What is a Lewis Diagram?

Lewis diagrams, also called electron-dot diagrams, are used to represent paired and unpaired valence (outer shell) electrons in an atom. For example, the Lewis diagrams for hydrogen, helium, and carbon are
where the symbol represents the element (in this case, hydrogen, helium, and carbon) and the dots represent the electrons in the outer shell (in this case, one, two, and four). These diagrams are based on the electron structures learned in the Atomic Structure and Periodic Table chapters.

What is a Lewis Structure?

The Lewis structure is used to represent the covalent bonding of a molecule or ion. Covalent bonds are a type of chemical bonding formed by the sharing of electrons in the valence shells of the atoms. Covalent bonds are stronger than the electrostatic interactions of ionic bonds, but keep in mind that we are not considering ionic compounds as we go through this chapter. Most bonding is not purely covalent, but is polar covalent (unequal sharing) based on electronegativity differences.
The atoms in a Lewis structure tend to share electrons so that each atom has eight electrons (the octet rule). The octet rule states that an atom in a molecule will be stable when there are eight electrons in its outer shell (with the exception of hydrogen, in which the outer shell is satisfied with two electrons). Lewis structures display the electrons of the outer shells because these are the ones that participate in making chemical bonds.

How to Build a Lewis Structure?

For simple molecules, the most effective way to get the correct Lewis structure is to write the Lewis diagrams for all the atoms involved in the bonding and adding up the total number of valence electrons that are available for bonding. For example, oxygen has 6 electrons in the outer shell, which are the pattern of two lone pairs and two singles. If the electrons are not placed correctly, one could think that oxygen has three lone pairs (which would not leave any unshared electrons to form chemical bonds). After adding the four unshared electrons around element symbol, form electron pairs using the remaining two outer shell electrons.

Incorrect Structure	Correct Structure

are two hydrogen atoms and one oxygen atom. The Lewis structure of each of these atoms would be as follows:
One good example is the water molecule. Water has the chemical formula of H₂O, which means there
We can now see that we have eight valence electrons (six from oxygen and one from each hydrogen). With few exceptions, hydrogen atoms are always placed on the outside of the molecule, and in this case the central atom would be oxygen. Each of the two unpaired electrons of the oxygen atom will form a bond with one of the unpaired electrons of the hydrogen atoms. The bonds formed by the shared electron pairs can be represented by either two closely places dots between two element symbols or more commonly by a straight line between element symbols:
Let us try another one.

Example: Write the Lewis structure for methane (CH₄).
Answer: Hydrogen atoms are always placed on the outside of the molecule, so carbon should be the central atom.
After counting the valence electrons, we have a total of 8 [4 from carbon + 4(1 from each hydrogen] = 8.
Each hydrogen atom will be bonded to the carbon atom, using two electrons. The four bonds represent the eight valence electrons with all octets satisfied, so your structure is complete.

Example: Write the Lewis structure for carbon dioxide (CO₂).
Answer: Carbon is the lesser electronegative atom and should be the central atom.
After counting the valence electrons, we have a total of 16 [4 from carbon + 2(6 from each oxygen)] = 16.
Each oxygen atom has two unshared electrons that can be used to form a bond with two unshared electrons of the carbon atom, forming a double bond between the two atoms. The remaining eight electrons will be place on the oxygen atoms, with two lone pairs on each.

Lewis Structures of Polyatomic Ions

Building the Lewis Structure for a polyatomic ion can be done in the same way as with other simple molecules, but we have to consider that we will need to adjust the total number of electrons for the charge on the polyatomic ion. If the ion has a negative charge, the number of electrons that is equal to the charge on the ion should be added to the total number of valence electrons. If the ion has a positive charge, the number of electrons that is equal to the charge should be subtracted from the total number of valence electrons. After writing the structure, the entire structure should then be placed in brackets with the charge on the outside of the brackets at the upper right corner.
Example: Write the Lewis structure for the ammonium ion (NH₄⁺).
Answer: Hydrogen atoms are always placed on the outside of the molecule, so carbon should be the central atom.

