Topic C – Intramolecular Bonding
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TOPIC C – INTRAMOLECULAR BONDING
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• Intermolecular attractions are relatively very strong, making these bonds much stronger than intermolecular forces.
• Ionic bonding• Ionic bonding involves the transfer of electrons between
atoms to form ions.
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• Metals have a tendency to lose electrons • forming positive ions
• Non-metals have tendency to gain electrons• Forming negative ions
• Ionic bonds are usually formed between metals & non-metals
• These strong electrostatic forces between the charged particles are called ionic bonds.
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• The electrostatic forces are large and the bond is a strong one
• Coulomb’s law predicts that force increases with increasing charge and decreasing distance between charges
• Therefore the strongest ionic bonds are formed between ions that are small (because they can get close to one another) and highly charged, i.e., one that have high charge densities.
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• Practice Problems:
• 1. List the ions present and hence the formula of the
following compounds; Sodium chloride, calcium chloride, iron(III) bromide, sodium oxide.
• 2. When comparing sodium chloride and sodium bromide, which compound would be expected to have the strongest ionic bonds? Explain.
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• The ions in an ionic solid are held rigidly in fixed positions in a giant, three-dimensional lattice.
• These rigid structures mean that ionic compounds are not malleable or ductile and tend to be brittle• When the ordered structure is disrupted, like charges repel and the
solid splits apart.
In the ionic structure the lattice of ions is held together by strong electrostatic interactions between them. The strong bonds give ionic solids high melting and boiling points and low volatility and subsequently low vapor pressures. Ionic substances can only conduct electricity when molten or in solution, since in the solid the ions are rigidly held and cannot move.
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• When ionic solids dissolve:• Polar water molecules are attracted to the oppositely charged ions• penetrate the lattice attaching themselves to the ions.
• The process is called hydration and the ions are said to be, hydrated.
• The ions become free to move when they are hydrated, and the solution will be a good conductor of electricity.
• A non-polar solvent will not be attracted to the ions in an ionic solid, • so the ionic bonds holding the solid together are not broken and the solid
will not dissolve.
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• Covalent bonding• Electrons are shared between atoms to achieve full s and
p sub-shells.• This forms discrete molecules.
• One shared pair of electrons represents a single covalent bond
• Two shared pairs represent a double bond, etc.
• These bonds usually occur between atoms that are non-metals.
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• It is possible to think of compounds being either 100% ionic or 100% covalent
• But they are actually they are two extremes of a sliding scale.
• Most bonds are intermediate between the two• ( a largely covalent substance actually having a degree of ionic
behavior, and vice-versa)
• Ionic Polar CovalentCovalent
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• Ionic substance with some covalent character
• Ionic Covalent Ionic w/covalent• character
• If the cation is small and highly charged it will distort the charge cloud around the anion
• Its possible to observe the ionic bond beginning to acquire some covalent character.
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• Fajans rules help to assess the degree of distortion (POLARIZATION).
• Distortion will be at a maximum when:• 1. The cation is small and highly charged,• (has a high charge density). • 2. When the anion is large and highly charged,• (electrons are more loosely held).
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• Covalent substance with some ionic character (polar covalent)
• If one atom in a covalent bond has a higher electronegativity than the other, then the electrons are attracted toward the more electronegative atom
• This leads to:• an electron cloud distortion and • a re-distribution of electron charge density
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• Using the Pauling scale of electronegativities• We can make predictions about the degree of ionic and covalent
character in a compound.
• Consider differences in electronegativity and using the table below, one can predict the % of covalent and ionic character present in a bond between two atoms.
• A difference of approximately 1.7 marks the boundary between predominately ionic or covalent character.
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• Metal structures (metallic bonding)
A metals’ structure can be considered to be a close packed lattice of positive atoms/ions surrounded by a “sea” of moving, delocalized electrons. These electrons and their movement cause metals to be good conductors of electricity. The close packed atoms/ions make them good conductors of heat. The metallic bond is the electrostatic attraction between the positive and negative charges. The flexibility of these bonds makes metals malleable & ductile.
