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How to Assign Oxidation Numbers

Assigning Oxidation Numbers

The oxidation number is the positive or negative number of an atom that indicates the electrical charge the atom has if its compound consists of ions. In other words, the oxidation number gives the degree of oxidation (loss of electrons) or reduction (gain of electrons) of the atom in a compound. Because they track the number of electrons lost or gained, oxidation numbers are a sort of shorthand for balancing charge in chemical formulas.

This is a list of rules for assigning oxidation numbers, with examples showing the numbers for elements, compounds, and ions.

Rules for Assigning Oxidation Numbers

Various texts contain different numbers of rules and may change their order. Here is a list of oxidation number rules:

  • Write the cation first in a chemical formula, followed by the anion. The cation is the more electropositive atom or ion, while the anion is the more electronegative atom or ion. Some atoms may be either the cation or anion, depending on the other elements in the compound. For example, in HCl, the H is H + , but in NaH, the H is H – .
  • Write the oxidation number with the sign of the charge followed by its value. For example, write +1 and -3 rather than 1+ and 3-. The latter form is used to indicate oxidation state .
  • The oxidation number of a free element or neutral molecule is 0. For example, the oxidation number of C, Ne, O 3 , N 2 , and Cl 2 is 0.
  • The sum of all the oxidation numbers of the atoms in a neutral compound is 0. For example, in NaCl, the oxidation number of Na is +1, while the oxidation of Cl is -1. Added together, +1 + (-1) = 0.
  • The oxidation number of a monatomic ion is the charge of the ion. For example, the oxidation number of Na + is +1, the oxidation number of Cl – is -1, and the oxidation number of N 3- is -3.
  • The sum of the oxidation numbers of a polyatomic ion is the charge of the ion. For example, the sum of the oxidation numbers for SO 4 2-  is -2.
  • The oxidation number of a group 1 (alkali metal) element in a compound is +1.
  • The oxidation number of a group 2 (alkaline earth) element in a compound is +2.
  • The oxidation number of a group 7 (halogen) element in a compound is -1. The exception is when the halogen combines with an element with higher electronegativity (e.g., oxidation number of Cl is +1 in HOCl).
  • The oxidation number of hydrogen in a compound is usually +1. The exception is when hydrogen bonds with metals forming the hydride anion (e.g., LiH, CaH 2 ), giving hydrogen an oxidation number of -1.
  • The oxidation number of oxygen in a compound is usually -2. Exceptions include OF 2 and BaO 2 .

Examples of Assigning Oxidation Numbers

Example 1: Find the oxidation number of iron in Fe 2 O 3 .

The compound has no electrical charge, so the oxidation numbers of iron and oxygen balance each other out. From the rules, you know the oxidation number of oxygen is usually -2. So, find the iron charge that balances the oxygen charge. Remember, the total charge of each atom is its subscript multiplied by its oxidation number. O is -2 There are 3 O atoms in the compound so the total charge is 3 x -2 = -6 The net charge is zero (neutral), so: 2 Fe + 3(-2) = 0 2 Fe = 6 Fe = 3

Example 2: Find the oxidation number for Cl in NaClO3.

Usually, a halogen like Cl has an oxidation number of -1. But, if you assume Na (an alkali metal) has an oxidation number of +1 and O has an oxidation number of -2, the charges don’t balance out to give a neutral compound. It turns out all of the halogens, except for fluorine, have more than one oxidation number. Na = +1 O = -2 1 + Cl + 3(-2) = 0 1 + Cl -6 = 0 Cl -5 = 0 Cl = -5

  • IUPAC (1997) “Oxidation Number”. Compendium of Chemical Terminology (the “Gold Book”) (2nd ed.). Blackwell Scientific Publications. doi: 10.1351/goldbook
  • Karen, P.; McArdle, P.; Takats, J. (2016). “Comprehensive definition of oxidation state (IUPAC Recommendations 2016)”.  Pure Appl. Chem .  88  (8): 831–839. doi: 10.1515/pac-2015-1204
  • Whitten, K. W.; Galley, K. D.; Davis, R. E. (1992).  General Chemistry  (4th ed.). Saunders.

