CHEMICAL BONDING Chapter 8-9.

CHEMICAL BONDING Chapter 8-9

States of Matter All matter has two distinct characteristics
Mass Volume Three states of matter—each with unique set of properties Solid Liquid Gas

States of Matter and Properties
Solids Definite shape Definite volume Particles packed tightly together Only motion allowed are vibrations in position Little movement relative to each particle. Largest amount of intermolecular forces (for given substance) Lowest energy (for given substance) Two categories Crystalline Regular, ordered, repeatable 3-D structure Amorphous Arrangement is not ordered or regular

Liquids Definite volume Fewer intermolecular bonds
Particles able to exchange positions. Greater freedom of movement Indefinite shape (takes shape of container) Relatively closely packed Similar volumes to solids Significantly greater amount of energy (but considerably less than gases) Energy input to convert solid to liquid used to break FEW intermolecular bonds Intermediate between extremely ordered solids and completely unordered gases.

Gases No definite shape No definite volume (limit)
Takes shape and volume of container No intermolecular bonds Highest Energy Energy input to convert liquid to gas used to break ALL intermolecular bonds RECALL: ∆Hvaporization is generally significantly HIGHER than ∆Hfusion Largely disordered RECALL: Perfectly elastic collisions Particles have sufficient energy to overcome intermolecular interactions

Properties of solids and liquids
Properties such as viscosity (ability to flow), surface tension (ability to acquire least surface area possible) , hardness, depends upon Arrangement of particles Extent of attractions between particles Ex: large chain hydrocarbons (Ex: C10H22, Decane, liquid. C20H42, Icosane, solid) get entangled and are highly viscous, even though the strength of individual bonds between particles is weak.

Types of Bonding

Intermolecular Forces
Based on Coulombic attractions between opposite charges Individual charges are usually relatively small Therefore inter bonding is a relatively VERY weak attraction compared to intra bonding

Background Electronegativity and dipoles
RECALL: electron cloud can be thought of a probabiliby map of where an electron may be found at any given time. In separate atoms, it can easily be represented as such

When atoms do form a bond, the probability map changes
When atoms do form a bond, the probability map changes. What are the possible combinations? Three possibilities based on Electronegativity Differences (defined as ability of an atom within a covalent bond to attract electrons to itself) Bond between SAME type of atoms. Ex: H2 Notice the GREATER density (probability) of electrons in the MIDDLE indicating EQUAL sharing or a non-polar bond Each nucleus has EXACTLY the same ability to attract the other atom’s electron Other examples: ALL diatomics

Bonds with similar types of atoms (or different BUT very SIMILAR electronegativity) also behave the same way. Show almost equal sharing Probability of each nucleus attracting the electrons of the other atom are relatively the same Example: BrCl Which atoms has greater pull? ASIDE: Why is Br larger, even though Cl has greater pull on electron? More shells

More likely to have one atom with much higher electronegativity than the other
i.e. the electrons are significantly more attracted to one atom than the other but NOT sufficiently enough to be completely pulled off and form ions Leads to electron cloud DISTORTION and RE-DISTRIBUTION of electron charge density Partial positive (δ+) --less electronegative atom Partial negative (δ-) --more electronegative atom --greater electron density Dipole: Creation of opposite charges at either end of the molecule Polar covalent bond

Intermolecular Forces
Attraction IN-BETWEEN molecules (RECALL: molecule is a discrete particle with covalent intramolecular bond) 3 types: depends on Presence or absence (or relative value) of dipoles, AND Intermolecular Coulombic attractions Greatly influence properties of the compound Ex: state of matter, boiling/melting points, solubility etc.

London Dispersion Forces (LDF’s)
RECALL: electrons are in constant motion Results in small electrostatic forces Due to TEMPORARY dipole (unequal distribution of charges) in molecules which otherwise have no permanent dipoles (Ex: I2—completely non-polar) One molecule approaches another  electrons of one or both get temporarily displaced due to mutual repulsion  movement results in temporary dipole to be set on surface  results in attraction for each other. Increases with surface area and with POLARIZABILITY. Polarizability (ability to become polarized or form dipole) increases with increased number of electrons Summary: larger molecules (larger surface area) have more electrons, therefore are more polarizable. Leads to more LDFs and greater attractions, and consequently higher boiling points. ALL (and we mean AAAALLLLLLLLLLLL) atoms and molecules exhibit LDF’s

Consider the boiling points of Noble gases
Consider the melting and boiling points of the Halogens Can LDF’s be used to explain the variations? On descending group 18 atoms of the elements get bigger, have more electrons and larger surface areas. This increases London dispersion forces between them, making them more difficult to separate and hence increasing their boiling points.

