READING: Chapter 10 sections 1 – 3 READING: Chapter 10 sections 1 – 3 HOMEWORK – DUE TUESDAY...

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READING: Chapter 10 sections 1 – 3 HOMEWORK – DUE TUESDAY 11/10/15

HW-BW 9.1 (Bookwork) CH 9 #'s 5, 6, 10, 51, 53, 54, 57 – 67 (odd), 86, 88, 89

HW-WS 16 (Worksheet) (from course website) HOMEWORK – DUE THURSDAY 11/12/15

HW-BW 9.2 (Bookwork) CH 9 #’s 27, 28, 29, 31, 33, 37, 38, 40, 47, 48, 49 HW-WS 17 (Worksheet) (from course website)

Lab Next Monday/Tuesday – EXP 13

Prelab

The lowercase Greek letter delta, d, is used to indicate a polar bond.

The MORE EN element has extra e-, so it is negative and is indicated by the symbol d–.

The LESS EN element is short of e-, so it is positive and is indicated by the symbol d+.

H – Cl d+ d–

3.02.1

Electronegativity and Polarity

Give delta notation and polarity arrows for the following:

C O Si OF H N Cd+ d– d+d– d+d– d+ d–

2.5 3.5 4.0 2.1 3.0 2.5 1.8 3.5

Electronegativity and Polarity

Lewis Structures

1) Determine the total number of valence electrons available in the chemical

If ion, add 1 electron for each negative charge and subtract 1 electron for each positive

2) Draw the skeletal structure of the molecule using single bonds to connect the atoms

Central atom(s) will be surrounded by other atomsCentral atom(s) tend to be the element that is the least electronegativeH and F always exterior atoms

3) Fill the octets for all atoms except hydrogen (2), beryllium (4) and, boron (6)

Count total electrons drawn and subtract this from the number of valence electrons available found in step #1.

If you have not drawn enough electrons, add the missing ones to the central atom

If you have drawn too many electrons, remove lone pair(s) and add multiple bonds

# of valence e- needed = # of bonds formed (guideline only)

Resonance

The Lewis structure of O3 can be drawn in 2 ways:

Neither structure depicts O3 accurately, because in reality the O-O bonds are identical in length and energy.

Resonance

Resonance structures have the same relative placement of atoms but different locations of bonding and lone electron pairs.

The structure of O3 is shown more correctly using both Lewis structures, called resonance structures.

A two-headed resonance arrow is placed between them.

ResonanceA species like O3, which can be depicted by more than one valid Lewis structure, is best represented as a resonance hybrid.

The real structure of the resonance hybrid for O3 is an average of its contributing resonance forms.

Lewis structures depict electrons as localized either on an individual atom (lone pairs) or in a bond between two atoms (shared pair).

In a resonance hybrid, electrons are delocalized: their density is “spread” over a few adjacent atoms.

Dotted lines are used to show delocalized electrons.

Formal charge on O = 6 - 2Formal charge on O = 6 - 2 - 4Formal charge on O = 6 - 2 - 4 = 0

Formal charge on C = 4Formal charge on C = 4 - 4Formal charge on C = 4 - 4 - 0Formal charge on C = 4 - 4 - 0 = 0

Formal Charge

Formal charge on an atom = # of valence e- - # of bonds - # of lone electrons

From periodic table Individual e-, NOT lone pair

Formal charge on C

Formal charge on O = 6Formal charge on O

Formal charge on O = 6 - 1Formal charge on O = 6 - 1 - 6Formal charge on O = 6 - 1 - 6 = -1

Formal charge on N = 5Formal charge on N = 5 - 4Formal charge on N = 5 - 4 - 0Formal charge on N = 5 - 4 - 0 = +1

Formal Charge

Formal charge on an atom = # of valence e- - # of bonds - # of lone electrons

From periodic table Individual e-, NOT lone pair

Formal charge on N

Formal charge on O = 6Formal charge on O N

O OFormal charge on O = 6 - 1 - 6 = -1

Formal charge on O = 6 - 2Formal charge on O = 6 - 2 - 4Formal charge on O = 6 - 2 - 4 = 0Formal charge on O = 6Formal charge on O