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AP®︎/College Chemistry

Course: ap®︎/college chemistry   >   unit 4.

  • Oxidation–reduction (redox) reactions

Worked example: Using oxidation numbers to identify oxidation and reduction

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Oxidation Number (Oxidation State)

What is oxidation number, oxidation number rules [1-6], how to find oxidation number.

An oxidation number is a number that is assigned to an atom to indicate its state of oxidation or reduction during a chemical reaction. Each atom in a redox reaction is assigned an oxidation number to understand its ability to donate, accept, or share electrons. It shows the total number of electrons that have been removed from or added to an element to get to its present state. For example, in Fe 2 O 3 , the oxidation number of Fe is +3, and in FeO, it is +2. A loss of electrons corresponds to an increase in oxidation number. On the other hand, a gain of electrons corresponds to a decrease in oxidation number [1-4] .

In order to assign oxidation numbers to atoms, we need to follow a set of rules.

1. The oxidation number of an element in its free state is zero.

Example : The oxidation number of Zn, Al, H 2 , O 2 , and Cl 2 is zero

2. The oxidation number of a monatomic ion is the same as the charge on the ion.

Example : The oxidation number of Na + is +1, Mg 2+ is +2, Al 3+ is +3, Cl -1 is -1, and O 2- is -2.

3. The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic ion is equal to the charge on the ion.

Example : In Fe 2 O 3 , the oxidation number of Fe is +3, and that of O is -2. The sum of all oxidation numbers is: +3 x 2 + (-2) x 3 = 0. The result is expected since Fe 2 O 3 is neutral.

4. The oxidation number of an alkali metal in a compound is +1, and the oxidation number of an alkaline earth metal in a compound is +2.

Example : In NaCl, Na is an alkali metal, and its oxidation number is +1. In MgO, Mg is an alkaline earth metal, and its oxidation number is +2.

5. The oxidation number of oxygen in a compound is usually -2. However, if the oxygen is in a category of compounds called peroxides, its oxidation number is -1. If the oxygen is bonded to fluorine, the number is +1 or +2, depending upon the compound.

Example : The oxidation number of O in H 2 O is -2, in H 2 O 2 is -1, in OF 2 is +2, and in O 2 F 2 is +1

6. The oxidation number of hydrogen in a compound is usually +1. In the case of a binary metal hydride, the oxidation number is -1.

Example : The oxidation number of H is +1 in H 2 O and -1 in NaH.

7. The oxidation number of fluorine is always -1.

Example : The oxidation number of F in NaF is -1.

8. Chlorine , bromine, and iodine usually have an oxidation number of -1 unless bonded to oxygen or fluorine.

Example : The oxidation number of Cl in NaCl is -1 and in ClO 2 is +4, and in FCl is +1.

to assign oxidation numbers rules

The oxidation number of an atom in an ion or compound can be determined using the above rules. Let us look at a few examples [1-6] .

1. Sulfuric Acid (H 2 SO 4 )

The oxidation number of hydrogen (H) and oxygen (O) are +1 and -2, respectively. Sulfuric acid is a neutral compound. Let x be the oxidation number of sulfur (S). Therefore,

(+1) x 2 + x + (-2) x 4 = 0

Or, 2 + x – 8 = 0

2. Nitric Acid (HNO 3 )

The oxidation numbers of hydrogen (H) and oxygen (O) are +1 and -2, respectively. Nitric acid is a neutral compound. Let x be the oxidation number of nitrogen (N). Therefore,

+1 + x + (-2) x 3 = 0

Or, +1 + x – 6 = 0

3. Potassium Permanganate (KMnO 4 )

The oxidation numbers of potassium (K) is +1 and oxygen (O) is -2. KMnO 4 is a neutral compound. Let x be the oxidation number of magnesium (Mn). Therefore,

+1 + x + (-2) x 4 = 0

Or, +1 + x – 8 = 0

4. Dichromate Ion (Cr 2 O 7 2- )

Dichromate is a complex ion. The oxidation number of oxygen (O) is -2. The charge of Cr 2 O 7 2- is -2. Let x be the oxidation number of chromium (Cr).

