Question:
\(CH_4(g)+4Cl_2(g)\rightarrow CCl_4(l)+4HCl(g)\). In the above reaction, the change of oxidation state of carbon is
\(CH_4(g)+4Cl_2(g)\rightarrow CCl_4(l)+4HCl(g)\). In the above reaction, the change of oxidation state of carbon is
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In \(CH_4\), carbon is \(-4\). In \(CCl_4\), carbon is \(+4\).
Updated On: Apr 29, 2026
- From \(+4\) to \(-4\)
- From \(+1\) to \(+4\)
- From \(-4\) to \(+4\)
- From \(-1\) to \(+1\)
- From \(-4\) to \(-1\)
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The Correct Option is C
Solution and Explanation
Concept:
Oxidation number is calculated by assigning usual oxidation states to hydrogen and halogens.
Step 1: Find oxidation state of carbon in \(CH_4\).
Hydrogen is \(+1\): \[ x+4(+1)=0 \] \[ x=-4 \]
Step 2: Find oxidation state of carbon in \(CCl_4\).
Chlorine is \(-1\): \[ x+4(-1)=0 \] \[ x=+4 \]
Step 3: Final change.
Carbon changes from: \[ -4 \rightarrow +4 \]
Step 1: Find oxidation state of carbon in \(CH_4\).
Hydrogen is \(+1\): \[ x+4(+1)=0 \] \[ x=-4 \]
Step 2: Find oxidation state of carbon in \(CCl_4\).
Chlorine is \(-1\): \[ x+4(-1)=0 \] \[ x=+4 \]
Step 3: Final change.
Carbon changes from: \[ -4 \rightarrow +4 \]
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