Oxidation States

Skills to develop

• Assign an oxidation state (or oxidation number) to an atom in a molecule or ion.
• Describe oxidation and reduction reaction in terms of oxidation state change.

Oxidation State

Oxidation state is a number assigned to an element in a compound according to some rules. This number enable us to describe oxidation-reduction reactions, and balancing redox chemical reactions. You are learning the skill to assign oxidation states (or oxidation numbers) to a variety of compounds and ions. When an oxidation number is assigned to the element, it does not imply that the element in the compound acquires this as a charge, but rather that it is a convenient number to use for balancing chemical reactions. The guidelines for assigning oxidation states (numbers) are given below:

1. The oxidation state of any element such as Fe, H₂, O₂, P₄, S₈ is zero (0).
2. The oxidation state of oxygen in its compounds is -2, except for peroxides like H₂O₂, and Na₂O₂, in which the oxidation state for O is -1.
3. The oxidation state of hydrogen is +1 in its compounds, except for metal hydrides, such as NaH, LiH, etc., in which the oxidation state for H is -1.
4. The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion.
5. The following elements usually have the same oxidation states in their compounds:
   • +1 for alkali metals - Li, Na, K, Rb, Cs;
   • +2 for alkaline earth metals - Be, Mg, Ca, Sr, Ba;
   • -1 for halogens except when they form compounds with oxygen or one another;

These rules are wordy because we have to point out the special cases such as H₂O₂ and Na₂O₂. Rule 3. deals with hydride. Other than these, you may simply remember the oxidation states for H and O are +1 and -2 respectively in a coumpound, and oxidation of other elements can be asigned by making the algebraic sum of the oxidation states equal to the net charge on the molecule or ion. For your practice, we provide some examples below. Please study the following examples and derive the oxidation state for all elements. The oxidation numbers of the key element are given in case you need help.

Element	Oxidation state	Compound or ion
H	+1	H+
Group 1	+1	H2O
	0	H2
	-1	NaAlH4
Cl	-1	Cl-
Group 7	0	Cl2
	+1	ClO-
	+3	ClO2-
	+4	ClO2
	+5	ClO3-
	+7	ClO4-
N	-3	NH3
Group 5	-2	N2H4
	-1	NH2OH
	0	N2
	+1	N2O
	+2	NO
	+3	NO2-
	+4	NO2
	+5	NO3-

The following are some common oxidants or reductants. Changes of oxidation states in redox reactions of the key elements are given in groups. Please justify the assigned oxidation state to your satisfaction as you read on, and assign the oxidation number to all element in the formulas.

Element	Oxidation state	Compound or ion
Fe	+2	Fe2+	Fe = Fe2+ + 2 e-
	+3	Fe3+	Fe2+ = Fe3++ e-
Zn	0	Zn	Zn is reducing agent
	+2	Zn2+
O	-1	H2O2	H2O2 = O2 + H2O
	0	O2
	-2	H2O
Cr	+6	Cr2O72-
	+3	Cr3+	Cr2O72- + 6 e = 2 Cr3+
	+6	CrO42-
Mn	+7	MnO4-
	+6	MnO42-
	+4	MnO2	MnO4- + 3 e = MnO2
	+2	Mn2+	MnO4- + 5 e = Mn2+
C	+3	H2C2O4
	+4	CO2
	+4	CO32-
	+2	CO

Oxidation and Reduction in Terms of Oxidation State

In the chemistry of battery, you have learned that oxidation is the loss of electron. A loss of one electron raises the oxidation state by one. We now have another definition for oxidatin and reduction.

A raise of oxidation state is an oxidation whereas a lower of oxidation state is a reduction.

For example, in the reaction (unbalanced),
MnO₄^- + H⁺ -> Mn²⁺ + H₂O the oxidation of Mn goes from +7 to +2. Thus, Mn is reduced. In the following reaction (unbalanced), H₂C₂O₄ + O₂ -> CO₂ + H₂O the element C is oxidized, because its oxidation state changes from +3 to +4 in the reaction.

