Transcript Basic Chemical Bonding - University of Waterloo

Basic Chemical Bonding
Molecules are artwork – just beautiful!
Cubane
Dodecahedrane
Side and top views of a single-wall
exohydrogenated
carbon nanotube
Basic Chemical
Bonding
1
Looking Back at Chemical Bonding
Bonding 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.
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Lewis Theory
G.N. Lewis (1875-1946) recognized
valence (outmost) electrons fundamental to bonding
electron transfer resulting in ionic bonds
sharing electrons resulting in covalent bonds
atoms tend to acquire a noble-gas electronic
configurations
The attraction between electrons of one atom to the nucleus of another
atom contribute to what is known as chemical bonds.
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Lewis Dot
Structure
Lewis wrote in a
memorandum
dated March 28,
1902
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Lewis Dot Structure – 2
Lewis' 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.
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Lewis Dot in Covalent Bond
Write the Lewis dot structures for these molecules:
H–H, H–Cl, H–O–H, NH3, H–, He, Cl–, Ne
H3O+, 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
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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.
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Periodic Table of Electronegativity
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Covalent and Ionic Bonds
The 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
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Electron Density of a Polar Bond Li–H
Li  H
dipole moment
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Writing Lewis Dot Structures
Show 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, HCN, H3PO4, O=N=O
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Formal Charge
The 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.
Basic Chemical Bonding
Explain & workout formal charge
judge stability of a formula
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Find Formal Charge
SO42–
Find FC in
these
structures
Confirm these FCs
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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.
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Draw Resonance Structures
Draw resonance structures for these:
CO2
NO2
NO2HCO2O3
SO3
NO3–
:O::C::O: (plus two more dots for each of O)
.NO (bent molecule due to the odd electron)
2
:NO2- (same number of valence electron as O3 & SO2)
H-CO2– ( ditto)
ozone
consider O-SO2, and the resonance structures
flat same number of valence electron as CO32-
Draw all resonance structures of all these
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Exceptions to the Octet Rule
Molecules with odd number of valence electrons, N=O
(compare to CO), CH3, OH, H, NO2 etc.
:Cl:
:Cl:
:Cl:
/
\
/
\ / \
M
M
M
M
Molecules with incomplete octets,
\
/
\
/ \ /
BeCl2 AlCl3, (gas and polymeric for both),
:Cl:
:Cl:
:Cl:
BF , compare with NH BF , BF –,
3
3
3
4
M = Al or Be
Expanded valence shells, PCl3, PCl5, SF6, H2SO4, H3PO4
Draw Lewis dot structures of all these molecules to see the exceptions
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Bond Properties
Bond length
bond angle
distance between the nuclei of bonded atoms
angles for any two bonds around an atom
bond energy energy required to break the bond
bond-dissociation energy
Bond
Length Energy
C – C 154
C = C 134
C  C 120
348
614
839
O – O 148
O = O 121
145
498
Compound
H2
HF
H2O
NH3
CH4
length energy
Bond (pm) (kJ/mol)
H–H
F–H
O–H
N–H
C–H
Chemical
Discuss the variations of bondBasic
length
andBonding
bond energy
74
92
96
101
109
436
565
464
389
414
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VSEPR Theory
Valence-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
Basic Chemical Bonding
HFE3
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Shape of Molecules
During the lecture, we will discuss structures of the following:
Make sure you can draw and name the
AX2 linear
geometrical shape of these structures.
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
AX4E, what’s my shape?
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Chemistry and Molecular Shapes
Neg Example
Descriptor
2
BeCl2, CO2 Linear
3
BF3, SO3
Trigonal planar
SO2E, OO2E Bent
4
CH4
NH3E
H2OE2
Tetrahedral
pyramidal
Bent
PF5
Trigonal
bypyramidal
Seesaw, butterfly
T-shape
5
SF4E
ClF3E2
Neg Example
Descriptor
6
Octahedral
Pyramidal
Square planar
SF6, OIF5
BrF5E
XeF4E2
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Structures with Multiple Covalent Bonds
We will talk about pi (p) 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
H
..
\
S
/
::O:
What is the Neg for
SO42–, COS, N2O?
C=O
\\
:O:
Basic Chemical Bonding
/
H
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Molecules with more than one central atom
Describe 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.
180o
109o
H-C
H
N=C=O
120o
H
What are the formal charges of all atoms in both structures?
Describe the structures of C2H5OH, CH3CO2H, and
H2NCH2CH2(OH)COOH. Basic Chemical Bonding
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Dipole Moment
The 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 distance
Symbol:
µ = e– x d = dq * dbond
For Cl+H, µ = 1.03 D, dH–Cl = 127.4 pm
Two ways of lookint at H+Cl,
dq = 1.03*3.34e–30 C m / 1.274e-12 m
= 2.70e-20 C (charge separation by H–Cl )
Ionic character = dq / e– = 0.17 = 17%
d = 3.44e-30 C m / 1.60e –19 C (e– charge)
– byChemical
Bonding
= 2.15E–11 m = 0.215 pm (+eBasic
0.215 pm)
mH–Cl = 1.03 D
mH–F = 1.9 D, find d and %
ionic character for them.
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Dipole moment of H2O
Verify please: The dipole moment of individual water molecules
measured by Shostak, Ebenstein, and Muenter (1991) is 6.18710–30 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 5.05310–30 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 = 5.05310–20 C, 32 % of the
charge of an electron (1.602210–19 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 6.18710–30 C m many be considered as
separation of charge of electron by 0.039 nm.
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Dipole moment and Molecular Shape
Dipole moments are vectors. The net dipole moment of a molecule is
the resultant (vector sum) of all bond-dipole-moment.
Answer & explain these:
mH–H = ____
mO=C=O = _____
mCH4 = _____
mCCl4 = _____
mBF3 = _____
mH2O = 1.84 D
mO3 = 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)
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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,
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Review 2
Apply 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 DHrxno.
Data:
What is the heat of reaction for
2 H2 (g) + O2 (g)  2 H2O (g),
DHrxno = 2 * D (H–H) + D(O=O) – 4 * D(H–O)
D(O=O) = 498 kJ mol–1
D(H–H) = 436;
D(H–O) = 464;
2 H (g)  H2 (g),
DH = – D (H–H)
H2 (g)  H (g),
= 2 * 436 + 489 – 4*464
DH = D (H–H)
= – 495 kJ compare to DHfo = –248 kJ mol–1 of H2O
Basic Chemical Bonding
Work on example 11-14 on page
423
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Review 3
What is the energy of reaction for
CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)?
Data:
D(C-H) = 414 kJ mol–1
D(Cl–Cl) = 243
D(C-Cl) = 339
D(H–Cl) = 431
Solution:
H3C – H + Cl – Cl  H3C – Cl + H – Cl
+ 414
+ 243
– 339 – 431 kJ
DHrxno = + 414 + 243 – 339 – 431 kJ
= – 113 kJ
Answer: 113 kJ is released in this reaction.
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