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Bonding and Molecular Structure. Goals: Understand the differences between ionic and covalent bonds . Draw Lewis electron dot structures for small molecules and ions.

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bonding and molecular structure

Bonding and Molecular Structure

Goals:

Understand the differences between ionic and covalent bonds.

Draw Lewis electron dot structures for small molecules and ions.

Use the valance shell electron-pair repulsion theory (VSEPR) to predict the shapes of simple molecules and ions and to understand the structures of more complex molecules.

Use electronegativity to predict the charge distribution in molecules and ions and to define the polarity of bonds.

Predict the polarity of molecules.

chemical bonding
Chemical Bonding

Problems and questions —

How is a molecule or polyatomic ion held together?

Why are atoms distributed at strange angles?

Why are molecules not flat?

Can we predict the structure?

How is structure related to chemical and physical properties?

There are 2 extreme forms of connecting or bonding atoms:

Ionic—complete transfer of ________________from one atom to another.

Covalent—some valence electrons _________ between atoms

Most bonds are somewhere in between!

chemical bonding1
Chemical Bonding
  • Wherever two atoms or ions are strongly attached to each other : chemical bond between them.
    • Ionic, covalent, metallic.
  • Ionic bond: electrostatic force between ions of opposite charge.
  • Covalent bond: sharing of electrons between two atoms.
lewis symbols
Lewis Symbols
  • Electrons involved in chemical bonding are the _______________ (those in the outermost occupied shell of an atom).
  • Lewis electron-dot symbols: simple way of showing the valance electrons.
  • Lewis symbol:
    • Chemical symbol for the element + a dot for each valance electron.
    • Dots are placed on the four sides of the atomic symbol. Each side can accommodate up to two electrons.
    • The number of valance electrons in any representative element is the same as the group number of the element.

Sulfur:

[Ne]3s23p4

••

S

••

the octet rule
The Octet Rule
  • Atoms tend to gain, lose, or share electrons until they are surrounded by eight valance electrons.
  • They have the same number of electrons as the ________ closest to them in the period table – stable electron arrangements (high ionization energies, low affinity for additional e-s, general lack of chemical reactivity.
  • An octet of electrons consists of full s and p subshells in an atom:
  • ns2np6
  • There are some exceptions…
ionic bonding
Ionic Bonding

Na (s) + 1/2 Cl2(g) ---> NaCl (s) DHfo = -410.9 kJ

Metal of low IE

Nonmetal of high EA

ionic bonding1
Ionic Bonding

Na (s) + 1/2 Cl2(g) ---> NaCl (s) DHfo = -410.9 kJ

  • An electron has been lost by a sodium atom and gained by a chlorine atom – electron transfer from the Na atom to the Cl atom.
  • NaCl is a typical ionic compound: it consists of a metal of low ionization energy and a nonmetal of high electron affinity.
  • Lewis electron-dot representation:
energetics of ionic bond formation
Energetics of Ionic Bond Formation
  • The principal reason that ionic compounds are stable is the attraction between ions of unlike charge.
  • This attraction draws the ions together, releasing energy and causing the ions to form a solid array – lattice.
  • A measure of how much stabilization results from arranging oppositely charged ions in an ionic solid:
  • Lattice energy – is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions.
  • NaCl  Na+ (g) + Cl- (g) DHlattice = + 788 kJ/mol
  • Highly endotermic process – the reverse process – the coming together is highly exothermic.
  • Eel = kQ1Q2 / d For a given arrangement of ions, the lattice energy increases as the charges on the ions increase and their radii decrease.
covalent bonding
Covalent Bonding
  • The bond arises from the mutual attraction of 2 nuclei for the same electrons.
  • Electron sharingresults.

Bond is a balance of attractive and repulsive forces.

bond formation
Bond Formation
  • A bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms.

Overlap of H (1s) and Cl (2p)

Note that each atom has a single, unpaired electron.

electron distribution in molecules

••

H

Cl

••

lone pair (LP)

shared or

bond pair

G. N. Lewis

1875 - 1946

Electron Distribution in Molecules
  • Electron distribution is depicted with Lewis electron dot structures
  • Valence electrons are distributed as shared or ______ PAIRS and unsharedor ______ PAIRS.
valance electrons
Valance Electrons

Electrons are divided between core and valence electrons:

Valance e-: e- in the outermost shell

B 1s2 2s2 2p1

Core = [He] , valence = 2s2 2p1

Br [Ar] 3d10 4s2 4p5

Core = [Ar] 3d10 , valence = 4s2 4p5

the octet rule1
The Octet Rule
  • Each atom (except H) has a share in four pairs of electrons, so each has achieved a noble gas configuration.
  • Each atom is surrounded by an octet of e-s.
  • Octet Rule: The tendency of molecules and polyatomic ions to ______________________________________ _______________.

