CNS 1- d. NO 2. The correct Lewis symbol for ground state carbon is a) b) c) d) e) 3. Which
More informationn 2 + 4n + 3. The answer in decimal form (for the Blitz): 0, 75. Solution. (n + 1)(n + 3) = n + 3 2 lim m 2 1
. Calculate the sum of the series Answer: 3 4. n 2 + 4n + 3. The answer in decimal form (for the Blitz):, 75. Solution. n 2 + 4n + 3 = (n + )(n + 3) = (n + 3) (n + ) = 2 (n + )(n + 3) ( 2 n + ) = m ( n
More informationIn part I of this two-part series we present salient. Practical Group Theory and Raman Spectroscopy, Part I: Normal Vibrational Modes
ELECTRONICALLY REPRINTED FROM FEBRUARY 2014 Molecular Spectroscopy Workbench Practical Group Theory and Raman Spectroscopy, Part I: Normal Vibrational Modes Group theory is an important component for understanding
More informationChapter 17. Orthogonal Matrices and Symmetries of Space
Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length
More information13 MATH FACTS 101. 2 a = 1. 7. The elements of a vector have a graphical interpretation, which is particularly easy to see in two or three dimensions.
3 MATH FACTS 0 3 MATH FACTS 3. Vectors 3.. Definition We use the overhead arrow to denote a column vector, i.e., a linear segment with a direction. For example, in three-space, we write a vector in terms
More informationContinuous Groups, Lie Groups, and Lie Algebras
Chapter 7 Continuous Groups, Lie Groups, and Lie Algebras Zeno was concerned with three problems... These are the problem of the infinitesimal, the infinite, and continuity... Bertrand Russell The groups
More informationA pure covalent bond is an equal sharing of shared electron pair(s) in a bond. A polar covalent bond is an unequal sharing.
CHAPTER EIGHT BNDING: GENERAL CNCEPT or Review 1. Electronegativity is the ability of an atom in a molecule to attract electrons to itself. Electronegativity is a bonding term. Electron affinity is the
More informationCovalent Bonding & Molecular Orbital Theory
Covalent Bonding & Molecular Orbital Theory Chemistry 754 Solid State Chemistry Dr. Patrick Woodward Lecture #16 References - MO Theory Molecular orbital theory is covered in many places including most
More informationBonding Models. Bonding Models (Lewis) Bonding Models (Lewis) Resonance Structures. Section 2 (Chapter 3, M&T) Chemical Bonding
Bonding Models Section (Chapter, M&T) Chemical Bonding We will look at three models of bonding: Lewis model Valence Bond model M theory Bonding Models (Lewis) Bonding Models (Lewis) Lewis model of bonding
More informationPhysics 235 Chapter 1. Chapter 1 Matrices, Vectors, and Vector Calculus
Chapter 1 Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. The main concepts that will be covered are: Coordinate transformations
More informationLMB Crystallography Course, 2013. Crystals, Symmetry and Space Groups Andrew Leslie
LMB Crystallography Course, 2013 Crystals, Symmetry and Space Groups Andrew Leslie Many of the slides were kindly provided by Erhard Hohenester (Imperial College), several other illustrations are from
More information11.1. Objectives. Component Form of a Vector. Component Form of a Vector. Component Form of a Vector. Vectors and the Geometry of Space
11 Vectors and the Geometry of Space 11.1 Vectors in the Plane Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. 2 Objectives! Write the component form of
More information1 2 3 1 1 2 x = + x 2 + x 4 1 0 1
(d) If the vector b is the sum of the four columns of A, write down the complete solution to Ax = b. 1 2 3 1 1 2 x = + x 2 + x 4 1 0 0 1 0 1 2. (11 points) This problem finds the curve y = C + D 2 t which
More information[1] Diagonal factorization
8.03 LA.6: Diagonalization and Orthogonal Matrices [ Diagonal factorization [2 Solving systems of first order differential equations [3 Symmetric and Orthonormal Matrices [ Diagonal factorization Recall:
More informationMolecular-Orbital Theory