After counting the valence electrons, we have a total of 9 [5 from nitrogen + 4(1 from each oxygen)] = 9. The charge of +1 means an electron should be subtracted, bringing the total electron count to 8.
Each hydrogen atom will be bonded to the nitrogen atom, using two electrons. The four bonds represent the eight valence electrons with all octets satisfied, so your structure is complete. (Do not forget your brackets and to put your charge on the outside of the brackets)

Example: Write the Lewis structure for the hydroxide ion (OH^-).
Answer: Since there are only two atoms, we can begin with just a bond between the two atoms. After counting the valence electrons, we have a total of 7 [6 from Oxygen + 1 from each Hydrogen)] = 7. The charge of -1 indicates an extra electron, bringing the total electron count to 8.
Oxygen will be bonded to the hydrogen, using two electrons. Place the remaining six electrons as three lone pairs on the oxygen atom. All octets are satisfied, so your structure is complete. (Do not forget your brackets and to put your charge on the outside of the brackets)

Lewis Structures for Resonance Structures

The existence of some molecules often involves two or more structures that are equivalent. Resonance can be shown using Lewis structures to represent the multiple forms that a molecule can exist. The molecule is not switching between these forms, but is rather an average of the multiple forms. This can be seen when multiple atoms of the same type surround the central atom. When all lone pairs are placed on the structure, all the atoms may still not have an octet of electrons. To deal with this problem, the atoms (primarily in a C, N, or O formula) form double or triple bonds by moving lone pairs to form a second or third bond between two atoms. The atom that originally had the lone pair does not lose its octet because it is sharing its lone pair. Double-headed arrows are placed between the multiple structures of the molecule or ion to show resonance. Let us look at how to build a nitrate ion (NO₃^-).
Nitrogen is the least electronegative atom and should be the central atom.
After counting the valence electrons, we have a total of 23[5 from nitrogen + 3(6 from each oxygen)] = 23. The charge of -1 indicates an extra electron, bringing the total electron count to 24.
Each oxygen atom will be bonded to the nitrogen atom, using a total of six electrons. We then place the remaining 18 electrons initially as 9 lone pairs on the oxygen atoms (3 pairs around each atom).
Although all 24 electrons are represented in the structure (two electrons for each of the three bonds and 18 for each of the nine lone pairs), the octet for the nitrogen atom is not satisfied. To satisfy the octet rule for the nitrogen atom, a double bond needs to be made between the nitrogen and one of the oxygen atoms. Because of the symmetry of the molecule, it does not matter which oxygen atoms is chosen. Because there are three different oxygen atoms that could form the double bond, there will be three different resonance structures showing each oxygen atom with a double bond to the nitrogen atom. Double-headed arrows will be placed between these three structures. (Do not forget your brackets and to put your charge on the outside of the brackets)

Example: What is the Lewis structure for the nitrite ion (NO₂⁻)?

Answer: Nitrogen is the least electronegative atom and should be the central atom.
After counting the valence electrons, we have a total of 17 [5 from nitrogen + 2(6 from each oxygen)] = 17. The charge of -1 indicates an extra electron, bringing the total electron count to 18.
Each oxygen will be bonded to the nitrogen, using two electrons. Place the remaining 16 electrons initially as nine lone pairs on the oxygen atoms (3 pairs around each atom) and the nitrogen (one pair).
Although all 18 electrons are represented in the structure (2 electrons for each of the two bonds and 14 for each of the seven lone pairs), the octet for the nitrogen atom is not satisfied. To satisfy the octet rule for the nitrogen atom, a double bond needs to be made between the nitrogen atom and one of the oxygen atoms. Because of the symmetry of the molecule, it does not matter which oxygen is chosen. Because there are two different oxygen atoms that could form the double bond, there will be two different resonance structures showing each oxygen atom with a double bond to the nitrogen atom. A double-headed arrow will be placed between these structures. (Do not forget your brackets and to put your charge on the outside of the brackets)

Lewis Structures for Electron-rich Compounds

Elements with atomic number greater than 13 often form compounds or polyatomic ions in which there are “extra” electrons. For these compounds we proceed as above. Once all of the octets are satisfied, the extra electrons are assigned to the central atom either as lone pairs or an increase in the number of bonds. (Never use multiple bonds with these compounds—you already have too many electrons.) Example: Draw the Lewis structure for phosphorus pentafluoride, PF₅.
Answer: The electronegativity of fluorine is greater than that of phosphorus—so the phosphorus atom is placed in the center of the molecule.
The total number of electros is 40 [5(7 from each fluorine) + 5 from the phosphorus] = 40. Using a single bond between the phosphorus atom and each of the fluorine atoms and filling the remaining electrons to satisfy the octet rule for the fluorine atoms accounts for all 40 electrons. Note that there are five bonds around the central atom.

