Profile
Settings
Refer your friends
Sign out
In chemistry, it is important to be able to calculate the oxidation number of atoms in a molecule. This number can help you to understand how the atoms are bonded together and what type of chemical reaction has taken place. In some cases, it can be difficult to determine the oxidation number of an atom that occurs more than once in a molecule. In this brief note, we will discuss how to calculate the oxidation number of atoms that have identical bonding.
Oxidation Number of Atoms Occurring more than Once in a Molecule
It is important to note that the oxidation number of atoms occurring more than once in a molecule and having identical bonding can be calculated by taking the average of all the oxidation numbers of that atom in the molecule. For example, consider the calculation of the oxidation number of chlorine in HClO₄. In this molecule, chlorine has an oxidation number of +VII. However, since there are two chlorine with identical bonding, we must take the average of their oxidation numbers. Therefore, the overall oxidation number for chlorine in HClO₄ is +VI½.
Similarly, when calculating the oxidation number of carbon in CH₃CHO, we must take into account that there are two carbons with identical bonding. Therefore, the overall oxidation number for carbon in CH₃CHO is +II½.
It is important to keep in mind that this method of calculation is only applicable when all atoms in the molecule have identical bonding. If there are atoms with different types of bonding, then a more sophisticated method of calculation must be used. Nevertheless, this technique can be useful in many situations and can help to quickly determine the oxidation number of atoms in a molecule.
How to calculate
The oxidation number of atoms occurring more than once in a molecule and having identical bonding can be calculated by taking the average of all the oxidation numbers of that atom in the molecule. For example, to calculate the oxidation number of chlorine in HClO₄, we would take the average of +VII and +VII, which would give us an overall oxidation number of +VII. Similarly, to calculate the oxidation number of carbon in CH₃CHO, we would take the average of +I and +I, which would give us an overall oxidation number of+I.
It is important to keep in mind that this method of calculation is only applicable when all atoms in the molecule have identical bonding. If there are atoms with different types of bonding, then a more sophisticated method of calculation must be used. Nevertheless, this technique can be useful in many situations and can help to quickly determine the oxidation number of atoms in a molecule.
Example:
Calculate the oxidation number of chlorine in HClO₄.
Since there are two chlorine with identical bonding, we must take the average of their oxidation numbers. Therefore, the overall oxidation number for chlorine in HClO₄ is +VI½.
Example:
Calculate the oxidation number of carbon in CH₃CHO.
Since there are two carbons with identical bonding, we must take the average of their oxidation numbers. Therefore, the overall oxidation number for carbon in CH₃CHO is +II½.
Example:
Calculate the oxidation number of oxygen in H₃PO₄.
Since there are four oxygens with identical bonding, we must take the average of their oxidation numbers. Therefore, the overall oxidation number for oxygen in H₃PO₄ is -II.
Conclusion
In Conclusion, the oxidation number of atoms that have identical bonding can be calculated by taking the average of all the oxidation numbers of that atom in the molecule. This method is only applicable when all atoms in the molecule have identical bonding. If there are atoms with different types of bonding, then a more sophisticated method of calculation must be used. Nevertheless, this technique can be useful in many situations and can help to quickly determine the oxidation number of atoms in a molecule. I hope this article helped you understand how to calculate the oxidation number of atoms that have identical bonding. If you have any questions, please feel free to leave a comment below. Thanks for reading!
