Question

Oxidation number of chlorine atoms in $CaOC{{l}_{2}}$ are
(A) 0, 0
(B) -1, -1
(C) -1, +1
(D) -2, +7

Answer 1

Hint: Oxidation number is the total number of electrons that an atom either gains or loses electrons in order to form a chemical bond with another atom. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element.

Complete step by step solution:
- Oxidation number, also called as oxidation state, can be described as the number that is allocated to elements in a chemical combination.
- It is the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.
- $CaOC{{l}_{2}}$ is a mixed salt of$HCl$ and $HOCl$ with $Ca{{(OH)}_{2}}$. It can be represented as
\[[Ca(OCl)Cl]\]
- The oxidation state of the compound can be represented as
\[[C{{a}^{+2}}{{(OCl)}^{-1}}C{{l}^{-1}}]\]
- The oxidation state of chlorine atom in $OCl$can be obtained as shown below
Oxidation number of oxygen atom is -2
Let oxidation number of Chlorine be x
$\begin{align}
& -2+x=-1 \\
& x=+1 \\
\end{align}$
- The oxidation state of other chlorine atom is -1

Therefore, answer to the question is (C) -1, +1

Additional information: Oxidation state helps us describe the transfer of electrons. It is also used to determine the changes that occur in redox reactions. It is quite similar to valence electrons.

Note: The oxidation number of atoms in their elemental state is taken as zero. The oxidation number of mono-atomic atoms like Na+ etc is taken as 1. The oxidation number of Hydrogen is +1 when present with non-metals and -1 when present with metals. In neutral compounds, the sum of all oxidation numbers is equal to zero.