Molecular-Orbital Theory 1 Introduction Orbitals in molecules are not necessarily localized on atoms or between atoms as suggested in the valence bond theory. Molecular orbitals can also be formed the
More informationCovalent Bonding and Molecular Geometry
Name Section # Date of Experiment Covalent Bonding and Molecular Geometry When atoms combine to form molecules (this also includes complex ions) by forming covalent bonds, the relative positions of the
More informationVector Math Computer Graphics Scott D. Anderson
Vector Math Computer Graphics Scott D. Anderson 1 Dot Product The notation v w means the dot product or scalar product or inner product of two vectors, v and w. In abstract mathematics, we can talk about
More information5.04 Principles of Inorganic Chemistry II
MIT OpenourseWare http://ocw.mit.edu 5.4 Principles of Inorganic hemistry II Fall 8 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.4, Principles of
More informationCH101/105, GENERAL CHEMISTRY LABORATORY
CH101/105, GENERAL CHEMITRY LABORATORY LABORATORY LECTURE 5 EXPERIMENT 5: LEWI TRUCTURE AND MOLECULAR HAPE Lecture topics I. LEWI TRUCTURE a) calculation of the valence electron numbers; b) choosing the
More informationGeometries and Valence Bond Theory Worksheet
Geometries and Valence Bond Theory Worksheet Also do Chapter 10 textbook problems: 33, 35, 47, 49, 51, 55, 57, 61, 63, 67, 83, 87. 1. Fill in the tables below for each of the species shown. a) CCl 2 2
More informationMATRIX ALGEBRA AND SYSTEMS OF EQUATIONS
MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a
More informationQuestion 2: How do you solve a matrix equation using the matrix inverse?
Question : How do you solve a matrix equation using the matrix inverse? In the previous question, we wrote systems of equations as a matrix equation AX B. In this format, the matrix A contains the coefficients
More informationSolving Simultaneous Equations and Matrices
Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering
More informationLecture L3 - Vectors, Matrices and Coordinate Transformations
S. Widnall 16.07 Dynamics Fall 2009 Lecture notes based on J. Peraire Version 2.0 Lecture L3 - Vectors, Matrices and Coordinate Transformations By using vectors and defining appropriate operations between
More informationChapter 9 Unitary Groups and SU(N)
Chapter 9 Unitary Groups and SU(N) The irreducible representations of SO(3) are appropriate for describing the degeneracies of states of quantum mechanical systems which have rotational symmetry in three
More informationNotes on Orthogonal and Symmetric Matrices MENU, Winter 2013
Notes on Orthogonal and Symmetric Matrices MENU, Winter 201 These notes summarize the main properties and uses of orthogonal and symmetric matrices. We covered quite a bit of material regarding these topics,
More informationForce on Moving Charges in a Magnetic Field
[ Assignment View ] [ Eðlisfræði 2, vor 2007 27. Magnetic Field and Magnetic Forces Assignment is due at 2:00am on Wednesday, February 28, 2007 Credit for problems submitted late will decrease to 0% after
More information1. Units of a magnetic field might be: A. C m/s B. C s/m C. C/kg D. kg/c s E. N/C m ans: D
Chapter 28: MAGNETIC FIELDS 1 Units of a magnetic field might be: A C m/s B C s/m C C/kg D kg/c s E N/C m 2 In the formula F = q v B: A F must be perpendicular to v but not necessarily to B B F must be
More informationThe excitation in Raman spectroscopy is usually. Practical Group Theory and Raman Spectroscopy, Part II: Application of Polarization
Electronically reprinted from March 214 Molecular Spectroscopy Workbench Practical Group Theory and Raman Spectroscopy, Part II: Application of Polarization In this second installment of a two-part series
More information9 MATRICES AND TRANSFORMATIONS
9 MATRICES AND TRANSFORMATIONS Chapter 9 Matrices and Transformations Objectives After studying this chapter you should be able to handle matrix (and vector) algebra with confidence, and understand the
