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OXIDATION NUMBERS CALCULATOR

To calculate oxidation numbers of elements in the chemical compound, enter it's formula and click 'Calculate' (for example: Ca2+ , HF2^- , Fe4[Fe(CN)6]3 , NH4NO3 , so42- , ch3cooh , cuso4*5h2o ).

Formula in Hill notation

The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The oxidation number is synonymous with the oxidation state. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it from the molecular formula (Figure 1b). The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons. Bonds between atoms of the same element (homonuclear bonds) are always divided equally.

When dealing with organic compounds and formulas with multiple atoms of the same element, it's easier to work with molecular formulas and average oxidation numbers (Figure 1d). Organic compounds can be written in such a way that anything that doesn't change before the first C-C bond is replaced with the abbreviation R (Figure 1c). Unlike radicals in organic molecules, R cannot be hydrogen. Since the electrons between two carbon atoms are evenly spread, the R group does not change the oxidation number of the carbon atom it's attached to. You can find examples of usage on the Divide the redox reaction into two half-reactions page.

Rules for assigning oxidation numbers

• The oxidation number of a free element is always 0.
• The oxidation number of a monatomic ion equals the charge of the ion.
• Fluorine in compounds is always assigned an oxidation number of -1.
• The alkali metals (group I) always have an oxidation number of +1.
• The alkaline earth metals (group II) are always assigned an oxidation number of +2.
• Oxygen almost always has an oxidation number of -2, except in peroxides (H 2 O 2 ) where it is -1 and in compounds with fluorine (OF 2 ) where it is +2.
• Hydrogen has an oxidation number of +1 when combined with non-metals, but it has an oxidation number of -1 when combined with metals.
• The algebraic sum of the oxidation numbers of elements in a compound is zero.
• The algebraic sum of the oxidation states in an ion is equal to the charge on the ion.

Assigning oxidation numbers to organic compounds

• cysteine: HO2CCH(NH2)CH2SH

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Oxidation Number Calculator

Oxidation Number Calculator is a free online tool that displays the oxidation number of the given chemical compound. BYJU’S online oxidation number calculator tool makes the calculation faster and it displays the oxidation number in a fraction of seconds.

How to Use the Oxidation Number Calculator?

The procedure to use the oxidation number calculator is as follows: Step 1: Enter the chemical compound in the respective input field Step 2: Now click the button “Calculate Oxidation Number” to get the result Step 3: Finally, the oxidation number of the given chemical compound will be displayed in the new window

What is Meant by the Oxidation Number?

In Chemistry, the oxidation number helps to keep track of the electrons in an atom. It is a number, which is generally assigned to the atoms of the chemical substance. The oxidation number can be positive, negative or it can be zero. It indicates that electrons lose or gain in the atom. In the chemical reaction, the change in oxidation number will help us to track the transfer of electrons. The oxidation number does not always correspond to the real charge of the molecule.

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Oxidation Number Calculator

Here is a simple online oxidation number calculator to calculate the oxidation number of any compound or element by just clicking on the respective compound name in the given elements table with ease.

Oxidation Number Table

Oxidation Number: The number that is assigned to an element to indicate the loss or gain of electrons by an atom of that element is called as the oxidation number. They are positive and negative numbers used for balancing the redox reaction. The oxidation number of diatomic and uncombined elements is zero. For a monatomic ion, it is the charge of that ion. And for alkali metals, the number is +1 and alkaline earth metal it is +2. For finding the number that is assigned to an element to indicate the loss or gain of electrons by an atom of that element, you can use this online oxidation number calculator . Just click on the multiple elements to know the oxidation number of all those elements.

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Oxidation State Calculator + Online Solver With Free Steps

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What Is an Oxidation State Calculator?

The Oxidation State Calculator is used to calculate the oxidation state or the oxidation number of a single element or all the elements in a compound.  The oxidation state defines the number of electrons lost or gained by the element in the chemical process. The ionic compounds contain elements that lose or gain electrons to form ionic bonds.

The element which loses electrons in the chemical reaction has a positive oxidation number and the element that gains electrons is assigned a negative oxidation number.

The oxidation state of a pure element is always zero as it didn’t react to form an ion. The positive ion is referred to as the cation and the negative ion as the anion .

