Nomenclature, the naming of compounds, is a fundamental skill that all chemists must master. Compound names are the basic language of chemistry. If you can’t speak the language, you don’t belong in the country. The following discussion is a comprehensive explanation of the rules that chemists follow to name the majority of inorganic compound. When you study organic chemistry you will learn additional rules for the naming of organic compounds. If you study biochemistry, additional rules for the naming of relevant biological macromolecules and their building blocks will be a required skill that you will also have to master. The initial ABC’s of the chemist’s language are introduced in this section.

Rules for naming compound:

First, determine if the compound is ionic or molecular

a) Recognizing an ionic compound from its formula:

Ionic compounds are recognized if the formula meets any of these criteria

1. Metal & non-metal examples: Ca3N2, NaCl, Fe2O3

2. Metal & polyatomic anion examples: Na2SO4, Ba3(PO4)2, Ni(OH)2

3. Polyatomic cation & poly atomic anion examples: (NH4)2CO3, Hg2(ClO)2

4. Polyatomic cation & group A anion examples: NH4Cl, Hg2S

b) Recognizing a molecular compound from its formula:

Molecular compounds contain 2 or more non-metals in their formula.

Examples: CH2O, N2O4, C5H7NO2

**** Be careful….. (NH4)2CO3 contains only non-metals but is an ionic compound. If you see the formula of poly-atomic ions in the compound it is ionic.

Naming ionic compounds:

General scheme: Name the cation, name the anion, put the two names together.

The first species in the formula of the ionic compound is always the cation, and the second species is always the anion

What you need to learn is how cations and anions are named. If you do, you can name any ionic compound.

a) Cation nomenclature:

There are three different classes of cations.

1. Group A cations (representative metals found in “A” groups of the periodic table).

2. Group B cations (transition and post-transition metals found in “B” groups of the periodic table) *****NOTE: the post-transition metals Pb, Sn, and Bi are in “A” groups.3. Polyatomic cation: The only 2 that you are responsible for are:

a. NH4+ - ammonium ion.

b. Hg22+- mercury (I) ion.

Group A cations are the easiest to name because they form only one stable ion. Group A cation nomenclature is simply the name of element + “ion”

Examples: a. The compound NaBr is an ionic compound (metal and non-metal). The first species in the compound (the Na) is a group A cation, (Na is found in group IA on the periodic table) Its name is sodium ionb. The compound Al2(SO4)3 is an ionic compound (metal and polyatomic ion). The first species in the

compound (the Al) is a group A cation, (Al is found in group IIIA on the periodic table)Its name is Aluminum ion.

Group B cations are named differently because they can form more than one type of cation. They can form multiple cations that have different charges. Example: Fe forms both the Fe2+ and the Fe3+ ions.We cannot call the cations in FeCl2 (the Fe2+) and FeCl3 (the Fe3+) both “iron ion” because they are different

Group B cations are named the following way: Name of element + (Roman numeral charge value) + “ion”.

This means that you must figure out the charge of the cation in the compound before it can be named.

The charge of a Group B cation in a compound can be determined the following way:

Example: The compound Cr2S3 is ionic (metal and non-metal). The cation (Cr) is a group B cation. The formula of the compound will be used to figure out the charge on the cation. “Go with what you know”. - The periodic table can be used to determine the charge of the anion (here the S). The stable ion of sulfur carries a 2- charge. There are 3 sulfur ions in the formula unit giving a total –6 (negative six) charge.The formula unit MUST be electronically neutral so there must be a total +6 (positive six) charge from the cations in the unit. There are two Cr cations that provide the +6 charge. Each one must carry a +3 charge. The name of the Cr cations in the compound Cr2S3 is the chromium (III) ion

Polyatomic cations: Their names come from memory.

In (NH4)2CO3 the name of the cation is ammonium ion

b) Anion nomenclature:

There are 2 classes of anions; group A anions and polyatomic anions.

Group A anions, as their name implies, are found in “A” groups of the periodic table. They are the non-metals.The anion is always the second species in the formula unit of the ionic compound.

Group A anions are named the following way:

a. Start with the name of the element,b. drop all the letters from the second vowel and on, add the ending “ide” + “ion”.

Note: There are some exceptions to this rule.

Examples: in Cr2S3, the S is a group A anion (it is found in group VIA). To name this anion we start with sulfur, drop the “ur” and add “ide ion” to get sulfide ion.

