assign oxidation number to each atom in hocl

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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 States (Oxidation Numbers)

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Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be useful to review and be familiar with the following concepts:

• oxidation and reduction in terms of electron transfer
• electron-half-equations

To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., \(\ce{V^{2+}}\) and \(\ce{V^{3+}}\)). The 2+ ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:

\[ \ce{V \rightarrow V^{2+} + 2e^{-}} \label{1}\]

The vanadium in the \( \ce{V^{2+}}\) ion has an oxidation state of +2. Removal of another electron gives the \(\ce{V^{3+}}\) ion:

\[ \ce{V^{2+} \rightarrow V^{3+} + e^{-}} \label{2}\]

The vanadium in the \(\ce{V^{3+} }\) ion has an oxidation state of +3. Removal of another electron forms the ion \(\ce{VO2+}\):

\[ \ce{V^{3+} + H_2O \rightarrow VO^{2+} + 2H^{+} + e^{-}} \label{3}\]

The vanadium in the \(\ce{VO^{2+}}\) is now in an oxidation state of +4.

Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations \ref{1} and \ref{2}), but not for the ion in Equation \ref{3}).

The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth electron to form another the \(\ce{VO_2^{+}}\) ion with the vanadium in a +5 oxidation state.

\[ \ce{VO^{2+} + H_2O \rightarrow VO_2^{+} + 2H^{+} + e^{-}}\]

Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of zero.

If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is common for nonmetals such as sulfur:

\[ \ce{S + 2e^- \rightarrow S^{2-}} \]

Here the sulfur has an oxidation state of -2.

The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.

• Oxidation involves an increase in oxidation state
• Reduction involves a decrease in oxidation state

Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations.

Determining oxidation states

Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules provide a simpler method.

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.
• 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 assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
• Some elements almost always have the same oxidation states in their compounds:

The reasons for the exceptions

Hydrogen in the metal hydrides : Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists 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, the sum of the oxidation states in a neutral compound is zero. Because 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, H 2 O 2 . This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.

Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.

Oxygen in F 2 O : The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.

Chlorine in compounds with fluorine or oxygen : Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference. An example of this situation is given below.

Example \(\PageIndex{1}\): Chromium

What is the oxidation state of chromium in Cr 2 + ?

For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion)

What is the oxidation state of chromium in CrCl 3 ?

This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let n equal the oxidation state of chromium:

n + 3(-1) = 0

The oxidation state of chromium is +3.

Example \(\PageIndex{2}\): Chromium

What is the oxidation state of chromium in Cr(H 2 O) 6 3+ ?

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, Cr 3 + . The oxidation state is +3.

What is the oxidation state of chromium in the dichromate ion, Cr 2 O 7 2 - ?

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

Example \(\PageIndex{3}\): Copper

What is the oxidation state of copper in CuSO 4 ?

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 sulfur) with variable oxidation states.

In cases like these, some chemical intuition is useful. Here are two ways of approaching this problem:

• Recognize CuSO 4 as an ionic compound containing a copper ion and a sulfate ion, SO 4 2 - . To form an electrically neutral compound, the copper must be present as a Cu 2+ ion. The oxidation state is therefore +2.
• Recognize the formula as being copper(II) sulfate (the (II) designation indicates that copper is in a +2 oxidation state, as discussed below).

Using oxidation states

In naming compounds.

You will have come across names like iron(II) sulfate 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 Fe 2 + and Fe 3 + ions.

This can also be extended to negative ions. Iron(II) sulfate is FeSO 4 . The sulfate ion is SO 4 2 - . The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion.

The sulfite ion is SO 3 2 - . The oxidation state of the sulfur is +4. This ion is more properly named the sulfate(IV) ion. The - ate ending indicates that the sulfur is in a negative ion.

FeSO 4 is properly named iron(II) sulfate(VI), and FeSO 3 is iron(II) sulfate(IV). Because of the potential for confusion in these names, the older names of sulfate and sulfite are more commonly used in introductory chemistry courses.

Using oxidation states to identify what has been oxidized and what has been reduced

This is the most common function of oxidation states. Remember:

In each of the following examples, we have to decide whether the reaction is a redox reaction, and if so, which species have been oxidized and which have been reduced.

