Question

part 1 of 2
what is the hybridization state of c in ch₄? select the single best answer.
○ sp³
○ sp³d²
○ sp²
○ sp³d
part 2 of 2
what is the hybridization state of c in h₃c - ch₃? select the single best answer.
○ sp³d
○ sp²
○ sp³
○ sp³d²

Explanation:

For \(CH_4\), carbon forms four sigma bonds with hydrogen atoms. The formula for hybridization is \(H=\frac{1}{2}(V + M - C + A)\), where \(V\) is the number of valence electrons of the central atom (\(V = 4\) for \(C\)), \(M\) is the number of monovalent atoms (\(M=4\) for \(H\)), \(C\) is the charge on cation ( \(C = 0\)), and \(A\) is the charge on anion (\(A=0\)). So, \(H=\frac{1}{2}(4 + 4)=4\). When \(H = 4\), the hybridization is \(sp^{3}\).

For \(H_3C - CH_3\) (ethane), each carbon atom forms four sigma bonds (three with \(H\) and one with the other \(C\)). Using the same hybridization formula \(H=\frac{1}{2}(V + M - C+A)\) (for each \(C\), \(V = 4\), \(M = 3\) (for \(H\)) + 1 (for the other \(C\)) \(= 4\), \(C = 0\), \(A = 0\)). So \(H=\frac{1}{2}(4 + 4)=4\), and the hybridization of each \(C\) is \(sp^{3}\).

Answer:

Part 1: \(sp^{3}\)
Part 2: \(sp^{3}\)