Lecture 20 © slg CHM 151

UNIT 4:

OXIDATION / REDUCTION

BONDING: ionic and covalent

LEWIS STRUCTURES: molecules polyatomic ions oxo acids

TOPICS

UNIT FOURCHAPTER 9: BONDING AND MOLECULAR

STRUCTURE

Now that we have examined the structure of the atom and the arrangement of its electrons, we are ready toturn to the molecules and compounds they form.

Our next studies will center on the bonds that holdthe atoms together in compounds, molecules andpolyatomic ions. We will also examine the threedimensional shape of these species, and their polar or non polar nature.

The interactions between atoms which lead to bondformation are all centered around the electrons in incomplete subshells and in incomplete outer shells:the valence electrons...

The atoms of the elements lose, gain or share theseelectrons to achieve, where possible, the noble gasconfigurations we have met.

BOND FORMATION

For the “main group elements”, the s and p block members, the electrons available for bonding, the “valence electrons”are the outer shell s and p electrons (except those of the noble gases!)

In forming compounds from these elements, only these electrons will be used, in two ways:

They may be transferred to form ions so thatincomplete subshells are completed or removed;

They may be shared so that two atoms together have complete subshells

For the transition metals, the valence electronsinclude both the s electrons from their outermostshell and also electrons from their inner, incomplete d subshell.

The PT column number of the main group and transition metals gives the sum of valence electrons for each element in the family.

Note that the column number indicates the maximum positive charge (or oxidation state) these metals can achieve in a compound through loss of e’s in achemical reaction.

Oxidation/Reduction (CH 5)SKIPPED EARLIER, READY NOW!

Before determining the exact type of bonding which exists in a particular molecule, ion or compound, let us brieflyconsider the type of reaction in which electron transfertakes place: the OXIDATION / REDUCTION reaction.

To date, the majority of reactions we have examined have been the “double replacement, ion exchange” type in which precipitates or molecules are formed but no charges change, no electrons transferred...

However, for a large number of reactions, electrontransfer is the driving force.

In these reactions, some substance(s) are made more positive through electron loss and are described as “oxidized.” Other substance(s) are made more negative through the gain of electrons and are said to be “reduced”.

These reactions are described as “redox” reactions, identified through change in charge in two or more species.

Identification of Electron Transfer in Reactions:

1. Combination of elements: Always “Redox”

2 Na + Cl2 ----> 2 NaCl

Na metal was transformed into Na+; each atom must have lost one electron for this to take place.Na was “oxidized”.

The neutral chlorine molecule broke apart into twoCl atoms, and each accepted an electron to become twoCl- ions. Cl was “reduced”.

for this reaction we can say: -2e2 Na -----> 2Na+ Sodium became more positive, oxidized

+2eCl2 -----> 2Cl- Chlorine became more negative, reduced

2 Na + Cl2 ----> 2 NaCl

In balancing the equation and giving sodium chloride itspredictable formula, we automatically insured that the electron count was equal: the number of e’s lost was equal to the number of e’s gained, a requirement of this kind of reaction.

for this reaction we can say: -2e2Cl- -----> Cl2 Chlorine became more positive, oxidized

+2e2Na+

-----> 2 Na Sodium became less positive, reduced

2 NaCl ----> 2 Na + Cl2 (Decomposition)

2. Decomposition of Compounds to Form Elements: always a redox reaction

3. Other Reactions Involving Redox:

Cu + AgNO3 Cu(NO3)2 + Ag

2 C2H2 + 5 O2 4CO2 + 2H2O

5 FeCl2 + KMnO4 + + 8 HCl

5 FeCl3 + MnCl2 + 4 H2O + KCl

To recognize that the reactions above are indeed “redox”,the system of “oxidation numbers” is employed. This system allows assignment of charge to every elementin every species based solely on its FORMULA.

Detection of REDOX: “Oxidation Numbers”

In order to recognize the occurrence of redox, you mustbe able to give an “OXIDATION NUMBER” to each element in each compound involved in a reaction.

Oxidation Numbers are assigned “charges” based on the guidelines described in the next several slides. Theyare not intended to represent true ionic type charges.

Oxidation Numbers are calculated from the formula of the compound, ion or molecule.

To distinguish between oxidation numbers and ionic charges, we represent true ionic charges with a value first, followed by its sign:

Al3+ S 2- Mg 2+ etc 3+ 2- 2+

On the other hand, when we assign oxidation numbers toelements we write them with “charge first, value second” to indicate that no ionic nature is implied.

Typical oxidation numbers: +3, -2, +5, +7

Guidelines for Determining Oxidation Numbers (“ox #’s”)

1. The oxidation number for each atom of an element is zero: Nao Cl2

o S8o ox # = 0

2. The algebraic sum of all ox#’s in a compound must equal 0;

Algebraic sum of all ox#’s in a polyatomic ion must equal the ionic charge on the ion.