Example \(\PageIndex{4}\):

This is the reaction between magnesium and hydrogen chloride:

\[ \ce{Mg + 2HCl -> MgCl2 +H2} \nonumber\]

Assign each element its oxidation state to determine if any change states over the course of the reaction:

The oxidation state of magnesium has increased from 0 to +2; the element has been oxidized. The oxidation state of hydrogen has decreased—hydrogen has been reduced. The chlorine is in the same oxidation state on both sides of the equation—it has not been oxidized or reduced.

Example \(\PageIndex{5}\):

The reaction between sodium hydroxide and hydrochloric acid is:

\[ NaOH + HCl \rightarrow NaCl + H_2O\]

The oxidation states are assigned:

None of the elements are oxidized or reduced. This is not a redox reaction.

Example \(\PageIndex{6}\):

The reaction between chlorine and cold dilute sodium hydroxide solution is given below:

\[ \ce{2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O} \nonumber\]

It is probable that the elemental chlorine has changed oxidation state because it has formed two ionic compounds. Checking all the oxidation states verifies this:

Chlorine is the only element to have changed oxidation state. However, its transition is more complicated than previously-discussed examples: it is both oxidized and reduced. The NaCl chlorine atom is reduced to a -1 oxidation state; the NaClO chlorine atom is oxidized to a state of +1. This type of reaction, in which a single substance is both oxidized and reduced, is called a disproportionation reaction.

Using oxidation states to determine reaction stoichiometry

Oxidation states can be useful in working out the stoichiometry for titration reactions when there is insufficient information to work out the complete ionic equation. Each time an oxidation state changes by one unit, one electron has been transferred. If the oxidation state of one substance in a reaction decreases by 2, it has gained 2 electrons.

Another species in the reaction must have lost those electrons. Any oxidation state decrease in one substance must be accompanied by an equal oxidation state increase in another.

Example \(\PageIndex{1}\):

Ions containing cerium in the +4 oxidation state are oxidizing agents, capable of oxidizing molybdenum from the +2 to the +6 oxidation state (from Mo 2 + to MoO 4 2 - ). Cerium is reduced to the +3 oxidation state (Ce 3 + ) in the process. What are the reacting proportions?

The oxidation state of the molybdenum increases by 4. Therefore, the oxidation state of the cerium must decrease by 4 to compensate. However, the oxidation state of cerium only decreases from +4 to +3 for a decrease of 1. Therefore, there must be 4 cerium ions involved for each molybdenum ion; this fulfills the stoichiometric requirements of the reaction.

The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.

Here is a more common example involving iron(II) ions and manganate(VII) ions:

A solution of potassium manganate(VII), KMnO 4 , acidified with dilute sulfuric acid oxidizes iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.

The oxidation state of the manganese in the manganate(VII) ion is +7, as indicated by the name (but it should be fairly straightforward and useful practice to figure it out from the chemical formula)

In the process of transitioning to manganese(II) ions, the oxidation state of manganese decreases by 5. Every reactive iron(II) ion increases its oxidation state by 1. Therefore, there must be five iron(II) ions reacting for every one manganate(VII) ion.

The left-hand side of the equation is therefore written as: MnO 4 - + 5Fe 2 + + ?

The right-hand side is written as: Mn 2 + + 5Fe 3 + + ?

The remaining atoms and the charges must be balanced using some intuitive guessing. In this case, it is probable that the oxygen will end up in water, which must be balanced with hydrogen. It has been specified that this reaction takes place under acidic conditions, providing plenty of hydrogen ions.

The fully balanced equation is displayed below:

\[ MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} \nonumber\]

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How can I assign oxidation numbers to each of the atoms?

The oxidation number of oxygen in HOCl is -2.

What is the oxidation number of chlorine in HOCl?

  -3     -2    -1    0    +1    +2    +3 

Assigning Oxidation States Example Problem

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The oxidation state of an atom in a molecule refers to the degree of oxidation of that atom. Oxidation states are assigned to atoms by a set of rules based on the arrangement of electrons and bonds around that atom. This means each atom in the molecule has its own oxidation state which could be different from similar atoms in the same molecule. These examples will use the rules outlined in Rules for Assigning Oxidation Numbers .

