What is the oxidation number of $ CuN{O_3} $ .
Answer
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Hint: Oxidation number is the total number of electrons gained or lost by an atom while it makes a bond with another atom. Oxidation is also known as the oxidation state of an element. The oxidation number of ionic compound is equal to net overall charge of the compound.
Complete answer:
The oxidation number of an ionic compound is the net overall charge of the compound. When an atom forms a cation or anion then the charge acquired by the element is also called oxidation number or oxidation state of the element. Here the overall charge of compound $ CuN{O_3} $ is zero. Thus we can say that it is electrically neutral. Also we know that the oxidation state of oxygen atoms is $ - 2 $ . Therefore for three moles of oxygen atom we have $ 3{\text{ }} \times {\text{ }}\left( { - 2} \right){\text{ = - 6}} $ . The given compound can be broken as, $ = {\text{ - 1 + 1 = 0}} $
$ CuN{O_3}{\text{ }}\xrightarrow{{}}{\text{ Cu + N}}{{\text{O}}_3} $
The sum of charge produced must be equal to zero. Therefore we will prove that as,
Total charge of $ {\text{N}}{{\text{O}}_3} $ will be calculated as, $ + 5{\text{ + 3}}\left( { - 2} \right) $ which is equal to $ - 1 $ .
Total charge on copper will be equal to $ = {\text{ + 1}} $
Thus the total charge on the ion formed is $ = {\text{ - 1 + 1 = 0}} $
Hence the total charge on reactants is equal to total charge on products. Thus the equation can be written as,
$ CuN{O_3}{\text{ }}\xrightarrow{{}}{\text{ C}}{{\text{u}}^{ + 1}}{\text{ + N}}{{\text{O}}_3}^{ - 1} $
Note:
The oxidation number for ionic compounds is the charge of ions. This is because the oxidation number of oxygen is taken as $ - 2 $ . Also for nitrogen it is taken as $ + 5 $ . Copper shows two oxidation states $ + 2 $ and $ + 1 $ . Here the oxidation state of copper is $ + 1 $ . When the oxidation state of copper is $ + 2 $ then two moles of nitrate are used.
Complete answer:
The oxidation number of an ionic compound is the net overall charge of the compound. When an atom forms a cation or anion then the charge acquired by the element is also called oxidation number or oxidation state of the element. Here the overall charge of compound $ CuN{O_3} $ is zero. Thus we can say that it is electrically neutral. Also we know that the oxidation state of oxygen atoms is $ - 2 $ . Therefore for three moles of oxygen atom we have $ 3{\text{ }} \times {\text{ }}\left( { - 2} \right){\text{ = - 6}} $ . The given compound can be broken as, $ = {\text{ - 1 + 1 = 0}} $
$ CuN{O_3}{\text{ }}\xrightarrow{{}}{\text{ Cu + N}}{{\text{O}}_3} $
The sum of charge produced must be equal to zero. Therefore we will prove that as,
Total charge of $ {\text{N}}{{\text{O}}_3} $ will be calculated as, $ + 5{\text{ + 3}}\left( { - 2} \right) $ which is equal to $ - 1 $ .
Total charge on copper will be equal to $ = {\text{ + 1}} $
Thus the total charge on the ion formed is $ = {\text{ - 1 + 1 = 0}} $
Hence the total charge on reactants is equal to total charge on products. Thus the equation can be written as,
$ CuN{O_3}{\text{ }}\xrightarrow{{}}{\text{ C}}{{\text{u}}^{ + 1}}{\text{ + N}}{{\text{O}}_3}^{ - 1} $
Note:
The oxidation number for ionic compounds is the charge of ions. This is because the oxidation number of oxygen is taken as $ - 2 $ . Also for nitrogen it is taken as $ + 5 $ . Copper shows two oxidation states $ + 2 $ and $ + 1 $ . Here the oxidation state of copper is $ + 1 $ . When the oxidation state of copper is $ + 2 $ then two moles of nitrate are used.
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