Question

1. ** what hybridization change does the carbon atom undergo in the following reaction?

ch₄(g) → c(s) + 2h₂(g)
a. sp³ → sp²
b. sp³ → sp
c. sp² → sp³
d. sp³ → no hybridization
e. none

Explanation:

• In \(CH_4\), the carbon atom forms 4 sigma bonds with hydrogen atoms. According to the hybridization rule (\(number\ of\ sigma\ bonds+number\ of\ lone - pairs\)), for \(CH_4\) (\(4 + 0\)), the hybridization of \(C\) is \(sp^{3}\).
• In elemental carbon (\(C(s)\)), if we consider the most common form (diamond has \(sp^{3}\) hybridization, but if we consider the reaction \(CH_4(g)\to C(s)+2H_2(g)\) in terms of the change from a compound to an element, in the elemental state (for example, in some interpretations of the formation of pure \(C\) from \(CH_4\) where we break the \(C - H\) bonds and the \(C\) atom reverts to its un - hybridized state. The ground - state electron configuration of \(C\) is \(1s^{2}2s^{2}2p^{2}\) and in the elemental form (not in a bonded state as in a compound like \(CH_4\)), there is no hybridization.

Answer:

D. \(sp^{3}\to\) no hybridization