More informationCHEM 101 Exam 4. Page 1
CEM 101 Exam 4 Form 1 (White) November 30, 2001 Page 1 Section This exam consists of 8 pages. When the exam begins make sure you have one of each. Print your name at the top of each page now. Show your
More information8 Square matrices continued: Determinants
8 Square matrices continued: Determinants 8. Introduction Determinants give us important information about square matrices, and, as we ll soon see, are essential for the computation of eigenvalues. You
More informationModule 3 : Molecular Spectroscopy Lecture 13 : Rotational and Vibrational Spectroscopy
Module 3 : Molecular Spectroscopy Lecture 13 : Rotational and Vibrational Spectroscopy Objectives After studying this lecture, you will be able to Calculate the bond lengths of diatomics from the value
More informationOrthogonal Projections
Orthogonal Projections and Reflections (with exercises) by D. Klain Version.. Corrections and comments are welcome! Orthogonal Projections Let X,..., X k be a family of linearly independent (column) vectors
More informationSelf Assessment_Ochem I
UTID: 2013 Objective Test Section Identify the choice that best completes the statement or answers the question. There is only one correct answer; please carefully bubble your choice on the scantron sheet.
More informationFigure 1.1 Vector A and Vector F
CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have
More informationIntroduction to Matrix Algebra
Psychology 7291: Multivariate Statistics (Carey) 8/27/98 Matrix Algebra - 1 Introduction to Matrix Algebra Definitions: A matrix is a collection of numbers ordered by rows and columns. It is customary
More informationMATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m
MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +
More informationSYSTEMS OF EQUATIONS AND MATRICES WITH THE TI-89. by Joseph Collison
SYSTEMS OF EQUATIONS AND MATRICES WITH THE TI-89 by Joseph Collison Copyright 2000 by Joseph Collison All rights reserved Reproduction or translation of any part of this work beyond that permitted by Sections
More informationChapter 10 Electronic Wavefunctions Must Also Possess Proper Symmetry. These Include Angular Momentum and Point Group Symmetries
Chapter 10 Electronic Wavefunctions Must Also Possess Proper Symmetry. These Include Angular Momentum and Point Group Symmetries I. Angular Momentum Symmetry and Strategies for Angular Momentum Coupling
More informationch9 and 10 practice test
1. Which of the following covalent bonds is the most polar (highest percent ionic character)? A. Al I B. Si I C. Al Cl D. Si Cl E. Si P 2. What is the hybridization of the central atom in ClO 3? A. sp
More informationCHAPTER 4: SYMMETRY AND GROUP THEORY
hapter 4 Symmetry and Group Theory 33 APTER 4: SYMMETRY AND GROUP TEORY 4. a. Ethane in the staggered conformation has 2 3 axes (the line), 3 perpendicular 2 axes bisecting the line, in the plane of the
More informationIntegrating algebraic fractions
Integrating algebraic fractions Sometimes the integral of an algebraic fraction can be found by first epressing the algebraic fraction as the sum of its partial fractions. In this unit we will illustrate
More informationReview Jeopardy. Blue vs. Orange. Review Jeopardy
Review Jeopardy Blue vs. Orange Review Jeopardy Jeopardy Round Lectures 0-3 Jeopardy Round $200 How could I measure how far apart (i.e. how different) two observations, y 1 and y 2, are from each other?
More informationNumber Sense and Operations
Number Sense and Operations representing as they: 6.N.1 6.N.2 6.N.3 6.N.4 6.N.5 6.N.6 6.N.7 6.N.8 6.N.9 6.N.10 6.N.11 6.N.12 6.N.13. 6.N.14 6.N.15 Demonstrate an understanding of positive integer exponents
More informationMolecular Geometry and VSEPR We gratefully acknowledge Portland Community College for the use of this experiment.
