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Chapter 1. mass. density =. volume. m. d =. V. Density – SI derived unit for density is kg/m 3. 1 g/cm 3 = 1 g/mL = 1000 kg/m 3. 0 F = x 0 C + 32. 5. 0 C = x ( 0 F – 32). 9. 9. 5. K = 0 C + 273.15. 273 K = 0 0 C 373 K = 100 0 C. 32 0 F = 0 0 C

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Chapter 1

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Chapter 1

Chapter 1


Chapter 1

mass

density =

volume

m

d =

V

Density – SI derived unit for density is kg/m3

1 g/cm3 = 1 g/mL = 1000 kg/m3


Chapter 1

0F = x 0C + 32

5

0C = x (0F – 32)

9

9

5

K = 0C + 273.15

273 K = 0 0C

373 K = 100 0C

32 0F = 0 0C

212 0F = 100 0C

Kelvin is the SI Unit of temperature: absolute temperature scale. 0K is the lowest temperature that can be achieved theoretically.

1.7


Chapter 1

Significant Figures

- The meaningful digits in a measured or calculated quantity

- The last digit is uncertain; 6.0±0.1ml

  • Any digit that is not zero is significant

    • 1.234 kg 4 significant figures

  • Zeros between nonzero digits are significant

    • 606 m 3 significant figures

  • Zeros to the left of the first nonzero digit are not significant

    • 0.08 L 1 significant figure

  • If a number is greater than 1, then all zeros to the right of the decimal point are significant

    • 2.0 mg 2 significant figures

  • If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant

    • 0.00420 g 3 significant figures

1.8


Chapter 1

89.332

+

1.1

two significant figures after decimal point

one significant figure after decimal point

90.432

round off to 0.79

round off to 90.4

3.70

-2.9133

0.7867

Significant Figures

Addition or Subtraction

The answer cannot have more digits to the right of the decimal

point than any of the original numbers.

3DP

1DP

1DP

2DP

4DP

2DP

Answer can have only as many decimal places (DP)as the number in the operation with the least number of decimal places:

1.8


Chapter 1

3 sig figs

round to

3 sig figs

2 sig figs

round to

2 sig figs

Significant Figures

Multiplication or Division

The number of significant figures in the result is set by the original number that has the smallest number of significant figures

4.51 x 3.6666 = 16.536366

= 16.5

3 sig figs

6.8 ÷ 112.04 = 0.0606926

= 0.061

2 sig figs

1.8


Chapter 1

Chapter 2


Chapter 1

A

X

Mass Number

Element Symbol

Z

Atomic Number

Atomic number, Mass number and Isotopes

Atomic number (Z) = number of protons in nucleus

Mass number (A) = number of protons + number of neutrons

= atomic number (Z) + number of neutrons

number of protons = number of electrons

2.3


Chapter 1

235

238

U

U

92

92

The Isotopes of Hydrogen

Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei

2.3


Chapter 1

Steps of naming ionic and binary molecular compounds

Naming acids and base

2.7


Chapter 1

Chapter 3


Chapter 1

7.42 x 6.015 + 92.58 x 7.016

100

Average atomic mass- average mass of naturally occurring mixture of isotopes.

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

Average atomic mass of lithium:

= 6.941 amu

3.1


Chapter 1

1S

32.07 amu

2O

+ 2 x 16.00 amu

SO2

SO2

64.07 amu

Molecular mass(or molecular weight) is the sum of

the atomic masses (in amu) in a molecule.

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

3.3


Chapter 1

1 12C atom

12.00 g

1.66 x 10-24 g

=

12.00 amu

6.022 x 102312C atoms

1 amu

= molar mass in g/mol

M

Relationship between amu and grams

x

1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

NA= Avogadro’s number

3.2


Chapter 1

Percent composition of compounds

Percent composition by mass is the percent of each element in a compound.