The oxidation state defines the electropositive or electronegative nature of an element in a compound. The charge on each of the elements defines how many electrons are lost or gained by the element by the ion.

The oxidation state or number is the charge on the individual ionic element.  For example, in the peroxide ion $ { O_{2} }^{-2} $, the oxidation state of each oxygen atom is -1 . In the superoxide ion $ { O_{2} }^{-1} $, the oxidation number of the individual oxygen atom is -1/2 .

In a monoatomic ion such as $ Na^{+1} $, the oxidation state is the same as the charge on the ion.  In an ionic compound such as $Mg Cl_2$, the oxidation state of magnesium and chlorine should be such that the net charge on the compound is zero .

As chlorine is a highly electronegative element, it has an oxidation state to be -1 . The two chlorine atoms make the oxidation number multiplied by 2 which becomes -2.

For the net charge to be zero, the alkaline earth metal magnesium should have the oxidation state to be +2 . The calculator computes the oxidation states for all the elements in the compound.

The Oxidation State Calculator is an online tool that is used to compute the oxidation states of all the elements in the compound by entering the chemical formula for the particular ionic compound.

It also computes the oxidation numbers of all the elements in different ions.

For example, the sulphate ion has the formula as $ {SO_{4} }^{-2} $. The calculator will compute the oxidation states of sulfur and oxygen in the sulfate ion.

It also displays the atomic structure of the compound with the highlighted oxidation states.

How To Use the Oxidation State Calculator

The user can use the Oxidation State Calculator by following the steps given below.

The user must first enter the chemical formula for the ion or compound whose oxidation states of the elements are required. It should be entered in the input block labeled as “ Type formula here ”.

After entering the chemical formula for the ionic compound or the particular ion, the user must press the “ Submit ” button.

The calculator processes the input formula and computes the oxidation states of all the elements in the formula.

The Oxidation State Calculator displays the output in the four windows given below.

Input Interpretation

The calculator interprets the input and displays the name of the compound whose chemical formula is entered by the user.

The calculator computes the oxidation states of all the elements and displays them in this window. It also shows the structure of the molecule or compound as entered by the user.

The molecule contains the same atoms of an element whereas a compound has a variety of elements.

Chemical Names and Formulas

This window shows the formula of the ionic compound entered by the user. It also shows Hill’s formula , name, and IUPAC name of the entered chemical formula.

Ion Equivalents

The calculator also shows the ion equivalents if the user enters the formula for an ionic compound. The ion equivalents show the number of cations and anions present in the ionic compound.

It also shows the individual oxidation states of all the ions in the ionic compound.

Solved Examples

The following examples are solved through the Oxidation State Calculator.

For the ionic compound magnesium hydroxide , calculate the oxidation states of magnesium, oxygen, and hydrogen in the compound.

The chemical formula for magnesium hydroxide is $ Mg (OH)_2 $. The user must first enter the chemical formula in the input window of the Oxidation State Calculator.

After submitting the input formula, the calculator computes the output in the following four windows.

The Input Interpretation window shows the name of the ionic compound as entered by the user. For this example, it shows “ magnesium hydroxide ”.

The calculator computes the oxidation states of magnesium, oxygen, and hydrogen to be +2 , -2 , and +1 respectively shown in the result window.

The molecular structure of magnesium hydroxide contains two hydroxide ions $ OH^{-1} $ having two ionic bonds with one magnesium ion $ Mg^{+2} $.

The calculator only shows the ions separately in this window.

Hill’s formula for magnesium hydroxide is $ H_2 Mg O_2 $ and the IUPAC name is magnesium dihydroxide.

The Ion Equivalents show that there is one magnesium cation and two hydroxide anions in magnesium hydroxide.

Calculate the oxidation states of phosphorous and oxygen in the phosphate ion .

The chemical formula entered by the user for phosphate ion is $PO_{4}$.

After submitting the input formula, the calculator shows interprets the input and displays the name for the entered ion. It shows “ phosphate anion ”.

The calculator computes the oxidation state of phosphorous as +5 and that of oxygen as -2 as displayed in the result window.

The structure of phosphate ion is shown in figure 1.

All the images are created using Geogebra.