Confidence Building Questions

• What is the oxidation state of Mo in MoO₄^2-?
  Skill:
  Note that (+6) + 4*(-2) = -2, the charge of the ion MoO₄^2-. An oxidation state is a signed number, in this case +6.
• What is the oxidation state of S in SO₃^2-?
  Discussion:
  Even though S is in group VIA: O, S, Se, Te, Bi, the sulfur in sulfite ions has an oxidation number +4.
• What is the oxidation state of As in AsO₄^3-?
  Discussion:
  This is the normal oxidation state of group VA: N, P, As, Sb, Bi.
• What is the oxidation state of Br in BrO₃^-?
  Discussion:
  Br has many oxidation states, e.g. +7 in BrO₄^-.
• What is the oxidation state of As in As₂S₃?
  Discussion:
  The oxidation for S is -2, since it is in the group VIA: O, S, Se, Te, Po. Super! S being in the same group as O, has an oxidation state of -2.
• What is the oxidation state of Cr in Cr(OH)₃?
  Discussion:
  There are three OH^- groups in the compound. Chromium hydroxide, Cr(OH)₃, is a greenish gelatinous solid.
• In the reaction, P₄(s) + 5O₂(g) = P₄O₁₀(s) What is element reduced?
  Discussion:
  What are the oxidation states of O in O₂ and P₄O₁₀? This equation represents the burning of phosphorus in air. P₄O₁₀ dissolves in water to form phosphoric acid.

 

Strand A: The Nature of Matter

 

Standard 2: The student understands the basic principles of atomic theory.

 

Benchmark SC.A.2.4.5: Knows that elements are arranged into groups and families based on similarities in electron structure and that their physical and chemical properties can be predicted.

 

Task Analysis: The student…

• explains the organization of the periodic table.
  
• interprets graphs that show the relationship between physical properties and electron structures of elements.
  
• compares and contrasts chemical properties of elements from different families.
  

 

Teacher Background

The Periodic Table was the outcome of several chemists working to make some sense out of the knowledge they were learning about the elements. John Newlands, Dmitri Mendeleev, and Henry Mosley all worked to give us the periodic table that we have today. John Newlands contribution to the periodic table was his "law of octaves" noted a pattern in the structures of atoms of elements with similar chemical properties. D. Mendeleev was a chemistry teacher who was eagerly trying to find a new way to show his students a better way to learn the properties of elements. In his attempts he wrote each element on a separate card and started to arrange these elemental cards based on similarities in physical and chemical properties and as a result ended up with a table that looked somewhat like our table today—except there were holes. According to Mendeleev, these holes represented elements that existed but were not yet found. The major factor Mendeleev used to arrange his elements was atomic mass (average mass of the atom). But there were problems where mass said elements should go one way and chemical properties said they should go another. He rationalized that the properties were more accurate than the masses since technology used to determine the mass was still improving. Later once technology improved and we learned that each atom has an integral positive charge (protons in the nucleus) in other words, each atom has a unique charge in its nucleus. Henry Mosley discovered that each element in Mendeleev's table was arranged in an order such that their integral positive charge increased numerically from left to right and top to bottom. Also, this new discovery fixed Mendeleev's problem spots. Now, the periodic table had a new basis of organization: atomic number (number of protons in the nucleus). This is basically still the organization except we have also added electron configuration (the placement of electrons into energy levels) to the organization. So now everything with the same valence shell electron configuration (outer shell electron configuration) is in the same family (column), where as properties are different from one period (row) to another.

Trends: the periodic table not only provides us with atomic numbers of elements and masses of elements, but it also provides us with a tool to determine general trends in chemical and physical properties.

