basic chemical bonding1 cubane dodecahedrane side and top views of a single-wall exohydrogenated...

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!


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.


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


Lewis Dot Structure

Lewis wrote in a memorandum dated March 28, 1902


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.


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


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.


Periodic Table of Electronegativity


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


Electron Density of a Polar Bond Li–H

Li Hdipole moment


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


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


Find Formal Charge

SO42–

Find FC in these structures

Confirm these FCs


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.


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


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


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


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


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?


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


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?


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.


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.


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.


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)


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,


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)


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

basic chemical bonding1 cubane dodecahedrane side and top views of a single-wall exohydrogenated...

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