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Oxidation Number Calculator

Oxidation Number Calculator is a free online tool that displays the oxidation number of a chemical compound. This calculator makes it easy for you to calculate the oxidation number of the given input chemical compound. Enter input and hit on the calculate button to get the oxidation number as output within no time.

Oxidation Number Calculator: Finding oxidation number is not going to be horror again for you with this easy to use and user friendly tool. This handy calculator gives the result i.e oxidation number in fraction of seconds. In the below sections, you will find the information on how to calculate the oxidation number of an element in a chemical compound. You can also see the solved examples here.

Procedure to Find Oxidation Number of an Element in Chemical Compound

Follow the simple and easy steps listed below to check the oxidation number of an element manually.

• Let us take any chemical compound.
• Go through the rules listed below to determine the oxidation numbers.
• According to those rules write the oxidation number of that element above the symbol and total number of atoms below the symbol.
• Evaluate the oxidation number and give the output answer.

Rules to Compute the Oxidation Numbers

1. Oxidation number of a free element is zero.

2. Oxidation number of a monatomic ion is equal to charge of that ion.

3. The oxidation number of Hydrogen H is +1, but it is -1 when combined with less electronegative elements.

4. Oxidation number of Oxygen O in compounds is -2, but it is -1 in peroxides.

5. Oxidation number of Group 1 element in compound is +1.

6. Oxidation number of Group 2 element in compound is +2.

7. Oxidation number of Group 17 element in a binary compound is -1.

8. The sum of oxidation numbers of all the atoms in a neutral compound is 0.

9. The sum of oxidation numbers in a poly atomic ion is equal to the charge of ion.

Question: What is the oxidation number of Cr in CrCl 3 ?

Given compound is CrCl 3

As per the rule 7, oxidation number of Chlorine Cl is -1.

                     -1

This gives CrCl 3

We have 3 Cl atoms. So its oxidation number becomes -3.

Rule 8 says that, sum of oxidation numbers of all atoms is zero.

We have only one Cr atom, so its oxidation number is +3.

Finally, the oxidation number of Cr in CrCl 3 is +3.

Have a look at Onlinecalculator.guru to get an idea on the various math concepts and its calculators. These calculators give instant results easily.

FAQs on Oxidation Number Calculator

1. What is the difference between oxidation state and oxidation number?

Oxidation state and oxidation number are the quantities that commonly equal the same value for atoms in a molecule and are often used interchangeably. Oxidation state refers to the degree of oxidation of an atom in a molecule. Oxidation numbers are used in coordination complex chemistry.

2. What does the Oxidation Number means?

The oxidation number is a number which is generally assigned to the atoms of chemical substance. it helps to keep track of electrons in an atom. It can be positive , negative or zero. Oxidation number indicates the lose or gain of electrons in an atom.

3. What is oxidation on the basis of oxidation number?

Oxidation is the loss of one or more electrons by an atom. If the oxidation number of an element increases, it says there is a loss of electrons and that element is being oxidized.

4. What is the oxidation number of Na 2 SO 4 ?

We know that oxidation number of group 1 element in the compound is +1.

So, Na oxidation number is +1.

A total of 2 elements are there in the given compound. That means +2.

The oxidation number of O in compounds is -2.

We have 4 O atoms in given compound. So it is -8.

   +1    -2

The sum of oxidation numbers of all atoms is zero. So Na number becomes +6.

  +2  +6  -8

Oxidation number of O is -8, Na is +2, S is +6 in the given compound.

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Oxidation numbers method of balancing redox reaction equations

Before we will get to explanation very important disclaimer: oxidation numbers don't exist. They were invented to help in charge accounting needed when balancing redox reaction equations, but they don't refer to any real life chemical concept.

The general idea behind the oxidation numbers (ON) method for balancing chemical equations is that electrons are transferred between charged atoms. These charges - assigned to individual atoms - are called oxidation numbers, just to remind you that they don't reflect real structure of the reagents.

