The oxidation number of Carbon in and are respectively: — Redox Reactions and Volumetric Analysis Chemistry Question
Question
The oxidation number of Carbon in $CH_4, CH_3Cl, CH_2Cl_2, CHCl_3,$ and $CCl_4$ are respectively:
Answer: B
💡 Solution & Explanation
Hydrogen has an O.S. of +1 and Chlorine has -1. Solving for C (x) in each: $CH_4 (x+4=0 \Rightarrow -4)$; $CH_3Cl (x+3-1=0 \Rightarrow -2)$; $CH_2Cl_2 (x+2-2=0 \Rightarrow 0)$; $CHCl_3 (x+1-3=0 \Rightarrow +2)$; $CCl_4 (x-4=0 \Rightarrow +4)$.
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