Question

identify the hybridization of the highlighted carbon present in 2-butyne

Explanation:

Step1: Analyze the highlighted carbon's bonds

The highlighted carbon (in 2 - butyne, the terminal carbons? Wait, no, the highlighted carbon here is the one with four single bonds? Wait, no, 2 - butyne has structure \( \text{CH}_3 - \text{C}\equiv\text{C}-\text{CH}_3 \). Wait, the highlighted carbon in the diagram: looking at the structure, the highlighted C is bonded to four single bonds? Wait, no, the diagram shows \( \text{H}-\text{C}(\text{H})(\text{H})-\text{C}\equiv\text{C}-\text{C}(\text{H})(\text{H})-\text{H} \)? Wait, no, 2 - butyne is \( \text{CH}_3\text{C}\equiv\text{CCH}_3 \). The carbon atoms in \( \text{CH}_3 \) groups: each \( \text{CH}_3 \) carbon is bonded to four single bonds (3 H and 1 C). For a carbon atom, the hybridization is determined by the number of sigma bonds (and lone pairs, but C has no lone pairs here).

- \( \text{sp}^3 \) hybridization: 4 sigma bonds (tetrahedral, bond angle ~109.5°), formed by mixing one s and three p orbitals.
- \( \text{sp}^2 \) hybridization: 3 sigma bonds and 1 pi bond (trigonal planar, bond angle ~120°), mixing one s and two p orbitals.
- \( \text{sp} \) hybridization: 2 sigma bonds and 2 pi bonds (linear, bond angle 180°), mixing one s and one p orbital.

The highlighted carbon (the one in the \( \text{CH}_3 \) - like group, bonded to four single bonds: 3 H and 1 C) has 4 sigma bonds. So its hybridization is \( \text{sp}^3 \)? Wait, no, wait 2 - butyne: the central carbons (the ones with the triple bond) are sp hybridized. The terminal carbons (the \( \text{CH}_3 \) carbons) are \( \text{sp}^3 \)? Wait, no, wait \( \text{CH}_3 - \text{C}\equiv\text{C}-\text{CH}_3 \): each \( \text{CH}_3 \) carbon is bonded to 3 H and 1 C (single bond), so 4 sigma bonds. So hybridization is \( \text{sp}^3 \)? Wait, but the options are \( \text{sp}^2 \), \( \text{sp} \), \( \text{sp}^3 \). Wait, maybe I misread the structure. Wait the diagram shows: H - C - H (top and bottom), and then C≡C, and another C with H - C - H. Wait, maybe the highlighted carbon is the one in the \( \text{CH}_3 \) group? Wait, no, maybe the highlighted carbon is the one with the triple bond? No, the triple bond carbons have 2 sigma bonds (one to C, one to C or H? Wait, no, in \( \text{CH}_3\text{C}\equiv\text{CCH}_3 \), each of the middle carbons (with triple bond) is bonded to one \( \text{CH}_3 \) carbon (single bond) and one triple bond (which has one sigma and two pi bonds). Wait, no: a triple bond has one sigma and two pi bonds. So for a carbon with a triple bond (e.g., \( \text{C}\equiv\text{C} \)), each carbon in the triple bond is bonded to one other carbon (sigma bond) and has a triple bond (sigma + 2 pi), so total sigma bonds: 2 (one to adjacent C, one to the other C in triple bond? No, wait \( \text{C}\equiv\text{C} \): each C is bonded to one C (triple bond: 1 sigma, 2 pi) and one other C (single bond? No, in 2 - butyne, it's \( \text{CH}_3 - \text{C}\equiv\text{C}-\text{CH}_3 \), so each middle C is bonded to one \( \text{CH}_3 \) C (single bond, sigma) and one triple bond (sigma + 2 pi) with the other middle C. So each middle C has 2 sigma bonds (one to \( \text{CH}_3 \) C, one to the other middle C (sigma part of triple bond)). So hybridization is sp (since 2 sigma bonds, so s + p (1 s, 1 p) hybridized, forming 2 sp orbitals for sigma bonds, and 2 p orbitals for pi bonds). But the highlighted carbon in the diagram: looking at the diagram, the highlighted C is bonded to four single bonds (H - C - H on top, H - C - H on bottom, and a single bond to the C≡C carbon). Wait, that would be a carbon with four single bonds, so 4 sigma bonds, so \( \text{sp}^3 \) hybridization? Wait, no, 2 - butyne's structure is \( \text{CH}_3\text{C}\equiv\text{CCH}_3 \), so each \( \text{CH}_3 \) carbon is \( \text{sp}^3 \) (four single bonds: 3 H, 1 C), and each middle C is sp (two sigma bonds: 1 to \( \text{CH}_3 \) C, 1 to the other middle C, and two pi bonds). So if the highlighted carbon is the \( \text{CH}_3 \) carbon, then hybridization is \( \text{sp}^3 \). Wait, but the options include \( \text{sp}^3 \) (orange), \( \text{sp} \) (blue), \( \text{sp}^2 \) (green). So the highlighted carbon: let's check the bonding. If it's bonded to four atoms (four single bonds), then it's \( \text{sp}^3 \). So the correct hybridization is \( \text{sp}^3 \)? Wait, no, wait maybe I made a mistake. Wait, in the diagram, the highlighted C is connected to four single bonds (H on top, H on bottom, H on left, H on right? No, the diagram shows: H - C - H (top and bottom, so two H), and a single bond to the C≡C carbon, and another H? Wait, the formula in the diagram: H - C (with H on left, H on right, H on bottom) - C≡C - C (with H on left, H on right, H on top) - H? Wait, no, the structure is \( \text{CH}_3 - \text{C}\equiv\text{C}-\text{CH}_3 \), so each \( \text{CH}_3 \) has three H and one C. So the carbon in \( \text{CH}_3 \) is bonded to four atoms (3 H, 1 C) with single bonds, so 4 sigma bonds, hence \( \text{sp}^3 \) hybridization. So the answer should be \( \text{sp}^3 \).

Answer:

\( \text{sp}^3 \) (the orange option)