Lewis Structures for Electron-poor Compounds

There is another type of molecule or polyatomic ion in which there is an electron deficiency of one or more electrons needed to satisfy the octets of all the atoms. In these cases, the more electronegative atoms are assigned as many electrons to complete those octets first and then the deficiency is assigned to the central atom. Example: Draw the Lewis structure for boron trifluoride, BF₃.
Answer: The electronegativity of fluorine is greater than that of boron—so the boron atom is placed in the center of the molecule.
The total number of electron is 24 [3(7 from each fluorine) + 3 from boron] = 24. Using a single bond between the boron and each of the fluorine atoms and filling the remaining electron as lone pairs around the fluorine atoms to satisfy the octets accounts for all 24 electrons.
The boron atom is two electrons shy of its octet. You may ask about the formation of a double bond (and even resonance). But, fluorine and boron are not in the list that can form double bonds (C, N, O, P, S) and so the compound is electron poor.

---

Try It Out!

Draw the Lewis structure for the following:

1. Hydronium ion (H₃O⁺)
2. Hypochlorite ion (ClO^-)
3. Carbonate ion (CO₃^-2)
4. Ammonia (NH₃)
5. Hydrogen fluoride (HF)
6. Ozone (O₃)
7. Xenon difluoride (XeF₂)

Lewis Symbols or Diagrams

Elemental properties and reactions are determined only by electrons in the outer energy levels. Electrons in completely filled energy levels are ignored when considering properties. Simplified Bohr diagrams which only consider electrons in outer energy levels are called Lewis Symbols. A Lewis Symbol consists of the element symbol surrounded by "dots" to represent the number of electrons in the outer energy level as represented by a Bohr Diagram. The number of electrons in the outer energy level is correlated by simply reading the Group number. Lewis symbols for oxygen, fluorine, and sodium are given in the diagram on the left. Lewis Symbols for the elements of the second period. Correlate the number of dots with the group number.

MOLECULE LEWIS DIAGRAMS A Lewis diagram depicts a mmolecule using an element symbol to represent the nucleus and core electrons of each atom. Valence electrons are represented by lines for electron pair bonds and dots for unbonded electrons. The following procedure can be followed to derive Lewis diagrams for most molecules. 1. Find the total number of electrons: Tabulate the total number of outer energy level electrons for all atoms in the molecule. For each atom, read the group number. 2. Draw a first tentative structure: The element with the least number of atoms is usually the central element. Draw a tentative molecular and electron arrangement attaching other atoms with single bonds as the first guess. Single bonds represented with a line represent 2 electrons 3. Add electrons as dots to get octets around atoms: When counting electrons for the octet around an atom, count both electrons in a bond for each atom and any lone pair electrons. Hydrogen, of course, gets only 2 electrons. 4. Count the total number of electrons in the final structure to see if the total agrees with the number tabulated in step #1. If not, then move a lone pair of electrons into a double bond. Or add more lone pairs of electrons. 5. Cycle through steps 3 and 4 several times until you get it right by trial and error.

Ionic radii and Radius ratio

(1) Ionic radii: X-ray diffraction or electron diffraction techniques provides the necessary information regarding unit cell. From the dimensions of the unit cell, it is possible to calculate ionic radii.

Let, cube of edge length 'a' having cations and anions say NaCl structure.

Then, rc + ra = a/2

where rc and ra are radius of cation and anion.

Radius of Cl– = √(a/2)2 + (a/2)2/2 = a/4

For bcc lattice say CsCl.rc + ra = √3a/2

(2) Radius ratio: Ionic compoundsoccur in crystalline forms. Ionic compounds are made of cations and anions. These ions are arranged in three dimensional array to form an aggregate of the type (A+B–)n. Since, the Coulombic forces are non-directional, hence the structures of such crystals are mainly governed by the ratio of the radius of cation (r+) to that of anion (r–). The ratio r+ to r– (r+/r–) is called as radius ratio.