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• Lewis structures of covalently bonded molecules • Lewis structures use dots to represent valence electrons
in atoms when they form molecules.
• Atoms form molecules and share electrons to achieve full s and p sub-shells. • In the case of hydrogen this is two (a duet)• In the case of the second period non-metal elements this is eight
(octet rule). • Some elements in the third period can have more than eight
electrons in their outer shells (expanded octet rule).
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Drawing Lewis structures • 1. Calculate the total number of valence electrons
• (taking into account any negative or positive charges).
• 2. Decide which atom is the central one, if > 2 atoms• (it will be the least electronegative atom but never hydrogen).
• Use a line to represent one pair of electrons forming a
covalent bond between each outer atom and the central atom.
• 3. Arrange the remaining electrons to complete the octets of the outer atoms
• Place any remaining electrons on the central atom• (if necessary expand the central atoms’ octet)
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• 4. If the total electrons used in the Lewis structure exceeds the total available form double or triple bonds.• (convert lone pair electrons into bonding pairs)
• 5. Any electron pairs that occur in the valence shell of an atom but do not form a bond with another atom are called non-bonding electrons or lone pairs.
• Draw Lewis diagrams for each of these species:• F2, O2, N2, HCl, HF, H2O, NH3, CBr4, PF5, PCl6-, NH4+
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Shapes of covalently bonded molecules and ions
• Consider the number of electron pairs around the central atom.
• Electron pairs repel one another and try to get as far apart as possible.
• Called - Valence Shell Electron Pair Repulsion theory or VSEPR.
• See handout or notes for the shapes, their names, bond angles and other relevant information.
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• A non-bonding pair will repel more strongly than a bonding pair.
• This tends to decrease the bond angles and change
the shape of molecules. (See next slide)
• For the purposes of predicting shape multiple bonds can be considered as single bonds.
• • Predict shape by:• 1. Drawing the Lewis dot structure for the molecule. • 2. Counting the electron pairs (both bonding and non-
bonding) around the central atom. • 3. Using the table on the handout to determine the correct
shape, name and bond angles that correspond to
the shape that arranges that minimizes repulsion.
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VSEPR Theory
Two electron pairs around the central atom - both bonded.
LINEAR
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VSEPR TheoryThree electron pairs around the central atom – all bonded.
TRIGONAL PLANAR
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VSEPR TheoryFour electron pairs around the central atom – all
bonded.
TETRAHEDRAL
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VSEPR TheoryFive electron pairs around the central atom – all
bonded.
Trigonal bipyramidal
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VSEPR TheorySix electron pairs around the central atom – all
bonded.
OCTAHEDRAL
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• Draw Lewis structures and sketch the shapes for the following:
• In each case identify the number of bonding & lone pairs around the central atom, and predict bond angles.
• PCl6-, ICl3, BrF5, SO32-,
• CH4, NH4+, ICl4-, SO2
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• Resonance structures & bond length
• Occasionally it is possible to draw a more than one valid Lewis structures.
• We call these resonance structures.
• For example, carbon dioxide has three possible Lewis structures.
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• The carbonate ion provides another example.
• The Lewis structure for the carbonate ion can be represented by the diagram below.
• It is also correct to draw the double bond on the right hand oxygen or the left hand oxygen creating three different resonance structures.
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• The structure is actually an “average” of all three possible resonance structures
• Each bond is between one and two covalent bonds• (not a single or double)
• ** If a multiple bond (double or triple) is created• bond length observed will be shorter than the corresponding single
bond.• This is because a double bond is stronger than a single bond and
hence pulls the atoms closer together.
• A triple bond is correspondingly shorter and stronger than a double bond.
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Formal charge• The formal charges of each atom within a structure can be
calculated using the formula below.
