Question

What is the change in oxidation number of carbon in the following reaction? 
\(CH_4(g) + 4Cl_2 (g) \rightarrow CCl_4(l) + 4HCl (g)\)

• A. +4 to +4
• B. 0 to +4
• C. -4 to +4
• D. 0 to -4

Answer 1

The Correct Option is C

To determine the change in oxidation number of carbon in the reaction:

\(CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)\)

1. Identify the initial oxidation number of Carbon in \(CH_4\):
   In methane, \(CH_4\), carbon is bonded to four hydrogen atoms. The usual oxidation state for hydrogen is +1. For a neutral molecule such as methane, the sum of the oxidation numbers must equal zero. Therefore, the equation becomes:

\(x + 4(+1) = 0\)

1. Solving for \(x\), the oxidation number of carbon:

\(x = -4\)

1. Thus, the oxidation number of carbon in methane is -4.
2. Determine the oxidation number of Carbon in \(CCl_4\):
   In carbon tetrachloride, \(CCl_4\), carbon is bonded to four chlorine atoms. The usual oxidation state of chlorine is -1. Thus, for the neutral molecule \(CCl_4\):

\(y + 4(-1) = 0\)

1. Solving for \(y\), the oxidation number of carbon:

\(y = +4\)

1. Hence, the oxidation number of carbon in carbon tetrachloride is +4.
2. Calculate the change in oxidation number:
   The oxidation number of carbon changes from -4 in \(CH_4\) to +4 in \(CCl_4\). Therefore, the change in oxidation number is:

\(+4 - (-4) = 8\)

1. However, in terms of the initial and final oxidation states, we describe it simply as:

-4 to +4

1. Conclusion:
   The correct option is -4 to +4, which demonstrates a significant increase in the oxidation state, indicating a transfer of electrons and oxidation process. Carbon undergoes an oxidation process whereby it loses electrons to chlorine, increasing its oxidation state.