Question

The oxidation states of carbon in $\text{CH}_4$ and $\text{CCl}_4$ respectively are:

• A. -4 and +4
• B. +4 and -4
• C. -4 and -4
• D. +4 and +4

Hint:

When bonded to less electronegative elements (like H), Carbon takes a negative oxidation state.
When bonded to more electronegative elements (like Halogens, O, N), Carbon takes a positive oxidation state.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to determine the oxidation state of Carbon (C) in Methane ($\text{CH}_4$) and Carbon Tetrachloride ($\text{CCl}_4$).
This requires evaluating electronegativity differences between carbon and the bonded atoms.

Step 2: Key Formula or Approach:
The oxidation state of an atom in a molecule is calculated by assigning formal charges based on electronegativity:
- Carbon (C) has a electronegativity of 2.55.
- Hydrogen (H) has a electronegativity of 2.20. (Hydrogen is less electronegative than Carbon, so it is assigned a $+1$ oxidation state).
- Chlorine (Cl) has a electronegativity of 3.16. (Chlorine is more electronegative than Carbon, so it is assigned a $-1$ oxidation state).

Step 3: Detailed Explanation:

• Case 1: Methane ($\text{CH}_4$)
Let the oxidation state of carbon in $\text{CH}_4$ be $x$.
Since hydrogen has an oxidation state of $+1$:
\[ x + 4(+1) = 0 \]
\[ x + 4 = 0 \]
\[ x = -4 \]

• Case 2: Carbon Tetrachloride ($\text{CCl}_4$)
Let the oxidation state of carbon in $\text{CCl}_4$ be $y$.
Since chlorine has an oxidation state of $-1$:
\[ y + 4(-1) = 0 \]
\[ y - 4 = 0 \]
\[ y = +4 \]

• This comparison shows that carbon can exhibit highly contrasting oxidation states (from $-4$ to $+4$) depending on the electronegativity of the atoms to which it is covalently bound.


Step 4: Final Answer:
The oxidation states of carbon in $\text{CH}_4$ and $\text{CCl}_4$ are $-4$ and $+4$, respectively.