Molecular Bonding Unit 5. Ionic bonding Cations (usually metals) lose electrons to anions...

Molecular Bonding

Unit 5

Ionic bonding

• Cations (usually metals) lose electrons to anions (monoatomic and polyatomic).

• Both cations and anions are stable like noble gases (octet rule)

• Strong electrostatic bonds

Example of an ionic bond

Covalent Bonds Sharing pairs of electrons Covalent bonds are the inter-atomic

attraction resulting from the sharing of electrons between atoms.

They result in ‘localized overlaps’ of orbitals of different atoms.

They also are the result of the attraction of electrons for the nucleus of other atoms.

Typical of molecular substances.

Example of a Covalent Bond

Covalent Bonds Cont.• Atoms bond together to form molecules

– molecules are electrically neutral groups of atoms joined together by covalent bonds

– strong attraction• Molecules attracted to each other weakly

form molecular compounds

Properties of Molecular Compounds

• Strong covalent bonds hold the atoms together within a molecule.

• The intermolecular forces that hold one molecule to another are much weaker.

• Properties vary depending on the strength of the intermolecular forces.

Lewis structures

• A Lewis structure is a representation of the valence e- in an atom, ion or molecule.

• Element symbol represents nucleus and core e-.

• Dots represent valence e-.

Electron pairs

• In covalent compounds electrons are shared between atoms creating electron pairs.

• Bonding pairs: e- that are shared between 2 atoms.•Lone or unshared pairs: e- that are NOT involved in bonding.

Covalent bonds

• Hydrogen follows the duet rule: sharing 2 electrons.

• Non-metals Carbon through Fluorine follow the octet rule: sharing 8 electrons.

Writing Lewis Structures of Molecules

1. Determine the central atom (atom in the middle)- usually is the “single” atom- least electronegative element- H never in the middle; C always in the middle

2. Count the total number of valence e- (group #)- add ion charge for “-”- subtract ion charge for “+”

3. Divide the total number of electrons by 2- sharing involves 2 electrons

4. Attach the atoms together with one pair of electrons

5. All remaining e- = LONE PAIRS! - lone pairs are NOT involved in bonding

Writing Lewis Structures Cont.6. Place lone pairs around non-central atoms to

fulfill the “octet rule” - some elements may violate this octet rule – (H=2, Be=4, B=6)

7. If more e- are still needed, create double or triple bonds around the central atom.

- single = 1 pair of shared electrons (2 e-) - double = 2 pair of shared electrons (4 e-) - triple = 3 pair of shared electrons (6 e-)

Practice

• Give the Lewis structure for:• HCl• NH3

• C2H6

• CO2

NAS• When drawing Lewis structures, remember: terminal atoms, atoms

that can only make one bond, must be on the outside.• N – A = S• #e- needed for – #e- available = #e- shared• octet or duet (valence e-) (bonds)

• S = the number of pairs of shared electrons• • S = the number of bonds (a dash may be used to • 2 represent a pair of shared electrons)

Resonance When there is more than one Lewis

structure for a molecule that differ only in the position of the electrons they are called resonance structures– Lone Pairs and Multiple Bonds in different

positions Resonance only occurs when there are

double bonds and when the same atoms are attached to the central atom

The actual molecule is a combination of all the resonance forms.

•••• •• ••••••••

•• ••O S O O S O•••••• ••••

••••

••••

• Some molecules are stable when the center atom has more than an octet. i.e. SF6, PCl5

sulfur has 12 electrons P has 10 electrons

Exceptions to the Octet Rule

• H, Be, B (stable with 2, 4, and 6 e-, respectively)

• Some molecules cannot be drawn with the Lewis structure rules, due to odd # of e-. i.e. NO & NO2 (There is no way for N to get an octet)

SAMPLE PROBLEM: Writing Resonance Structures

PLAN:

SOLUTION:

PROBLEM: Write resonance structures for the nitrate ion, NO3

-.

After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms.

Nitrate has 1(5) + 3(6) + 1 = 24 valence e-

N

O

O O

N

O

O O

N

O

O O

N does not have an octet; a pair of e- will move in to form a double bond.

N

O

O O

N

O

O O

N

O

O O

1. Smaller formal charges (either positive or negative) are preferable to larger charges;

2. A more negative formal charge should exist on an atom with a larger EN value.

3. Get unlike charges as close together as possible

4. Avoid like charges (+ + or - - ) on adjacent atoms

Four criteria for choosing the more important resonance structure:

Resonance and Formal Charge

Formal charge of atom =

# valence e- =

(# unshared electrons +

1/2 the # shared electrons)

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• We will calculate the formal charge for each atom in this Lewis structure.

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• Hydrogen shares 2 electrons with oxygen.