• No. of valence electrons of an atom = group number

• For Groups 1A-4A, no. of bond pairs = group number

• For Groups 5A -7A, BP’s = 8 – group number

• Except for H (and sometimes atoms of 3rd and higher periods),

BP’s + LP’s = 4

building an electron dot structure
Building an Electron Dot Structure

Ammonia, NH3

1. Decide on the central atom: Central atom is atom of lowest affinity for electrons; never H.

Therefore, N is central

2. Count total valence electrons

H = 1 and N = 5

Total = (3 x 1) + 5

= 8 electrons / 4 pairs

The trick is to count the e-s!

Students should be familiar with writing Lewis structures.

ammonia nh 3

H

H

N

H

••

H

H

N

H

Ammonia, NH3

3. Form a single bond between the central atom and each surrounding atom. Each line represent 1 pair of shared e-.

4. Remaining electrons form LONE PAIRS to complete octet as needed.

N: 3 BOND PAIRS and 1 LONE PAIR.

Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair.

Practice with the Lab Experiment # 11

sulfite ion so 3 2
Sulfite ion, SO32-

Step 1. Central atom

= S

Step 2. Count total valence electrons

Notice charge in ion: add or substract e- according to charge!

Negative charge = 2 e- more

sulfite ion so 3 21
Sulfite ion, SO32-

Step 3. Form bonds

Step 4. Assign as lone pairs to outside atoms (never to H), if some left, then to the center atom.

Step 5. Check all atoms have an octet.

If the central atom as fewer than eight e-, move one or more of the lone pairs on the terminal atoms and form multiple bonds. – See next example.

carbon dioxide co 2
Carbon Dioxide, CO2

Step 1. Central atom

= C

Step 2. Count total valence electrons

Step 3. Form bonds

Step 4. Assign as lone pairs to outside atoms (never to H), if some left, then to the center atom.

Step 5. Check all atoms have an octet.

carbon dioxide co 21
Carbon Dioxide, CO2

Step 5. Check all atoms have an octet. Form multiple bonds if needed.

So that C has an octet, we shall form DOUBLE BONDS between C and O.

The second bonding pair forms a pi (π) bond.

sulfur dioxide so 2
Sulfur Dioxide, SO2

Step 1. Central atom

= S

Step 2. Count total valence electrons

Step 3. Form bonds

Step 4. Assign as lone pairs to outside atoms (never to H), if some left, then to the center atom.

Step 5. Check all atoms have an octet.

sulfur dioxide so 21
Sulfur Dioxide, SO2

Step 5. Check all atoms have an octet. Form double bond so that S has an octet — but note that there are two ways of doing this.

This leads to the following structures.

These equivalent structures are called

RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two.

formal atom charges
Formal Atom Charges
  • Atoms in molecules often bear a charge (+ or -).
  • The predominant resonance structure of a molecule is the one with charges as close to 0 as possible.
  • Formal charge = Number of valance e- (group number) – 1/2 (no. of bonding electrons) - (no. of LP electrons)
formal atom charges1
Formal Atom Charges

Formal charge = Valance e- – 1/2 (BP e-) – LP e-

For C,

= 4 – ½ (8) – 0 = 0

For O,

= 6 – ½ (4) – 4 = 0

Yellow= negative&red= positive

Relative size = relative charge

formal atom charges2

S

C

N

S

C

N

S

C

N

Formal Atom Charges

Formal charge = Group number – 1/2 (BP e-) – LP e-

For S,

6 - ½ (2) - 6 = -1

For C,

= 4 – ½ (8) – 0 = 0

For N,

5 - ½ (6) - 2 = 0

Thiocyanate Ion, SCN-

Which is the most important resonance form?

All atoms are negative, but most negative is S.

violations of the octet rule
Violations of the Octet Rule

Usually occurs with B and elements of higher periods.