3.5


Chapter 1

Percent Composition and Empirical Formulas

3.5


How to read chemical equations

IS NOT

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

2 grams Mg + 1 gram O2 makes 2 g MgO

2 Mg(s) + O2 (g) 2 MgO(s)

Indicate physical state

2 HgO(s) O2 (g) + 2Hg(l)

3.7


Chapter 1

C2H6 + O2

CO2 + H2O

NOT

2C2H6

C4H12

Balancing Chemical Equations

  • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

  • Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

3.7


Chapter 1

1 carbon

on right

6 hydrogen

on left

2 hydrogen

on right

2 carbon

on left

C2H6 + O2

C2H6 + O2

C2H6 + O2

CO2 + H2O

2CO2 + H2O

2CO2 + 3H2O

Balancing Chemical Equations

  • Start by balancing those elements that appear only once on each side of the equation and balance them.

start with C or H but not O

multiply CO2 by 2

multiply H2O by 3

3.7


Chapter 1

multiply O2 by

4 oxygen

(2x2)

+ 3 oxygen

(3x1)

2 oxygen

on left

C2H6 + O2

2CO2 + 3H2O

C2H6 + O2

2CO2 + 3H2O

2C2H6 + 7O2

4CO2 + 6H2O

7

7

2

2

Balancing Chemical Equations

  • Balance those elements that appear in two or more reactants or products.

= 7 oxygen

on right

remove fraction

multiply both sides by 2

3.7


Chapter 1

4 C (2 x 2)

4 C

Reactants

Products

12 H (2 x 6)

12 H (6 x 2)

14 O (7 x 2)

14 O (4 x 2 + 6)

4 C

4 C

12 H

12 H

14 O

14 O

2C2H6 + 7O2

4CO2 + 6H2O

Balancing Chemical Equations

  • Check to make sure that you have the same number of each type of atom on both sides of the equation.

3.7


Chapter 1

Amounts of Reactants and Products

  • Write balanced chemical equation

  • Convert quantities of known substances into moles

  • Use coefficients in balanced equation to calculate the number of moles of the sought quantity

  • Convert moles of sought quantity into desired units

3.8


Chapter 1

2NO + 2O2 2NO2

Limiting Reagents

The reactant used up first in a reaction is called limiting reagent.

7 mole NO and 8 mole O2

NO is the limiting reagent

O2 is the excess reagent

The maximum amount of products formed depends on the amount of the limiting reagent.

3.9


Chapter 1

Chapter 4


Molecular equation ionic equation and net ionic equation

precipitate

Pb(NO3)2(aq) + 2NaI (aq) PbI2(s) + 2NaNO3(aq)

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3-

Pb2+ + 2I- PbI2 (s)

Molecular equation, Ionic equation and net Ionic equation

molecular equation

ionic equation

net ionic equation

Na+ and NO3- are spectator ions: ions that are not involved in the overall reaction

4.2


Writing net ionic equations

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

Write the net ionic equation for the reaction of silver

nitrate with sodium chloride.

Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3-

Ag+ + Cl- AgCl (s)

Writing Net Ionic Equations

  • Write the balanced molecular equation.

  • Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.Weak and non electrolytes are written as molecules

  • Cancel the spectator ions on both sides of the ionic equation

  • Check that charges and number of atoms are balanced in the net ionic equation

4.2


Oxidation number

Oxidation number

The charge the atom would have in a molecule (or an

ionic compound) if electrons were completely transferred.

  • Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

  • In monatomic ions, the oxidation number is equal to the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

  • The oxidation number of oxygen isusually–2. In H2O2 and O22- it is –1.

4.4


Chapter 1

Oxidation numbers of all the elements in HCO3- ?

  • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds (e.g. LiH, CaH2). In these cases, its oxidation number is –1.

  • Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

  • The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. For example,NH4+ , the sum of oxidation numbers is

  • -3+4(+1)=+1

7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2-, is -½.