Dipole-Dipole forces RECALL: a dipole is formed when bonded atoms have significant electronegativity differences but not high enough for electrons to be transferred. Results in permanent dipole (greater pull of electrons towards one atom making it partially -, and making the other partially +) Results in molecules arranging themselves in such a way that negative and positive ends of the molecules attract one another

In a 3-dimensional plane, the complete picture is more complicated
Molecules come together in such a way that attractions are maximized Repulsions are minimized Eventual alignment with best compromise between attraction and repulsion Results in dipole-dipole forces Generally stronger than London Dispersion Forces.

Hydrogen Bonding Special case of dipole-dipole forces
Hydrogen: Exceptional element Extremely small Only 1 electron In covalent bond, when electron is attracted to the nucleus of the OTHER atom, it is “held” to one side. The other side is completely exposed (and δ+) Any approaching negatively charged group can get VERY CLOSE to the hydrogen nucleus Results in unexpectedly large electrostatic attraction Attractions are further exaggerated when H is bonded to a MORE electronegative element that is ALSO very small. Allows for a SIGNIFICANTLY strong intermolecular interaction, than would be predicted based on mass/polarizability alone.

Hydrogen Bonding Exists between
elements F, O, N connected to H in one molecule, AND the un-bonded lone pairs of electrons on F, O, N in another molecule (**more of this later)

Consequences of Hydrogen Bonding
Substances with intermolecular hydrogen bonding Have unusually high boiling points Tend to be more viscous Both explained by Increased attraction between molecules due to Hydrogen bonding More difficult to separate molecules due to exceptionally large intermolecular attractions

Boiling Points of hydrides of groups 14, 15, 16, 17
Group 14: notice the increase in boiling points (as expected. Why?) Due to increase in size and hence London Dispersion Forces Group 15: similar patter to group 14. Exception? Why? Ammonia displays Hydrogen bonding. Results in a much HIGHER boiling point than would normally be predicted if only LDF’s were present. Which Group 16 and 17 elements do you expect to have higher than predicted b.p?

Repeated in Group 16 (water) and 17 (hydrogen fluoride) due to Hydrogen bonding displayed by these elements.

THINK: Why doesn’t CH4 display hydrogen bonding?
No lone pair of electrons to be attracted to Hydrogen Equal sharing on all sides, entirely non-polar compound in terms of bond polarity AND geometry (huh??? More later) Why are NH3 and HF gases at room temperature but H2O is a liquid? NH3—not “heavy” enough HF—too small H2O—just right in terms of molar mass and size to allow sufficient interactions

Properties affected by Intermolecular Bonds
Miscibility (ability to mix) When intermolecular forces of two substances are similar, they tend to mix. “like dissolves like”. Non-polar dissolves in non-polar. Polar in does NOT dissolve in non-polar.

Vapor Pressure Pressure exerted by the vapor phase of liquid (or solid) above the liquid (or solid) as it vaporizes. Consider a liquid in a sealed container Even at temperatures BELOW b.p  a few molecules at surface possess sufficient energy to overcome attractive forces holding them together  escape into the vapor phase above liquid. Weaker intermolecular forces  little energy needed to vaporize  easier to vaporize  more molecules enter vapor phase  HIGH vapor pressure  LOW boiling point volatile liquids—easily vaporized, low b.p, high vapor pressure THINK about it: why do you think people confuse volatility with flammability? Many organic molecules such as hydrocarbons, alcohols, acetone etc. tend to have LDF’s as their intermolecular forces, hence tend to also have low b.p, high vapor pressure and are volatile. They also happen to be extremely flammable (but NOT due to low b.p). Since our common experience is with these molecules which are BOTH flammable AND volatile, it is easy to confuse the two.

