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Basic Chemical Bonding 1
Basic Chemical Bonding
Cubane Dodecahedrane
Side and top views of a single-wall exohydrogenated carbon nanotube
Molecules are artwork – just beautiful!
Basic Chemical Bonding 2
Looking Back at Chemical BondingBonding must be electric nature.
1852, E. Frankland proposed the valence concept, using “–” for valence.
1857, F.A. Kekule figured out the structure of benzene C6H6.
1874 J.H. van't Hoff and le Bel postulated the tetrahedral arrangement of 4 bonds around carbon.
1916 G.N. Lewis propsed the dot symbol for valence electrons
1923 G.N. Lewis wrote Valence and the structure of atoms and molecules.
1939 L. Pauling wrote The nature of chemical bond
1940 N.V. Sidgwick and H.E. Powell studied the lone pairs of valence electrons.
Basic Chemical Bonding 3
Lewis Theory
The attraction between electrons of one atom to the nucleus of another atom contribute to what is known as chemical bonds.
G.N. Lewis (1875-1946) recognized valence (outmost) electrons fundamental to bondingelectron transfer resulting in ionic bondssharing electrons resulting in covalent bondsatoms tend to acquire a noble-gas electronic configurations
Basic Chemical Bonding 4
Lewis Dot Structure
Lewis wrote in a memorandum dated March 28, 1902
Basic Chemical Bonding 5
Lewis Dot Structure – 2Lewis' Paper of 1916
In this paper, Lewis begins by using cubes, but he moves away from them by the end of the paper. Here is how he visualized the elements lithium through fluorine:
Please illustrate modern Lewis dot structures of periods 2 and 3 elements. Chieh does that during lecture.
Basic Chemical Bonding 6
Lewis Dot in Covalent Bond
Write the Lewis dot structures for these molecules:
H–H, H–Cl, H–O–H, NH3, H–, He, Cl–, NeH3O+, NH4
+, OH –, (coordinate covalent)Cl2, O2, (multiple bonds) N2, CO2
Explain the types in each line and write the dot structures.
Define: bond pair, lone pair, single bond, double bond, triple bond
Basic Chemical Bonding 7
Polar Covalent Bond & Electronegativity
Discuss the nature of these bonds:
H–F, H–Cl
H–O–H (including lone pairs)
Electronegativity: the ability of an element competing for bonding electrons.
The variation as a function of atomic number and its trends on the Periodic Table has been discussed previously, and the Periodic Table showing electronegativity is shown next.
Basic Chemical Bonding 8
Periodic Table of Electronegativity
Basic Chemical Bonding 9
Covalent and Ionic BondsThe ionicity of a bond depends on the difference in electronegativity.
A difference of 1.7 is given as 50% ionic, and usually considered ionic.
Analyze these
Basic Chemical Bonding 10
Electron Density of a Polar Bond Li–H
Li Hdipole moment
Basic Chemical Bonding 11
Writing Lewis Dot StructuresShow all valence electrons.
Each bond represents two electrons.
All electrons are paired, usually (exceptions).
Each atom acquires 8 valence electrons, usually (exceptions).
Multiple bonds are needed sometimes.
Show class how to write Lewis structure for
CF4, (CX4, SiX4), NH3, H2O, HF
C2H5OH, HCN, H3PO4, O=N=O
Basic Chemical Bonding 12
Formal ChargeThe formal charge on any atom in a Lewis structure is a number assigned to it according to the number of valence electrons of the atom and the number of electrons around it.
The formal charge of an atom is equal to the number of valence electrons, Nv.e. subtract the number of unshared electrons, Nus.e. and subtract half of the bonding electrons, ½ Nb.e..
Formal charge = Nv.e. - Nus.e. - ½ Nb.e.
Stability rules:Formulas with the lowest magnitude of formal charges are more stable.More electonegative atoms should have negative formal charges.Adjacent atoms should have opposite formal charges.
Explain & workout formal charge judge stability of a formula
Basic Chemical Bonding 13
Find Formal Charge
SO42–
Find FC in these structures
Confirm these FCs
Basic Chemical Bonding 14
Resonance When several structures with different electron distributions among the bonds are possible, all structures contribute to the electronic structure of the molecule. These structures are called resonance structures.
