oxidation number assignment rules

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How to Assign 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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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.

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].

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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Oxidation Numbers

Key Questions

The valence electrons determine how many electrons an atom is willing to give up or how many spaces need to be filled in order to satisfy the rule of octet.

Lithium (Li), Sodium (Na) and Potassium (K) all have an electron configuration that ends as #s^1# . Each of these atoms would readily release this electron to have a filled valence shell and become stable as #Li^+1# , #Na^+1# and #K^+1# . Each element having an oxidation state of +1.

Oxygen (O) and Sulfur (S) all have an electron configuration that ends as #s^2 p^4# . Each of these atoms would readily take on two electrons to have a filled valence shell and become stable as #O^-2# , and #S^-2# . Each element having an oxidation state of -2.

There are exceptions to the rules and the transition metals usually have more than one oxidation state.

I hope this was helpful. SMARTERTEACHER

And so we can interpret a given chemical reaction with the use of oxidation numbers. Consider the oxidation of ammonia to give nitrate ion..........in terms of formal oxidation state this is the transition, #stackrel(-III)N# to #stackrel(+V)N# , an 8 electron oxidation, which we formally represent in the equation.......

#NH_3(aq) +3H_2O rarr NO_3^(-) +9H^(+) + 8e^(-)# #(i)#

Is this balanced with respect to mass and charge? It must be if we purport to represent physical reality.

And, inevitably, something must be reduced to effect the oxidation; let's say it is oxygen.

#stackrel(0)O_2+4e^(-) rarr2O^(2-)# #(ii)#

We add the individual redox equations together in a way to eliminate the electrons, which are particles of convenience......And so we take #(i) + 2xx(ii)# :

#NH_3(aq) +3H_2O +2O_2+8e^(-)rarr NO_3^(-) +underbrace(9H^(+) +4O^(2-))_(4H_2O+H^+) + 8e^(-)#

And so we cancel out what we can.....

#NH_3 +cancel(3H_2O) +2O_2+cancel(8e^(-))rarr NO_3^(-) +underbrace(9H^(+) +4O^(2-))_(cancel(4)H_2O+H^+) + cancel(8e^(-))#

....to give finally.........

#NH_3 +2O_2rarr NO_3^(-) +H_2O+H^+#

....or........

#stackrel(-III)NH_3 +2stackrel(0)O_2rarr Hstackrel(+V)NO_3 +H_2stackrel(-II)O#

I acknowledge that is a lot of pfaff.......but your question was rather open-ended.

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Assigning oxidation numbers
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Oxidation Number| Rules Assigning Oxidation no Class 11 |1st year Chemistry Lecture Smart Syllabus
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