O = -2

H = +1

3x(-2) + 1 + ? = -1

HCO3-

C = +4

4.4


Chapter 1

5. Disproportionation reaction


Chapter 1

The Activity Series for Metals

The Activity Series

* Arranges metals according to their ease of oxidation

*The higher the metal on the Activity Series, the more active that metal (the easier it is oxidized.)Any metal can be oxidized by the metal ions below it.

*Any metal above hydrogen will displace it from water or from an acid.


Chapter 1

3. Halogen Displacement

F2 > Cl2 > Br2 > I2

F2 is the greatest oxidizing halogen

I2 is the least oxidizing halogen

Example:

-1 0 -1 0

2 Br- + Cl2 --> 2 Cl- + Br2

Br2 + Cl- --> no reaction (NR)


Chapter 1

moles of solute

mole

n

liters of solution

liters

V

Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =

=

=

• Read 2.5 M NaCl as 2.5 molar sodium chloride

Often use square bracket [ ] to indicate the concentration

Example:

What is the molarity of a solution made from 4.00 g of NaOH diluted to a final volume of 250 mL?

First find the number of moles of NaOH.

Then divide by the volume in Liters.

4.5


Chapter 1

Chapter 5


Ideal gas equation

Boyle’s law: V a (at constant n and T)

Va

nT

nT

nT

P

P

P

V = constant x = R

1

P

Ideal Gas Equation

Charles’ law: VaT(at constant n and P)

Avogadro’s law: V a n(at constant P and T)

R is the gas constant

PV = nRT

Ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.

5.4


Chapter 1

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

R =

(1 atm)(22.414L)

PV

=

nT

(1 mol)(273.15 K)

Molar volume of gas

• 1 mole of gas at STP = 22.4 Liters

• Example:

2 moles of gas at STP = 44.8 L

PV = nRT

R = 0.082057 L • atm / (mol • K)=8.314J/(K·mol)

= 8.314 L·kPa/(K·mol)

5.4

In calculation, the units of R must match those for P,V,T and n.


Chapter 1

The combined gas law

P1V1

n1T1

P2V2

n2T2

R=

R=

P1V1

n1T1

P1V1

T1

P2V2

n2T2

P2V2

T2

=

=

If n1=n2


Chapter 1

Dalton’s Law of Partial Pressures

Partial pressure is the pressure of the individual gas in the mixture.

V and T are constant

P1

P2

Ptotal= P1 + P2

5.6


Chapter 1

PA =

nART

nBRT

V

V

PB =

nA

nB

nA + nB

nA + nB

XB =

XA =

ni

mole fraction (Xi) =

nT

Consider a case in which two gases, A and B, are in a container of volume V.

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB

PA = XAPT

PB = XBPT

Pi = XiPT

5.6


Chapter 1

Chapter 6


Chapter 1

Another form of the first law for DEsystem

DE = q + w

DE is the change in internal energy of a system

q is the heat exchange between the system and the surroundings

w is the work done on (or by) the system

w = -PDVwhen a gas expands against a constant external pressure

6.3


Chapter 1

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ/mol

x

H2O (l) H2O (g)

H2O (s) H2O (l)

3013 kJ

1 mol P4

x

DH = 6.01 kJ/mol

DH = 44.0 kJ/mol

1 mol P4

123.9 g P4

Thermochemical Equations

  • The physical states of all reactants and products must be specified in thermochemical equations.

= 6470 kJ

266 g P4

6.4


Chapter 1

Consider the reaction below: 

2CH3OH(l) + 3O2(g)  4H2O(l) + 2CO2(g)

∆H = -1452.8 kJ/mol 

What is the value of ∆H for the reaction of 

8H2O(l) + 4CO2(g)  4CH3OH (l) + 6O2 (g)?

If you reverse a reaction, the sign of ΔH changes.

If you multiply both sides of the equation by a factor n, then ∆H must change by the same factor n.