• Assign 1 electron to H and 1 to O.• A neutral hydrogen atom has 1 electron.• Therefore, the formal charge of H in

nitric acid is 0.

Formal charge of H

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• Oxygen has 4 electrons in covalent bonds.• Assign 2 of these 4 electrons to O.• Oxygen has 2 unshared pairs. Assign all 4

of these electrons to O.• Therefore, the total number of electrons

assigned to O is 2 + 4 = 6.

Formal charge of O

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• Electron count of O is 6.• A neutral oxygen has 6 electrons.• Therefore, the formal charge of oxygen

is 0.

Formal charge of O

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• Electron count of O is 6 (4 electrons from unshared pairs + half of 4 bonded electrons).

• A neutral oxygen has 6 electrons.• Therefore, the formal charge of oxygen

is 0.

Formal charge of O

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• Electron count of O is 7 (6 electrons from unshared pairs + half of 2 bonded electrons).

• A neutral oxygen has 6 electrons.• Therefore, the formal charge of oxygen

is -1.

Formal charge of O

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• Electron count of N is 4 (half of 8 electrons in covalent bonds).

• A neutral nitrogen has 5 electrons.• Therefore, the formal charge of N is +1.

Formal charge of N

–

Nitric acid

.. :

..H O

O

O

N

:

:..

..

• A Lewis structure is not complete unless formal charges (if any) are shown.

Formal charges

–

+

Formal Charge

Formal charge =

Number of

valence

electrons

number ofbonds

number ofunshared electrons

– –

An arithmetic formula for calculating formal charge.

–

"Electron Counts" and Formal Charges in NH4

+ and BF4-

1

4

N

H

H H

H

+7

4

..

BF

F

F

F

..

......: :

: :

: :

..

Formal Charge: Selecting the Best Resonance Structure

An atom “owns” all of its nonbonding electrons and half of its bonding electrons.

Formal charge of atom =

# valence e-

- # unshared electrons

- 1/2 # shared electrons

OO O

A

B

CFor OA # valence e- = 6

# nonbonding e- = 4

# bonding e- = 4 X 1/2 = 2

Formal charge = 0

For OB

# valence e- = 6

# nonbonding e- = 2

# bonding e- = 6 X 1/2 = 3

Formal charge = +1For OC

# valence e- = 6

# nonbonding e- = 6

# bonding e- = 2 X 1/2 = 1

Formal charge = -1

EXAMPLE: NCO- has 3 possible resonance forms -

Resonance and Formal Charge

N C O

A

N C O

B

N C O

C

N C O N C O N C O

Now Determine Formal Charges

-2 0 +1 -1 0 0 0 0 -1

Forms B and C have smaller formal charges; this makes them more important than form A. (rule 1)

Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid. (rule 2)

Coordinate Covalent Bond• A covalent bond in which one atom

contributes both bonding electrons.

Predicting Molecular Geometry

VSEPR Theory– Valence Shell Electron Pair Repulsion

The shape around the central atom(s) can be predicted by assuming that the areas of electrons on the central atom will repel each other

• Each Bond counts as 1 area of electrons– single, double or triple all count as 1 area

• Each Lone Pair counts a 1 area of electrons– Even though lone pairs are not attached to

other atoms, they do “occupy space” around the central atom

– Lone pairs generally “push harder” than bonding electrons, affecting the bond angle

Shapes

• Straight Line– molecule made up of only 2 atoms

Shapes- Linear

– 2 atoms on opposite sides of central atom, no lone pairs around CA

– 180° bond angles Trigonal Planar

– 3 atoms form a triangle around the central atom, no lone pairs around CA

– Planar– 120° bond angles

180°

120°

• Tetrahedral– 4 surrounding atoms form a tetrahedron

around the central atom, no lone pairs around the CA

– 109.5° bond angles

109.5°

Shapes

• Trigonal Pyramidal– 3 bonding areas and 1 lone pair around the CA– Bond angle = 1070

• V-shaped or Bent– 2 bonding areas and 2 lone pairs around the CA– bond angle = 104.50

Polar Bonds: Electronegativity

Measure of the ability of an atom to attract shared electrons– Larger electronegativity means atom attracts more

strongly– Values 0.7 to 4.0

Increases across period (left to right) on Periodic Table

Decreases down group (top to bottom) on Periodic Table

• Larger difference in electronegativities means more polar bond– negative end toward more electronegative

atom• Covalent bonding between unlike atoms

results in unequal sharing of the electrons– One end of the bond has larger electron

density (more electronegative) than the other– Polar covalent – unequal sharing– Nonpolar covalent – equal sharing

