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Oxidation number of carbon in $C{{H}_{2}}C{{l}_{2}}$
$\begin{align}
 & A)\text{ 0} \\
 & B)\text{ +1} \\
 & C)\text{ +2} \\
 & D)\text{ +4} \\
\end{align}$

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: Oxidation state is termed as Oxidation number which describes the degree of oxidation. Oxidation state of an atom is the hypothetical charge that it would have when all the bonds are treated as purely ionic.

Complete answer:
The IUPAC name of $C{{H}_{2}}C{{l}_{2}}$ is dichloromethane, often abbreviated as DCM. Carbon in $C{{H}_{2}}C{{l}_{2}}$ is \[sp3\] hybridized leading to a Tetrahedral geometry of molecule as shown below-

Let oxidation state of carbon in $C{{H}_{2}}C{{l}_{2}}$ be $x$
Oxidation state of $H$ in $C{{H}_{2}}C{{l}_{2}}$ is $+1$
Oxidation state of $Cl$ in $C{{H}_{2}}C{{l}_{2}}$ is $-1$
Since, there is no net charge on the DCM molecule, the sum of the oxidation state of all the atoms constituting it will be zero.
Sum of oxidation state = zero
$\begin{align}
 & \text{ }C+2(H)+2(Cl)=0 \\
 & \Rightarrow x+2(+1)+2(-1)=0 \\
 & \Rightarrow \text{}x+2-2=0 \\
 & \Rightarrow \text{}x=0 \\
\end{align}$
The oxidation state of carbon in $C{{H}_{2}}C{{l}_{2}}$ is Zero. Hence the option A is correct.

Additional information:
The word Oxidation was firstly used by Antoine Lavoiser. Oxidation state of an element in a compound represents the charge on that element if the charge is completely ionized. Loss of electrons, addition of Hydrogen or addition of oxygen are all known as Oxidation. Increase in oxidation state is termed as Oxidation and decrease in oxidation state is termed as Reduction. In the redox reaction the process of oxidation and reduction occurs simultaneously.

Note: Sum of oxidation state of a constituting atom of a molecule or compound is equal to the net charge present on that molecule. Hydrogen shows two oxidation states $(+1,-1)$. It shows normally $+1$ oxidation state when attached to Carbon, Oxygen, halogen etc. When attached to metals it shows $-1$ oxidation state because metals are electropositive.


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