
What are some examples of Oxidation numbers ?
Answer
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Hint:To solve the given question, we should have knowledge about oxidation numbers and how to calculate it. Oxidation Number, is the number of electrons gained or lost or shared to form a compound from an element.
Some general rules that follow oxidation number are that the oxidation number of a free element is always $ 0 $, the oxidation number of a Group $ 1 $ element in a compound is $ +1 $ , a Group $ 2 $ element in a compound is $ +2 $ and the sum of oxidation numbers of all of the atoms in a compound is of total charge of the molecule.
Complete answer:
The first example we could look forward to was the oxidation number of $ Cr $ in $ CrCl_3 $ . Let the oxidation number of Chromium be $ x $ . We can say that the total charge or molecule is zero.
$ x-3=0 $
$ x=+3 $
Therefore, the oxidation number of $ Cr $ in $ CrCl_3 $ is $ +3 $ .
Next one can be $ KCl $ , so, oxidation number of $ Cl $ is $ -1 $ . While in the case of $ Cl_2 $ , it is $ 0 $ . In $ KClO_2 $ , the oxidation number of $Cl $ is $ +3 $.
In case of $ KMnO_4 $ , the oxidation number of $ Mn $ is ;
Let the oxidation number of $ Mn $ be $ x $ . The oxidation number of the molecule as a whole is $ 0 $ .
So, it can be calculated as :
$ +1+x-8=0 $
$ x=+7 $
So, the oxidation number of $ Mn $ in $ KMnO_4 $ is $ +7 $ .
Note:
While calculating the oxidation number the charge on the whole molecule should be duly noted and identified. This is a crucial step which should not be messed up to find the correct oxidation state.
Some general rules that follow oxidation number are that the oxidation number of a free element is always $ 0 $, the oxidation number of a Group $ 1 $ element in a compound is $ +1 $ , a Group $ 2 $ element in a compound is $ +2 $ and the sum of oxidation numbers of all of the atoms in a compound is of total charge of the molecule.
Complete answer:
The first example we could look forward to was the oxidation number of $ Cr $ in $ CrCl_3 $ . Let the oxidation number of Chromium be $ x $ . We can say that the total charge or molecule is zero.
$ x-3=0 $
$ x=+3 $
Therefore, the oxidation number of $ Cr $ in $ CrCl_3 $ is $ +3 $ .
Next one can be $ KCl $ , so, oxidation number of $ Cl $ is $ -1 $ . While in the case of $ Cl_2 $ , it is $ 0 $ . In $ KClO_2 $ , the oxidation number of $Cl $ is $ +3 $.
In case of $ KMnO_4 $ , the oxidation number of $ Mn $ is ;
Let the oxidation number of $ Mn $ be $ x $ . The oxidation number of the molecule as a whole is $ 0 $ .
So, it can be calculated as :
$ +1+x-8=0 $
$ x=+7 $
So, the oxidation number of $ Mn $ in $ KMnO_4 $ is $ +7 $ .
Note:
While calculating the oxidation number the charge on the whole molecule should be duly noted and identified. This is a crucial step which should not be messed up to find the correct oxidation state.
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