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: Set up the rule for assigning oxidation states in covalent molecules.
For each bond, the more electronegative atom is assigned the shared electrons; carbon (2.5) is more electronegative than hydrogen (2.1) but less electronegative than chlorine (3.0).
Step 2: Work out \( \text{CH}_4 \).
Carbon is bonded to 4 hydrogens, and since carbon is more electronegative than H, each bond contributes -1 to carbon:
\[ x + 4(+1) = 0 \Rightarrow x = -4 \]
Step 3: Work out \( \text{CCl}_4 \).
Carbon is bonded to 4 chlorines, and since chlorine is more electronegative than carbon, each bond contributes +1 to carbon here:
\[ x + 4(-1) = 0 \Rightarrow x = +4 \]
So carbon flips from its most reduced state in methane to its most oxidised state in carbon tetrachloride.
\[ \boxed{-4 \text{ and } +4} \]