Radius ratio = r+/r–

Solved example 1: If the radius ratio is in the range of then the coordination number will be

(a) 2 (b) 4

(c) 6 (d) 8

Ans: (d)

Solved example 2: If the radius ratio is in the range of then the coordination number will be Na2O

(a) 2 (b) 4

(c) 6 (d) 8

Ans: (c)

Solved example 3: What is the coordination number of sodium in

(a) 6 (b) 4

(c) 8 (d) 2

Ans: (b)

Solved example 4: The ratio of cationic radius to anionic radius in an ionic crystal is greater than 0.732. Its coordination number is

(a) 6 (b) 8

(c) 1 (d) 4

Ans: (b)
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I. ANALYTICAL CHEMISTRY — 15%
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II. INORGANIC CHEMISTRY — 25%
1. General Chemistry — Periodic trends, oxidation states, nuclear chemistry
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III. ORGANIC CHEMISTRY — 30%
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Nuclear Chemistry

An Introduction

Traditional chemical reactions occur as a result of the interaction between valence electrons around an atom's nucleus (see our Chemical Reactions module for more information). In 1896, Henri Becquerel expanded the field of chemistry to include nuclear changes when he discovered that uranium emitted radiation. Soon after Becquerel's discovery, Marie Sklodowska Curie began studying radioactivity and completed much of the pioneering work on nuclear changes. Curie found that radiation was proportional to the amount of radioactive element present, and she proposed that radiation was a property of atoms (as opposed to a chemical property of a compound). Marie Curie was the first woman to win a Nobel Prize and the first person to win two (the first, shared with her husband Pierre and Becquerel for discovering radioactivity; the second for discovering the radioactive elements radium and polonium).

Radiation and nuclear reactions

In 1902, Frederick Soddy proposed the theory that "radioactivity is the result of a natural change of an isotope of one element into an isotope of a different element." Nuclear reactions involve changes in particles in an atom's nucleus and thus cause a change in the atom itself. All elements heavier than bismuth (Bi) (and some lighter) exhibit natural radioactivity and thus can "decay" into lighter elements. Unlike normal chemical reactions that form molecules, nuclear reactions result in the transmutation of one element into a different isotope or a different element altogether (remember that the number of protons in an atom defines the element, so a change in protons results in a change in the atom). There are three common types of radiation and nuclear changes:

1. Alpha Radiation (α) is the emission of an alpha particle from an atom's nucleus. An α particle contains two protons and two neutrons (and is similar to a He nucleus: ). When an atom emits an a particle, the atom's atomic mass will decrease by four units (because two protons and two neutrons are lost) and the atomic number (z) will decrease by two units. The element is said to "transmute" into another element that is two z units smaller. An example of an a transmutation takes place when uranium decays into the element thorium (Th) by emitting an alpha particle, as depicted in the following equation:

238 92

U

4 2

He

+

234 90

Th

(Note: in nuclear chemistry, element symbols are traditionally preceded by their atomic weight (upper left) and atomic number (lower left).

2. Beta Radiation (â) is the transmutation of a neutron into a proton and an electron (followed by the emission of the electron from the atom's nucleus: ). When an atom emits a β particle, the atom's mass will not change (since there is no change in the total number of nuclear particles), however the atomic number will increase by one (because the neutron transmutated into an additional proton). An example of this is the decay of the isotope of carbon named carbon-14 into the element nitrogen:

14 6

C

0 -1

e

+

14 7

N

3. Gamma Radiation (ã) involves the emission of electromagnetic energy (similar to light energy) from an atom's nucleus. No particles are emitted during gamma radiation, and thus gamma radiation does not itself cause the transmutation of atoms, however γ radiation is often emitted during, and simultaneous to, α or β radioactive decay. X-rays, emitted during the beta decay of cobalt-60, are a common example of gamma radiation.

Half-life

Radioactive decay proceeds according to a principal called the half-life. The half-life (T_½) is the amount of time necessary for one-half of the radioactive material to decay. For example, the radioactive element bismuth (210Bi) can undergo alpha decay to form the element thallium (206Tl) with a reaction half-life equal to five days. If we begin an experiment starting with 100 g of bismuth in a sealed lead container, after five days we will have 50 g of bismuth and 50 g of thallium in the jar. After another five days (ten from the starting point), one-half of the remaining bismuth will decay and we will be left with 25 g of bismuth and 75 g of thallium in the jar. As illustrated, the reaction proceeds in halfs, with half of whatever is left of the radioactive element decaying every half-life period.

