Bonding
Ionic Bonds
• An attraction between anions and cations.
• Stronger than covalent bonds.
• Commonly called salts.
Coulomb’s Law
• The strength of an ionic bond can be
determined by 2 factors:
– The charge of the ions, the higher the
numerical charge value the stronger the
attraction.
– The distance of the ions, the closer they are
the stronger the bond.
Lattice Energy
• The energy required to break an ionic
bond into its gaseous foundational
elements:
Covalent Bonding
• When two atoms have a small difference in their
tendencies to lose or gain electrons, we observe
electron sharing and covalent bonding.
• This type of bonding most commonly occurs
between nonmetal atoms (although a pair of metal
atoms can sometimes form a covalent bond).
• Each nonmetal atom holds onto its own electrons
tightly (high IE) and tends to attract other electrons
as well (highly negative EA).
• The attraction of each nucleus for the valence
electrons of the other draws the atoms together.
• A shared electron pair is considered to be localized
between the two atoms because it spends most of
its time there, linking them in a covalent bond of a
particular length and strength.
Bond Energy
• Atoms will position themselves so that the
system will achieve the lowest possible
energy.
• The distance where this energy is minimal
is the bond length (distance between 2
nuclei, usually in nm or pm).
• The energy required to break a bond is
called the bond energy (usually in kJ/mol).
– The shorter the bond length (the stronger the
attraction) the higher the bond energy.
– Rank in order of increasing bond energy:
• H—F , H—Br, H—Cl
Diatomic Hydrogen
Polarity
• The distribution of bonding electrons in a
molecule maybe be equal or not depending on
the surrounding atoms (and their
electronegativities) and lone pairs in a molecule.
This leads to a discussion on polarity:
– Polar – the unsymmetrical distribution of electrons will
cause a negative and positive end creating an overall
dipole moment in the molecule.
– Nonpolar – the symmetrical distribution of electrons
will cause the dipole moments in the molecule to
cancel out, and leaving the molecule with no net
dipole.
Dipole Moments & Polarity
Dipole moments maybe symbolized by either Greek letter delta (lowercase),
δ , or arrows crossed at the positive end, or both.
Lewis Structures
• Lewis Dot Diagrams represent the number
of valence electrons present in an atom.
• Lewis structures represent the covalent
bonds in a molecule. Lines are used to
represent bonds (1 line = single bond, 2
lines = double bond, 3 lines = triple bond)
and 2 dots represent lone pairs of
electrons
Gilbert Newton Lewis (1875–1946). As early as 1902, nearly a
decade before Rutherford proposed the nuclear model of the atom,
Lewis’s notebooks show a scheme that involves the filling of outer
electron “shells” to explain the way elements combine. His electrondot symbols and associated structural formulas have become
standards for representing bonding. Among his many other
contributions is a more general model for the behavior of acids and
bases.
Drawing Lewis Structures
1. Count the total number of valence electrons in the molecule. Usually,
its easier to divide by 2 and work with pairs of electrons.
2. Create a “skeleton” structure by connecting surrounding (terminal)
atoms to the central atom. The central atom is the one that there is
one of or the least electronegative (generally the first element in the
formula is the central).
3. Place electrons on surrounding atoms so they each may have an
octet (remember hydrogen only needs 2). Place the remaining
electrons (if any) on the central atom.
4. Once all electrons are used check to see that all atoms have an
octet.
5. If the central (C, N, O, P, S, Se) does not have an octet must use
lone pairs on the surrounding to make multiple bonds (only possible
are double and triple).
Examples
• Draw a Lewis Structure for:
– The hypochlorite ion:
– A water molecule:
– Ammonia:
– Ammonium:
– Carbon dioxide:
– Methane:
– Ethane, C2H6:
Exceptions to the Octet Rule
• Electron Deficient
– The central atom has less than 8 electrons
• Boron Trifluoride
• Expanded Octet
– The central atom is sharing more than 8 electrons
Phosphorus pentachloride
• Odd Electron
– A few molecules contain a central atom with an odd number of
valence electrons, so they cannot possibly have all their
electrons in pairs. Such species, called free radicals, contain a
lone (unpaired) electron, which makes them paramagnetic and
extremely reactive.
• Nitrogen monoxide and nitrogen dioxide
Resonance
• Resonance structures express the various
plausible locations of a double/triple bond
location on a molecule.
• Resonance does not imply different kinds
of molecules but just reflects plausible
Lewis structures.
Examples
• Draw a Lewis Structure for:
– The nitrite ion:
– Sulfur dioxide:
– The nitrate ion:
– Ozone:
– Nitrogen gas:
– Benzene*, organic molecule, that forms a
hexagonal ring structure.