Bond Polarity The result is bond polarity

– The end with the larger electron density gets a partial negative charge

– The end that is electron deficient gets a partial positive charge

H F••+d -d

Dipole Moment Bond polarity results in an unequal electron

distribution, resulting in areas of partial positive and partial negative charge

Any molecule that has a center of positive charge and a center of negative charge in different points is said to have a dipole moment

• If a molecule has more than one polar covalent bond, the areas of partial negative and positive charge for each bond will partially add to or cancel out each other

• The end result will be a molecule with one center of positive charge and one center of negative charge

• The dipole moment affects the attractive forces between molecules and therefore the physical properties of the substance

Charge distribution in the water molecule

Polarity of Molecules

• Molecule will be NONPOLAR if:– the bonds are nonpolar (Br-Br, F-F)– there are no lone pairs around the central atom

and all the atoms attached to the central atom are the same

• Molecule will be POLAR if:– the central atom has lone pairs

Bond Length

• Distance between the nuclei of two bonded atoms.

• Average value• Same bond get shorter when bond doubles or

triples (C-C: 154 pm; C=C: 134 pm)

Bond Dissociation Energy

• Energy required to break a single bond• Ex H-H: + 435 kJ H● + H●

– 435 kJ of energy/mole of H2

• Shorter bond = Stronger bond = higher energy• Calculate the amount of energy needed to

dissociate 1 mole of C2H6.• C-C: 348 kJ/mol; C-H: 398 kJ/mol

Enthalphy of reaction

• Breaking bonds: endothermic• Formation of bonds: exothermic• Calculate the enthalpy (ΔrH) of the combustion

of 1 mol CH4

• Break: 4 x C-H, 2 x O=O: + 2648 kJ/mol• Form: 2 x C=O, 4 x O-H: - 3342 kJ/mol• ΔrH = -694 kJ/mol

Bondlength (pm) and bond energy (kJ/mol)

Bond Length Energy Bond Length Energy

H--H 74 435 H--C 109 393

C--C 154 348 H--N 101 391

N--N 145 170 H--O 96 366

O--O 148 145 H--F 92 568

F--F 142 158 H--Cl 127 432

Cl-Cl 199 243 H--Br 141 366

Br-Br 228 193 H--I 161 298

I--I 267 151

C--C 154 348

C--C 154 348 C=C 134 614

C--N 147 308 CºC 120 839

C--O 143 360 O--O 148 145

C=O 121 736 O=O 121 498

C--F 135 488 N--N 145 170

C--Cl 177 330 NºN 110 945

Overlap of Orbitals

The point at which the potential energy is a minimum is called the equilibrium bond distance

The degree of overlap is determined by the system’s potential energy

equilibrium bond distance

2s

These new orbitals are called hybrid orbitals

The process is called hybridization

What this means is that both the s and one p orbital are involved in bonding to the connecting

atoms

Formation of sp hybrid orbitals

The combination of an s orbital and a p orbital produces 2 new orbitals called sp orbitals.

Hybridization

• refers to a mixture or a blending• Biology – refers to genetic material• Chemistry – refers to blending of orbitals• Remember, orbitals can only predict an area in

space where an e- may be located.• Sometimes blending orbitals can produce a

lower, more stable bonding opportunity.• Orbital hybridization occurs through e-

promotion in orbitals that have similar energies (i.e. same energy level).

Hybridization Cont.

• Hybridization occurs WITHIN the atom to enhance bonding possibilities.

• Do not confuse this concept with orbital overlap (bonding).

• Hybridization is a concept used to explain observed phenomenon about bonding that can’t be explained by dot structures.

• EXAMPLES – draw box diagrams for Be, B, and C (use noble gas core).

How do I know if my CA is hybridized?

• If your CA is B, Be, C, Si, or Al then it is hybridized.

• If your molecule has multiple bonds in it then it is hybridized.– Double bonds – sp2 hybridized– Triple bonds – sp hybridized

Formation of sp2 hybrid orbitals

Formation of sp3 hybrid orbitals

Hybrid orbitals can be used to explain bonding and molecular geometry

Multiple Bonds

 Everything we have talked about so far has only dealt

with what we call sigma bonds Sigma bond (s) A bond where the line of electron density is concentrated symmetrically along the line connecting the two atoms.

Pi bond (p) A bond where the overlapping regions exist above and below the internuclear axis (with a nodal plane along the internuclear axis).

Example: H2C=CH2

Example: H2C=CH2

Example: HCCH

Delocalized p bonds

 When a molecule has two or more resonance structures,

the pi electrons can be delocalized over all the atoms that have pi bond overlap.

In general delocalized p bonding is present in all molecules where we can draw resonance structures with

the multiple bonds located in different places.

Benzene is an excellent example.  For benzene the p orbitals all overlap leading to a very delocalized electron

system

Example: C6H6 benzene