2x + (-2) x 7 = -2

Or, 2x -14 = -2

5. Carbonate (CO 3 2- )

The oxidation number of oxygen (O) is -2 and the charge on CO 3 2- is -2. Let x be the oxidation number of carbon (O). Therefore,

x + (-2) x 3 = -2

Or, x – 6 = -2

6.  Phosphite (PO 3 3- )

The oxidation number of oxygen (O) is -2 and the charge of PO 3 3- is -3. Let x be the oxidation number of phosphorous (P). Therefore,

x + (-2) x 3 = -3

Or, x – 6 = -3

7. Potassium Perchlorate (KClO 4 )

The oxidation numbers of potassium (K) and oxygen (O) are +1 and -2, respectively. Let x be the oxidation number of chlorine (Cl). KClO 4 is a neutral compound. Therefore,

Or, 1 + x – 8 = 0

8. Potassium Nitrate (KNO 3 )

The oxidation number of potassium (K) and oxygen (O) are +1 and -2, respectively. KNO 3 is a neutral compound. Let x be the oxidation number of nitrogen (N). Therefore,

Or, 1 + x – 6 = 0

The following image shows a chart consisting of the oxidation numbers of the periodic table elements [7].

to assign oxidation numbers rules

Ans. The difference between valency and oxidation number is that valency is the maximum number of electrons an atom can donate, accept, or share to become stable. In contrast, the oxidation number is the number of electrons an atom can donate or accept to form a bond with another atom.

Ans. The d-block or translational elements have incomplete d- and s-subshells. The valence electrons are present in both these subshells. It is for this reason that they can form variable oxidation states.

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Write the rules for assigning oxidation number .

(1) all atoms in the elemental or molecular state have 'zero'oxidation state. (2)elements of group $$ i $$ and $$ ii $$ in the periodic table always have $$ '+1'$$ and $$ '+2' $$ oxidation states respectively. (3). hydrogen $$ '+1'$$ oxidation state in all its compounds expect metal hydrides where it is $$ '-1'$$ (4) oxygen has been assigned an oxidation number of $$ '-2' $$ in all its compounds except peroxides and oxygen fluoride. in peroxide it is $$ '-1'$$and in oxygen fluoride it is $$ '+2'$$. (5) halogens generally have $$ '-1'$$ oxidation state. except for fluorine, other halogens may have positive oxidation states in their oxides and inter halogen compounds (flurione always has $$ '-1'$$ state). (6) the algebraic sum of oxidation numbers of all the elements in a compound is zero and that in ion is equal to the net charge on the ion..

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3.4: Oxidation States

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Learning Objectives

  • To identify oxidation–reduction reactions in solution.

The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O 2− ). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction.

Any oxidation must ALWAYS be accompanied by a reduction and vice versa.

Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). The reaction is as follows:

\[ Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g) \label{3.4.1}\]

Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O 2 or H 2 , are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is

\[ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s) \label{3.4.2} \]

Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al 2 O 3 , electrons are transferred as follows (the superscript 0 emphasizes the oxidation state of the elements):

\[ 4 Al^0 + 3 O_2^0 \rightarrow 4 Al^{3+} + 6 O^{2-} \label{3.4.3}\]

Equation 3.4.1 and Equation 3.4.2 are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation 3.4.3 , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:

\[ electrons \, lost = 4 \, Al \, atoms \times {3 \, e^- \, lost \over Al \, atom } = 12 \, e^- \, lost \label{3.4.4a}\]

\[ electrons \, gained = 6 \, O \, atoms \times {2 \, e^- \, gained \over O \, atom} = 12 \, e^- \, gained \label{3.4.4a}\]

The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained.

An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in Figure \(\PageIndex{1}\) .

In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.