In BaSe, the Se is a group A anion. To name the anion we start with selenium, drop the ium and add ide ion to get the selenide ion (exception to 2nd vowel). Iodide ion is also an exception.

Some names of common group A anions:

C4- Carbide ion (rare) S2- Sulfide ion

N3- Nitride ion Cl- Chloride ion

O2- Oxide ion Se2- Selenide ionF- Fluoride ion Br- Bromide ionP3- Phosphide ion I- Iodide ion

Polyatomic anions: Polyatomic ions can be divided into two classes.

There are diatomic anions (usually containing 2 atoms) :

CN- cyanide ion, OH- hydroxide ion, O22- peroxide ion, and SCN- thiocyanide.

***Note- Their names all have the suffix “ide”

The second class is collectively called the oxy-anions.As their name implies their formula contain (multiple) oxygen atoms.

Common oxy-anions: (commit to memory)

1- ions 2- ions 3- ions

NO3- - nitrate ion

MnO4- -permanganate ion

C2H3O2- - acetate ion

ClO3- - chlorate ion

SO42- -sulfate ion

CO32- -carbonate ion

C2O42- -oxalate ion

CrO42- -chromate ion

Cr2O72- -dichromate ion

PO43- - phosphate ion

*** Note- The names of all the anions in the above table have the suffix “ate”.

There are many more oxy-anions that are not in the above table. Some of them are related to those in this table. Oxy-anions that have one oxygen atom less than the “ate” versions are given the suffix “ite”.

When you see one, you change the “ate” suffix to “ite”.

Examples: The anion NO3- is named nitrate ion. The anion NO2

- also exists. Its name is the nitrite ion.

The polyatomic ion SO32- has one less oxygen than SO4

2-. SO42- is the sulfate ion so SO3

2- is the sulfite ion.

Chlorate (ClO3-) is a special case. Not only is there the chlorite ion (ClO2

-), there is also the hypochlorite ion (ClO-), and the perchlorate ion (ClO4

-). Hypo is a prefix meaning below (like hypodermic, which means below

the skin). Hypochlorite has one less oxygen (below) chlorite. The prefix per (in naming) comes from hyper which means above. Perchlorate has one more oxygen (above) chlorate.

With these rules for naming cations and anions you can name all ionic compounds. First, name the cation, then name the anion. Finally, you put the names together, dropping the word ion from each, to get the name of the compound.

Examples:

Name the following compounds:

a. FeBr3

1. Identify that the compound is ionic (metal and a non-metal).2. Name the cation first:Use the periodic table to determine if the cation is group A, or group B, or identify it as a polyatomic cation.Fe is a group B cation. Follow the rule for naming a group B cation (figure out the charge of the cation using the charge of the anion). Br is a group A anion whose charge can be predicted with the periodic table. The charge of the Br ion is –1. There are three Br ions in the formula unit of the compound. The total negative charge is –3. The unit must be neutral so the total positive charge must be +3. Since only one iron is providing all the positive charge it must be the 3+ ion. The name of the cation is the “iron (III) ion”.

3. Next, you name the anion. As already stated, Br is a group A anion. The rule for naming a group A anion is to drop the 2nd vowel an on then add ide ion. The Br- ion is called the “bromide ion”

4. Put the names together dropping the word ion from each name:

“iron (III) bromide” is the name of the compound.b. Al2(SO3)3

1. The compound is ionic (metal and polyatomic ion)2. Al is a group A cation…….Its name is Aluminum ion3. The Polyatomic anion (oxy-anion) table shows SO4

2- is the sulfate ion. SO32- is 1 oxygen less than the sulfate

ion so its name is the sulfite ion4. Aluminum sulfite is the name of the compound.

Naming binary molecular compounds:

As their name implies these compounds are composed of TWO different kinds of non-metals

Examples: N3O5, CO2, CCl4

To name binary molecular compounds, first pretend that the compound is ionic.

The first species in the formula would be considered the cation and the second species would be considered the anion.

Each is named according to the rules for naming group A cations and anions. Once the compound is given an ionic name, Latin numerical prefixes are included to indicate the number of each species in the formula of the compound

Examples:

a. N3O5 –First, consider N the group A cation and give it the name nitrogen ion. The O is named oxide ion after its group A anion name. put the names together to get nitrogen oxide and then add the appropriate Latin numerical prefixes to the name.