Surface tension Ability to liquid to have the smallest surface area possible In body of a liquid, all particles (molecules) feel attractive forces in three dimensions  result: no net force On surface of liquid no liquid particles above  feel “inward” net force into the body of the liquid  cohesion  causes liquid to contract to smallest size possible (a sphere)  creates internal pressure at surface  can resist external pressure Effect: water “beads” into droplets. Possible to float small objects of density higher than water. Insects to walk on water

Capillary action: When placed in a very thin tube, combination of cohesive (within) forces and adhesive (forces between liquid and walls of the tube) forces can ADD up to overcome forces of gravity Liquid is drawn up the tube without external forces being applied Recall: in a glass tube NOT all liquids form a downward (concave) meniscus like water Ex: mercury has a convex meniscus What other substances could have convex meniscus, in a glass tube?

Intermolecular forces in biological molecules
Ex: enzymes in order for a substrate to interact with an enzyme, an intermolecular attraction must exist 1o, 2o, 3o and 4o, structures of proteins (enzymes)  leads to different 3D shapes  different orientation and functional groups  results in attraction (hydrophilic) or repulsion (hydrophobic) to polar water molecules

INTRAMOLECULAR BONDING
Intra (within) bonding Also based upon Coulombic attractions Significantly stronger attractions RESULT—intra bonds are MUCH stronger than inter bonds Types Ionic Covalent (Discrete) Polar Non-polar Giant Covalent Metals (Alloys)

Ionic Bonds Transfer of electrons between atoms to form ions
Between metals and nonmetals, due to tendency of Metals  lose  cation  positive Nonmetals  gain  anion  negative Atoms have equal number of protons and electrons No overall charge. Neutral In ionic compound Total electrons lost MUST equal total electrons gained Overall charge is O (zero) or neutral formula Elements that are generally widely separated on periodic table Results in VERY large electrostatic forces  i.e ionic bonds are very STRONG.

Q1Q2 E =  r Coulomb’s Law where Predicts Lattice Energy
E = energy (J)  = Coulomb’s constant (2.31 X J·nm) Q1 and Q2 = numerical ion charge r = distance between the ion centers Predicts Lattice Energy Recall: Lattice energy-amount of energy required to convert a solid ionic compound into its gaseous ions Because the strength of a bond is measured in terms of the amount of energy required to break the bond, a measure of lattice energy is an indication of the strength of the FORCE (or bond strength) for a given ionic compound. E =  Q1Q2 r

Summary of Coulomb’s Law
Predicts that force (of attraction or strength of bond) increases with Increasing charge Decreasing distance between charges Therefore, the strongest ionic bonds are formed between ions that are small (because they can get close to each other) and highly charged (ions that have high charge densities Explain in terms of ionic radius and energy transitions

Practice When comparing sodium chloride and sodium bromide, which compound would be expected to have stronger ionic bonds, i.e. the greatest coulombic attraction? Explain For each pair, indicate the stronger ionic bond MgO vs MgS NaCl vs. MgS

Structure and Properties of Ionic compounds
Ions held in rigid, fixed positions Giant, 3-D lattice Result  not malleable or ductile Brittle Why? When ordered structure is disrupted, charges repel, solid splits apart Strong electrostatic forces result in high melting points and boiling points Low volatility Low vapor pressure Electrical Conductivity Do NOT conduct as solid (ions not free to move) DO conduct when molten (liquid) or in solution (ions free to move

Polar water molecules are attracted to oppositely charged ions.
Dissolution in water Polar water molecules are attracted to oppositely charged ions. Penetrates lattice  attach themselves to ions (like dissolves like) Process known as hydration  ions are “hydrated” when they “dissociate” Ions become free to move around (able to conduct electricity) Non-polar solvent  no “charged” ends to allow attraction and penetration of ions  ions stay together  hence ionic compound do not dissolve in non-polar solvents.

Covalent Bonding Valence electrons shared between atoms
Achieve full s and p sub-shells (noble gas configuration) Forms discrete molecules Each shared pair of electrons represents a single bond Two shared pair = double bond Three paired pairs = triple bond Usually occurs between non-metal atoms Elements close to each other on top right hand of periodic table

Could it really be so simple?
In simplest term, possible to think of compounds as either 100% ionic or 100% covalent Actually 100% ionic or 100% covalent are extreme ends of a sliding scale Most bonds are intermediate between the two Ex: particle with largely covalent properties, actually has some degree of ionic behavior and vice-versa.