When two or more plausible Lewis structures can be written but the “correct” structure cannot be written is called resonance. For example:
. . . . . . O O O
:O: :O : :O: :O: :O: :O :
Please complete the dot structure and find the formal charge for the above structures.
Basic Chemical Bonding 15
Draw Resonance Structures
Draw resonance structures for these:
CO2 :O::C::O: (plus two more dots for each of O)NO2 .NO2 (bent molecule due to the odd electron)NO2
- :NO2- (same number of valence electron as O3 & SO2)
HCO2- H-CO2
– ( ditto)O3 ozone
SO3 consider O-SO2, and the resonance structuresNO3
– flat same number of valence electron as CO32-
Draw all resonance structures of all these
Basic Chemical Bonding 16
Exceptions to the Octet Rule
Molecules with odd number of valence electrons, N=O (compare to CO), CH3, OH, H, NO2 etc.
Molecules with incomplete octets, BeCl2 AlCl3, (gas and polymeric for both), BF3, compare with NH3BF3, BF4
–,
Expanded valence shells, PCl3, PCl5, SF6, H2SO4, H3PO4
:Cl: :Cl: :Cl: / \ / \ / \M M M M \ / \ / \ / :Cl: :Cl: :Cl:
M = Al or Be
Draw Lewis dot structures of all these molecules to see the exceptions
Basic Chemical Bonding 17
Bond PropertiesBond length distance between the nuclei of bonded atoms
bond angle angles for any two bonds around an atom
bond energy energy required to break the bond bond-dissociation energy
length energy Compound Bond (pm) (kJ/mol)
H2 H – H 74 436 HF F – H 92 565
H2O O – H 96 464NH3 N – H 101 389CH4 C – H 109 414
Bond Length Energy
C – C 154 348C = C 134 614C C 120 839
O – O 148 145O = O 121 498
Discuss the variations of bond length and bond energy
Basic Chemical Bonding 18
VSEPR TheoryValence-Shell Electron Repulsion Theory: The VSEPR model counts both bonding and nonbonding (lone) electron pairs (E), and call the total number of pairs number of electron groups (Neg). If the element A has m atoms bonded to it and n nonbonding pairs (E), then
Neg = m + n
Discuss the electronic and molecular structures of CH4, ENH3, & E2OH2. All have Neg = 4.
Bond angles in these structure indicates that E E repulsion is stronger than that of bonding electrons.
CH4 ENH3 H2OE2 HFE3
Basic Chemical Bonding 19
Shape of MoleculesDuring the lecture, we will discuss structures of the following:
AX2 linear
BeCl2
AX3, AX2E triangular planar, bent
BF3, SO2E
AX4, AX3E AX2E2 tetrahedral, pyramidal, bent
AX5, AX4E, AX3E2, AX2E3 triangular pyramidal, butterfly
PCl5, SF4E, ClF3E2, XeF2E3 T-shape, linear
AX6, AX5E, AX4E2 octahedral, square pyramidal, square planar
SF6, BrF5E, XeF4E2, ICl4E2
Make sure you can draw and name the geometrical shape of these structures.
AX4E, what’s my shape?
Basic Chemical Bonding 20
Chemistry and Molecular Shapes
Neg Example Descriptor
2 BeCl2, CO2 Linear
3 BF3, SO3 Trigonal planar
SO2E, OO2E Bent
4 CH4 TetrahedralNH3E pyramidalH2OE2 Bent
5 PF5 Trigonal bypyramidal
SF4E Seesaw, butterflyClF3E2 T-shape
Neg Example Descriptor
6 SF6, OIF5 OctahedralBrF5E PyramidalXeF4E2 Square planar
Basic Chemical Bonding 21
Structures with Multiple Covalent BondsWe will talk about pi () bonding later.
At this stage, you may consider all electrons in a multiple bond are confined around the lines connecting the two atoms. Thus the number of electron groups Neg for a multiple bond is 1.
For example, Neg = 3 for . .S
/ \\ ::O: :O:
H \ C = O /H
What is the Neg forSO4
2–, COS, N2O?
Basic Chemical Bonding 22
Molecules with more than one central atomDescribe the structure of methyl isocyanate, CH3NCO.