Key Takeaways: Assigning Oxidation States

• An oxidation number refer to the quantity of electrons that may be gained or lost by an atom. An atom of an element may be capable of multiple oxidation numbers.
• The oxidation state is the positive or negative number of an atom in a compound, which may be found by comparing the numbers of electrons shared by the cation and anion in the compound needed to balance each other's charge.
• The cation has a positive oxidation state, while the anion has a negative oxidation state. The cation is listed first in a formula or compound name.

Problem: Assign oxidation states to each atom in H 2 O According to rule 5, oxygen atoms typically have an oxidation state of -2. According to rule 4, hydrogen atoms have an oxidation state of +1. We can check this using rule 9 where the sum of all oxidation states in a neutral molecule is equal to zero. (2 x +1) (2 H) + -2 (O) = 0 True The oxidation states check out. Answer: The hydrogen atoms have an oxidation state of +1 and the oxygen atom has an oxidation state of -2. Problem: Assign oxidation states to each atom in CaF 2 . Calcium is a Group 2 metal. Group IIA metals have an oxidation of +2. Fluorine is a halogen or Group VIIA element and has a higher electronegativity than calcium. According to rule 8, fluorine will have an oxidation of -1. Check our values using rule 9 since CaF 2 is a neutral molecule: +2 (Ca) + (2 x -1) (2 F) = 0 True. Answer: The calcium atom has an oxidation state of +2 and the fluorine atoms have an oxidation state of -1. Problem: Assign oxidation states to the atoms in hypochlorous acid or HOCl. Hydrogen has an oxidation state of +1 according to rule 4. Oxygen has an oxidation state of -2 according to rule 5. Chlorine is a Group VIIA halogen and usually has an oxidation state of -1 . In this case, the chlorine atom is bonded to the oxygen atom. Oxygen is more electronegative than chlorine making it the exception to rule 8. In this case, chlorine has an oxidation state of +1. Check the answer: +1 (H) + -2 (O) + +1 (Cl) = 0 True Answer: Hydrogen and chlorine have +1 oxidation state and oxygen has -2 oxidation state. Problem: Find the oxidation state of a carbon atom in C 2 H 6 . According to rule 9, the sum total oxidation states add up to zero for C 2 H 6 . 2 x C + 6 x H = 0 Carbon is more electronegative than hydrogen. According to rule 4, hydrogen will have a +1 oxidation state. 2 x C + 6 x +1 = 0 2 x C = -6 C = -3 Answer: Carbon has a -3 oxidation state in C 2 H 6 . Problem: What is the oxidation state of the manganese atom in KMnO 4 ? According to rule 9, the sum total of oxidation states of a neutral molecule equal zero. K + Mn + (4 x O) = 0 Oxygen is the most electronegative atom in this molecule. This means, by rule 5, oxygen has an oxidation state of -2. Potassium is a Group IA metal and has an oxidation state of +1 according to rule 6. +1 + Mn + (4 x -2) = 0 +1 + Mn + -8 = 0 Mn + -7 = 0 Mn = +7 Answer: Manganese has an oxidation state of +7 in the KMnO 4 molecule. Problem: What is the oxidation state of the sulfur atom in the sulfate ion - SO 4 2- . Oxygen is more electronegative than sulfur, so the oxidation state of oxygen is -2 by rule 5. SO 4 2- is an ion, so by rule 10, the sum of the oxidation numbers of the ion is equal to the charge of the ion. In this case, the charge is equal to -2. S + (4 x O) = -2 S + (4 x -2) = -2 S + -8 = -2 S = +6 Answer: The sulfur atom has an oxidation state of +6. Problem: What is the oxidation state of the sulfur atom in the sulfite ion - SO 3 2- ? Just like the previous example, oxygen has an oxidation state of -2 and the total oxidation of the ion is -2. The only difference is the one less oxygen. S + (3 x O) = -2 S + (3 x -2) = -2 S + -6 = -2 S = +4 Answer: Sulfur in the sulfite ion has an oxidation state of +4.

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