Question

Oxidation state of C and H in CH$_4$ are respectively

• A. +4, -1
• B. -4, +1
• C. 0, +1
• D. -2, 0

Hint:

In hydrocarbons, Hydrogen is always $+1$ because Carbon is more electronegative than Hydrogen.
Consequently, you can quickly find the oxidation state of Carbon by simply taking the negative of the number of Hydrogen atoms attached to it.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The goal is to determine the individual oxidation states of Carbon (C) and Hydrogen (H) in a neutral molecule of methane ($CH_4$).

Step 2: Key Rules or Approach:
The following standard rules for assigning oxidation states are applied:
1. The oxidation state of Hydrogen is $+1$ when it is bonded to non-metals (Carbon is a non-metal).
2. For any neutral molecule, the algebraic sum of the oxidation states of all its constituent atoms must be zero.

Step 3: Detailed Explanation:
Let the unknown oxidation state of Carbon be represented as $x$.
Methane ($CH_4$) consists of one Carbon atom and four Hydrogen atoms.
As per the rules, each Hydrogen atom is assigned an oxidation state of $+1$.
Applying the rule for the sum of oxidation states in a neutral molecule:
\[ 1 \times (\text{Oxidation state of } C) + 4 \times (\text{Oxidation state of } H) = 0 \]
Substituting the assigned value for Hydrogen:
\[ x + 4 \times (+1) = 0 \]
\[ x + 4 = 0 \]
\[ x = -4 \]
This calculation confirms that Carbon has an oxidation state of $-4$ and each Hydrogen atom has an oxidation state of $+1$.

Step 4: Final Answer:
The oxidation states of C and H in $CH_4$ are $-4$ and $+1$ respectively, which corresponds to option (B).