 

 

 

 

 

 

 

 

 

• Atomic Radius—the atomic radius is defined to be the distance from the nucleus to the outer energy level where the outermost electrons are. The atomic radius decreases as you move from left to right across a period and increases as you move from top to bottom in a family. But why? For a period, you are in the same energy level which itself describes the size of the electron cloud (the area where the electrons are most probably located), so based on this elements should have the same radius in the same period. But as you move across the period, the total positive charge in the nucleus increases thereby increasing the total force attracting the electrons in the outer energy level thereby making the cloud contract. As you go down a family it is much easier to understand the trend. As you go down a family the energy level increases thereby increasing the radius of the cloud so the atomic radius increases.
  

 

• Ionization Energy—this is defined as the amount of energy needed to remove an electron from the outer energy level of an atom. Here the trend is to increase from left to right across a period and to decrease from top to bottom in a family. Why? As previously stated, the size decreases as you move from left to right as the total positive charge increases. By the inverse square law (F=kQ₁Q₂/r²) this means the force attracting the electron increases as size decreases and as the positive charge increases. As you move from top to bottom again using the inverse square law, you get farther from nucleus thus reducing the attractive force on the electron.
  

 

• Electron Affinity—this is defined as the amount of energy required to add an electron. For all but one column this value is either negative or zero. The 2^nd family is the only one that is consistently positive. Here the negative energy implies that energy is released rather than required when an electron is added. The trends here are exactly the same as for ionization energy. From left to right the energy is more negative—high affinity--and from top to bottom it is less negative—lower affinity. The reasoning is very similar to those for ionization energy.
  

 

 

• Electro negativity—this is defined as the attraction of one atom in a molecule for the shared electrons. The trend is exactly the same as for electron affinity and ionization energy. The reasoning is also the same. Fluorine is set at the highest at 4.0 and all other atoms are lower than this.
  

  
   

• Reactivity—this is defined as how readily an atom reacts. For metals this is based on ionization energy where the most reactive metals are those with the lowest ionization energies. For nonmetals this is based on electron affinity where the most negative electron affinity is the most reactive. This has to do with the fact that metals form positive ions (lose electrons) and nonmetals for negative ions (gains electrons).
  

  
   

Materials

• graph paper
  
• activity series for metals and nonmetals (found in any chemistry text or on the internet)
  

 

Description

Note: This activity is intended to be an activity used before the students are taught the trends. It is to be used to have the students recognize the repeating patterns in the graphs. After the completion of the activity, the explanations of the trends may be addressed depending on the level of the students.

 

For the following data, plot the atomic number on the x-axis and the property on the y-axis. Each property should be graphed on a separate sheet of graph paper.

 

Element	Atomic Number	Atomic Radius (pm)	1st Ionization Energy (kJ/mol)	Electron Affinity (kJ/mol)	Electro negativity (Pauling Units)
H	1	53	1312	-73	2.1
He	2	31	2372	0	NA
Li	3	167	520	-60	1.0
Be	4	112	899	240	1.5
B	5	87	801	-27	2.0
C	6	67	1086	-122	2.5
N	7	56	1402	0	3.0
O	8	48	1314	-141	3.5
F	9	42	1681	-328	4.0
Ne	10	38	2081	0	NA
Na	11	190	496	-53	0.9
Mg	12	145	738	230	1.2
Al	13	118	578	-44	1.5
Si	14	111	787	-134	1.8
P	15	98	1012	-72	2.1
S	16	88	1000	-200	2.5
Cl	17	79	1251	-349	3.0
Ar	18	71	1521	0	NA

Graphs are for teacher to check students 'work:(note on the Electro negativity graph, the 3 noble gases are omitted because values are not available)

 

Analysis

1. Based on your graphs, what is the trend in atomic radius across a period? down a family?
   

   Atomic radius decreases from left to right and increases from top to bottom.
   

   
    

2. Based on your graphs, what is the trend in ionization energy across a period? down a family?
   

   Ionization energy increases from left to right and decreases from top to bottom.
   