There are several simple rules used for assigning oxidation numbers to every atom present in any compound:

• First of all, charged mono atomic ion has oxidation number equal to its charge. Thus Na + has oxidation number +1, Fe 3+ has oxidation number +3, F - has oxidation number of -1 and S 2- has oxidation number of -2.
• Second rule says that the oxidation number of a free element is always 0. Thus oxidation number of solid, metallic Cu is 0, oxidation number of O in O 2 is 0, the same holds for S in S 8 and so on.
• Oxygen in almost all compounds has oxidation number -2.
• Hydrogen in almost all compounds has oxidation number +1.
• Some elements usually have the same oxidation number in their compounds:
• alkali metals - Li, Na, K, Rb, Cs - oxidation numbers are +1
• alkaline earth metals - Be, Mg, Ca, Sr, Ba - oxidation numbers are +2
• halogens (except when they form compounds with oxygen or one another) - oxidation numbers are -1 (always true for fluorine)
• Last rule says that the charge of the ion or molecule equals sum of oxidation numbers of all atoms.

There are some exceptions to the rules 3 and 4 - for example oxygen in peroxides has oxidation number of -1, it is also not -2 in compounds with fluorine (where F is always -1), hydrogen in hydrides has oxidation number -1.

Before we will try to balance any equations let's use above rules to assign oxidation numbers to atoms in several substances.

For example - what is oxidation number of sulfur in SO 2 ? Particle is not charged, so oxidation number of sulfur must equal sum of oxidation numbers of oxygens, but with the opposite sign. Oxygen oxidation number is -2, there are two oxygens - that gives -4 together, so sulfur must have ON=+4.

What is oxidation number of atoms in CrO 4 2- ? Oxygen is -2 and there are 4 oxygens - that gives overall of -8, ion has charge of -2, so central atom must have ON=+6.

How do we use oxidation numbers for balancing? First of all, we have to understand that oxidation means increase of oxidation number, while reduction means decrease of oxidation number. In both cases change of oxidation number is due to electrons lost (oxidation) or gained (reduction). We calculate oxidation numbers for all atoms present in the reaction equation (note that it is not that hard as it sounds, as for most atoms oxidation numbers will not change) and we look for a ratio that makes the number of electrons lost equal to the number of electrons gained. That gives us additional information needed for reaction balancing.

Let's try with following reaction:

KIO 3 + KI + H 2 SO 4 → K 2 SO 4 + H 2 O + I 2

First of all - we don't need any spectators here, as they are only making things look more difficult then they are in reality. Quick glance tells us that the net ionic reaction is

IO 3 - + I - + H + → H 2 O + I 2

Looks like IO 3 - is oxidizing agent here and I - is reducting agent. I - has oxidation number of -1, iodine in IO 3 - has oxidation number of +5. On the right side in I 2 both iodine atoms have oxidation number 0. It means that iodine in IO 3 - must gain 5 electrons. These electrons come from I - - one for every I - ion. Assuming (just like we do in the inspection method) that IO 3 - is the most complicated molecule and it's coefficient is 1 we will need five I - for the redox process to complete:

1 IO 3 - + 5 I - + H + → H 2 O + I 2

Now that the ratio between oxidizer and reducing agent is known we use simple techniques we know from the inspection method to balance remaining elements. There are six atoms of iodine on the left, so we need three I 2 molecules to balance iodine:

1 IO 3 - + 5 I - + H + → H 2 O + 3 I 2

And the final, trivial step is balancing oxygen, hydrogen and water:

IO 3 - + 5 I - + 6 H + → 3 H 2 O + 3 I 2

Other case we can try is oxidation of Mn 2+ with NaBiO 3 in acidic conditions:

Mn 2+ + BiO 3 - + H + → MnO 4 - + Bi 3+ + H 2 O

Using methods for oxidation numbers calculation we can easily check that manganese is oxidized from +2 to +7 (freeing five electrons) and bismuth is reduced from +5 to +3 (accepting two electrons). To balance electrons transferred we can put coefficients 2 and 5 on the left side of reaction equation:

2 Mn 2+ + 5 BiO 3 - + H + → MnO 4 - + Bi 3+ + H 2 O

Rest can be balanced by inspection and is not difficult to do, yielding:

2 Mn 2+ + 5 BiO 3 - + 14 H + → 2 MnO 4 - + 5 Bi 3+ + 7 H 2 O

Now the same equation can be also easily balanced as a full (non net-ionic) version:

4 MnSO 4 + 10 NaBiO 3 + 14 H 2 SO 4 → 4 NaMnO 4 + 5 Bi 2 (SO 4 ) 3 + 14 H 2 O + 3 Na 2 SO 4

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