∆H = 2*(+1452.8 kJ/mol)

= +2905.6 KJ/mol

Practice problem of chap6:Q2


Chapter 1

Calorimetry- the measurement of heat.

The specific heat(s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity(C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = m x s

Heat (q) absorbed or released:

q = m x s x Dt

q = C x Dt

Dt = tfinal - tinitial

6.5


Chapter 1

Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions.

f

Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. “0”-standard state at 1atm; “f”-formation

f

DH0 (O2) = 0

DH0 (O3) = 142 kJ/mol

DH0 (C, graphite) = 0

DH0 (C, diamond) = 1.90 kJ/mol

f

f

f

f

Standard enthalpy of formation and reaction

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

The standard enthalpy of formation of any element in its most stable form is zero.

6.6


Chapter 1

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

-

[

+

]

[

+

]

=

-

S

S

=

DH0

DH0

rxn

rxn

mDH0 (reactants)

dDH0 (D)

nDH0 (products)

cDH0 (C)

aDH0 (A)

bDH0 (B)

f

f

f

f

f

f

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

6.6


Chapter 1

Chapter 7


Chapter 1

Maxwell (1873), proposed that visible light consists of electromagnetic waves.– provides a mathematical description of the general behavior of light.

Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves, which has an electrical field component and a magnetic field component .

Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiation

l x n = c

7.1


Chapter 1

Energy of light:

Quantum = packet of energy

Photon = packet of light

• E = h n

Planck’s constant (h)

h = 6.63 x 10-34 J•s


Chapter 1

ni = 3

ni = 3

f

i

ni = 2

nf = 2

DE = RH

i

f

( )

( )

( )

Ef = -RH

Ei = -RH

nf = 1

nf = 1

1

1

1

1

n2

n2

n2

n2

Transition between energy states

Ephoton = DE = Ef - Ei

= h x n

if ni > nf, then emission occurs

7.3


Chapter 1

Quantum Mechanical Description of the distribution of electrons in the atom:

• 4 quantum numbers:

n (principal quantum number)

l (angular momentum quantum number)

ml (magnetic quantum number)

ms (spin quantum number)

Allowed values:

• n = 1,2,3 …,n

• l = 0,1,2,…,(n-1)

• ml = (-l, …, 0. …+l)

• ms = +1/2, -1/2


Chapter 1

Electron Configurations

Periods 1, 2, and 3

Three rules:

1.Electrons fill orbitals starting with lowest n and moving upwards (Aufbau principle: Fill up electrons in lowest energy orbitals )

2. No more than two electrons can be placed in each orbital.

No two electrons can fill one orbital with the same spin (Pauli exclusion principle: no two electrons in an atom

can have the same four quantum numbers.)

3. For degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins )

Period 4 and Beyond

the d orbitals begin to fill


Chapter 1

Chapter 8


Chapter 1

Electron Configurations of Cations and Anions

Of Representative Elements

Atoms lose electrons so that cation has a noble-gas outer electron configuration.

Na [Ne]3s1

Na+ [Ne]

Ca [Ar]4s2

Ca2+ [Ar]

Remember: electrons are first removed from orbitals with the highest principal quantum number.

Al [Ne]3s23p1

Al3+ [Ne]

H 1s1

H- 1s2 or [He]

Atoms gain electrons so that anion has a noble-gas outer electron configuration.

F 1s22s22p5

F- 1s22s22p6 or [Ne]

O 1s22s22p4

O2- 1s22s22p6 or [Ne]

N 1s22s22p3

N3- 1s22s22p6 or [Ne]

8.2


Chapter 1

Electron Configurations of Cations of Transition Metals

When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals.

Fe: [Ar]4s23d6

Mn: [Ar]4s23d5

Fe2+: [Ar]4s03d6 or [Ar]3d6

Mn2+: [Ar]4s03d5 or [Ar]3d5

Fe3+: [Ar]4s03d5 or [Ar]3d5

8.2


Chapter 1

Periodic Variation in Physical Properties


Chapter 1

8.3


Chapter 1

Cation is always smaller than atom from which it is formed. This is because the nuclear charge remains the same but the reduced electron repulsion resulting from removal of electrons make the electron clouds shrink.