   
    

3. Based on your graphs, what is the trend in electron affinity across a period? down a family?
   

   Electron affinity increases (becomes more negative) as you move from left to right and decreases (becomes less negative) as you move from top to bottom).
   

   
    

4. Based on your graphs, what is the trend in electro negativity across a period? down a family?
   

   Electro negativity increases as you move from left to right and decreases as you move from top to bottom.
   

   
    

5. Using an activity series, what can you deduce about the relationship between ionization energy and reactivity of metals?
   

   The lower the ionization energy the more reactive the metal.
   

 

1. Using an activity series, what can you deduce about the relationship between electron affinity and reactivity of nonmetals?
   

   The more negative the electron affinity is for the nonmetal, the more reactive the nonmetal.
   

 

1. Explain in 3-4 paragraphs the organization and usefulness of the modern periodic table based on what you have learned in class and what you have learned from this activity. (See teacher notes).
   

 

Periodic Trends

Student Sheet

 

Description

For the following data, plot the atomic number on the x-axis and the property on the y-axis. Each property should be graphed on a separate sheet of graph paper.

 

Element	Atomic Number	Atomic Radius (pm)	1st Ionization Energy (kJ/mol)	Electron Affinity (kJ/mol)	Electro negativity (Pauling Units)
H	1	53	1312	-73	2.1
He	2	31	2372	0	NA
Li	3	167	520	-60	1.0
Be	4	112	899	240	1.5
B	5	87	801	-27	2.0
C	6	67	1086	-122	2.5
N	7	56	1402	0	3.0
O	8	48	1314	-141	3.5
F	9	42	1681	-328	4.0
Ne	10	38	2081	0	NA
Na	11	190	496	-53	0.9
Mg	12	145	738	230	1.2
Al	13	118	578	-44	1.5
Si	14	111	787	-134	1.8
P	15	98	1012	-72	2.1
S	16	88	1000	-200	2.5
Cl	17	79	1251	-349	3.0
Ar	18	71	1521	0	NA

 

Analysis

1. Based on your graphs, what is the trend in atomic radius across a period? down a family?
   

   
    

2. Based on your graphs, what is the trend in ionization energy across a period? down a family?
   

 

1. Based on your graphs, what is the trend in electron affinity across a period? down a family?
   

 

1. Based on your graphs, what is the trend in electro negativity across a period? down a family?
   

 

1. Using an activity series, what can you deduce about the relationship between ionization energy and reactivity of metals?
   

 

Using an activity series, what can you deduce about the relationship between electron affinity and reactivity of nonmetals? Explain in 3-4 paragraphs the organization and usefulness of the modern periodic table based on what you have learned in class and what you have learned from this activity.

Periodic Trends

A

Atomic Radii

1) As you move down a group, atomic radius increases.

WHY? - The number of energy levels increases as you move down a group as the number of electrons increases.  Each subsequent energy level is further from the nucleus than the last.  Therefore, the atomic radius increases as the group and energy levels increase.

2) As you move across a period, atomic radius decreases.

WHY? - As you go across a period, electrons are added to the same energy level.  At the same time, protons are being added to the nucleus.  The concentration of more protons in the nucleus creates a "higher effective nuclear charge."  In other words, there is a stronger force of attraction pulling the electrons closer to the nucleus resulting in a smaller atomic radius.

 

Ionic Radii

1) Anions (negative ions) are larger than their respective atoms.

WHY?

Electron-electron repulsion forces them to spread further apart.
Electrons outnumber protons; the protons cannot pull the extra electrons as tightly toward the nucleus.

2) Cations (positive ions) are smaller than their respective atoms.

WHY?

There is less electron-electron repulsion, so they can come closer together.
Protons outnumber electrons; the protons can pull the fewer electrons toward the nucleus more tightly.
If the electron that is lost is the only valence electron so that the electron configuration of the cation is like that of a noble gas, then an entire energy level is lost.  In this case, the radius of the cation is much smaller than its respective atom.