Trinitrogen pentoxide

CO2: Carbon ion, oxide ion – carbon oxide - carbon dioxideB2P4: Boron ion, Phosphide ion - boron phosphide – diboron tetraphosphide

Hydrocarbon nomenclature:

There is a unique class of binary molecular compounds collectively called hydrocarbons. These are compounds that contain only carbon and hydrogen. C4H10, C6H12, and C3H4 are examples of hydrocarbons. The naming of simple (un-branched) hydrocarbons follows its own set of nomenclature rules.

Straight chain hydrocarbons consist of a carbon backbone with the hydrogen atoms attached to the carbons of the backbone chain.

Simple hydrocarbons are classified based on the carbon-carbon bonds present in the carbon backbone.

If all the carbon-carbon bonds are single covalent bonds, the compound is classified as an alkane.If there is a carbon-carbon double covalent bond in the backbone, the compound is classified as an alkene.If there is a carbon-carbon triple covalent bond in the backbone, the compound is classified as an alkyne.

The formula of alkanes follow the general scheme CnH2n+2, where n is the number of carbons in the chain

CH4, C2H6, and C4H10 all follow the scheme CnH2n+2, and are all alkanes.

The formula of alkenes follow the general scheme CnH2n, where n is the number of carbons in the chain

C2H4, C6H12, and C8H16 all follow the scheme CnH2n, and are all alkenes.

The formula of alkynes follow the general scheme CnH2n-2, where n is the number of carbons in the chain.

C3H4, and C7H12 follow the scheme CnH2n-2, and are alkynes.

Hydrocarbon nomenclature is based on the compounds class (alkane, alkene, or alkyne) and number of carbon atoms in the chain.

To name a hydrocarbon, the compound is first classified as being an alkane, alkene, or alkyne based on which scheme its formula adheres to.

Examples:

C4H10 is classified as an alkane because if n = 4 (the number of carbons), the number of hydrogen atoms (10) follows the scheme 2n + 2, (2 x 4 + 2 = 10).

C6H12 is classified as an alkene because if n = 6 (the number of carbons), the number of hydrogen atoms (12) follows the scheme 2n, (2 x 6 = 12).

C7H12 is classified as an alkyne because if n = 7 (the number of carbons), the number of hydrogen atoms (12) follows the scheme 2n - 2, (2 x 7 - 2 = 12).

The suffix used for the name of the hydrocarbon comes from the class of the hydrocarbon. Alkane all have the suffix “ane”, alkenes all use the suffix “ene” and alkynes all use the suffix “yne”The prefix used in the name of a hydrocarbon indicates the number of carbons in the formula. The numerical prefixes for hydrocarbons are as follows:

# C Prefix # C Prefix # C Prefix

1 meth 5 pent 9 non

2 eth 6 hex 10 dec

3 prop 7 hept

4 but 8 oct

The appropriate prefix is added to the appropriate suffix to get the name of the hydrocarbon:

C4H10 – alkane with 4 carbons - 4 = but, alkane = ane – compound = butane

C6H12 – alkene with 6 carbons – 6 = hex, alkene = ene – compound = hexane

C7H12 – alkyne with 7 carbons – 7 = hept, alkyne = yne – compound = heptyne

Reverse process – generating the formula from the name:

1. Generating the formula for an ionic compound:

It is rather easy to recognize that a compound is ionic from its name. Generally names of ionic compounds lack Latin numerical prefixes that you find in the names of binary molecular compounds. The only ionic compounds that have Latin numerical prefixes in their name are those that have a polyatomic anion whose name contains a numerical prefix (like dichromate).

When you recognize the compound is ionic, simply reinsert the word ion back into the name and generate the formula of the cation and anion. Then, criss-cross the charges to generate the subscripts. Lastly, reduce the formula if necessary as the formula unit of an ionic compound is always empirical.

Examples: Magnesium bromide-- magnesium ion, bromide ion-- Mg2+ Br - MgBr2

Lead (IV) sulfite—lead (IV) ion, sulfite ion—Pb4+ SO32--- Pb2(SO3)4 – Pb(SO3)2

The formulas of binary molecular compounds are generated simply by interpreting the name.

Example: Tetraphoshorus decoxide--- tetra = 4, deca = 10---- P4O10

The formulas of hydrocarbons are obtained by interpreting the prefix to get the number of carbon atoms (n) and the suffix to get the appropriate hydrocarbon scheme (i.e. CnH2n).