Formal Charge
• Formal charge is used to determine the one structure that exists in
nature when more than one arrangement of atoms in the Lewis
structure is possible.
– i.e.
A—A—B
or
A—B—A
• You must determine the formal charge for all central atoms. This is
done using the following formula:
Cf = X - (Y + Z/2)
X = the number of valence electrons on the atom
Y = the # of unshared electrons around the atom
Z = the # of bonding electrons
• The more likely Lewis structure is the one in which:
– The formal charges are closest to zero
– Any negative formal charge is located on the stronger electronegative
atom.
• Formal charges must sum to the actual charge on the species: zero for a
molecule and the ionic charge for an ion.
Formal Charge in Resonance
Three criteria help us choose the more important resonance
structures:
•
Smaller formal charges (positive or negative) are
preferable to larger ones.
•
The same nonzero formal charges on adjacent atoms are
not preferred.
•
A more negative formal charge should reside on a more
electronegative atom.
Example
• Decide which is the more relevant Lewis
structure for the cyanate ion.
» Deadly Free-Radical Activity Free
radicals can be extremely dangerous to
biological systems because they rupture
bonds in cells’ biomolecules. If a free
radical reacts with a biomolecule, it
typically forms a covalent bond to one of
the H atoms and removes it, and the
biomolecule is left with an unpaired
electron, thereby becoming a new free
radical. That species repeats the process
and creates other species with lone
electrons that proliferate to disrupt
chromosomes and cell membranes. We
benefit from this activity when we apply the
disinfectant hydrogen peroxide to a cut
because it forms free radicals that destroy
bacterial membranes. (The photo shows
H2O2 reacting vigorously with a drop of
blood.) On the other hand, recent studies
have suggested that several disease states,
including certain forms of cancer, may be
attributable to free radicals. Also, vitamin E
is believed to interrupt free-radical
proliferation.
VSEPR
• Molecular geometry can be predicted on the basis of
VSEPR (valence shell electron pair repulsion theory).
The idea that valence electron pairs around an atom
repel one another as far apart as possible thus impacting
the geometry (shape) of the molecule.
• To determine molecular geometry, first determine the
type of molecule from the Lewis structure and the rest is
really memorization.
– You must know the name, 3-D shape, bond angles, the
hybridization, polarity, and bond order (to be discussed
later this section) for each type of molecule, use the Lewis
Structure to determine the type of molecule:
• A = central atom
• X = surrounding atoms
• E = lone pairs of electrons on central atom
Molecules With No Lone Pairs on Central
•
•
•
•
•
AX2
AX3
AX4
AX5
AX6
-
linear
triangular planar
tetrahedron
triangular bipyramid
octahedron
-
sp
sp2
sp3
sp3d
sp3d2
Molecules With Lone Pairs on Central
• AX2E
• AX3E
• AX2E2
-
bent
triangular pyramid
bent
-
sp2
sp3
sp3
•
•
•
•
•
-
see-saw
T-shaped
linear
square pyramidal
square planar
-
sp3d
sp3d
sp3d
sp3d2
sp3d2
AX4E
AX3E2
AX2E3
AX5E
AX4E2
Molecular Shape & Polarity
• If there is no lone pair on the central and all
of the surrounding atoms are the same then
the molecule will be nonpolar:
– linear 1, triangular planar, tetrahedral, triangular
bipyramidal, & octahedral.
• If there are lone pairs around the central and
unsymmetrical distribution of the
surrounding atoms the molecule is polar:
– bent 1 and bent 2, triangular pyramidal, square
pyramidal, t-shaped, & seesaw,
• If the polar A—X bonds in a molecule AXmEn
are symmetrical around the central atom
then the molecule is nonpolar:
– square planar and linear 2.
Polarity of Tetrahedral
Hybridization
• The sharing of electrons leads to the
formation of hybrid orbitals:
Bond Order
As the bond order increases bond length will decrease thus the bond energy increases.
In resonance structures the bond order is calculated by taking the average bond order.
Sigma & Pi Bonds
• Sigma bonds – basically all single bonds
in a molecule
• Pi bonds – one of the electron pairs in a
multiple bond is a sigma bond, the rest are
pi bonds.
– Example 7.11 – page 187
σ(sigma) and π (pi) bonding
MC #1
• Which of the following molecules has the
shortest bond length?
(A) N2
(B) O2
(C) Cl2
(D) Br2
(E) I2
MC #2
• Which of the following has a zero dipole
moment?