First Ionization Energy

Definition:  The energy required to remove the outermost (highest energy) electron from a neutral atom in its ground state.

1) As you move down a group, first ionization energy decreases.

WHY?

Electrons are further from the nucleus and thus easier to remove the outermost one.
"SHIELDING" - Inner electrons at lower energy levels essentially block the protons' force of attraction toward the nucleus.  It therefore becomes easier to remove the outer electron

2) As you move across a period, first ionization energy increases.

WHY? - As you move across a period, the atomic radius decreases, that is, the atom is smaller.  The outer electrons are closer to the nucleus and more strongly attracted to the center.  Therefore, it becomes more difficult to remove the outermost electron.

Exceptions to First Ionization Energy Trends

1) Xs2 > Xp1  e.g. 4Be > 5B WHY? - The energy of an electron in an Xp orbital is greater than the energy of an electron in its respective Xs orbital.  Therefore, it requires less energy to remove the first electron in a p orbital than it is to remove one from a filled s orbital.  2) Xp3 > Xp4  e.g.  7N > 8O WHY? - After the separate degenerate orbitals have been filled with single electrons, the fourth electron must be paired.  The electron-electron repulsion makes it easier to remove the outermost, paired electron. (See Hund's Rule)

Veiw a periodic table with first ionization energies.

Second and Higher Ionization Energies

Definition:  Second Ionization Energy is the energy required to remove a second outermost electron from a ground state atom.
Subsequent ionization energies increase greatly once an ion has reached the state like that of a noble gas.  In other words, it becomes extremely difficult to remove an electron from an atom once it loses enough electrons to lose an entire energy level so that its valence shell is filled.

Ionization Energies (kJ/mol)
Element Na Mg Al		1st IE 495.8 737.7 577.6		2nd IE 4562.4 1450.6 1816.6		3rd IE 6912 7732.6 2744.7		4th IE 9543 10,540 11,577

Electron Affinity

Definition:  The energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion.

1) As you move down a group, electron affinity decreases.

2) As you move across a period, electron affinity increases.

	Exceptions Among nonmetals, however, the elements in the first period have lower electron affinities than the elements below them in their respective groups.  Elements with electron configurations of Xs2, Xp3, and Xp6 have electron affinities less than zero because they are unusually stable.  In other words instead of energy being given off, these elements actually require an input of energy in order to gain electrons.  e.g.  Be, N, Ne  WHY? - Electron affinities are all much smaller than ionization energies.  Xs2 < 0:  Stable, diamagnetic atom with no unpaired electrons.  Xp3 < 0: Stable atom with 3 unpaired p-orbital electrons each occupying its own subshell.  Xp6 < 0: Stable atom with filled valence (outermost) shell.

Lattice Energy

Definition:  The energy given off when oppositely charged ions in the gas phase come together to form a solid.

The strength of a bond between ions of opposite charge can be calculated using Coulomb's Law.

Coulomb's Law - The force of attraction between oppositely charged particles is directly proportional to the product of the charges of the particles (q₁ and q₂) and inversely proportional to the square of the distance between the particles.

1) As you move down a group, lattice energy decreases.

WHY? - The atomic radius increases as you move down a group.  Since the square of the distance is inversely proportional to the force of attraction, lattice energy decreases as the atomic radius increases.

2) As you increase the magnitude of the charge (becomes more positive or more negative), lattice energy increases.

WHY? - The product of the charges of the particles is directly proportional to the force of attraction. Therefore, lattice energy increases as the charges increase.

Lattice Energies of Alkali Metals with Halides (kJ/mol) Li+ Na+ K+ Rb+ Cs+ F- 1036 923 821 785 740 Cl- 853 787 715 689 659 Br- 807 747 682 660 631 I- 757 704 649 630 604

Lattice Energies of Salts of OH- and O2- with Cations of varying charge (kJ/mol) Na+ Mg2+ Al3+ OH- 900 3006 5627 O2- 2481 3791 15916