Atomic radii increase right to left across the period, and top to bottom down the group. Opposite is true for ionization energy. Covalent bonds are made when difference in electronegativity of the elements is less than 1.9, i.e. when ions cannot be formed.Covalent bonding gives all atoms apparent octet, i.e. the configuration of the closest noble gas. Only valence electrons are involved.

CO32-

CO O

O

(4 + 3x6 +2)e- = 24e-

[24 - (3x2)]e- = 18e-

CO O

O

: :

: :. .. .. .

. .

. . Octet rule not satisfied for C

in lone pairs

total

CO O

O

: :

: :. .. . . .

2-. .The ion has three resonance structures.

Step 1: Find the total number of valence e- by adding up the group numbers of all atoms. For ions, adjust the dot count accordingly (subtract e- for cation, add for anion).Step 2: Assume that the first non-hydrogen atom in the formula of the group is the central atom. Connect atoms with single bonds.

Step 4: Put in the remaining electrons, two at a time, as lone pairs. Start with the terminal atoms, and continue with the central atom if there are any electron pairs left.Step 5: Check that each atom has octet satisfied (doublet for H). If not, move electron pair(s) from the adjacent atom to form multiple bonds.

Step 3: Subtract 2e- for each bond from total #e- to get #e- in lone pairs.

Step 6. Count the number of electron pairs around the central atom, including lone pairs (i.e. the steric number, SN). A multiple bond counts as a single electron group.Step 7. Pretend the lone pairs are invisible and describe the resulting shape of the molecule.

SN = 3, no lone pairs all atoms in a plane, 120o.

SN=4, tetrahedral, 109.5o

SN=3, trigonal planar, 120o

SN=2, linear, 180o

Chapter 11Chemical Bonds

Chapter 12

Pressure (P) – atmospheresTemperature (T) – KelvinVolume (V) – litersAmount of Gas (n) – moles

PV = nRT

Behavior of an ideal gas can be totally described by:

The total pressure of a mixture of gases is the sum of the partial pressures exerted by each gases:

Ptotal = PA + PB + PC + …

Experimentally determined that 1 mol of ANY gas occupies 22.4 L at STP (Avogadro’s law).If the mass and volume of the gas at STP are known, molar mass M is:

Standard temp. and pressure (STP) is 1 atm and 0oC.

M = mRT PV

Volume A

Mass A

Atoms A

Moles A

Volume B

Mass B

Atoms B

Moles B

Gaseous State

Mass (g) x 1 mol molar mass (g)

# atoms x 1 mol 6.022 x 1023 atoms

Volume (L) x 1 mol22.4 L

What volume of O2 at STP can you make from 10. g KClO3 (p.c.)? 2 KClO3 2 KCl + 3 O2

10. g p.c.x 1 mol KClO3

122.55 g p.c.x

3 mol O2

2 mol KClO3x

22.4 L1 mol O2

= 2.7 L

g - mol A mol A - mol B mol - Liters B

GasStoichiometry:

Partial pressure: What is the volume of the dry oxygen if it was collected over water at 23oC and 760. torr in a 500 mL container? Water vapor pressure at 23oC is 21.2 torr.

Step 1: determine the pressure of dry O2: PO2 = Ptot – PH2O PO2 = 760 torr – 21.2 torr = 739 torr.

V2 =P1V1

P2=

739 torr x 500 mL 760. torr = 486 mL dry O2.

P1 = 739 torrV1 = 500 mL

P2 =760 torrV2 = ?

Conversion factors

If two parameters are constant, the relationship simplifies:

P1V1 = P2V2

P1

T1

P2

T2=

V1

T1

V2

T2=

n1T1 = n2T2P1

n1

P2

n2= etc.

One constant parameter:

P1V1

n1

P2V2

n2

=

Liquid State Chapter 13

Rate of vaporization is proportional to vapor pressure.

meltingsolid liquidfreezing

evaporationliquid gas

condensation

Heating curve for a pure substance

Temperature is constant during changes of phase.

Qfusion = (mass) (spec.heat of fusion)

Qvaporization = (mass) (spec.heat of vaporization)

Qheating = (mass) (spec.heat) (temp.change)

Qtot = Qfusion + Qheating + Qvaporization

H – O :. .

|H H

H – O :|

. .. . .

H bondHydrogen bond exists if H is directly bonded to either N, O, or F. Very strong intermolecular forces.

CuSO4 5 H2O.Copper(II) sulfate pentahydrate.

Hydrate is a true compound, water is an integral part of it.

. .

. .