Energetics of Ionic Bonding The ionization energy of the metal is endothermic

Na(s) → Na+(g) + 1 e ─ DH° = +496 kJ/mol

The electron affinity of the nonmetal is exothermic ½Cl2(g) + 1 e ─ → Cl─

(g) DH° = −349 kJ/mol

Generally, the ionization energy of the metal is larger than the electron affinity of the nonmetal, therefore the formation of the ionic compound should be endothermic

Yet the heat of formation of most ionic compounds is exothermic and generally large. Why? Na(s) + ½Cl2(g) → NaCl(s) DH°f = −411 kJ/mol

The extra energy that is released comes from the formation of a structure in which every cation is surrounded by anions, and vice versa

This structure is called a crystal lattice held together by the electrostatic attraction of the cations for all the

surrounding anions maximizes the attractions between cations and anions, leading to

the most stable arrangement Electrostatic attraction is nondirectional!!

no direct anion–cation pair There is no ionic molecule

the chemical formula is an empirical formula, simply giving the ratio of ions based on charge balance – Formula Unit

Crystal Lattice

The extra stability that accompanies the formation of the crystal lattice is measured as the lattice energy

Lattice energy is the energy released when the solid crystal forms from separate ions in the gas state always exothermic

Lattice energy depends directly on size of charges and inversely on distance between ions

Crystal Lattice

Lattice Energy: The Born–Haber Cycle The Born–Haber Cycle is a hypothetical series of reactions

representing the formation of an ionic compound from its constituent elements

Reactions are chosen so that the change in enthalpy of each is known except for the last one, which is the lattice energy

Hess’s Law returns!! DH°f(salt) = DH°f(metal atoms, g) + DH°f(nonmetal atoms, g) + DH°f(cations, g) + DH°f(anions, g) + DH°(crystal lattice)

DH°(crystal lattice) = Lattice Energy

for metal atom(g) cation(g), DH°f = 1st ionization energy

for nonmetal atoms (g) anions (g), DH°f = electron affinity

heat of formation, Na(g)

bond energy Cl–Cl OR heat of formation Cl(g)

1st ionization energy Na(g)

electron affinity Cl(g)

lattice energy NaCl(s)

heat of formation NaCl(s)

Lattice Energy: The Born–Haber Cycle

Na(s) → Na(g) DHf (Na(g))

½ Cl2(g) → Cl(g) DHf (Cl(g))

Na(g) → Na+(g) DHf (Na+

Cl(g) → Cl−(g) DHf (Cl −

Na+ (g) + Cl−

(g) → NaCl(s) DH (NaCllattice)

Na(s) + ½ Cl2(g) → NaCl(s) DHf (NaCl(s))

Lattice Energy: The Born–Haber CycleNa(s) → Na(g) DHf (Na(g))

½ Cl2(g) → Cl(g) DHf (Cl(g))

Na(g) → Na+(g) DHf (Na+

Cl(g) → Cl−(g) DHf (Cl −

Na+ (g) + Cl−

(g) → NaCl(s) DH (NaCllattice)

Na(s) + ½ Cl2(g) → NaCl(s) DHf (NaCl(s))

Na(s) → Na(g) 108 kJ

½ Cl2(g) → Cl(g) ½ (244 kJ)

Na(g) → Na+(g) 496 kJ

Cl(g) → Cl−(g) −349 kJ

Na(s) + ½ Cl2(g) → NaCl(s) −411 kJ

NaClLattice = − 411 kJ − 108 kJ − 122 kJ − 496 kJ + 349 kJ

= −788 kJ

Lattice Energy: The Born–Haber CycleCalculate the lattice energy of MgCl2 given:

heat of formation Mg(g) DH = +147.1 kJ/molheat of formation Cl(g) DH = +122 kJ/mol1st ionization magnesium DH = +738 kJ/mol2nd ionization magnesium DH = +1450 kJ/mol1st electron affinity chlorine DH = −349 kJ/molheat of formation MgCl2(s) DH = −641 kJ/mol