Assigning Oxidation States

Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previosuly, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure \(\PageIndex{1}\) ), magnesium oxide (MgO), and calcium chloride (CaCl 2 ). In covalent compounds, in contrast, atoms share electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.

bc779aea262f6991d515f6f8928bd425.jpg

A set of rules for assigning oxidation states to atoms in chemical compounds follows.

Rules for Assigning Oxidation States

  • The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
  • The oxidation state of a monatomic ion is the same as its charge—for example, Na + = +1, Cl − = −1.
  • The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
  • Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
  • Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
  • The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.

Nonintegral oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states .

In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.

Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6 . Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H − ion, whereas HCl forms H + and Cl − ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF 2 but −½ in KO 2 . Note that an oxidation state of −½ for O in KO 2 is perfectly acceptable.

The reduction of copper(I) oxide shown in Equation 3.4.5 demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H 2 and Cu. From rule 4, hydrogen in H 2 O has an oxidation state of +1, and from rule 5, oxygen in both Cu 2 O and H 2 O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu 2 O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:

\[ \overset {+1}{Cu_2} \underset {-2}{O} (s) + \overset {0}{H_2} (g) \rightarrow 2 \overset {0}{Cu} (s) + \overset {+1}{H_2} \underset {-2}{O} (g) \label{3.4.5} \]

Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:

\[ electrons \, lost = 2 \, H \, atoms \times {1 \, e^- \, lost \over H \, atom } = 2 \, e^- \, lost \label{3.4.6a}\]

\[ electrons \, gained = 2 \, Cu \, atoms \times {1 \, e^- \, gained \over Cu \, atom} = 2 \, e^- \, gained \label{3.4.6b}\]

Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.

Example \(\PageIndex{1}\)

Assign oxidation states to all atoms in each compound.

  • sulfur hexafluoride (SF 6 )
  • methanol (CH 3 OH)
  • ammonium sulfate [(NH 4 )2SO 4 ]
  • magnetite (Fe 3 O 4 )
  • ethanoic (acetic) acid (CH 3 CO 2 H)

Given : molecular or empirical formula

Asked for : oxidation states

Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.

a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF 6 ), the oxidation state of sulfur must be +6:

[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0

b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:

[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0

c. Note that (NH 4 ) 2 SO 4 is an ionic compound that consists of both a polyatomic cation (NH 4 + ) and a polyatomic anion (SO 4 2− ) (see Table 2.4 ). We assign oxidation states to the atoms in each polyatomic ion separately. For NH 4 + , hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:

[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH 4 + ion

For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:

[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion

d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:

Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe 3 O 4 can be viewed as having two Fe 3 + ions and one Fe 2 + ion per formula unit, giving a net positive charge of +8 per formula unit. Fe 3 O 4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”

e. Initially, we assign oxidation states to the components of CH 3 CO 2 H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of

[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0

So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH 3 ) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO 2 H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of

[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3

To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus

\[ \underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \]

Thus the sum of the oxidation states of the two carbon atoms is indeed zero.

Exercise \(\PageIndex{1}\)

  • barium fluoride (BaF 2 )
  • formaldehyde (CH 2 O)
  • potassium dichromate (K 2 Cr 2 O 7 )
  • cesium oxide (CsO 2 )
  • ethanol (CH 3 CH 2 OH)
  • Ba, +2; F, −1
  • C, 0; H, +1; O, −2
  • K, +1; Cr, +6; O, −2
  • Cs, +1; O, −½
  • C, −3; H, +1; C, −1; H, +1; O, −2; H, +1

Redox Reactions of Solid Metals in Aqueous Solution

A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure \(\PageIndex{1}\)). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl − ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:

\(Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \label{3.4.5}\)

In subsequent steps, FeCl 2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH) 3 .

a1821e16fd102a37cb6a5029132c9516.jpg

Many metals dissolve through reactions of this type, which have the general form

\[metal + acid \rightarrow salt + hydrogen \label{3.4.6}\]

Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:

\[ Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \label{3.4.7}\]

Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!