(A) HCN
(B) NH3
(C) SO2
(D) NO2
(E) PF5
MC #3
• Pi bonding occurs in each of the following
species EXCEPT
(A) CO2
(B) C2H4
(C) CN¯
(D) C6H6
(E) CH4
MC #4
• The SbCl5 molecule has trigonal bipyramid
structure. Therefore, the hybridization of
Sb orbitals should be
(A) sp2
(B) sp3
(C) dsp2
(D) dsp3
(E) d2sp3
MC #5
• CCl4, CO2, PCl3, PCl5, SF6 Which of the
following does not describe any of the
molecules above?
(A) Linear
(B) Octahedral
(C) Square planar
(D) Tetrahedral
(E) Trigonal pyramidal
FRQ #1
• Discuss briefly the relationship between
the dipole moment of a molecule and the
polar character of the bonds within it. With
this as the basis, account for the
difference between the dipole moments of
CH2F2 and CF4.
FRQ #2
(a) Draw the Lewis electron-dot structures
for CO32-, CO2, and CO, including
resonance structures where appropriate.
(b) Which of the three species has the
shortest C-O bond length? Explain the
reason for your answer.
(c) Predict the molecular shapes for the
three species. Explain how you arrived at
your predictions.
FRQ #3
CF4
XeF4
ClF3
(a) Draw a Lewis electron-dot structure for
each of the molecules above and identify
the shape of each.
(b) Use the valence shell electron-pair
repulsion (VSEPR) model to explain the
geometry of each of these molecules.
FRQ #4
• Use simple structure and bonding models to
account for each of the following.
(a) The bond length between the two carbon
atoms is shorter in C2H4 than in C2H6.
(b) The H-N-H bond angle is 107.5º, in NH3.
(c) The bond lengths in SO3 are all identical and
are shorter than a sulfur-oxygen single bond.
(d) The I3- ion is linear.
FRQ #5
• Answer the following questions, which pertain to binary
compounds.
(a)
In the box provided below, draw a complete Lewis
electron-dot diagram for the IF3 molecule.
(b)
On the basis of the Lewis electron-dot diagram that
you drew in part (a), predict the molecular geometry of
the IF3 molecule.
(c)
In the SO2 molecule, both of the bonds between
sulfur and oxygen have the same length. Explain this
observation, supporting your explanation by drawing in
the box below a Lewis electron-dot diagram (or
diagrams) for the SO2 molecule.
(d)
On the basis of your Lewis electron-dot diagram(s)
in part (c), identify the hybridization of the sulfur atom in
the SO2 molecule.
FRQ #6
• Answer the following questions about the structures of
ions that contain only sulfur and fluorine.
(a) The compounds SF4 and BF3 react to form an ionic
compound according to the following equation.
SF4 + BF3  SF3BF4
(i) Draw a complete Lewis structure for the SF3+ cation
in SF3BF4.
(ii) Identify the type of hybridization exhibited by sulfur
in the SF3+ cation.
(iii) Identify the geometry of the SF3+ cation that is
consistent with the Lewis structure drawn in part (a)(i).
(iv) Predict whether the F—S—F bond angle in the SF3+
cation is larger than, equal to, or smaller than 109.50˚.
Justify your answer.
FRQ #6 (cont.)
(b)
The compounds SF4 and CsF react to form an ionic
compound according to the following equation.
SF4 + CsF  CsSF5
(i) Draw a complete Lewis structure for the SF5– anion
in CsSF5.
(ii) Identify the type of hybridization exhibited by sulfur
in the SF5– anion.
(iii) Identify the geometry of the SF5– anion that is
consistent with the Lewis structure drawn in part (b)(i).
(iv) Identify the oxidation number of sulfur in the
compound CsSF5.
FRQ #7
Answer the following questions that relate to chemical bonding
(a) In the boxes provided, draw the complete Lewis structure (electron-dot
diagram) for each of the three molecules represented below.
CF4
PF5
SF4
(b) On the basis of the Lewis structures drawn above, answer the following
questions about the particular molecule indicated.
(i)
What is the F-C-F bond angle in CF4?
(ii) What is the hybridization of the valence orbitals of P in PF5?
(iii) What is the geometric shape formed by the atoms in SF4?
(c) Two Lewis structures can be drawn for the OPF3 molecule, as shown below.
Structure 1
Structure 2
(i)
How many sigma bonds and how many pi bonds are in structure 1?
(ii) Which one of the two structures best represents a molecule of OPF3?
Justify your answer in terms of formal charge.
Credits
• (Silberberg, Martin S.. Chemistry: The
Molecular Nature of Matter and Change,
5th Edition. McGraw-Hill, 012009. 9.4.2).