Draw an enthalpy diagram for the above process

Lattice Energy: The Born–Haber CycleCalculate the lattice energy of MgCl2 given:

heat of formation Mg(g) DH = +147.1 kJ/molheat of formation Cl(g) DH = +122 kJ/mol1st ionization magnesium DH = +738 kJ/mol2nd ionization magnesium DH = +1450 kJ/mol1st electron affinity chlorine DH = −349 kJ/molheat of formation MgCl2(s) DH = −641 kJ/mol

1881.1 + MgCl2(lattice) = −641

MgCl2(lattice) = −2522.1 kJ/mol

147.1 kJ147.1 kJ + 2(122 kJ)147.1 kJ + 2(122 kJ) + 738 kJ147.1 kJ + 2(122 kJ) + 738 kJ + 1450 kJ147.1 kJ + 2(122 kJ) + 738 kJ + 1450 kJ + 2(−349 kJ)147.1 kJ + 2(122 kJ) + 738 kJ + 1450 kJ + 2(−349 kJ) + MgCl2(lattice)147.1 kJ + 2(122 kJ) + 738 kJ + 1450 kJ + 2(−349 kJ) + MgCl2(lattice) = −641 kJ147.1 kJ + 2(122 kJ)147.1 kJ + 2(122 kJ) + 738 kJ + 1450 kJ + 2(−349 kJ)

Mg2+(g) + 2 Cl(g) + 2 e–

MgCl2(s)

Mg(s) + Cl2(g)

Mg(g) + ½ Cl2(g) + Cl(g)

Mg(g) + 2 Cl(g)

Mg+(g) + 2 Cl(g) + 1 e–

Mg2+(g) + Cl –

(g) + Cl(g) + 1 e–

Mg2+(g) + 2 Cl–

Mg(g) + Cl2(g)

Lattice energy MgCl2(s)

heat of formation Mg(g) DH = +147.1 kJheat of formation Cl(g) DH = +122 kJ1st ionization magnesium DH = +738 kJ2nd ionization magnesium DH = +1450 kJ1st electron affinity chlorine DH = −349 kJheat of formation MgCl2(s) DH = −641 kJ

DH = −2522 kJ

Trends in Lattice Energy: Ion Size Lattice energy is a measure of the strength of an ionic bond. The force of attraction between charged particles is

inversely proportional to the distance between them Larger ions mean the center of positive charge (nucleus of

the cation) is farther away from the negative charge (electrons of the anion) larger ion = weaker attraction = smaller lattice energy

Electrostatic energy charge A x charge Bdistance

Electrostatic energy cation charge x anion chargecation radius + anion radius DHo

lattice

Trends in Lattice Energy: Ion Size

The force of attraction between oppositely charged particles is directly proportional to the product of the charges

Larger charge means the ions are more strongly attracted larger charge = stronger attraction stronger attraction = larger lattice energy

Of the two factors, ion charge is generally more important

Lattice Energy = −910 kJ/mol

Trends in Lattice Energy: Ion Charge

Lattice Energy = −3414 kJ/mol

Trends in Lattice Energy

Order the following ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO

Based on charge: (KBr = KCl) < (CaO = SrO)

Based on size: KBr < KCl and SrO < CaO

Overall order: KBr < KCl < SrO < CaO

Trends in Lattice Energy

Order the following ionic compounds in order of increasing magnitude of lattice energy: MgS, NaBr, LiBr, SrS

Based on charge: (NaBr = LiBr) < (MgS = SrS)

Based on size: NaBr < LiBr and SrS < MgS

Overall order: NaBr < LiBr < SrS < MgS

Electron pairs surrounding an atom repel each other. This is referred to as Valence Shell Electron Pair Repulsion (VSEPR) theory.

VSEPR Theory

NOT based on how many electron pair there are, but where they are.Where they are: electron groups (charge clouds)

HUH?!?!?!

VSEPR Theory

HO C O

What counts as an electron group?

1) A connection between the atom of interest and any other atom

a) single, double, and triple bonds each count one electron group

2) A lone pair on the atom of interest

O HH4 electron groups 4 electron groups

4 electron groups

2 electron groups

3 electron groups2 electron groups

3 electron groups

4 electron groups

VSEPR Theory

NOT based on how many electron pair there are, but where they are.Where they are: electron groups (charge clouds)

Each group of valence electrons around a central atom is located as far as possible from the others, to minimize repulsions.

