Question

Determine the oxidation number of underlined atoms in the following”
 $ 8K\underline {Cl} {O_3} + 24HCl \to 8K\underline {Cl} + 12{H_2}O + 9\underline {C{l_2}} + 6\underline {Cl} {O_2} $
 $ A. $ $ + 5, - 1,0, + 4 $
 $ B. $ $ + 5, - 1,0, + 2 $
 $ C. $ $ + 4, - 1,0, + 5 $
 $ D. $ None of these

Answer 1

Hint: Oxidation number is referred to as oxidation state. Oxidation number is defined as the charge that atoms appear to have on forming ionic bonds with other heteroatoms. It is also defined as the total number of electrons present that an atom either gains or loses in order to form a chemical bond.

Complete step by step solution:
In the given question we have determined the oxidation number of the underlined atom, let’s calculate the oxidation number one by one.
Oxidation number of underlined atom in $ K\underline {Cl} {O_3} $ ,
Let $ x $ the oxidation number of underlined atom $ \left( {Cl} \right) $ in $ K\underline {Cl} {O_3} $
 $ 1 + x + 3\left( { - 2} \right) = 0 $
 $ \Rightarrow $ $ x - 5 = 0 $
 $ x = + 5 $
Or, oxidation number of underlined atom in $ K\underline {Cl} $
Let $ x $ the oxidation number of underlined atoms $ \left( {Cl} \right) $ in $ K\underline {Cl} $ .
 $ x + 1 = 0 $
 $ x = - 1 $
Or, oxidation number of underlined atom in $ \underline {C{l_2}} $
Let $ x $ the oxidation number of underlined atoms $ \left( {Cl} \right) $ in $ \underline {C{l_2}} $ .
 $ 2x = 0 $
 $ x = 0 $
And, Or, oxidation number of underlined atom in $ \underline {Cl} {O_2} $
Let $ x $ the oxidation number of underlined atoms $ \left( {Cl} \right) $ in $ \underline {Cl} {O_2} $ .
 $ x + 2\left( { - 2} \right) = 0 $
 $ \Rightarrow $ $ x - 4 = 0 $
 $ x = + 4 $
Hence the oxidation number of underlined atoms is $ + 5, - 1,0, + 4 $ .
So, the correct option is $ A. $

Note:
There are certain rules for finding the oxidation number of an atom. These rules are stated below;
The oxidation number of a free element is always zero, the oxidation number of a monatomic ion equal to the charge of that ion, the oxidation number of hydrogen is $ + 1 $ , but when it combine with more electropositive element the oxidation number of hydrogen is $ - 1 $ , the oxidation number of oxygen in compound is usually $ 2 $ but it is $ 1 $ when there is a peroxide linkage and the sum of the oxidation number of all the atom in a neutral compound is $ 0 $ whereas in case of charged compound the sum is equal to that charge.