Single-Displacement Reactions

Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 3.4.8) and the reduction of silver salts by copper (Equation 3.4.9 and Figure \(\PageIndex{2}\)):

\[ Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \label{3.4.8}\]

\[ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \label{3.4.9}\]

The reaction in Equation 3.4.8 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.

Precipitation_of_Silver_on_Copper.jpg

The Activity Series

By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn 2 + . Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:

\[ Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \label{3.4.10}\]

\[ Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \label{3.4.11}\]

Magnesium has a greater tendency to be oxidized than zinc does.

Pairwise reactions of this sort are the basis of the activity series (Figure \(\PageIndex{4}\)), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1) , the alkaline earth metals (group 2) , and Al (group 13) . In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).

bc6148e2d176471fe6f7f1b70fe6ebb6.jpg

When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series . Because magnesium is above zinc in Figure \(\PageIndex{4}\), magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H 2 . Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example \(\PageIndex{2}\) demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.

Example \(\PageIndex{2}\)

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

  • A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
  • A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
  • Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.

Given: reactants

Asked for: overall reaction and net ionic equation

  • Locate the reactants in the activity series in Figure .3.4.4 and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
  • Write the net ionic equation for the redox reaction.

\[ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s) \]

Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.

  • A Mercury lies below lead in the activity series, so no reaction will occur.

\[ Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) \]

Lead(II) sulfate is the white solid that forms on corroded battery terminals.

19ed64d0a413643429e781853dacedcc.jpg

Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.

Exercise \(\PageIndex{2}\)

  • A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
  • A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
  • A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).
  • \(no\: reaction\)
  • \(3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)\)
  • \(2Al(s) + 6CH_3CO_2H(aq) \rightarrow 2Al^{3+}(aq) + 6CH_3CO_2^-(aq) + 3H_2(g)\)
  • Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation.

In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table \(\PageIndex{1}\)), in which the overall reaction is separated into an oxidation equation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (Figure \(\PageIndex{3}\)), which arranges metals and H 2 in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series.

to assign oxidation numbers rules

Create a form in Word that users can complete or print

In Word, you can create a form that others can fill out and save or print.  To do this, you will start with baseline content in a document, potentially via a form template.  Then you can add content controls for elements such as check boxes, text boxes, date pickers, and drop-down lists. Optionally, these content controls can be linked to database information.  Following are the recommended action steps in sequence.  

Show the Developer tab

In Word, be sure you have the Developer tab displayed in the ribbon.  (See how here:  Show the developer tab .)

Open a template or a blank document on which to base the form

You can start with a template or just start from scratch with a blank document.

Start with a form template

Go to File > New .

In the  Search for online templates  field, type  Forms or the kind of form you want. Then press Enter .

In the displayed results, right-click any item, then select  Create. 

Start with a blank document 

Select Blank document .

Add content to the form

Go to the  Developer  tab Controls section where you can choose controls to add to your document or form. Hover over any icon therein to see what control type it represents. The various control types are described below. You can set properties on a control once it has been inserted.

To delete a content control, right-click it, then select Remove content control  in the pop-up menu. 

Note:  You can print a form that was created via content controls. However, the boxes around the content controls will not print.

Insert a text control

The rich text content control enables users to format text (e.g., bold, italic) and type multiple paragraphs. To limit these capabilities, use the plain text content control . 

Click or tap where you want to insert the control.

Rich text control button

To learn about setting specific properties on these controls, see Set or change properties for content controls .

Insert a picture control

A picture control is most often used for templates, but you can also add a picture control to a form.

Picture control button

Insert a building block control

Use a building block control  when you want users to choose a specific block of text. These are helpful when you need to add different boilerplate text depending on the document's specific purpose. You can create rich text content controls for each version of the boilerplate text, and then use a building block control as the container for the rich text content controls.

building block gallery control

Select Developer and content controls for the building block.

Developer tab showing content controls

Insert a combo box or a drop-down list

In a combo box, users can select from a list of choices that you provide or they can type in their own information. In a drop-down list, users can only select from the list of choices.

combo box button

Select the content control, and then select Properties .