VSEPR Theory

VSEPR TheoryThe electron pair geometry gives the arrangement

of atoms AND the lone pair electrons around the central atom.There are five basic arrangements of electron groups

around a central atom.

VSEPR Theory The electron pair geometry gives the arrangement of atoms

AND the lone pair electrons around the central atom. There are five basic arrangements of electron groups around a

central atom.

The molecular shape is the three-dimensional arrangement of nuclei joined by the bonding groups. This is defined only by the relative positions of the nuclei. Classified by AXE

A = central atomX = surrounding atomE = nonbonding valence-electron group

For CO2, the central C atom has two atoms attached (the two oxygen atoms) and has no lone pair.

2 electron groups – AX2

Hybridization – sp The electron pair geometry is linear. The molecular shape is linear. Bond angle is 180o

Linear Molecules

Cl Be Cl

For H2CO, the central C atom has three atoms attached and has no lone pair.

Hybridization – sp2

The electron pair geometry is trigonal planar. The molecular shape is trigonal planar. Bond angle is 120o

Trigonal Planar

In ozone, O3, the central O atom has two atoms attached (the other 2 oxygens) and has one lone pair.

3 electron groups – AX2E

The electron pair geometry is trigonal planar. The molecular shape is bent. Bond angle is 120o

Trigonal Planar

OO O O

In SiCl4, the central Si atom has four atoms attached (four chlorine atoms) and has no lone pair.

The electron pair geometry is tetrahedral. The molecular shape is tetrahedral. Bond angle is 109.5o

Tetrahedral

In NH3, the central N atom has three atoms attached (three hydrogen atoms) and has one lone pair.

The electron pair geometry is tetrahedral. The molecular shape is trigonal pyramidal. Bond angle is ~109.5o (~107o actually)

Tetrahedral

In SCl2, the central S atom has two atoms attached (two chlorine atoms) and has two lone pair.

4 electron groups – AX2E2

The electron pair geometry is tetrahedral. The molecular shape is bent. Bond angle is ~109.5o (~104.5o actually)

Tetrahedral

In PCl5, the central P atom has five atoms attached (five chlorine atoms) and has no lone pair.

Hybridization – sp3d The electron pair geometry is trigonal bipyramidal. The molecular shape is trigonal bipyramidal. Bond angles are 120o (equatorial) and 90o (axial)

Trigonal Bipyramidal

In SF4, the central S atom has four atoms attached (four fluorine atoms) and has one lone pair.

Hybridization – sp3d The electron pair geometry is trigonal bipyramidal. The molecular shape is seesaw. Bond angles are 120o (equatorial) and 90o (axial)

In BrF3, the central Br atom has three atoms attached (three fluorine atoms) and has two lone pair.

Hybridization – sp3d The electron pair geometry is trigonal bipyramidal. The molecular shape is T-shaped. Bond angles are 120o (equatorial) and 90o (axial)

In XeF2, the central Xe atom has two atoms attached (two fluorine atoms) and has three lone pair.

Hybridization – sp3d The electron pair geometry is trigonal bipyramidal. The molecular shape is linear. Bond angle is 180o

In SF6, the central S atom has six atoms attached (six fluorine atoms) and has no lone pair.

Hybridization – sp3d2

The electron pair geometry is octahedral. The molecular shape is octahedral. Bond angles are 90o (equatorial) and 90o (axial)

Octahedral

In BrF5, the central Br atom has five atoms attached (five fluorine atoms) and has one lone pair.

The electron pair geometry is octahedral. The molecular shape is square pyramidal. Bond angles are 90o (equatorial) and 90o (axial)

Octahedral

In XeF4, the central Xe atom has four atoms attached (four fluorine atoms) and has two lone pair.

The electron pair geometry is octahedral. The molecular shape is square planar. Bond angles are 90o (equatorial) and 90o (axial)

Octahedral

READING: Chapter 10 sections 1 – 3 READING: Chapter 10 sections 1 – 3 HOMEWORK – DUE TUESDAY...

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