3. Ox #’s, First Element in Formula

• if fixed charge metal (no Roman Numeral in name), charge = ox # column 1A= +1; column 2A = +2; Ag +1 Zn, Cd +2 Al +3

• if Hydrogen, ox # = +1 (anywhere except after metal)

• if variable charge metal, charge on ion is determined by formula, and charge per ion = ox #

4. Ox #: Last Element in Formula

• if written last, all halides (F, Cl, Br, I) are assigned an ox # = -1;

• if H follows an active metal, ox # = -1; otherwise +1 even if at end of formula

• if written last in the formula, assign O or S an ox # of -2

• in general, only the last element in a formula will be assigned a negative oxidation number

Let’s do Oxidation Numbers (ox#’s) for the following:

HNO3 N2O5 NO2- Cr(NO3)3

In all cases:

Ox# is the charge assigned per individual atom

Determine in “variable cases” by adding up all the known charges in the species and equating totaloverall charge of species to 0 or to ion charge value.

H N O

assigned ox # +1 ? -2 total charge +1 ? -6

since sum of all charges must = 0 for compound:( +1 ) + ( - 6 ) + (charge, N) = 0 charge, N = +6 -1 = +5

Ox#’s for HNO3 are: H, +1; O, -2; N, +5

HNO3

Ox #’s HNO3

N O

assigned ox #/atom ? -2 total charge: ? -10 balancing charge, N: +10 -10 calculated ox #, per N: +5

N2O5

5 X (-2)

+10 / 2

Ox #’s N2O5

Ox#’s: N, +5; O, -2

(N O)-

assigned ox # / atom: ( ? -2)-

total charge: ( ? -4)-

balancing charge, N: ( +3 -4)-

(?) + (-4) = -1 (?) = 4 -1 = +3

Ox#’s: N, +3; O, -2

(NO2)- Ox #’s, NO2

-

Ox #’s, Cr(NO3)3

( N O)-

assigned ox # / atom: ( ? -2)-

total charge: ( ? -6)-

balancing charge, N: ( +5 -4)-

(?) + (-6) = -1 (?) = 6 -1 = +5

First step: Use knowledge that nitrate ion, NO3- , has

a (1-) charge and therefore Cr has a (3+) charge. The ox # for Cr is +3, and the ox # of the N in the nitrate can be determined from the O:

NO3-

GROUP WORK

Determine the OX #’s for elements in colored reactantsand products, and decide if reaction is a “redox” type”

Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag

5 FeCl2 + KMnO4 + + 8 HCl

5 FeCl3 + MnCl2 + 4 H2O + KCl

0 +1 +5 -2 +2 +5 -2 0

Cu(s) + 2 AgNO3 Cu(NO3)2 + 2 Ag(s)

Cu, Ag: Ox # = 0

AgNO3 Ag Ox # +1 (always 1+ ion)Cu(NO3)2 Cu Ox# +2 (from nitrate 1- charge)

NO31-:

O: Ox # -2 ( assigned, guidelines) N: Ox # +5 (as below)

? -2 ( N O3 )1-

total ( ? -6)1- (+5 -6)1-

+2 -1 +1 +7 -25 FeCl2 + KMnO4 + + 8 HCl +3 -1 +2 -1 5 FeCl3 + MnCl2 + 4 H2O + KCl

FeCl2 Fe: +2 Cl: -1FeCl3 Fe: +3 Cl: -1MnCl2 Mn: +2 Cl: -1 KMnO4 K: +1 O: -2 Mn: +7 (as below): +1 +7 -2 K Mn O4

+1 (+7) -8

Fe+2 oxidized to Fe+3; Mn+7 reduced to Mn+2

In the language of “redox” reactions:

Any substance which receives electrons and is reduced is called “ the oxidizing agent”(the one who makes another more positive!)

Any substance which gives away electrons and is “oxidized” is termed “the reducing agent”(the one who makes another less positive!)

0 +1 +5 -2 +2 +5 -2 0

Cu(s) + 2 AgNO3 Cu(NO3)2 + 2 Ag(s)

Cu0 was oxidized to Cu+2, reducing the Ag+1 to Ag0. Cu0 was the reducing agent ; Ag+1 was the oxidizing agent.

+2 -1 +1 +7 -25 FeCl2 + KMnO4 + + 8 HCl +3 -1 +2 -1 5 FeCl3 + MnCl2 + 4 H2O + KCl

Mn+7 is the oxidizing agent, and Fe+2 is the reducing agent.

For all main group elements (the columns 1A-8A), it is convenient for bonding purposes, to represent the elements as “Lewis Dot Symbols”, which include one dot for each of their valence electrons.

The tendency of these elements to achieve an outershell configuration of eight electrons (the “octet rule”) is easily visualized through use of these symbols.

The valence electrons for transition elements (columns 3B-8B, 1B, 2B) are not represented by dot symbols.