Examples: Pentene – pent means n = 5, ene means CnH2n -- C5H10

Octyne – oct ; n = 8, yne; CnH2n-2 : C8H14

Propane – prop: n = 3, ane; CnH2n+2 : C3H8

ACID NOMENCLATURE:

There are a set of compounds that are ionic-like called acids. These compounds are molecular in there pure state but dissociate into ions when put into solution. The class of acids that you are required to recognize and name fall under the Arrhenius definition of acids; any species that increases the H+ concentration when put into solution is an Arrhenius acid. You will recognize Arrhenius acids as they all have cation-like hydrogen(s) in their formula, followed by either a non-metal or a poly-atomic anion. An (aq) physical state must also be present in the written formula.

Examples: H2S(aq), HCl(aq), HNO3(aq), H2SO3(aq)

Acids are named on the basis of their anion’s name. To name an acid simply name the anion and use the appropriate scheme.

a. For acids that contain anions whose name ends with the suffix “ide” take the anion name, drop the ide and put what is left into the blank:

Hydro______________ic acid

Examples: HCl(aq), HCN(aq), H2S(aq) Anions: Chloride, cyanide, sulfide

Roots for nomenclature: Chlor, cyan, sulfur***** (for sulf you must add back the ur)

Acid name: Hydrochloric acid, Hydrocyanic acid, Hydrosulfuric acid

b. For acids that contain anions whose name ends with the suffix “ate” take the anion name, drop the ate and put what is left into the blank:

_________________ic acid

Examples: HClO3 (aq), H2CO3 (aq), H2SO4 (aq) Anions: Chlorate, carbonate, sulfate

Roots for nomenclature: Chlor, carbon, sulfur***** (for sulf you must add back the ur)

Acid name: Chloric acid, Carbonic acid, Sulfuric acid

c. c. For acids that contain anions whose name ends with the suffix “ite” take the anion name, drop the ite and put what is left into the blank:

_________________ous acid

Examples: HClO2 (aq), HNO2 (aq), H2SO3 (aq) Anions: Chlorite, nitrite, sulfite

Roots for nomenclature: Chlor, nitr, sulfur***** (for sulf you must add back the ur)

Acid name: Chlorous acid, nitrous acid, Sulfurous acid

CONTINUE TO THE NEXT PAGE

Balancing chemical equations:

Chemists use chemical equations to represent reactions. The generic form of a chemical equation is the following:

Reactants products

The chemical formulas of the starting materials are listed (the reactants), an arrow is used to represent a process, and then the chemical formulas of the materials that form (the products) are listed. This is how chemists generate skeletal chemical equations.

Example: lead (II) nitrate reacts with potassium chloride to produce lead (II) chloride and potassium nitrate.

To convert this into a chemical equation, we first write the formula of each reactant, then the arrow, and then the formula of each product

Pb(NO3)2 + KCl PbCl2 + KNO3

While this skeletal equation shows the species that react and the species that form, skeletal equations do not obey the law of conservation of matter. The number of each type of atom on the reactant side does not match the number of each type of atom on the product side. In order to get skeletal equations to obey the law of conservation of matter chemists balance the equation by altering its coefficients.

If the coefficient 2 is placed in front of the KCl and the KNO3 the equation becomes balanced.

Pb(NO3)2 + 2KCl PbCl2 + 2KNO3

Now there are equal numbers of each atom on both sides of the equation

(1 Pb, 2 N, 6 O, 2 K, and 2 Cl)

General rules for balancing equations:

1) Balance element at a time2) Add coefficients determined by: what you need/what you have3) Never change a subscript or alter the formula of any compound4) Coefficients are added only before the formula of a compound, never in the middle of a chemical formula

Specific rules for balancing equations

1. Identify any and all elements existing in the equation as an elemental state. Save these species for the end2. Balance non-hydrogen, non-oxygen first3. Balance hydrogen and/or oxygen4. Balance the species from step 15. Multiply equation by largest denominator present to get rid of any fractions IF NEEDED.

