Question

What is the oxidation number of $ H $ and $ Cl $ in $ HCl{O_4} $ respectively?
(A) $ + 1, + 7 $
(B) $ + 2, + 6 $
(C) $ + 3, + 7 $
(D) $ + 2, + 7 $

Answer 1

Hint :Perchloric acid has the formula $ HCl{O_4} $ and is a mineral acid. This colourless compound, which is usually contained in an aqueous solution, is a stronger acid than sulfuric and nitric acids.

Complete Step By Step Answer:
The oxidation state which is also known as the oxidation number, determines the degree of oxidation of an atom in a chemical compound. In simple terms, it's the number that's assigned to each element in a chemical mixture. The total number of electrons that atoms in a molecule will share, lose, or gain when forming chemical bonds with atoms of another substance is known as the oxidation number.
We know that the oxidation number of $ H = + 1 $
Now we have to find the oxidation number of $ Cl $ .
Let the oxidation number of $ Cl = x $
The oxidation number of $ O = - 2 $
That is, $ \left( { + 1} \right) \times 1 + \left( x \right) \times 1 + \left( { - 2} \right) \times 4 = 0 $
 $ \Rightarrow 1 + x - 8 = 0 $
 $ \Rightarrow x = 8 - 1 = 7 $
Therefore, the oxidation number of $ Cl = + 7 $
So, the correct answer is Option A.

Additional Information:
The oxidation state of an atom has little to do with its "real" formal charge or any other atomic property. This is especially true at high oxidation states, where the ionisation energy needed to generate a multiply positive ion exceeds the energies available in chemical reactions. Furthermore, depending on the electronegativity scale used in their measurement, the oxidation states of atoms in a given compound can differ.

Note :
Oxidation is the process of raising an atom's oxidation state through a chemical reaction; reduction is the process of decreasing an atom's oxidation state. In such reactions, the formal transfer of electrons is involved: a reduction is a net gain of electrons, and an oxidation is a net loss of electrons.