Molecular and VSEPR We gratefully acknowledge Portland ommunity ollege for the use of this experiment. Objectives To construct molecular models for covalently bonded atoms in molecules and polyatomic ions
More information5.3 The Cross Product in R 3
53 The Cross Product in R 3 Definition 531 Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] Then the vector given by [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] is called the cross product (or
More informationLecture L26-3D Rigid Body Dynamics: The Inertia Tensor
J. Peraire, S. Widnall 16.07 Dynaics Fall 008 Lecture L6-3D Rigid Body Dynaics: The Inertia Tensor Version.1 In this lecture, we will derive an expression for the angular oentu of a 3D rigid body. We shall
More informationChemistry Workbook 2: Problems For Exam 2
Chem 1A Dr. White Updated /5/1 1 Chemistry Workbook 2: Problems For Exam 2 Section 2-1: Covalent Bonding 1. On a potential energy diagram, the most stable state has the highest/lowest potential energy.
More informationState of Stress at Point
State of Stress at Point Einstein Notation The basic idea of Einstein notation is that a covector and a vector can form a scalar: This is typically written as an explicit sum: According to this convention,
More informationLinear Algebra Review. Vectors
Linear Algebra Review By Tim K. Marks UCSD Borrows heavily from: Jana Kosecka kosecka@cs.gmu.edu http://cs.gmu.edu/~kosecka/cs682.html Virginia de Sa Cogsci 8F Linear Algebra review UCSD Vectors The length
More informationEigenvalues and Eigenvectors
Chapter 6 Eigenvalues and Eigenvectors 6. Introduction to Eigenvalues Linear equations Ax D b come from steady state problems. Eigenvalues have their greatest importance in dynamic problems. The solution
More informationVector Notation: AB represents the vector from point A to point B on a graph. The vector can be computed by B A.
1 Linear Transformations Prepared by: Robin Michelle King A transformation of an object is a change in position or dimension (or both) of the object. The resulting object after the transformation is called
More informationDecember 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS
December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in two-dimensional space (1) 2x y = 3 describes a line in two-dimensional space The coefficients of x and y in the equation
More information5.61 Physical Chemistry 25 Helium Atom page 1 HELIUM ATOM
5.6 Physical Chemistry 5 Helium Atom page HELIUM ATOM Now that we have treated the Hydrogen like atoms in some detail, we now proceed to discuss the next simplest system: the Helium atom. In this situation,
More informationChemistry 105, Chapter 7 Exercises
hemistry 15, hapter 7 Exercises Types of Bonds 1. Using the periodic table classify the bonds in the following compounds as ionic or covalent. If covalent, classify the bond as polar or not. Mg2 4 i2 a(3)2
More informationAP Chemistry A. Allan Chapter 8 Notes - Bonding: General Concepts
AP Chemistry A. Allan Chapter 8 Notes - Bonding: General Concepts 8.1 Types of Chemical Bonds A. Ionic Bonding 1. Electrons are transferred 2. Metals react with nonmetals 3. Ions paired have lower energy
More informationLINEAR ALGEBRA. September 23, 2010
LINEAR ALGEBRA September 3, 00 Contents 0. LU-decomposition.................................... 0. Inverses and Transposes................................. 0.3 Column Spaces and NullSpaces.............................
More informationChapter 2: Crystal Structures and Symmetry
Chapter 2: Crystal Structures and Symmetry Laue, ravais December 28, 2001 Contents 1 Lattice Types and Symmetry 3 1.1 Two-Dimensional Lattices................. 3 1.2 Three-Dimensional Lattices................