Ionic substance with some covalent character
Small, highly charged cation Ability to distort the charge cloud around anion Compare to diagram with charge distribution of 100% ionic or 100% covalent. Electron IS completly transfer Resultant electron cloud around anion is distorted Translation:- ionic bond with some acquired covalent character Fajans rule—helps assess degree of distortion (polarization). Distortion will be maximum when Cation is small and highly charged (high charge density) Anion is large and highly charged (electrons NOT held tightly)

Covalent Substance with some Ionic character (polar covalent)
Based on electronegativity differences Incomplete electron transfer (sharing) Electrons attracted towards more electronegative atom Electron cloud distortion and re-distribution of electron charge density

Pauling Scale Pauling scale of electronegativities
Allows predictions about degree (or percent) of ionic and covalent character in a compound

Metallic bonding Metal structures
Considered to be a close packed lattice of positive atoms/ions surrounded by a “sea” of moving (mobile), delocalized electrons The mobile electrons and their movement allow metals to be good conductors of electricity Close packed lattice of atoms allow metals to be good conductors of heat Metallic bond due to electrostatic attraction between positive and negative charges All charges are the same, so any repulsion cancels out Allows metals to be malleable and ductile Close packed lattice allows very smooth surface Light reflected off surface allows luster (shiny)

Lewis Structures of Covalent Molecules
Use dots to represent valence electrons in atoms in a bonded molecule RECALL: covalent bonds form through sharing of electrons to achieve full s and p sublevels (octet) Exception: Hydrogen—forms duet Boron—incomplete octet Period 3 onward non-metals—MAY have expanded octet Empty d sublevels become available, and may participate in bonding

Rules for Drawing Lewis Structures
Calculate total number of valence electrons Add 1 for each negative charge on a polyatomic anion Subtract 1 for each positive charge on a polyatomic cation Determine CENTRAL atom Least electronegative, EXCEPT Hydrogen Generally more towards left on periodic table Place terminal atoms around central Form single bonds by using one pair of electrons per bond Ex: PCl3 Valence electrons on P: 5 Cl: 7 X 3 Total: 26 Central atom: P Draw Molecule Count electrons placed Each bond=2 electrons 6 electrons placed Cl—P—Cl Cl

:Cl—P—Cl: :Cl: :Cl—P—Cl: :Cl: . . . . Electrons remaining
Subtract electrons placed from total Arrange remaining electrons in PAIRS around terminal atoms until each has an octet Each pair of electron + 2 electrons for the bond Electrons remaining? Arrange remaining electrons on central atom Check whether EACH atom has an octet Count bonded electrons for BOTH atoms 26 – 6 = 20 electrons remaining 20-(3X6) = 2 :Cl—P—Cl: :Cl: . . :Cl—P—Cl: :Cl: . .

Practice NH3 CH4 CCl4 H2O

Multiple Bonds If central atom lacks octet and there are no electrons remaining Form multiple bonds (double or triple) by converting NON-BONDING PAIRS of electrons from terminal atoms into bonding pairs Ex: HCN Total electrons: 1+4+5 = 10 Central atom: C (recall H can’t be central) Draw molecule 4 electrons placed 6 remaining electrons placed on N. H has duet, N has octet, no remaining electrons for C Make double bond by “shifting” non-bonded pair from N C still lacks octet Make triple bond

Practice N2 O  CO2 C2H4

Exceptions—Incomplete octet
Some atoms remain electron deficient Generally limited to Boron—too small to support an octet Ex: BCl3 Total electron 3+7(3) = 24 Central atom: B 3 bonds = 6 electrons 6 unbonded electrons around each Cl = 18 Total electrons placed = 24 No electrons left over for B Forms an INCOMPLETE octet :Cl—B—Cl: :Cl: . .