Examples:

C3H8O + O2 CO2 + H2O

Step 1: Save the oxygen for the end because it appears in the equation as O2, an elemental state

Step 2: Balance the carbon first. There are 3 carbons on the reactant side and only 1 on the product side.We need the product side to be 3 (what you need), and there is 1 on the product side (what you have). TheFraction 3/1 (which is just 3) is what we put in front of the CO2 to balance the carbon. This 3 also affects the number of oxygen in the CO2 but we do not care at this time.Because there are no other species that is not hydrogen or oxygen step 2 is over

C3H8O + O2 3CO2 + H2O

Step 3: Balance the hydrogen. Oxygen is being saved for the end so we balance the hydrogen now. There are 8 hydrogen on the reactant side and 2 on the product side. We need the product side to be 8, we have 2. 8/2 = 4 is the coefficient that we add to the front of the H2O

C3H8O + O2 3CO2 + 4H2O

Step 4: Balance the oxygen. The last thing that is not balanced is the oxygen. On the product side there is a total of 10 oxygen (6 from the 3CO2 and 4 from the 4H2O). There is one O in the C3H8O and 2 from the O2. We do not want to change the C3H8O because that will change the carbon and the hydrogen. That means that the O2

needs to be 9 to add to the one from the C3H8O to make a total of 10. We need the O2 to be 9, we have 2 the fraction 9/2 is placed in front of the O2.

C3H8O + 9/2 O2 3CO2 + 4H2O

The equation is numerically balanced but we cannot leave the fraction. All coefficients (on the reactant and product side) are multiplied by 2 (the largest denominator) to get rid of the fraction.

2[C3H8O + 9/2 O2 3CO2 + 4H2O]

2C3H8O + 9 O2 6CO2 + 8H2O

Lets do another by the rules

Fe2O3 + H2 Fe + H2O

Step 1: save hydrogen and iron for the endStep 2: balance the O by putting a 3 in front of the H2O

Fe2O3 + H2 Fe + 3H2O

Balance the Fe by putting a 2 in front of the Fe

Fe2O3 + H2 2Fe + 3H2O

Balance the H by putting a 3 (6/2) in front of the H2

Fe2O3 + 3H2 2Fe + 3H2O

The equation is balanced.

Redox Reactions: Oxidation – Reduction Reactions.

Reactions that involve the transfer of electrons from one species to another are referred to as oxidation – reduction reactions or redox reactions. The mneumonic OIL RIG is useful in remembering the electron transfer process. OIL- Oxidation Is Loss. When a species is losing electrons it is undergoing oxidation. RIG- Reduction Is Gain. When a species is gaining electrons it is undergoing reduction.

Certain reactions are easily identified as redox reactions due to the presence (and changes) of charge values in the balanced equation. Here is an example of a redox reaction that is recognizable by merely looking at the equation:

2Al(s) + 3Fe2+(aq) 2Al3+

(aq) + 3Fe(s)

By simple inspection it is easy not only to identify this reaction as a redox reaction, but also to identify the species that is undergoing oxidation and the species that is undergoing reduction. Aluminum is going from the neutral metalic state to a 3+ ion during the reaction. This can only occur if each aluminum atom loses 3 electrons (OIL- oxidation is loss). We say that the aluminum atoms are undergoing oxidation. When you focus on the iron (II) ions you see that they are becoming neutral, metalic iron atoms during the reaction. This can only occur if each ion gains 2 electrons (RIG- reduction is gain). We say that the iron (II) ions are undergoing reduction.

While it is easy to identify the above reaction as a redox reaction, many redox reactions are not this obvious. Look at the following reaction:

NaI + 3HOCl NaIO3 + 3HCl

Although this is a redox reaction, it is not obvious like the first reaction. How can this reaction be identified as a redox reaction and the how can the species undergoing oxidation and reduction be identified as well?

We define a value called an oxidation number (also known as an oxidation state) as the charge atoms in a molecule would have if shared electrons (between bonded atoms) are given to the more electronegative atom.

To identify reactions as redox reactions, oxidation numbers are assigned to each atom and the changes in oxidation numbers during the reaction are identified and used to show that the reaction is in fact a redox reaction and what species are involved in the redox.

The rules for assigning oxidation numbers are as follows:

1. Fluorine is the most electronegative element. It has an oxidation number of -1 except in the diatomic elemental state F2 where its oxidation number is zero.

2. Oxygen is the second-most electronegative element. It has an oxidation number of -2 except:a. as the peroxide ion (O2

2-) where its oxidation number is -1.b. in the elemental states O2 and O3 where its oxidation number is zero.c. when bonded to fluorine (see summation rule below).

