Test 3 Review

NaMass #

Atomic #

Electric charge

# of atoms

Also referred to as a “salt”

Formation involves a transfer of electrons

Usually made up of a metal and a non-metal

Are good conductors when they can be melted or dissolved

Typically have extremely high melting points

Ionic Compounds

Formation of an Ionic Bond

Electron acceptor (Cl)

meetselectron donor

(Na)

Ions attract to

form a neutral

pair

e- jumps from Na to Cl

Smallest building blocks are ions, NOT MOLECULES

Large numbers of ions can attract to form clusters and eventually crystals

Structure

Ion pair Ion cluster Crystal

lattice

Cations – positively charged ions◦ Na+ Ca2+ Al3+

Anions – negatively charged ions◦ Cl- O2-

Polyatomic ions – ions made up of more than one type of atom◦ NO3

- SO4-2 PO4

-3

Ions

The number of e- gained, lost or shared ub compound formations◦ Alkali metals +1◦ Alkaline earth metals +2◦ Oxygen group -2◦ Halogens -1

Oxidation Number

K+ and N3-

◦ K3N

Ca2+ and N3-

◦ Ca3N2

Ba2+ and NO3-

◦ Ba(NO3)2

Criss-cross rule

Write the formulas – ALWAYS put cation first

Binary – made of 2 ions

Write cation first Change anion ending to –ide

Na+ and Cl-◦ Sodium chloride

H+ and F-

◦ Hydrogen fluoride

CaBr2◦ Calcium bromide

Naming Binary Ionic Compounds

Name the cation Polyatomic ion name is unchanged

NaNO3◦ Sodium nitrate

Zinc carbonate◦ ZnCO3

Naming Polyatomic Ionic Compounds

Also called covalent compounds

A molecule is a neutral group of atoms that are held together by covalent bonds

The valence e- are shared by the atoms

Covalent bonding usually occurs between 2 non-metals◦ H2O, CO2, O2, NO

Molecular Compounds

Naming Molecular Compounds Use prefixes 1 mono-

2 di-3 tri-4 tetra-5 penta-6 hexa-7 hepta-8 octa-9 nona-10 deca-

P4O10

N2O3

As2O5

OF2

ExamplesTetraphosphorous decoxide

Dinitrogen trioxide

Diarsenic pentoxide

Oxygen difluoride

H2O2N2Cl2Br2 I2F2

Diatomic Molecules 7 diatomic molecules No noble gases Halogens and N, O, H They are all gases (not

noble gases) except for Br and I

“Honcl brif”

H2SO4

HF

H3PO4

H2SO3

H2CO3

HNO3

Try these. . .Sulfuric AcidHydrofluoric AcidPhosphoric Acid

Sulfurous Acid

Carbonic Acid

Nitric Acid

More Practice. . . Calcium bromide

Chromium (III) acetate

Barium sulfate

Copper (I) sulfide

Sulfur hexafluoride

CaBr2

Cr(C2H3O2)3

BaSO4

Cu2SSF6

Cr2(C2O4)3

Hg(CN)2

Cu(ClO4)2

ZnC4H4O6

More Practice. . . Chromium (III) oxalateMercury (II) cyanide

Copper (II) perchlorateZinc tartrate

The mass of a compound

In order to calculate molar mass (also called molecular weight) you add up the masses of each element in the compound◦ Be aware of subscript numbers that designate the

amount of atoms per element

You get the masses from the periodic table

**be careful when rounding the mass

How to Calculate Molar Mass

NaCl◦ Na = 23 g/mol◦ Cl = 35.5 g/mol

H2O◦ H = 1 g/mol (but there are 2) = 2 g/mol◦ O = 16 g/mol

HNO3◦ H = 1 g/mol◦ N = 14 g/mol◦ O = 16 g/mol (but there are 3) = 48 g/mol

Ba(NO3)2◦ Ba = 137.3 g/mol◦ N = 14 g/mol (but there are 2) = 28 g/mol◦ O = 16 g/mol (but there are 6) = 96 g/mol

Examples 58.5 g/mol

18 g/mol

63 g/mol

261.3 g/mol

All metal atoms in a metallic solid contribute their valence e- to form a “sea” of e-◦ These e- move easily and freely because they are

not tied to a specific atom Delocalized electrons

◦ Metallic cation is formed

Electron Sea Model

All empty space is evenly distributed v.e-

The attraction of a metallic cation for delocalized electrons

This accounts for a lot of theproperties of metals◦ Range of melting points◦ Malleability◦ Ductile◦ Durable

Hard to remove metallic cation because of the strong e- attraction

◦ Mobile e- Explains why they are good conductors

Metallic Bonds

Find the difference in electronegativities of the two elements

Electronegativity and Bond Type

0.5 1.7PureCovalent-share e- evenly-2 non metals and/or metalloids

Non-polar

PolarCovalent-Share e- but not evenly-One element holds e- more

Polar

Ionic-Metal and non-metal

Count total valence electrons available

Place electrons around atoms

Ensure each atom has an octet (8)◦ Or a pair for H (2)

To create a Lewis Dot Diagram

Draw the Lewis Structure for the molecule

Count the total number of . . .◦ Bonded regions around the central atom

DOUBLE and TRIPLE bonds count as ONE REGION◦ Unshared e- pair

Count as ONE REGION

VSEPR Rules

Molecular Lewis Dot electron pairs around central atom

Structure structure total shared unshared

H CH4 H-C-H 4 4 0 “tetrahedral” H NH3 H-N-H 4 3 1 “trigonal H pyramidal”

H2O H-O-H 4 2 2 “bent”

Molecule Total no. of

electron pairs

No. of shared pairs

No. of unshared pairs

Molecular shape

A molecule is polar if◦ There is a polar bond◦ It is ASSYMETRICAL (not symmetric)

Polarity

O

HH

(-)

(+)(+)H

H

H

H

C

(+)

(+)

(+)

(+)

PolarNon-Polar

Symmetric (non-polar)◦ Linear◦ Tetrahedral◦ Trigonal planar

If all elements around the center atom are the same

Asymmetric (polar)◦ Bent◦ Trigonal pyramidal

Typically. . .

Van der Waals forces (London Dispersion forces)◦ Weak forces between non-polar molecules◦ These forces determine volatility

Doesn’t take much nrg to break apart (liquid gas) Most likely to be a gas

Like playing red rover and only holding pinkies together

Intermolecular Forces

Dipole-Dipole◦ Attraction between polar molecules

Most likely to be a liquid

Play red rover and hold hands

Intermolecular Forces

Hydrogen Bonding (H-Bonds)◦ Between hydrogen (H) and a highly

electronegative element F, O, N

◦ Extreme case of dipole-dipole◦ Strongest of the intermolecular forces

Play red rover and link elbows Needs A LOT of nrg to break bonds

Intermolecular Forces

Carbon has a mass of 12 g

Oxygen has a mass of 16 g

H2O molecules has a mass of 18 g

How do these #’s relate to the atom or compound?◦ Atomic mass

1 mole of . . .

Amedeo Avogadro (1776-1856) 1 mole = 6.0221415 x 1023

◦ Particles◦ Molecules◦ Atoms◦ Ions◦ Formula units◦ Etc, etc

Avogadro’s Number

Determine the mass percentage of each element in the compound.

Percent Composition

100____

compoundofmasselementofmass

Gives the lowest whole # ratio of elements in a compound.

The empirical formula for C6H12O6 is

The empirical formula for C2H6 is

* most basic ratio of elements in the compound

Empirical Formula

CH2O

CH3