• Formal Charge = • [# of valence electrons on atom] – [electrons in lone pairs + ½ the # of bonding electrons]
• The formal charge of an atom is used in one of two ways. • (1) To suggest where charges may most reasonably lay (for example where the
2- charge of the carbonate ion actually resides)
• Using the Lewis structure for the carbonate ion (CO32-) it is
possible to determine the formal charges on the atoms and therefore the most likely separation of charge.
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• Carbon atom Formal charge = 4 – 0 – ½ (8) = 0 • Oxygen atom in C=O Formal charge = 6 – 4 – ½ (4) = 0• Each oxygen atom in C-O Formal charge = 6 – 6 – ½ (2) = -1
• Note that the sum of the formal charges adds up to the total charge on the ion or molecule.
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• (2) To help select the most plausible structure from a set of resonance structures
• Using the two possible Lewis structures for CH2O with formal charges added, we can determine the most likely Lewis structure for the compound.
• The sum of the formal charges adds up to the total charge on the species.
• To determine which structure is most likely,• choose the structure with zero formal charges, • and/or formal charges with absolute values as low as possible, • and/or keep any negative formal charges on the most electronegative
atoms. • Since the first structure has no formal charges it is considered to be the most
plausible.
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Dative or Co-ordinate bonding (electron deficient species)
• Some molecules have a central atom that does not have a complete octet of electrons.
• Example = BF3.
• In this molecule the boron atom has only six electrons surrounding it rather than the usually required eight.
• The boron atom is said to be electron deficient. • It can complete the octet by forming bonds with other
compounds that have non-bonding pairs of electrons, [ ammonia (NH3) ].
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• The new covalent bond formed by using both the electrons from nitrogen (one species).
• This type of covalent bond is called a dative or co-ordinate bond.
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Polar bonds and polar molecules• In order for a substance to be “polar”• (1) a dipole moment must exist • (2) and the dipoles must NOT cancel out due to
symmetry.
• The dipole moment is indicated by an arrow that points toward the negative charge and the tail of the arrow indicating the positive charge.
• using + and - to indicate small areas of positive and negative charge.
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• For example:• 2-pentanone and trifluoromethane (both polar)• Opposite ends of the molecules carry different charges
and there is no canceling of the dipoles.
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• In Hexane (C6H14):
• carbon and hydrogen have similar electronegativities and the bonds are non-polar.
• Carbon tetrachloride (CCl4):
• has four polar C-Cl bonds but has a symmetrical shape
• The dipoles cancel and it is non-polar overall.
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• Other non-polar compounds as a result of symmetry include CO2 and BCl3.
• Three geometries that are symmetrical:• Tetrahedral• Linear• Trigonal Planar
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Polyatomic molecules and hybridization of orbitals
• For polyatomic molecules the orbitals of the central atoms are said to undergo hybridization (a process of mixing).
• Consider the example in the table below of the sp3 hybridization of carbon in the methane (CH4) molecule.
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• In its ground state – carbon has two unpaired electrons• To form the stable CH4 molecule one of the electrons in
the 2s orbital is promoted into the empty 2pz orbital
• This forms - [He] 2s1 2px1 2py
1 2pz1.
• These four orbitals then ‘hybridize’ to form the four, sp3 equivalent orbitals.
• The sp3 represents the one “s” orbital and three “p” orbitals that hybridized.
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• These four hybrid orbitals then form simple sigma bonds
(4 covalent bonds) and the familiar, tetrahedral shape.
The examples on the next slide use simple sigma bonds to form the molecule.
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Predicting Hybridized Orbitals• The type of hybridization present is simple to predict.• • The total number of electron pairs around the central atom
determines the number of orbitals that need to be hybridized.
• By taking (in order), one s, and as many p orbitals as required (up to 3), we can determine the correct number of orbitals needed.
• For example:• Four electron pairs, require four orbitals and have a
hybridization of (s + p + p + p) or sp3.