Draw the skeleton and add all valence electrons
H3C – N – C – O
Draw the Lewis dot structure that satisfy the octet rule.
N = C = OH - C
H H
120o109o
180o
What are the formal charges of all atoms in both structures? Describe the structures of C2H5OH, CH3CO2H, and H2NCH2CH2(OH)COOH.
Basic Chemical Bonding 23
Dipole MomentThe product of magnitude of charge on a molecule and the distance between two charges of equal magnitude with opposite sign is equal to dipole moment; D (unit is debye, 1 D = 3.34E–30 C m (coulumb.metre); representation Cl+H, a vector )
Dipole moment = charge x distanceSymbol: µ = e– x d = q * dbond
For Cl+H, µ = 1.03 D, dH–Cl = 127.4 pmTwo ways of lookint at H+Cl, q = 1.03*3.34e–30 C m / 1.274e-12 m = 2.70e-20 C (charge separation by H–Cl )Ionic character = q / e– = 0.17 = 17%
d = 3.44e-30 C m / 1.60e –19 C (e– charge) = 2.15E–11 m = 0.215 pm (+e– by 0.215 pm)
H–Cl = 1.03 DH–F = 1.9 D, find d and % ionic character for them.
Basic Chemical Bonding 24
Dipole moment of H2OVerify please: The dipole moment of individual water molecules measured by Shostak, Ebenstein, and Muenter (1991) is 6.18710–30 C m (or 1.855 D). This quantity is a vector resultant of two dipole moments of due to O–H bonds. The bond angle H–O–H of water is 104.5o. Thus, the dipole moment of a O–H bond is 5.05310–30 C m. The bond length between H and O is 0.10 nm, and the partial charge at the O and the H is therefore q = 5.05310–20 C, 32 % of the charge of an electron (1.602210–19 C). Of course, the dipole moment may also be considered as separation of the electron and positive charge by a distance 0.031 nm. For the water molecule, a dipole moment of 6.18710–30 C m many be considered as separation of charge of electron by 0.039 nm.
Basic Chemical Bonding 25
Dipole moment and Molecular ShapeDipole moments are vectors. The net dipole moment of a molecule is the resultant (vector sum) of all bond-dipole-moment.
Answer & explain these: H–H = ____O=C=O = _____ CH4 = _____ CCl4 = _____ BF3 = _____ H2O = 1.84 D O3 = 0.534 D (implication of long pair)
Which are polar and non-polar, SF6, H2O2, C2H4, Cl3CCH3, PCl5, I-Cl, NO, SO2, CH2Cl2, NH3, (put your skill to tell molecular shape at work)
Basic Chemical Bonding 26
Review 1
Predict the molecular geometry of the polyatomic anion ICl4–
Hint:
Draw the Lewis dot structure for Cl and I (figure out the valence e–s)
Drew the Lewis dot structure for ICl4–
What is the number of unshared e– of the above?
Drew the ion, and describe this shape in proper term.
Do the same for NCl3, POCl3, COS, H2CO,
Basic Chemical Bonding 27
Review 2Apply bond energy for thermochemistry calculation
In a chemical reaction, add (–ve) energy released from bonds formed and (+ve) energy required to break the bonds is the energy of the reaction Hrxn
o.
What is the heat of reaction for 2 H2 (g) + O2 (g) 2 H2O (g),
Hrxno = 2 * D (H–H) + D(O=O) – 4 * D(H–O)
= 2 * 436 + 489 – 4*464 = – 495 kJ compare to Hf
o = –248 kJ mol–1 of H2O
Work on example 11-14 on page 423
Data: D(O=O) = 498 kJ mol–1
D(H–H) = 436; D(H–O) = 464;
2 H (g) H2 (g), H = – D (H–H) H2 (g) H (g), H = D (H–H)
Basic Chemical Bonding 28
Review 3What is the energy of reaction for CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)?
Solution:H3C – H + Cl – Cl H3C – Cl + H – Cl + 414 + 243 – 339 – 431 kJ
Hrxno = + 414 + 243 – 339 – 431 kJ
= – 113 kJ
Answer: 113 kJ is released in this reaction.
Data: D(C-H) = 414 kJ mol–1
D(Cl–Cl) = 243 D(C-Cl) = 339 D(H–Cl) = 431