3. Hydrogen has an oxidation number of +1 except:a. When bonded to a metal in metal hydrides (i.e. NaH) where its oxidation number is -1b. As the diatomic elemental state H2 where its oxidation number is zero.

4. The oxidation number of a pure element (monatomic, diatomic, or polyatomic state) is zero.5. The oxidation number of a free ion is the charge of the ion.6. The oxidation number of an ion in an ionic compound is the charge of the ion.

7. Summation rule- the total contribution of oxidation numbers in a compound must add up to the charge of the compound.

With these rules, most of the oxidation numbers in a reaction can be determined.

Example:

Assigning oxidation numbers to the species in the reaction: NaI + 3HOCl NaIO3 + 3HCl

NaI is ionic (metal and non-metal). Rule 6- The oxidation number of an ion in an ionic compound is the charge of the ion. Na carries a 1+ charge and I carries a 1- charge. The oxidation numbers of Na and I are +1 and -1 respectively.

HOCl- Rule 2- Oxygen is the second-most electronegative element. It has an oxidation number of -2 Rule 3- Hydrogen has an oxidation number of +1Application of these two rule gives us the following states:

The above rules determine the single state oxidation number for both the oxygen and the hydrogen. The total contribution is calculated by multiplying the single state oxidation number by the subscript of each species in the chemical formula.

The final rule used for the chlorine is the summation rule. Rule 7- Summation rule- the total contribution of oxidation numbers in a compound must add up to the charge of the compound.

Application of this rule allows us to determine the total contribution of chlorine. The compound is neutral so the sum of the total contribution must add up to zero. +1 + -2 + total contribution of Cl = 0. The total contribution of Cl in HOCl is +1. The total contribution is then divided by the subscript of Cl in the formula to give the single state oxidation number of Cl in the compound (+1).

NaIO3 is ionic (IO3- is the iodate ion).

Rule 2- Oxygen is the second-most electronegative element. It has asingle state oxidation number of -2

The total contribution of oxygen is calculated by multiplying the single state -2 by the subscript of oxygen in the formula (3) to get -6.

Rule 6- The oxidation number of an ion in an ionic compound is the charge of the ion. Na carries a 1+ charge. The oxidation numbers of Na is +1.

The oxidation number of iodine is determined with the summation rule. The total contribution of iodine must be +5 so that all the total contributions sum to zero (the overall charge on NaIO3). Since there is only one iodine in the formula its single state oxidation number is +5.

HCl - Rule 3- Hydrogen has an oxidation number of +1 Rule 7- Summation rule- the total contribution of oxidation numbers in a compound must add up to the charge of the compound.

Once all single states have been assigned, changes in the single state values are identified and interpreted in terms of redox.

Iodine as a reactant has an oxidation number of -1. During the reaction its oxidation number changes to +5. The oxidation number becomes more positive which corresponds to less negative and a loss of electrons (OIL- oxidation is loss). The iodine is undergoing oxidation (by losing 6 electrons).

Chlorine as a reactant has an oxidation number of +1. During the reaction its oxidation number changes to -1. The oxidation number becomes more negative which corresponds to a gain of electrons (RIG- reduction is gain). The chlorine is undergoing reduction (by each chlorine gaining 2 electrons).

Sometimes the rules for assigning oxidation numbers do not work. There are many compound that contain elements that are not covered by the limited rules for assigning oxidation number. Once you cover the chapter on drawing Lewis Dot Structures, the original definition of an oxidation number can be used along with the preferred structure to determine the oxidation number of any species. The following examples will illustrate how oxidation numbers can be determined from the preferred Lewis dot structure of a compound.

Example 1: OF2 oxygen difluoride.

The set of rules for assigning oxidation numbers works for this compound. However, we will use both methods to illustrate that they yield the same result for the oxidation numbers of O and F. Then we will apply this technique to a compound whose oxidation numbers cannot be determined by the rules but can be determined with the Lewis structure.

Rule 1- Fluorine has an oxidation number of -1, and the total contribution for fluorine will be -2

Rule 2 states that when oxygen is bonded to fluorine, the summation rule must be used to determine its oxidation number. Application of the summation rule yields a +2 for the single state oxidation number fot the oxygen in OF2.