Exceptions—Expanded Octet
RECALL: starting with non-metals of period 3 onwards, empty d subshells are available, and MAY hold shared electrons in covalently bonded atoms. Ex: SF6 Total electrons: 6 + 7(6) = 48 Central atom: S 2 e- = 1 bond

Practice SF4 PCl5 BF3 ClF5 IF5 XeF4

Charged particle Ex: NO3-1 Central atom Ex: NH4+1 Draw Molecule
Total electrons 5 + 6(3) + 1 = 24 Central atom N Draw Molecule Multiple correct structures resonance Ex: NH4+1 Total electrons – 1 = 8 Central atom N Draw molecule Put [brackets] around molecule with charge shown outside [] as superscript

Practice O2-2 NO2-1 OH- SO3-2

REMEMBER: each atom must get octet
Exceptions: Hydrogen-- duet Boron—incomplete octet Non-metals of period 3 onwards—expanded octet possible Hydrogen and Fluorine are ALWAYS terminal Can only form single bonds Each pair of electrons = 1 bond Pairs of electrons in valence shell NOT in a bond Always shown in pairs Called non-bonding pairs of electrons or lone pairs of electrons

RESONANCE and BOND LENGTH
As in previous examples, multiple correct lewis structure may be possible Ex: SO3-2 The double bond on oxygen could be in ANY three possible positions Therefore ALL are correct Known as resonance In actual molecule, it is noticed that actual measured bond length on ALL three sides are the same and inbetween (average) what would be predicted for a single bond (longer, weaker) vs. a double bond (shorter, stronger) Shown as RESONANCE for ease of use Ex: Ozone, O3 Actual observed bond length is 1.5 (or average) of the single and double bonds between oxygen atoms.

FORMAL CHARGE When multiple structures are possible, use FORMAL charge to determine BEST structure To calculate formal charge Total valence electrons around atom Number of non-bonding electrons in Lewis strucure ½ number of bonding electrons (or bonds) _____________________________________________________________ = Formal charge The BEST structure is the one with formal charge as LOW as possible, preferably zero

Determining Formal Charge
Ex: CO3-2 Three possible lewis structures Formal charges Carbon atom formal charge = 4 – 0 – ½ (8) = 0 Oxygen in C=O formal charge = 6 – 4 – ½ (4) = 0 Oxygen in C—O formal charge = 6 – 6 – ½(2) = -1 NOTE: The sum of formal charges adds up to the total charge on the species

Note: Incorrect structure for CO3-2 due to formal charges NOT being lowest possible
Total electrons= 4 + 6(3) + 2 = 24 3 bonds = 6 electrons. Remaining 18 6 electrons around EACH oxygen = 18 Problem? Carbon lacks octet Formal charges C: – 0 – ½ (6) = +1 O in each C—O bond: 6 – 6 – ½ (2) = -1 Total adds up to -2 charge, HOWEVER, each atom DOES NOT have lowest formal charge possible, therefore not accurate [ ] -2 . . . . :O—C—O: :O: . .

Practice

Practice Determine best structure based on formal charge

Dative or Coordinate Covalent bonds (electron deficient species)
RECALL: Boron has incomplete octet Said to be electron deficient Can make up the octet by forming bonds with other compounds with non-bonding pairs of electrons: ex: ammonia NH3 New, shared pair (covalent bond) is made up by using BOTH electrons from one species (in this case, the non-bonded electron pair on Nitrogen in ammonia), rather than one from each species as in a normal covalent bond Called DATIVE or CO-ORDINATE bond =

Electron Domain Geometry and Molecular Geometry (shapes of molecules) VSEPR
Electron domain geometry: indicates the “groups” or “pairs” of bonding and non-bonding electrons around central atom Helps identify bond angles Molecular geometry: indicates specific shape of molecule based on actual configuration. **NOTE: both use SAME terminology. Don’t confuse the two terms. Shapes of covalently bonded molecules and ions can be determined by considering the number of electron pairs around central atom Electron pairs repel one another to get as far apart as possible Known as Valence Shell Electron Pair Repulsion (VSEPR) Standard shapes and deviations can be predicted Rules to keep in mind Non-bonding pairs will REPEL MORE STRONGLY than bonding pairs of electrons For purposes of predicting shape multiple bonds are treated as single bonds

Predicting shapes Draw the lewis structure for molecule
Count the electron pairs (both bonding and non-bonding) around central atom Equivalent to electron domain geometry Use table to recall correct shape, name and bond angles that correspond to the shape that arranges electrons such that it minimizes repulsion Equivalent to molecular geometry. Specific consideration for bonded vs. non-bonded electrons.