Radioactive Decay of Bismuth-210 (T½ = 5 days)

The fraction of parent material that remains after radioactive decay can be calculated using the equation:

Fraction remaining =

1  2n

(where n = # half-lives elapsed)

The amount of a radioactive material that remains after a given number of half-lives is therefore:

Amount remaining = Original amount * Fraction remaining

The decay reaction and T_½ of a substance are specific to the isotope of the element undergoing radioactive decay. For example, Bi210 can undergo a decay to Tl206 with a T_½ of five days. Bi215, by comparison, undergoes b decay to Po215 with a T_½ of 7.6 minutes, and Bi208 undergoes yet another mode of radioactive decay (called electron capture) with a T_½ of 368,000 years!

Stimulated nuclear reactions

While many elements undergo radioactive decay naturally, nuclear reactions can also be stimulated artificially. Although these reactions also occur naturally, we are most familiar with them as stimulated reactions. There are two such types of nuclear reactions:

1. Nuclear fission: reactions in which an atom's nucleus splits into smaller parts, releasing a large amount of energy in the process. Most commonly this is done by "firing" a neutron at the nucleus of an atom. The energy of the neutron "bullet" causes the target element to split into two (or more) elements that are lighter than the parent atom.

The Fission Reaction of Uranium-235

The Fission of U235

Concept simulation - Illustrates a nuclear fission reaction.

(Flash required)

During the fission of U235, three neutrons are released in addition to the two daughter atoms. If these released neutrons collide with nearby U235 nuclei, they can stimulate the fission of these atoms and start a self-sustaining nuclear chain reaction. This chain reaction is the basis of nuclear power. As uranium atoms continue to split, a significant amount of energy is released from the reaction. The heat released during this reaction is harvested and used to generate electrical energy.

Two Types of Nuclear Chain Reactions

Concept simulation - Reenacts controlled and uncontrolled nuclear chain reactions.

(Flash required)

2. Nuclear fusion: reactions in which two or more elements "fuse" together to form one larger element, releasing energy in the process. A good example is the fusion of two "heavy" isotopes of hydrogen (deuterium: H2 and tritium: H3) into the element helium.

Nuclear Fusion of Two Hydrogen Isotopes

Nuclear Fusion

Concept simulation - Reenacts the fusion of deuterium and tritium inside of a tokamak reactor.

(Flash required)

Fusion reactions release tremendous amounts of energy and are commonly referred to as thermonuclear reactions.  Although many people think of the sun as a large fireball, the sun (and all stars) are actually enormous fusion reactors.  Stars are primarily gigantic balls of hydrogen gas under tremendous pressure due to gravitational forces.  Hydrogen molecules are fused into helium and heavier elements inside of stars, releasing energy that we receive as light and heat.  

Oxidation States Ref

Oxidation States

Remember that elements can have various oxidation states depending on the number of electrons that they have compared to the number of protons that are in the nucleus.

	If the number of electrons is the same as the number of protons, then we say that the atom is neutral and that it has an oxidation state of 0.
	If the element has reacted in a way to have gained electrons, then it has a negative oxidation state, because the gained electrons have negative charges.
	If the element has reacted in a way that has caused it to lose electrons, then it has a positive oxidation state, because it has lost negative charges.

Guidelines

Here are some guidelines that you can use to figure oxidation states of chemicals.

If you have a pure element, the oxidation state is 0.	For example, K is a pure element in its 0 state.
If you have an isolated ion that consists of one atom that has either gained or lost electrons, the oxidation state is equal to the charge on that ion.	Na+ is in a +1 state. Chloride ion (Cl-) has a -1 state for the chlorine.
In a compound more figuring has to be done. Remember that all of the oxidation states have to add up to zero. First find the most electronegative element and give it the negative state that it wants. That usually works. Then figure out what the other element has to be to bring the total to zero.	In Cl2O the oxygen gets to be -2, then the chlorine must be +1 in order for the oxidation states of all the atoms to add up to zero.In Fe2O3 the oxygen gets to be -2, then the iron must be +3 in order for the oxidation states of all the atoms to add up to zero.
If there is more than one other element, start with the ones you can count on being +1 because they are in group IA or +2 because they are in group IIA. Then figure out the others.	In HNO3 the oxygen gets to be -2, and we can count on the hydrogen to be +1, then the nitrogen must be +5 in order for the oxidation states of all the atoms to add up to zero.
If you are dealing with a polyatomic ion by itself, the only change is that all of the oxidation states have to add up to the charge on the ion.	In PO43- the oxygen gets to be -2, then the phosphorus has to be +5 so that the oxidation states of all the atoms adds up to -3.
If you recognize a polyatomic ion in the formula, it is often helpful to note it and its charge because that can help you figure out the oxidation state of the element it is combined with. Then continue with the rest as before.	In FeSO4 the sulfate ion has a -2 charge, therefore the iron has to be +2 in this compound. The oxygen gets to be -2, then the sulfur has to be +6 in order for the oxidation states of all the atoms to add up to zero.CoCO3, because carbonate is -2 the cobalt has to be +2. The oxygen is -2 and the carbon is +4.