Using the Lewis dot structure of OF2 and the definition of an oxidation number the same result will be obtained.Here is the application and explanation:

The preferred Lewis dot structure for OF2 is

Since fluorine is more electronegative than oxygen the shared electrons are given to each fluorine and a charge is determined for each atom by comparison to the normal valence of each atom.

Fluorine normally has 7 valence electrons. When the shared electrons are given to the fluorine atoms, each one winds up with 8 valence electron, This results in a -1 charge on the fluorine atoms. Recall the definition of oxidation number. It’s the charge the atoms would have if shared electrons are given to the more electronegative atom. The oxidation number of fluorine in OF2 is -1.

The oxygen winds up with only 4 valence electrons when the shared electrons are given to the fluorine atoms. Oxygen normally has 6 valence electrons. This results in the oxygen having a charge of 2+ and an oxidation number of +2. These results correspond to the results of using the appropriate rules to assign oxidation numbers to the atoms of OF2.

The rules for assigning oxidation numbers fail to produce results for (CN)2. This is due to the fact that for both C and N there aren’t any rules to assign oxidation numbers. The oxidation numbers for both C and N can be determined from the preferred Lewis dot structure of (CN)2.

The preferred Lewis dot structure for (CN)2 is :

The shared electrons of each triple bond is given to each nitrogen as it is more electronegative than carbon. The pair of electrons shared between the two carbons is split between them as each has the same electronegativity. The resulting electron distribution is as follows:

Nitrogen normally has 5 valence electrons. When the electrons are distributed on the basis of electronegativity, each nitrogen winds up with a 3- charge. Hence, the oxidation number of the nitrogen atoms is -3.

Carbon normally has 4 valence electrons. When the electrons are distributed on the basis of electronegativity, each carbon atom winds up with a 3+ charge. Hence, the oxidation number of the carbon atoms is +3.

The Balancing of Redox Reactions:

Like all reactions, redox reactions must obey the law of conservation of matter. Therefore, balanced equations are required when they are represented. Compared to other types of reactions there are additional considerations of the law of conservation of matter when balancing redox reactions.

As with all reactions the number of each atom on the reactant and product sides of the equation must be equal. Since electrons are forms of matter, the electron transfer during the redox reaction must also obey the law of conservation of matter. That is, the total number of electrons that are lost by the species undergoing oxidation must be equal to the total number of electrons being gained by the species undergoing reduction. Finally, since the electron transfer is conserved, the total charge on the reactant side of the equation must equal the total charge on the product side of the equation. While charges on individual species may change, there can be no net generation of charge during the reaction.

The rules used to balance redox reactions take into consideration all three aspects of conservation; the balancing of atoms, and the balancing of electrons, while preventing charge generation.

Skeletal unbalanced redox reactions typically show the compounds containing the species undergoing oxidation and reduction on the reactant side, and on the product side, the compounds containing the species in their new oxidation state. Usually, skeletal unbalanced redox reactions do not obey the law of conservation of matter with respect to hydrogen and/or oxygen. Often, these species are missing from either the product or reactant side. This is not an initial concern because the steps of balancing redox reactions will incorporate H2O, and either H+ or OH- into the equation, balancing the missing hydrogen and/or oxygen atoms.

Example: C2O42- + HNO2 CO2 + NO

By assigning oxidation numbers the atoms undergoing oxidation and reduction are identified.

The carbon in the oxalate ion is going from a +3 oxidation state to a +4 oxidation state. The change to a more positive state corresponds to a less negative state and the loss of electrons (OIL). Carbon is undergoing oxidation during the reaction. The nitrogen in nitrous acid is going from an oxidation state of +3 to a state of +2. A less positive state corresponds to a more negative state and the gain of electrons (RIG). Nitrogen is undergoing reduction during the reaction.

Once the species undergoing redox are identified, the skeletal equation is split into oxidation and reduction half reactions. Half reactions show the oxidation and reduction reactions as distinct entities (even though they are occurring together). Each half reaction will be balanced and then recombined to yield the final balanced redox equation.

Oxidation half-reaction: C2O42- CO2

Reduction half-reaction: HNO2 NO

The atom that is undergoing the redox is then numerically balanced in each half-reaction. In the example below the carbon needs to be numerically balanced. Nitrogen in the reduction half-reaction is already numerically balanced.