BACK TO BONDING, CH 9:

1A 2A 3A 4A 5A 6A 7A

ns1 ns2 ns2np1 ns2np2 ns2np3 ns2np5ns2np4

Li Be B C N O F

8A

ns2np6

Ne

LEWIS DOT STRUCTURES FOR PERIOD 2

All elements, same column: same dot structure

Bond Formation: Ionic vs. Covalent

The elements come together to form compoundsso that each element can achieve a more satisfactoryouter shell electronic configuration.

Elements may lose or gain electrons resulting in cation and anion formation and the attractionbetween the two which we call the “ionic bond”

Elements may share one or more pairs of electrons.The attraction of both nuclei for the same pair of electrons results in the force we call the “covalent bond”.

The “ionic bond”: attraction of opposite chargeswhen transfer of electrons cause formation ofpositive and negative species: cations and anions.

The individual ions radiate charge in all directions and cluster in geometric patterns which are describedas crystal lattices.

Note in the following slide that each ion has many neighbors and the compound itself is not molecularin nature: no discrete “formula units” exist.

Ionic compounds are all solids at room temperaturewith elevated melting points. Their melting pointsreflect the very high degree of attraction exhibited bythese fully charged particles, which depends on themagnitude of their charge and the ionic size:

The larger each charge and the smaller each ion, thegreater the attraction.

Energy = n (+) X n (-) d

n= magnitude of charge d=distance between ions

Ion formation and the resulting ionic bondoccurs when metals of sufficiently low electronegativity (X) react with non-metals ofsufficiently high X values.

The most ionic of compounds are those formed between the active s block metals (X < 1)with the non-metals whose X values are 3.0 or larger.

All compounds we have met containing “polyatomic” anions or ammonium are also of course truly ionic type compounds.

ELECTRONEGATIVITY

INCREASES

>2

.8-1.4

1.5-1.9

Most active metals

Most active non-metals

The second mode of bond formation occurs when elements share one or more pairs of electrons to achieve where possible an outer shell octet.

The attraction of both nuclei for the same pair of shared electrons is the basis of the covalentbond.

THE COVALENT BOND

Covalent bonds are directed between two atoms, sharing together one or more pairs of electrons. This type of bonding leads to formation of discreet molecules, individual units made up of two or more atoms covalently bonded together.

Any formula consisting solely of nonmetals and metalloids can assumed to molecular and covalent in nature. Covalent bonds hold together the atoms within a polyatomic ion. Occasionally, metals with higher electronegativity values will form a compound more covalent than ionic in nature.

LEWIS DOT STRUCTURES: MOLECULES ANDCOMPOUNDS

We are next going to use the Lewis Dot Symbols forthe “main group elements” to represent the bonding and structure for various species, molecules, compounds and polyatomic ions.

• all valence electrons for every atom will be included

• all shared pairs of e’s will be indicated by a “dash”

• all unshared pairs of e’s will be indicated by a dot

We will use the “octet rule” as our guiding principle.

Diatomic Elements: H2 Cl2 N2 O2

H H H H H H

H H

H H

Cl ClCl Cl ClCl

ClCl ClCl

ClCl

Cl Cl

Check octets!

Chlorine, Cl2 (same for Br2, F2, I2)

N N N N N N

NO OCTET STILL NO

OCTET

N2, Nitrogen:

N N N N N N

N N

Covalent “triple bond”

O O O O O OO O

O OO O

OXYGEN ACTSLIKE THIS*

CORRECTLEWIS STRUCTURE,Covalent double bond

*Required whole new bonding theory to explain...

Lewis Structures: Compounds and Polyatomic IonsGUIDELINES

Decide on arrangement of atoms. For most species, the element written first in the molecule or ion is the centralatom and the remainder of the atoms are grouped around it.

Hydrogen is a problem in “oxo acids” where it is writtenfirst in the formula. Ignore H, start with the next atom informula and place the H or H’s on the O or O’s.

First step:

NH3

CH2O

SO42-

HNO3

CHBr 3

PCl 3

NH H

H

PCl Cl

Cl

C

H

Br

Br

BrC

O

HH

N

O

OO H S

O

O

O

O

Second Step

•Add up all available valence electrons. • If species is cation, subtract positive charge from total.• If species is anion, add negative charge to total.

•Divide total by two to determine available number of electron pairs

Third Step

Place a pair of electrons between each pair of bondedatoms to represent a single bond (use a “dash”!)

NH3

NH H

H

N: 1 x 5= 5 valence electrons

3H: 3 x 1= 3 valence e's

8 valence e's

8 e's / 2 = 4 electron pairs

NH H

H

Step 1Step 2

Step 3

Fourth Step

Place leftover electron pairs around “terminal” atomsto achieve their octet (except H). Do central atom last.

Fifth Step

Examine central atom to determine if a double or triple bond is required to achieve the central atom’s octet.Do so using unshared pairs, IF central atom is: C, N, P, O, S

NH H

H

NH H

H

8 e's / 2 = 4 electron pairs

NH H

H

N, octet

H, duetStep 4:

No Step 5 needed