To create a list of choices, select Add under Drop-Down List Properties .

Type a choice in Display Name , such as Yes , No , or Maybe .

Repeat this step until all of the choices are in the drop-down list.

Fill in any other properties that you want.

Note:  If you select the Contents cannot be edited check box, users won’t be able to click a choice.

Insert a date picker

Click or tap where you want to insert the date picker control.

Date picker button

Insert a check box

Click or tap where you want to insert the check box control.

Check box button

Use the legacy form controls

Legacy form controls are for compatibility with older versions of Word and consist of legacy form and Active X controls.

Click or tap where you want to insert a legacy control.

Legacy control button

Select the Legacy Form control or Active X Control that you want to include.

Set or change properties for content controls

Each content control has properties that you can set or change. For example, the Date Picker control offers options for the format you want to use to display the date.

Select the content control that you want to change.

Go to Developer > Properties .

Controls Properties  button

Change the properties that you want.

Add protection to a form

If you want to limit how much others can edit or format a form, use the Restrict Editing command:

Open the form that you want to lock or protect.

Select Developer > Restrict Editing .

Restrict editing button

After selecting restrictions, select Yes, Start Enforcing Protection .

Restrict editing panel

Advanced Tip:

If you want to protect only parts of the document, separate the document into sections and only protect the sections you want.

To do this, choose Select Sections in the Restrict Editing panel. For more info on sections, see Insert a section break .

Sections selector on Resrict sections panel

If the developer tab isn't displayed in the ribbon, see Show the Developer tab .

Open a template or use a blank document

To create a form in Word that others can fill out, start with a template or document and add content controls. Content controls include things like check boxes, text boxes, and drop-down lists. If you’re familiar with databases, these content controls can even be linked to data.

Go to File > New from Template .

New from template option

In Search, type form .

Double-click the template you want to use.

Select File > Save As , and pick a location to save the form.

In Save As , type a file name and then select Save .

Start with a blank document

Go to File > New Document .

New document option

Go to File > Save As .

Go to Developer , and then choose the controls that you want to add to the document or form. To remove a content control, select the control and press Delete. You can set Options on controls once inserted. From Options, you can add entry and exit macros to run when users interact with the controls, as well as list items for combo boxes, .

Adding content controls to your form

In the document, click or tap where you want to add a content control.

On Developer , select Text Box , Check Box , or Combo Box .

Developer tab with content controls

To set specific properties for the control, select Options , and set .

Repeat steps 1 through 3 for each control that you want to add.

Set options

Options let you set common settings, as well as control specific settings. Select a control and then select Options to set up or make changes.

Set common properties.

Select Macro to Run on lets you choose a recorded or custom macro to run on Entry or Exit from the field.

Bookmark Set a unique name or bookmark for each control.

Calculate on exit This forces Word to run or refresh any calculations, such as total price when the user exits the field.

Add Help Text Give hints or instructions for each field.

OK Saves settings and exits the panel.

Cancel Forgets changes and exits the panel.

Set specific properties for a Text box

Type Select form Regular text, Number, Date, Current Date, Current Time, or Calculation.

Default text sets optional instructional text that's displayed in the text box before the user types in the field. Set Text box enabled to allow the user to enter text into the field.

Maximum length sets the length of text that a user can enter. The default is Unlimited .

Text format can set whether text automatically formats to Uppercase , Lowercase , First capital, or Title case .

Text box enabled Lets the user enter text into a field. If there is default text, user text replaces it.

Set specific properties for a Check box .

Default Value Choose between Not checked or checked as default.

Checkbox size Set a size Exactly or Auto to change size as needed.

Check box enabled Lets the user check or clear the text box.

Set specific properties for a Combo box

Drop-down item Type in strings for the list box items. Press + or Enter to add an item to the list.

Items in drop-down list Shows your current list. Select an item and use the up or down arrows to change the order, Press - to remove a selected item.

Drop-down enabled Lets the user open the combo box and make selections.

Protect the form

Go to Developer > Protect Form .