Oxidation half-reaction: C2O42- 2CO2

Reduction half-reaction: HNO2 NO

Electrons are then added to each half-reaction to account for the change in oxidation numbers of the redox species, taking into account the number of atoms that are undergoing the change in oxidation state. In the example below the carbon atom’s oxidation number is changing from +3 to +4, a loss of 1 electron. However, 2 carbon atoms must undergo this change in oxidation state requiring a total of 2 electrons lost in the half reaction. Lost electrons are shown on the product side for oxidation half-reactions.

Oxidation half-reaction: C2O42- 2CO2 + 2e-

The nitrogen atom in the reduction half-reaction is gaining 1 electron which accounts for the change in its oxidation state from +3 to +2. Gained electrons are shown on the reactant side for reduction half reactions.

Reduction half-reaction: 1e- + HNO2 NO

Redox reactions usually occur in either acidic or basic solutions. In acidic solutions H+ is present in abundance. In basic solutions, OH- is an abundant species. H+ is incorporated into half-reactions to balance charge when redox reactions are occurring in acidic solutions. OH- is incorporated into half-reactions to balance charge when redox reactions occur in basic solutions. The conditions of the solution are usually indicated along with the skeletal unbalanced equation.

C2O42- + HNO2 CO2 + NO (in acidic solution)

The balancing of redox reactions in acidic solutions will be discussed first.

Into each half reaction, H+ will be added to the appropriate side to balance the charges on the product and reactant sides.

Oxidation half-reaction: C2O42- 2CO2 + 2e-

For the oxidation half-reaction, the total charge on the reactant side is a -2 (from the charge on the oxalate ion) and the total charge on the product side is -2 (due to the charge of the 2 e_ incorporated during the previous step).

Reduction half-reaction: 1e- + HNO2 NO

For the reduction half-reaction, the total charge on the reactant side is -1 (from the 1 e-) and zero on the product side. An H+ added to the reactant side of the equation gives both sides of the half-reaction a zero charge, balancing the charge for the half reaction.

Reduction half-reaction: H+ + 1e- + HNO2 NO

Once the charge has been balanced H2O is added to the half-reactions to balance the oxygen atoms (if needed) in each half-reaction. In addition, this usually balances the hydrogen on both sides of each half reaction.

Oxidation half-reaction: C2O42- 2CO2 + 2e-

The oxygen atoms are balanced in the oxidation half-reaction, so H2O will not be used.

Reduction half-reaction: H+ + 1e- + HNO2 NO

The reactant side of the reduction half-reaction has 2 oxygen atoms while the product side only has 1 oxygen atom. One H2O will be added to the product side to bring the number of oxygen atoms on that side to 2. When this is done, the hydrogen atoms will balance as well as the added water molecule adds hydrogen to the side it is added to.

Reduction half-reaction: H+ + 1e- + HNO2 NO + H2O

Each half reaction is now balanced: The atoms are numerically balanced, electrons account for the redox, and charge is balanced. The final aspect of redox balancing will now be addressed; the electron transfer must be balanced.

The entire reduction half-reaction will be multiplied by 2 so that the number of elections gained will equal the number of electrons lost in the oxidation half-reaction.

Reduction half-reaction: 2H+ + 2e- + 2HNO2 2NO + 2H2O

The oxidation half-reaction and the reduction half-reaction will now be added together. All species that appear on both sides of the equations will be canceled (in this example only the electrons cancel from the equations).

The combined equation is the complete balanced equation for the redox reaction.

Balancing redox reactions in basic solution:

The steps involved in balancing redox reactions in basis solution are nearly identical to balancing redox reactions in acidic solution. In fact, there is only one additional step.

Consider the skeletal unbalanced redox reaction in basic solution:

ClO3- + N2H4 NO + Cl- (in basic solution)

After oxidation numbers are assigned the species undergoing the redox are identified:

The reaction is then split into half reactions:

The atom(s) undergoing the redox are balanced:

Electrons are added to account for the change in oxidation number:

Even though the reaction is occurring in basic solution, pretend that it is in acidic solution and use H+ to balance charge.

The H+ used to balance charge are converted to the same number of H2O, and the same number of OH- is added to the other side.

H2O is then added to balance the oxygen atoms:

The equations are then multiplied so that the electrons cancel:

Electrons, and all like species on opposite sides of the equations are canceled and the half-reactions are added together.

Check that atoms are numerically balanced and that the total charges on the product and reactant sides are equal.