Electron Domain Geometry and Molecular Geometry—MEMORIZE THIS

Practice Draw lewis structure and sketch the shapes for the following molecules. In each case, identify the number of bonding and non-bonding pairs of electrons around central atom, predict electron domain geometry, molecular shape and bond angle. PCl6- ICl3- BrF5 SO3 CH4 NH4+1 ICl4- SO2

Polar Bonds and Polar Molecules
Recall: Polarity = unequal sharing of electrons. Difference in charges at different ends of a particle. Consider Bond Polarity Comparison of Electronegativity values between TWO atoms Ex: CCl4: ∆EN 3.0 – 2.5 = 0.5  polar covalent Consider Molecular Polarity Comparison of charge differences (or distribution) on ENTIRE molecule Electron pair geometry: tetrahedral. Lone pairs on Central atom: none. Molecular geometry: tetrahedral. All BONDs are the SAME and EQUALLY distributed. Therefore MOLECULE is non-polar

In order for a substance to be “polar”, consider two factors
Bonds within the molecule must carry different charges (dipole must exist). AND Dipoles that are present must NOT cancel out due to symmetry RECALL: Dipole moments may be represented two ways +--> an arrow pointing towards negative charge center with the tail indicating positive charge center Using δ+ or δ- to indicate small areas of positive and negative charge centers

Examples: consider 3-pentanone (C2H5COC2H5) and trifluoromethane (CHF3). Draw molecule. Determine dipole moments. Predict molecular polarity. ̈ : 3-pentanone CH bonds: non-polar CO bond: Dipole moment due to non-bonding pairs on oxygen Nothing to CANCEL the dipole moment. Therefore: polar molecule Trifluoromethane CH bond: non-polar CF bonds: polar (dipole moments) Molecule NOT symmetrical. Dipoles DO NOT cancel out Therefore: polar molecule

Now consider hexane C6H14 and Carbon tetrachloride CCl4
Draw Lewis structure. Determine dipole moments. Determine molecular geometry Hexane CH Bonds: non-polar Dipole moments: none Symmetrical molecule Therefore: NON-polar molecule Carbon tetrachloride CCl bonds: Polar Dipole moments: each CCl bond Symmetrical molecule. Dipoles CANCEL each other out due to partial negative charges being EQUALLY spread around partially positive central Carbon Therefore: NON-polar molecule

Draw Lewis structures and Predict molecular geometry of the following compounds
BCl3 NH3 OF2 H2O

Hybridization Process of mixing of atomic orbitals or subshells
RECALL: in bonded atoms electrons exist in PAIRS (bonded or non-bonded) Consider CH2 (a molecule that does NOT exist) In the ground state Carbon has the following valence shell configuration 2s22p2 Hydrogen has the following valence shell configuration 1s1 Based on this, one may predict that the two electrons in 2p will participate in bonding and the 2s electrons that are already paired will not participate in bonding However, CH2 is an extremely unstable molecule and does not exist

The simplest hydrocarbon that does form is CH4. HOW?
By addition of a very small amount of energy, one of the 2s electrons gets promoted to the empty 2pz orbital to form excited state C with electron configuration of [He] 2s1 2px1 2py1 2pz1 The four orbitals are then “mixed together” to form four, sp3 hybrid equivalent orbitals. Each of the four hybrid orbitals can from simple sigma (single) bonds When they overlap with each of the 4 atomic s orbitals of 4 hydrogen atoms four bonds around central Carbon

Examples Let’s consider the following covalent molecules and predict their hybridization by filling in the table below Example Molecule Ground state electron configuration Number & Type of Hybrid orbitals Number & Type of empty, unhybridized orbitals Type of Hybridization BeCl2 BF3 CH4

Determining Type of Hybridization
Determining type of hybridization is simple to predict Consider total number of electron pairs (including BOTH bonding and non-bonding pairs) around central atom RECALL: each electron pair needs one orbital Therefore total number of electron pairs equal total number of orbitals needed RECALL: count of each type of orbital possible (s=1, p=3, d=5, f=7) Starting with 1 s type orbital, add as many p, d and so on orbitals as there are electron pairs. Example 2 pairs = sp 3 pairs = sp2 4 pairs = sp3 5 pairs = sp3d (**generally not tested on ap) 6 pairs = sp3d2 (**generally not tested on ap) RECALL: multiple bonds are considered a SINGLE electron pair