OH H

+

-

Vapor pressure of any gas at the boiling point is equal to the atmospheric pressure.

Liquids have intermediate properties between solids and gases. Liquids are almost incompressible, have definite volume and assume the shape of the container.Vapor pressure is the pressure exerted by a liquid at equilibrium with its liquid.

Water is unique because it is liquid at room temp, its solid form (ice) has lower density than the liquid, and is an excellent solvent.

…is a homogeneous mixture of two or more substances with the same composition throughout, thus the concentration of the substances is the same. Consists of solvent (present in greatest amount), and solute(s).

Solution

all Na+, K+, NH4+

all NO3-, C2H3O2

-

most CO32-, PO4

3-,OH-, S2-.

most SO42-, Cl-, Br-, I-

Solubility of ions in watersoluble insoluble

(except Na+,K+,NH4+)Solubility, or the maximum mass of solute in 100. g of

solvent, increases with temperature for most solids. For gases, solubility increases with P and decreases with T.Solution can be saturated, unsaturated, and supersaturated. The latter is unstable and precipitates upon disturbance.

% composition = Amount of soluteAmount of solution x 100

Mass%Vol%Mass/vol%

Molarity is the number of moles Molarity (M) =

Moles of solute1 L of solution

To prepare 1 L of NaCl, measure molar mass of NaCl, add it to a 1L volumetric flask and add water to the mark.

Vstock soln. = Mdiluted soln. x Vdiluted soln.

Mstock soln.

To prepare 1 L of NaCl, calculate Vstock soln of NaCl, add it to a 1L volumetric flask and add water to 1L.

Molality (m) is the number of moles of solute per kilogram of solvent. ∆tf = m x Kf ∆tb = m x Kb

Freezing Point Depression and Boiling Point Elevation depend on molality.

What is the freezing point (F.p.) of solution of 100.0 g ethylene glycol in 200. g H2O?

100.0 g eg x 1 mol eg62.05 g eg= 8.05 m

∆tf = m x Kf = 8.05 x 1.86 = 15 oCmol egkg H2O

oC kg H2O mol eg

F.p. = -15 oC

Prepare 500. mL of 0.15 M NaCl using solid NaCl.

0.5 L x 0.15 mol NaCl 1 L soln.

58.4 g NaCl1 mol NaCl

x = 4.4 g NaCl

Prepare above using stock soln of 0.50 M NaCl. M1V1

M2

Vstock = = = 0.15 M x 500 mL 0.5 M

= 150 mL stock soln. 150 mL stock soln. + 350 mL H2O.

of solute in 1 L of solution

Chapter 14

Electrolytes, Acids and Bases Chapter 15

Acid (Latin acidus - sour): sour taste; turns plant dye litmus red; dissolves metals producing H2.

Base: bitter taste; turns plant dye litmus blue; aqueous solutions feel slippery to touch.

An electrolyte is a solute that dissolves in water and dissociates into ions, yielding a solution that conducts electricity.

Ionic compounds are usually metal plus nonmetal or group of nonmetals.

Exceptions: HX (X-halide) - polar covalent, produce acids in H2O; ammonium salts.

Find electrolytes from the list: Al(NO3)3, (CH3)2O, (NH4)2SO4, CH3OH, HBr

Dissociation constant, Keq, shows how many protons can be obtained from 1 mol of acid.

H2SO4 + H2O H3O+ + HSO4-

HSO4- + H2O H3O+ + SO4

2-

Keq > 1.0 x 103

Keq = 1.2 x 10-2

Strong acids #H3O+ Weak acidsHydrochloric (HCl) 1 1 Hydrofluoric (HF)Hydrobromic (HBr) 1 1 Hypochlorous (HClO)Hydroiodic (HI) 1 1 Acetic (HC2H3O2)Nitric (HNO3) 1 2 Carbonic (H2CO3)Sulfuric (H2SO4) 2 3 Phosphoric (H3PO4)

Base - anything that accepts a proton.Acid - anything that donates a proton.

Kw = Keq x [H2O]2 = [H3O+] x [OH-]

Keq =[H3O+] x [OH-][H2O] x [H2O]

Water auto-ionization.

pH = - log [H3O+][H3O+] = 10-(pH)

Kw = 1 x 10-14

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l)

H+(aq) + OH-(aq) H2O(l)

Formula eq.Total ionic eq.Net ionic eq.spectator ions cancel

Buffer, a solution of a weak acid and its conjugate base, maintains pH. Addition of a strong acid or base to a buffer changes pH only slightly.Ionic Equations