Protect form button on the Developer tab

Note:  To unprotect the form and continue editing, select Protect Form again.

Save and close the form.

Test the form (optional)

If you want, you can test the form before you distribute it.

Protect the form.

Reopen the form, fill it out as the user would, and then save a copy.

Creating fillable forms isn’t available in Word for the web.

You can create the form with the desktop version of Word with the instructions in Create a fillable form .

When you save the document and reopen it in Word for the web, you’ll see the changes you made.

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IMAGES

  1. Assigning oxidation numbers

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  2. Rules to assign oxidation numbers

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  3. How to Assign Oxidation Numbers

    to assign oxidation numbers rules

  4. PPT

    to assign oxidation numbers rules

  5. Assigning oxidation numbers

    to assign oxidation numbers rules

  6. PPT

    to assign oxidation numbers rules

VIDEO

  1. \" Oxidation number\")

  2. Oxidation Number| Rules Assigning Oxidation no Class 11 |1st year Chemistry Lecture Smart Syllabus

  3. Rules for Assigning Oxidation Numbers

  4. Oxidation Number //Oxidation State//Rules To Assign Oxidation Number

  5. Oxidation state

  6. Determining Oxidation Number 3D Animation

COMMENTS

  1. Rules for Assigning Oxidation Numbers

    Updated on August 16, 2019 Electrochemical reactions involve the transfer of electrons. Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom.

  2. Rules for Assigning Oxidation Numbers to Elements

    Rule 1: The oxidation number of an element in its free (uncombined) state is zero — for example, Al (s) or Zn (s). This is also true for elements found in nature as diatomic (two-atom) elements and for sulfur, found as: Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion, for example:

  3. 22.6: Assigning Oxidation Numbers

    Assigning Oxidation Numbers. The oxidation number is a positive or negative number that is assigned to an atom to indicate its degree of oxidation or reduction. In oxidation-reduction processes, the driving force for chemical change is in the exchange of electrons between chemical species. ... Six rules for determining oxidation numbers are ...

  4. Oxidation States (Oxidation Numbers)

    Rules to determine oxidation states. The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl 2, S 8, and large structures of carbon or silicon each have an oxidation state of zero.; The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

  5. How to Assign Oxidation Numbers

    Here is a list of oxidation number rules: Write the cation first in a chemical formula, followed by the anion. The cation is the more electropositive atom or ion, while the anion is the more electronegative atom or ion. Some atoms may be either the cation or anion, depending on the other elements in the compound.

  6. 4.3: Oxidation Numbers and Redox Reactions

    Rules for Assigning Oxidation Numbers. The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl 2, S 8, and large structures of carbon or silicon each have an oxidation state of zero.(Since atoms of the same element always form pure covalent bonds, they share electrons equally, neither losing nor gaining, e.g., Cl 2.)

  7. Using oxidation numbers to identify oxidation and reduction (worked

    By assigning oxidation numbers to the atoms of each element in a redox equation, we can determine which element is oxidized and which element is reduced during the reaction. In this video, we'll use this method to identify the oxidized and reduced elements in the reaction that occurs between I⁻ and MnO₄⁻ in basic solution. Created by Sal ...

  8. Assigning Oxidation Numbers ( Read )

    Practice All Modalities Assigning Oxidation Numbers [Figure1] Iron what? Once we move from the element iron to iron compounds, we need to be able to designate clearly the form of the iron ion. An example of this is iron that has been oxidized to form iron oxide during the process of rusting.

  9. Oxidation Number (State): Definition, Rules, How to Find, and Examples

    A loss of electrons corresponds to an increase in oxidation number. On the other hand, a gain of electrons corresponds to a decrease in oxidation number [1-4]. Oxidation Number Rules [1-6] In order to assign oxidation numbers to atoms, we need to follow a set of rules. 1. The oxidation number of an element in its free state is zero.

  10. Assigning Oxidation Numbers

    Assigning Oxidation Numbers - Chemistry Tutorial TheChemistrySolution 61.2K subscribers Subscribe Subscribed 385K views 12 years ago This chemistry tutorial discusses how to assign oxidation...