BF3

SF4

boron trifluoride bf 3

Calc’d partial charges in BF3

Boron Trifluoride, BF3

The B atom has a share in only 6 pairs of electrons (or 3 pairs). B atom in many molecules is electron deficient.

F is negative and B is positive

sulfur tetrafluoride sf 4
Sulfur Tetrafluoride, SF4

5 pairs around the S atom. A common occurrence outside the 2nd period.

slide29

Which of the following species will have a Lewis structure most like that of a sulfate ion, SO42-? Assume that the Lewis structure has no double bonds.

a) NH3b) CBr4c) SO3d) H2CO

free radicals
Free Radicals
  • Free radicals – atomic or molecular specie with ______ ____________.
  • Free radicals are generally _____________.
  • H, Cl
  • O2 - Oxygen itself is a free radical, but one with two unpaired electrons; reduction by adding one electron to give superoxide involves the pairing of two of the electron spins, leaving one unpaired O2 + 1e-  O2·-
  • NO – vascular system (11 valance electrons – odd number)
  • Vitamin C, Vitamin E – Antioxidants
    • They produce very stable radicals (stabilized by resonance structures)

Vit E

molecular geometry
MOLECULAR GEOMETRY

VSEPR

  • Valence Shell Electron Pair Repulsion theory.
  • Most important factor in determining geometry is ________________________ electron pairs.
  • Molecule adopts the shape

that ___________ the

electron pair repulsions.

exp 11 molecular geometry
Exp 11: Molecular Geometry
  • I suggest you practice with:

3. SO2 13. CS2 29. NO2+

7. NO3- 22. CH3- 36. PO33-

9. NH4+ 27. H3O+ 43. H2O

10.CH2Cl2 28. O3 46. CH3+

48. HCN

1. Give the total number of valance electrons.

2. Draw the correct Lewis dot structure.

3. Give the geometry using VSEPR theory.

4. Give the polarity of the MOLECULE or ION.

structure determination by vsepr

••

H

H

N

H

Structure Determination by VSEPR

Ammonia, NH3

1. Draw electron dot structure

2. Count BP’s and LP’s = 4

3. The 4 electron pairs are at the corners of a tetrahedron.

The ELECTRON PAIR GEOMETRY is tetrahedral.

structure determination by vsepr1
Structure Determination by VSEPR

Ammonia, NH3

The electron pair geometry is tetrahedral.

The MOLECULAR GEOMETRY — the positions of the atoms — is PYRAMIDAL.

structure determination by vsepr2
Structure Determination by VSEPR

Water, H2O

1. Draw electron dot structure

2. Count BP’s and LP’s = 4

3. The 4 electron pairs are at the corners of a tetrahedron.

The electron pair geometry is TETRAHEDRAL.

structure determination by vsepr3
Structure Determination by VSEPR
  • Water, H2O

The electron pair geometry is tetrahedral.

The molecular geometry is BENT.

geometries for four electron pairs
Geometries for Four Electron Pairs

Students should be familiar with predicting molecular geometries.

the hydronium ion
The Hydronium Ion

H3O+

Draw the Lewis dot structure and predict the geometry.

molecular geometries for five electron pairs
Molecular Geometries for Five Electron Pairs

All based on trigonal bipyramid.

sulfur tetrafluoride sf 41

9

0

o

F

F

1

2

0

o

S

F

F

Sulfur Tetrafluoride, SF4

Lone pair is in the equator because _____ __________________.

molecular geometries for six electron pairs
Molecular Geometries for Six Electron Pairs

All are based on the 8-sided octahedron

sulfur hexafluoride sf 6

O

c

t

a

h

e

d

r

o

n

9

0

o

F

F

F

S

9

0

o

F

F

F

Sulfur Hexafluoride, SF6

6 electron pairs

electronegativity
Electronegativity
  • Concept proposed by Linus Pauling (1901-1994).

The only person to receive two unshared Nobel prizes (for Peace and Chemistry).

Chemistry areas: bonding, electronegativity, protein structure.

electronegativity1
Electronegativity
  •  Is the ability of _________________to ____________________.
  • It is related to its ionization energy and electron affinity (properties of isolated atoms).
  • Electronegativity values are unitless.
  • ____________, the most electronegative, has a value of 4.0
  • __________, the least electronegative has a value of 0.7.
  • The values for all other elements lie between these two extremes.
electronegativity2
Electronegativity
  • Used for atoms in molecules.
  • It is related to its ionization energy and electron affinity (properties of isolated atoms).
bond polarity
Bond Polarity

HCl is ______ because it has a positive end and a negative end.