Practice

Now practice by determining the oxidation states of the elements in these chemicals (which are also listed in exercise 11 in your workbook). Check your answers below.

Au3+

N2

FeCl3

N2O3

NO3-

PCl5

OF2

H2SO4

HNO2

MnO4-

Answers

Au3+	N2	FeCl3	N2O3	NO3-	PCl5	OF2	H2SO4	HNO2	MnO4-
Au is +3	N is 0	Cl is -1 Fe is +3	O is -2 N is +3	O is -2 N is +5	Cl is -1 P is +5	F is -1 O is +2	O is -2 H is +1 S is +6	O is -2 H is +1 N is +3	O is -2 Mn is +7

OXIDATION STATES (OXIDATION NUMBERS)

This page explains what oxidation states (oxidation numbers) are and how to calculate them and make use of them. Oxidation states are straightforward to work out and to use, but it is quite difficult to define what they are in any quick way. Explaining what oxidation states (oxidation numbers) are Oxidation states simplify the whole process of working out what is being oxidised and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be helpful if you knew about: oxidation and reduction in terms of electron transfer electron-half-equations
	Note:  If you aren't sure about either of these things, you might want to look at the pages on redox definitions and electron-half-equations. It would probably be best to read on and come back to these links if you feel you need to.
We are going to look at some examples from vanadium chemistry. If you don't know anything about vanadium, it doesn't matter in the slightest. Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons: The vanadium is now said to be in an oxidation state of +2. Removal of another electron gives the V3+ ion: The vanadium now has an oxidation state of +3. Removal of another electron gives a more unusual looking ion, VO2+. The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one). The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element. It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5. Every time you oxidise the vanadium by removing another electron from it, its oxidation state increases by 1. Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero. What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur. The sulphur has an oxidation state of -2. Summary Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state. Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state Recognising this simple pattern is the single most important thing about the concept of oxidation states. If you know how the oxidation state of an element changes during a reaction, you can instantly tell whether it is being oxidised or reduced without having to work in terms of electron-half-equations and electron transfers. Working out oxidation states You don't work out oxidation states by counting the numbers of electrons transferred. It would take far too long. Instead you learn some simple rules, and do some very simple sums! The oxidation state of an uncombined element is zero. That's obviously so, because it hasn't been either oxidised or reduced yet! This applies whatever the structure of the element - whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon. The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion. The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second. Some elements almost always have the same oxidation states in their compounds: element usual oxidation state exceptions Group 1 metals always +1   Group 2 metals always +2   Oxygen usually -2 except in peroxides and F2O (see below) Hydrogen usually +1 except in metal hydrides where it is -1 (see below) Fluorine always -1   Chlorine usually -1 except in compounds with O or F (see below) The reasons for the exceptions Hydrogen in the metal hydrides Metal hydrides include compounds like sodium hydride, NaH. In this, the hydrogen is present as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion - in this case, -1. Alternatively, you can think of it that the sum of the oxidation states in a neutral compound is zero. Since Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0). Oxygen in peroxides Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound and so the sum of the oxidation states of the hydrogen and oxygen must be zero. Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it. Oxygen in F2O The problem here is that oxygen isn't the most electronegative element. The fluorine is more electronegative and has an oxidation state of -1. In this case, the oxygen has an oxidation state of +2. Chlorine in compounds with fluorine or oxygen There are so many different oxidation states that chlorine can have in these, that it is safer to simply remember that the chlorine doesn't have an oxidation state of -1 in them, and work out its actual oxidation state when you need it. You will find an example of this below. Warning! Don't get too bogged down in these exceptions. In most of the cases you will come across, they don't apply! Examples of working out oxidation states What is the oxidation state of chromium in Cr2+? That's easy! For a simple ion like this, the oxidation state is the charge on the ion - in other words: +2 (Don't forget the + sign.) What is the oxidation state of chromium in CrCl3? This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n: n + 3(-1) = 0 n = +3 (Again, don't forget the + sign!) What is the oxidation state of chromium in Cr(H2O)63+? This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia. The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3. What is the oxidation state of chromium in the dichromate ion, Cr2O72-? The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present. 2n + 7(-2) = -2 n = +6