  11. Oxidation Numbers

    Updated: 11/21/2023 Table of Contents Oxidation Numbers Rules for Assigning Oxidation Numbers Examples of Assigning Oxidation Numbers Lesson Summary Frequently Asked Questions How...

  12. Oxidation Number Meaning, Rules & Examples

    Table of Contents What is an Oxidation Number? Assigning Oxidation Numbers Using Rules Examples of How to Calculate the Oxidation Number Lesson Summary Frequently Asked Questions...

  13. 19.1 How to Assign Oxidation Numbers

    Chad begins a chapter on Electrochemistry with a lesson on How to Assign Oxidation Numbers (i.e. Oxidation States). Six rules for determining oxidation numb...

  14. 4.7: Oxidation-Reduction Reactions

    The rules for assigning oxidation numbers to atoms are as follows: Atoms in their elemental state are assigned an oxidation number of 0. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from ...

  15. PDF RULES FOR ASSIGNING OXIDATION NUMBERS

    RULES FOR ASSIGNING OXIDATION NUMBERS http://www.csun.edu/~hcchm001/IntroChemHandouts.html All elements in their free state (uncombined with other elements) have an oxidation number of zero (for example, Na, Cu, Mg, H , O 2 , Cl 2 , N 2 ). 2. H is +1, except in metal hydrides, where it is -1 (for example, NaH, CaH ) 2 Na : H

  16. PDF Oxidation Numbers: Rules

    7) The oxidation number of Group 1A elements is always +1 and the oxidation number of Group 2A elements is always +2. 8) The oxidation number of oxygen in most compounds is -2. 9) Oxidation numbers for other elements are usually determined by the number of electrons they need to gain or lose to attain the electron configuration of a noble gas.

  17. How to Find Oxidation Numbers (Rules and Examples)

    Using a list of simple rules you'll learn how to find the oxidation numbers for elements and compounds. For each rule there are examples and practice calcul...

  18. Oxidation Number Notes:Definition,Rules, Redox Reaction &Examples

    Home Chemistry Oxidation Numbers Rules to Identify and Assign Oxidation Numbers to Atoms, Molecules and Ions Last updated on May 15, 2023 Download as PDF Overview Test Series The oxidation number of an atom is the charge that an atom would have if the compound was composed of ions.

  19. 11.1: Oxidation Numbers

    The sum of all oxidation numbers in the sulfate ion would be \(1 \left( +6 \right) + 4 \left( -2 \right) = -2\), which is the charge of the ion. An examination of the rules for assigning oxidation numbers reveals that there are many elements for which there are no specific rules, such as nitrogen, sulfur, and chlorine.

  20. Rules for Assigning Oxidation Number

    The sum of the oxidation numbers in an electrically neutral substance is zero; For example, in hematite (Fe2O3), the oxidation number of the two iron atoms (+6 in total) balances the oxidation number of the three oxygen atoms (6) in an electronically neutral substance.

  21. How to Calculate Oxidation Numbers Introduction

    We'll learn how to determine the oxidation numbers or oxidation states for a the elements in a chemical compound. The oxidation numbers tell us how electrons...

  22. Write the rules assigning oxidation number

    Question Write the rules for assigning oxidation number . Solution Verified by Toppr (1) All atoms in the elemental or molecular state have 'zero'oxidation state. (2)Elements of group I and I I in the periodic table always have ′ + 1 ′ and ′ + 2 ′ oxidation states respectively. (3).

  23. 3.4: Oxidation States

    Rules for Assigning Oxidation States. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero. The oxidation state of a monatomic ion is the same as its charge—for example, Na + = +1, Cl − = −1. The oxidation state of fluorine in chemical compounds is always −1.

  24. Create a form in Word that users can complete or print

    Show the Developer tab. If the developer tab isn't displayed in the ribbon, see Show the Developer tab.. Open a template or use a blank document. To create a form in Word that others can fill out, start with a template or document and add content controls.