MHT CET · Chemistry · Redox Reactions
The oxidation number of \(\mathrm{C}\) atom in \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{CCl}_{4}\) are respectively
- A \(-2\) and \(-4\)
- B 0 and \(-4\)
- C 0 and 4
- D 2 and 4
Answer & Solution
Correct Answer
(C) 0 and 4
Step-by-step Solution
Detailed explanation
Let the oxidation number of \(C\) is \(x\)
\(\mathrm{CH}_{2} \mathrm{Cl}_{2}\)
\(x+2(+1)+2(-1)=0\)
\(x+2-2=0\)
\(x=0\)
\(\mathrm{CCl}_{4}\)
\(x+4(-1)=0\)
\(x-4=0\)
\(x=4\)
\(\mathrm{CH}_{2} \mathrm{Cl}_{2}\)
\(x+2(+1)+2(-1)=0\)
\(x+2-2=0\)
\(x=0\)
\(\mathrm{CCl}_{4}\)
\(x+4(-1)=0\)
\(x-4=0\)
\(x=4\)
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