Practice Determine the hybridization on each carbon in the following molecules a) d) count carbons from left (1) to right (5) b) c) count carbons from left (1) to right (4)

Sigma (σ) and Pi(π) Bonds
Consider two hydrogen atoms forming H2 Each atom has 1 electron in its spherical 1s orbital In H2, the two 1s orbitals overlap to share a pair of electrons Form a single sigma (σ) covalent bond

Consider two oxygen atoms forming O2
Each atom has the electronic configuration [He]2s22p4 Indicating there is 1 filled 2p orbital and 2 half-filled 2p orbitals One of the two half filled, incomplete 2p orbital can participate in bonding by Simple end on end overlap Electron density is located the axis of the bond Sigma (σ) covalent bond The other half filled, incomplete 2p orbital will overlap sideways Hence electron density is located above and below the axis of the bond Pi (π) covalent bond Results in oxygen atoms forming a double bond

Summary Whenever a double or triple covalent bond is formed
The first (and strongest) bond is AlWAYS a sigma (σ) bond Hence, single bonds are sigma bonds All bonds after the first are considered to be pi (π) bonds Pi bonds lead to delocalized electron clouds due to overlap of unhybridized p orbitals Potential for some electron movement (and occasionally, rare ability of molecular substances to conduct electricity) Sigma bonds can freely rotate Pi bonds prevent rotation

Ex: alkenes (double bonded C=C hydrocarbons) prevent rotation along double bond
Leads to existence of cis- and trans- isomers. Example: cis and trans but-2-ene. The C=C bond cannot rotate because of presence of pi bonds Result: CH3- groups remain fixed on the same side of double bond (cis) or on opposite sides of the double bonds (trans) In Alkanes (single bonded carbons in a hydrocarbon), all carbons are connected via single sigma bonds Rotation can occur No difference in actual structure or connectivity

Bonding and Properties of Solids
Alloys Mixtures of metals No metallic compounds Because by definition a compound has a fixed ratio of elements and fixed, unique, specific properties Metallic mixtures do NOT need to be in a fixed proportion due to sharing of valence electrons and type of bonding that exists

Two common types of alloys
Interstitial Alloy Additional, smaller atoms of different element fill the spaces in the metallic lattice Ex: steel-carbon atoms fill spaces between iron atoms Properties Less malleable and ductile than pure metals Presence of smaller atoms makes structure more rigid and less flexible Substitutional Alloy One metal’s atoms are replaced by another metal’s. Usually of similar radius Ex: Brass—copper atoms replaced with zinc atoms Similar to interstitial—reduced malleability and ductility Densities typically between densities of component metals

Common properties Sea of mobile electrons is maintained
Alloys remain good conductors In some cases, surface of alloy or metal may take on a different property than remainder of solid Due to oxide layer forming, following reaction with oxygen in the air

Practice Problems Draw a cross section illustration of the exterior surface and interior of a sample of steel given the following information. Steel is composed of primarily Iron with a small amount of carbon. Additionally approximately 15% by mass of chromium is added to the alloy The chromium atoms on the surface form an oxide layer which is primarily non-reactive, making the steel stainless Which of the following diagrams best depicts an alloy of Ni and B? Element Molar Mass (g/mol) Atomic Radius (pm) Fe 55.85 125 Cr 52.00 127 C 12.01 77 O 16.00 73

Giant covalent network solids
RECALL: group 14 elements can make four covalent bonds Allows them to bond together in very large, continuous networks. Few examples Diamonds Graphite Silicon dioxide Silicon carbide

Diamonds and graphite Allotropes of carbon
Different forms (due to differences in bonding) of the same substance Made entirely of carbon atoms Covalently bonded in a giant continuous network

Diamond Three dimensional tetrahedral unit
All carbons atoms bonded to one another with very strong covalent bonds. HUGE macro structure Makes diamond very strong, hard Fixed, rigid covalent bond angles High melting and boiling points

Graphite Two dimensional layered structure
Each carbon bonded to three others in each plane Covalent bonds within layers are strong—high melting point Conducts electricity in ONLY one plane Due to one “free” outer electron on EACH carbon Spread over each layer (delocalized) leading to a sea of electrons along one plane only Electrons cannot move between layers  no conduction in the vertical plane Used as a lubricant Weak london dispersion forces hold the layers together Able to slide over one another, making graphite soft and a good lubricant