Anion is always larger than atom from which it is formed. This is because the nuclear charge remains the same but electron repulsion resulting from the additional electron enlarges the electron clouds.

Ionic Radius

8.3


Chapter 1

  • From top to bottom, both the atomic and ionic radius increase within group.

  • Across a period the anions are usually larger than cations.

  • For ions derived from different groups, size comparison is meaningful only if the ions are isoelectronic.

  • E.g. Na+ (Z=11) is smaller than F- (Z=9).

  • The larger effective charge results in a smaller radius.

  • Radius of tripositive ions < dipositive ions<unipositive ions

  • Al3+ <Mg2+ <Na+

  • Radius of uninegative ions < dinegative ions

  • O2->F-


Chapter 1

Increasing First Ionization Energy

Increasing First Ionization Energy

General Trend in First Ionization Energies

8.4


Chapter 1

Variation of Electron Affinity With Atomic Number (H – Ba)

Overall trend: increases from left to right across the period. The values varied little in the group. The EA of metals are generally lower than those of nonmetals.

8.5


Chapter 1

Chapter 9


Chapter 1

The Octet Rule

Atoms tend to gain, lose, or share electrons until they

are surrounded by 8 valence electrons.

An octet = 8 electrons

= 4 pairs of electrons

= noble gas arrangement

= very stable

Exception: H = 2 electrons only

Example:

Na [Ne]3s1 loses 1 e----> Na+ [Ne]

Cl [Ne]3s23p5 gains 1 e- --> Cl- [Ar]


Chapter 1

Writing Lewis Structures

  • Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. Arrange atoms & connect with single bonds

  • – Single bond = 2 electrons

  • 2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.

  • 3. Complete an octet for atoms bonded to the central atom except hydrogen (H=2). Then add leftover electrons to central atom. Electrons belonging to the central or surrounding atoms must be shown as lone pairs if they are not involved in bonding.

  • 4. If the central atom has fewer than 8 electrons, try adding double and triple bonds on central atom as needed.

9.6


Chapter 1

formal charge on an atom in a Lewis structure

total number of valence electrons in the free atom

total number of nonbonding electrons

(

total number of bonding electrons

)

1

-

-

=

2

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

FC = [ Val. e-] -[(nonbonded e-) + 1/2(bonded e-)]

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

• Remember: formal charges are NOT real charges

9.7


Chapter 1

Which is the most likely Lewis structure for CH2O?

H

C

O

H

H

C

O

H

-1

+1

0

0

Formal Charge and Lewis Structures

  • For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

  • Lewis structures with large formal charges are less plausible than those with small formal charges.

  • Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

9.7


Chapter 1

N – 5e-

S – 6e-

N

O

6F – 42e-

O – 6e-

48e-

11e-

F

6 single bonds (6x2) = 12

F

F

Total = 48

S

18 lone pairs (18x2) = 36

F

F

F

Exceptions to the Octet Rule

Odd-Electron Molecules

Result: unpaired electron

NO

The Expanded Octet (too many e- on central atom; central atom with principal quantum number n > 2)

SF6

9.9


Chapter 1

F

B

F

F

  • Which of the following is correct concerning the lone pairs on the underlined atoms in compounds?

  • ICl, 3(b) H2S, 6(c) CH4, 4

  • (d) CaH2, 2(e) SCl2, 4

  • Which of the following obeys the octet rule?