More informationIntroduction to Group Theory with Applications in Molecular and Solid State Physics
Introduction to Group Theory with Applications in Molecular and Solid State Physics Karsten Horn Fritz-Haber-Institut der Max-Planck-Gesellschaft 3 84 3, e-mail horn@fhi-berlin.mpg.de Symmetry - old concept,
More informationPCV Project: Excitons in Molecular Spectroscopy
PCV Project: Excitons in Molecular Spectroscopy Introduction The concept of excitons was first introduced by Frenkel (1) in 1931 as a general excitation delocalization mechanism to account for the ability
More informationSection 5.3. Section 5.3. u m ] l jj. = l jj u j + + l mj u m. v j = [ u 1 u j. l mj
Section 5. l j v j = [ u u j u m ] l jj = l jj u j + + l mj u m. l mj Section 5. 5.. Not orthogonal, the column vectors fail to be perpendicular to each other. 5..2 his matrix is orthogonal. Check that
More informationDCI for Electronegativity. Data Table:
DCI for Electronegativity Data Table: Substance Ionic/covalent EN value EN Value EN NaCl ionic (Na) 0.9 (Cl) 3.0 2.1 KBr (K) 0.8 (Br) 2.8 MgO (Mg) 1.2 (O) 3.5 HCl (H) 2.1 (Cl) 3.0 HF (H) 2.1 (F) 4.0 Cl
More informationCHEM 340 CHEMICAL BONDING - in General Lect-07 IONIC COVALENT METAL COVALENT NETWORK
CHEM 340 CHEMICAL BONDING in General Lect07 BONDING between atoms classified as belonging to one of the following types: IONIC COVALENT METAL COVALENT NETWORK or each bond type, the valence shell electrons
More informationWhich substance contains positive ions immersed in a sea of mobile electrons? A) O2(s) B) Cu(s) C) CuO(s) D) SiO2(s)
BONDING MIDTERM REVIEW 7546-1 - Page 1 1) Which substance contains positive ions immersed in a sea of mobile electrons? A) O2(s) B) Cu(s) C) CuO(s) D) SiO2(s) 2) The bond between hydrogen and oxygen in
More informationAntenna Measurement 1 Antenna Ranges antenna range
Antenna Measurement 1 Antenna Ranges An antenna range is a facility where antenna radiation characteristics are measured. An antenna range includes the following typical components: 1. A substantial space
More informationEXPERIMENT # 17 CHEMICAL BONDING AND MOLECULAR POLARITY
EXPERIMENT # 17 CHEMICAL BONDING AND MOLECULAR POLARITY Purpose: 1. To distinguish between different types of chemical bonds. 2. To predict the polarity of some common molecules from a knowledge of bond
More information7.14 Linear triatomic: A-----B-----C. Bond angles = 180 degrees. Trigonal planar: Bond angles = 120 degrees. B < B A B = 120
APTER SEVEN Molecular Geometry 7.13 Molecular geometry may be defined as the three-dimensional arrangement of atoms in a molecule. The study of molecular geometry is important in that a molecule s geometry
More informationComputing Orthonormal Sets in 2D, 3D, and 4D
Computing Orthonormal Sets in 2D, 3D, and 4D David Eberly Geometric Tools, LLC http://www.geometrictools.com/ Copyright c 1998-2016. All Rights Reserved. Created: March 22, 2010 Last Modified: August 11,
More informationThe Matrix Elements of a 3 3 Orthogonal Matrix Revisited
Physics 116A Winter 2011 The Matrix Elements of a 3 3 Orthogonal Matrix Revisited 1. Introduction In a class handout entitled, Three-Dimensional Proper and Improper Rotation Matrices, I provided a derivation
More informationRotation Matrices and Homogeneous Transformations
Rotation Matrices and Homogeneous Transformations A coordinate frame in an n-dimensional space is defined by n mutually orthogonal unit vectors. In particular, for a two-dimensional (2D) space, i.e., n
More informationCHEMISTRY BONDING REVIEW
Answer the following questions. CHEMISTRY BONDING REVIEW 1. What are the three kinds of bonds which can form between atoms? The three types of Bonds are Covalent, Ionic and Metallic. Name Date Block 2.
More informationRecall that two vectors in are perpendicular or orthogonal provided that their dot
Orthogonal Complements and Projections Recall that two vectors in are perpendicular or orthogonal provided that their dot product vanishes That is, if and only if Example 1 The vectors in are orthogonal
More information