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In terms of energyIn terms of energyE
nerg
y
2p
2s
Hybridization sp3
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• We know the geometry from experiment.• We know the orbitals of the atom• hybridizing atomic orbitals can explain the geometry.
• So if the geometry requires a tetrahedral shape, it is sp3 hybridized
• This includes bent and trigonal pyramidal molecules because one of the sp3 lobes holds the lone pair.
How we get to hybridization
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The Bonding in MethaneH
CH H
HC
H
H
H
H
C:2s 2p
1s
4 x H:
+
C forms 4 bonds to H
Electron configuration of C shows only 2 places to bond
but . . .
1) How does carbon form 4 bonds?2) At 109.5o angles?
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sp3 Hybridization of Atomic Orbitals
+
+ +
4 atomic orbitals2s + 2px + 2py + 2pz
yield4 hybrid atomic orbitals
4 sp3 orbitals with 2e- per orbital
Electron density distributed to the corners of a tetrahedron
C:s p 4 sp3
109.5o4 e-
domains
C MIX
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Electrons From Hydrogen Share With sp3 Electrons
Pairs of electrons keeps nuclei from
repelling
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Similar for NH3
8 Electrons in sp3, one is a lone pair
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• C2H4
• Double bond acts as one pair.• trigonal planar• Have to end up with three blended orbitals.
• Use one s and two p orbitals to make sp2 orbitals.
• Leaves one p orbital perpendicular.
sp2 hybridization
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In terms of energyIn terms of energyE
nerg
y
2p
2s
sp2Hybridization
2p
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• Where is the P orbital?• Perpendicular• The overlap of orbitals makes a sigma bond (s bond)
• Two Types of bonds• Sigma bonds from overlap of orbitals.• Between the atoms.
• Pi bond (p bond) above and below atoms• Between adjacent p orbitals.
• The two bonds of a double bond. CCH
H
H
H
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When an sp2 Orbital is Used a p Orbital is Left Over for Pi Bonding
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• When three total bonded or unbonded electrons pairs on the central atom.
• trigonal planar• 120º• one s one p bond
When do we get sp2 hybridization?
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• When two things come off.• One s and one p hybridize.• linear
What about two
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• When two bonded or unbonded pairs of electrons are present around the central atom.
• One s and one p hybridize.• linear
• End up with two lobes 180º apart.• p orbitals are at right angles• Makes room for two p bonds and two sigma bonds.• A triple bond or two double bonds.
sp hybridization
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In terms of energyIn terms of energyE
nerg
y
2p
2s
Hybridizationsp
2p
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sp Hybrid Orbitals
BeCl2 or CO2
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• C can make two s and two p• O can make one s and one p
CO2
CO O
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• PCl5• The model predicts that we must use the d orbitals.
• dsp3 hybridization• There is some controversy about how involved the d orbitals
are.
•Dsp3
• Trigonal bipyrimidal• has only s bonds.• no p bonds.• basic shape for five things.
Breaking the octet
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dsp3
d2sp3
Hybridization Using d Orbitals
PCl5
SF6
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Sigma () and Pi () bondsConsider two hydrogen atoms forming H2.
Each hydrogen atom has one electron that occupies a spherical 1s orbital.
On forming H2 these 1s orbitals overlap to share a pair of electrons forming a molecule and a covalent bond.
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• Whenever a double or triple covalent bond is formed:• the first (and strongest) bond is always a sigma bond.
• All bonds after that are considered to be pi bonds.
• Pi bonds lead to delocalized electron clouds via the overlap p orbitals
• Sigma bonds can freely rotate, but pi bonds in alkenes prevent the rotation of carbon-carbon bonds, leading to the existence of cis and trans isomers.
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• The C=C bond cannot rotate because of the presence of the pi bond, so the CH3 groups remain fixed on the same side of the double bond (cis), or on opposites sides of the double bond (trans).
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• In butane:• The sigma bond between the central carbons can rotate
(no pi bonds) so these two conformations are 100% interchangeable and are the same compound (not isomers).