Cl has a greater share in bonding electrons than does H.

Cl has slight negative charge (-d) and H has slight positive charge (+ d)

bond polarity1
Bond Polarity

Due to the bond polarity, the H—Cl bond energy is __________ than expected for a “pure” covalent bond.

BOND ENERGY

“pure” bond 339 kJ/mol calc’d

real bond 432 kJ/mol measured

Difference = 92 kJ. This difference is proportional to the difference in ELECTRONEGATIVITY,.

bond polarity2
Bond Polarity
  • Three molecules with polar, covalent bonds.
  • Each bond has one atom with a slight negative charge (-d) and and another with a slight positive charge (+ d)
  • Relative values ofdetermine BOND POLARITY (and point of attack on a molecule).
  • Dipole moment – _________ ______________________ ______________________ ______________________.
  • They are added as VECTORS (in math)
which bond is more polar
Which bond is more polar?

O—H O—F

 3.5 - 2.1 3.5 - 4.0

 ____ _____

______ is more polar than ____

and polarity is “reversed.”

net dipole moment
Net Dipole Moment

Molecules—such as HI and H2O— can be POLAR (or dipolar).

They have a DIPOLE MOMENT. The polar HCl molecule will turn to align with an electric field.

net dipole moment1
Net Dipole Moment

Compare CO2 and H2O. Which one is polar?

The magnitude of the dipole is given in Debye units. Named for Peter Debye (1884 - 1966). Rec’d 1936 Nobel prize for work on x-ray diffraction and dipole moments.

molecular polarity
Molecular Polarity

Molecules will be polar if

a) bonds are polar

AND

b) the molecule is NOT “symmetric”

Students should be familiar with identifying polar vs. nonpolar molecules.

molecular polarity1
Molecular Polarity

C—F bond is very polar. Molecule is not symmetrical and so, it is polar.

C-H bond is polar.

Methane is symmetrical and is NOT polar.

bond properties
Bond Properties
  • What is the effect of bonding and structure on molecular properties?
  • Bond Order – Number of bonding e- pairs between a pair of atoms (1, 2, 3… fractional bond orders).
    • Bond Length
    • Bond Strength

Free rotation around C–C single bond

No rotation around C=C double bond

bond order
Bond Order

______ bond

______ bond

Acrylonitrile

______ bond

Fractional bond ordersoccur in molecules with _______________ structures (book example: O3).

Nitrite ion, NO2-

The N—O bond order = _____

bond order1

414 kJ

123 pm

745 kJ

110 pm

Bond Order

Bond order is proportional to two important bond properties:

(a) bond length

(b) bond strength

bond length
Bond Length
  • Bond length is the distance between _____ ____________ ___________.
bond length1

H—F

H—Cl

H—I

Bond Length

Bond length depends on size of ___________.

Bond length depends on _____________.

Bond distances measured in Angstrom units where

1 A = 10-2 pm.

bond strength
Bond Strength
  • —measured by the energy req’d to break a bond. See Tables 9.9 and 9.10

BOND STRENGTH (kJ/mol)

H-H 436

C-C 346

C=C 602

CC 835

NN 945

The GREATER the number of bonds (bond order) the ______ the bond strength and the ________ the bond.

bond strength1
Bond Strength

Bond Order Length Strength

HO—OH

O=O

142 pm

210 kJ/mol

1

2

121

498

1.5

128

?

using bond energies
Using Bond Energies

Estimate the energy of the reaction

H—H(g) + Cl—Cl(g) ----> 2 H—Cl(g)

Net energy = ∆Hrxn =

= energy required to break bonds

- energy evolved when bonds are made

H—H = 436 kJ/mol

Cl—Cl = 242 kJ/mol

H—Cl = 432 kJ/mol

using bond energies1
Using Bond Energies
  • Estimate the energy of the reaction
  • H—H + Cl—Cl ----> 2 H—Cl

H—H = 436 kJ/mol

Cl—Cl = 242 kJ/mol

H—Cl = 432 kJ/mol

remember
Remember
  • Go over all the contents of your textbook.
  • Practice with examples and with problems at the end of the chapter.
  • Practice with OWL tutor.
  • Work on your assignment for Chapter 9.
  • Practice with the quiz on CD of Chemistry Now.