Silicon and semi-conductors
Silicon forms structure very similar to diamonds Not surprising (same group, similar valence configuration) Pure semi-conductors (like silicon) are generally poor conductors of electricity When doped (deliberate introduction of an impurity), conductivity increases n-type (negative) conductor: doping carried out with an element with “extra” valence electrons compared to silicon (i.e. group 15 element, such as phosphorous—similar size) p-type (positive) conductor: doping carried out with an element with one less valence electron compared to silicon (i.e group 13 element, such as Boron) Result—disrupting valence shells of silicon atoms allows effective flow of electrons Previous insulator becomes a good conductor Important to electronics—boundaries between p and n known as p-n junctions Allows controlled flow of electrons Conductivity increases with increasing temperature as electrons are promoted from valence band to conduction band

Molecular solids Non-metal solids Low melting point
Ex: iodine Polymers Low melting point Primary intermolecular bonding: LDFs Structure appears similar to NaCl (unit cell) but inter bonds are very weak Do not conduct electricity No charged particles (ions or electrons)

Review Based upon the data given in the table below, deduce the type of bonding present in solids A, B and C.

BOND ENERGIES and ENTHALPIES
RECALL: Atoms are attracted to each other when the outer electrons of one atom are electrostatically attracted to the nuclei of another atom Attraction between the two atoms makes them increasingly stable, giving lower and lower potential energies However, if atoms continue to approach one another, and get increasingly close, there comes a point at which the two nuclei (and/or the more densely packed core electrons) will start to repel each other As they start to repel, potential energy is raised, and the atoms become less stable Happy medium (i.e. bond formed) at a distance where the force of attraction and repulsion results in the LOWEST potential energy Distance between the nuclei at this point is the bond length Potential energy at this point is the bond strength

Plotting potential energy vs. bond length
Typical plot of potential energy vs. bond length appears as follows Ex: Hydrogen, H2 bond length: 74 pm bond strength: 436 kJ Since forces of attraction stabilize atoms, the greater the number of electrons involved, the stronger the attraction Result: triple bonds tend to be stronger than double bonds, and double bonds are stronger than single bonds Shorter bonds also tend to be stronger than longer bonds

Bond dissociation energy
RECALL: Bond breaking is always an endothermic process (ΔH is +) and bond making is always an exothermic process (ΔH is -) The strength of a bond can be measured by measuring the amount of energy required to break a bond Ex: H2  2H Bond Energy (ΔH) = +436 kJ

Consider a polyatomic molecule, such as methane, CH4
Composed of 4 identical C-H bonds. However, each bond dissociation energy is actually quite different CH4(g)  CH3(g) + H(g) 435 kJ/mol CH3(g)  CH2(g) + H(g) 453 kJ/mol CH2(g)  CH (g) + H(g) 425 kJ/mol CH(g)  C(g) + H(g) 339 kJ/mol TOTAL: kJ/mol TO overcome complexity of accounting for each individual bond dissociation energy, we define bond energy as the AVERAGE of each of the four, separate C-H bonds. Therefore 1652/4 = 413 kJ/mol

Determining ΔH values based on Bond strength
The energy change in a reaction can be calculated by Summing up the energy required to break each of the bonds in the reactants (a positive value) Summing up the energy released by making each of the bonds in the products (a negative value) Summing the totals of the two values above If the total energy needed to break the reactant bonds is greater than the total energy released in making the product bonds, then the reaction (system) is endothermic Energy is absorbed by surrounding Temperature of surrounding decrease If the total energy needed to break the reactant bonds is less than the totoal energy released in making the product bonds then the reaction (system) is exothermic Energy is released to the surrounding Temperature of surrounding rises

Common Bond Enthalpy Values

Practice Calculate the standard enthalpy change of reaction below
CH3CH=CH2 + H2  CH3CH2CH3 CH3CH=CH2 + Br2  CH2BrCHBrCH3 CH Cl F2  CF2Cl HF + 2HCl CH3CH2OH + CH3COOH  CH3CO2CH2CH3 + H2O

CHEMICAL BONDING Chapter 8-9.

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CHEMICAL BONDING Chapter 8-9.

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Presentation on theme: "CHEMICAL BONDING Chapter 8-9."— Presentation transcript:

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