  • BF3(b) SF4(c) NO

  • (d) PF5(e) NO3-

Practice problem of chap9:Q3,9


Chapter 1

Bond Energies (BE) and Enthalpy changes in reactions

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

DH0 = total energy input – total energy released

= SBE(reactants) – SBE(products)

Endothermic

reaction

Exothermic

reaction

9.10


Chapter 1

Chapter 10


Chapter 1

Molecules in which the central atom has no lone pairs

No Lone Pairs on Central Atom

• Arrangement of electron domains is very symmetrical

• The molecular geometry is the same as the electron domain geometry

10.1


Chapter 1

Steps to Assign Molecular Geometry

• Draw Lewis structure

• Count # of electron domains around the Central atom(treat double and triple bonds as though there were single bond)

• Assign electron domain geometry

• Finally assign the molecular shape

– based on the arrangement of the atoms

– not the arrangement of the domains

No Lone Pairs on Central Atom

• Arrangement of electron domains is very symmetrical

• The molecular geometry is the same as the electron domain geometry


Chapter 1

10.1


Chapter 1

Molecules containing net dipole moments are called polar molecules. Otherwise they are called nonpolar molecules because they do not have net dipole moments.

Diatomic molecules: Determined by the polarity of bond

Polar molecules:HCl, CO,NO

Nonpolar molecules: H2,F2,O2

Molecules with three or more atoms: determined by the polarity of the bond and the molecular geometry

Dipole moment is a vector quantity, which has both magnitude and direction.


Chapter 1

Chloroform, CCl3H, is

asymmetric. The vector

sum of the bond dipoles

is nonzero giving the

molecule a net dipole.

• A molecule will be nonpolar if: (a) the bonds are nonpolar, or (b) there are no lone pairs in the valence shell of the

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


Chapter 1

10.4


Chapter 1

How do I predict the hybridization of the central atom?

  • Draw the Lewis structure of the molecule.

  • Count the number of lone pairs AND the number of atoms bonded to the central atom. Predict the overall arrangement of electron pairs using VSEPR model ( Table 10.1)

  • Deduce hybridization of central atom by matching the arrangement of the electron pairs with those of hybrid orbitals shown in Table 10.4.

# of Lone Pairs

+

# of Bonded Atoms

Hybridization

Examples

2

sp

BeCl2

3

sp2

BF3

4

sp3

CH4, NH3, H2O

5

sp3d

PCl5

10.4

6

sp3d2

SF6


Chapter 1

Molecular Orbital (MO) Configurations

-MOs follow the same filling rules as atomic orbitals

  • The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.

  • The more stable the bonding MO, the less stable the corresponding antibonding MO.

  • The filling of MOs proceeds from low to high energies.

  • Each MO can accommodate up to two electrons.

  • Use Hund’s rule when adding electrons to MOs of the same energy.(Electrons spread out as much as possible, with spins unpaired, over orbitals that have the same energy)

  • The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms.

10.7


Chapter 1

The approximate relative energies of molecular orbitals in

second period diatomic molecules. (a) Li2 through N2, (b) O2

through Ne2.


Chapter 1

10.7


Chapter 1

bond order =

Number of electrons in bonding MOs

Number of electrons in antibonding MOs

(

)

-

1

2

bond order

½

1

½

0

10.7


Chapter 1

Chapter 11


Chapter 1

E = k

Q+Q-

r

Types of Intermolecular Forces (IMF)

• Should actually be called Interparticulate Forces (molecules, ions, and/or atoms)

• Ion - ion forces

• Ion-dipole forces

• Dipole-dipole forces

• Dispersion Forces

• Hydrogen Bonds

Ion - ion forces: (lattice energy-ionic compound)

• Remember Chapter 9

• Force depends on the charge on the ions and

the distance separating the ions

• (STRONG FORCES)


Chapter 1

4 atoms/unit cell

2 atoms/unit cell

1 atom/unit cell

(8 x 1/8 + 6 x 1/2 = 4)

(8 x 1/8 + 1 = 2)

(8 x 1/8 = 1)

11.4


Chapter 1

11.4


Chapter 1

phase diagram

A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas.

Phase Diagram of Water

11.9


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