	Warning:  Because these are simple sums it is tempting to try to do them in your head. If it matters (like in an exam) write them down using as many steps as you need so that there is no chance of making careless mistakes. Your examiners aren't going to be impressed by your mental arithmetic - all they want is the right answer! If you want some more examples to practice on, you will find them in most text books, including my chemistry calculations book.
What is the oxidation state of copper in CuSO4? Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulphur) whose oxidation states can both change. The only way around this is to know some simple chemistry! There are two ways you might approach it. (There might be others as well, but I can't think of them at the moment!) You might recognise this as an ionic compound containing copper ions and sulphate ions, SO42-. To make an electrically neutral compound, the copper must be present as a 2+ ion. The oxidation state is therefore +2. You might recognise the formula as being copper(II) sulphate. The "(II)" in the name tells you that the oxidation state is 2 (see below). You will know that it is +2 because you know that metals form positive ions, and the oxidation state will simply be the charge on the ion. Using oxidation states In naming compounds You will have come across names like iron(II) sulphate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions. This can also be extended to the negative ion. Iron(II) sulphate is FeSO4. There is also a compound FeSO3 with the old name of iron(II) sulphite. The modern names reflect the oxidation states of the sulphur in the two compounds. The sulphate ion is SO42-. The oxidation state of the sulphur is +6 (work it out!). The ion is more properly called the sulphate(VI) ion. The sulphite ion is SO32-. The oxidation state of the sulphur is +4 (work that out as well!). This ion is more properly called the sulphate(IV) ion. The ate ending simply shows that the sulphur is in a negative ion. So FeSO4 is properly called iron(II) sulphate(VI), and FeSO3 is iron(II) sulphate(IV). In fact, because of the easy confusion between these names, the old names sulphate and sulphite are normally still used in introductory chemistry courses.
	Note:  Even these aren't the full name! The oxygens in the negative ions should also be identified. FeSO4 is properly called iron(II) tetraoxosulphate(VI). It all gets a bit out of hand for everyday use for common ions.
Using oxidation states to identify what's been oxidised and what's been reduced This is easily the most common use of oxidation states. Remember: Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced. Example 1: This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas: Have the oxidation states of anything changed? Yes they have - you have two elements which are in compounds on one side of the equation and as uncombined elements on the other. Check all the oxidation states to be sure:. The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced. Example 2: The reaction between sodium hydroxide and hydrochloric acid is: Checking all the oxidation states: Nothing has changed. This isn't a redox reaction. Example 3: This is a sneaky one! The reaction between chlorine and cold dilute sodium hydroxide solution is: Obviously the chlorine has changed oxidation state because it has ended up in compounds starting from the original element. Checking all the oxidation states shows: The chlorine is the only thing to have changed oxidation state. Has it been oxidised or reduced? Yes! Both! One atom has been reduced because its oxidation state has fallen. The other has been oxidised. This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced. Using oxidation states to work out reacting proportions This is sometimes useful where you have to work out reacting proportions for use in titration reactions where you don't have enough information to work out the complete ionic equation. Remember that each time an oxidation state changes by one unit, one electron has been transferred. If one substance's oxidation state in a reaction falls by 2, that means that it has gained 2 electrons. Something else in the reaction must be losing those electrons. Any oxidation state fall by one substance must be accompanied by an equal oxidation state increase by something else. This example is based on information in an old AQA A' level question. Ions containing cerium in the +4 oxidation state are oxidising agents. (They are more complicated than just Ce4+.) They can oxidise ions containing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). In the process the cerium is reduced to the +3 oxidation state (Ce3+). What are the reacting proportions? The oxidation state of the molybdenum is increasing by 4. That means that the oxidation state of the cerium must fall by 4 to compensate. But the oxidation state of the cerium in each of its ions only falls from +4 to +3 - a fall of 1. So there must obviously be 4 cerium ions involved for each molybdenum ion. The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.