Draw the Lewis structure of CH
CCBr and then choose the appropriate set of hybridization states for the three central atoms. Your answer choice is independent of the orientation of your drawn structure. A) sp
/ sp / sp
B) sp
/ sp
/ sp
C) sp
/ sp / sp D) sp
/ sp
/ sp
E) sp / sp
/ sp
The Correct Answer and Explanation is:
To determine the correct Lewis structure and hybridization for the molecule CH≡C–CBr (commonly called bromo-propyne), we must:
Step 1: Determine the Molecular Structure
The molecular formula CH≡C–CBr consists of:
- A terminal hydrogen atom (H)
- A triple bond between two carbon atoms (C≡C)
- A single bond between the second carbon and a third carbon atom
- The third carbon is bonded to a bromine atom (Br)
This gives the following condensed structural formula:
H–C≡C–C–Br
Step 2: Draw the Lewis Structure
- The first carbon (C1) is bonded to hydrogen and forms a triple bond with the second carbon (C2). This gives C1 a linear shape with 2 electron domains.
- C2 is in the middle of the triple bond (with C1) and a single bond to the third carbon (C3). It also has 2 electron domains, so it is linear.
- C3 is bonded to C2 and to Br and has two lone pairs if it’s saturated. So it has 4 electron domains, giving it a tetrahedral electron geometry.
Step 3: Assign Hybridization
- C1 (≡CH): 2 regions of electron density (H and C2) → sp hybridization
- C2 (≡C–C): 2 regions of electron density (C1 and C3) → sp hybridization
- C3 (C–Br): 4 regions of electron density (C2, Br, and 2 lone pairs) → sp³ hybridization
Correct Answer: E) sp / sp / sp³
Explanation
To determine the hybridization of atoms in CH≡C–CBr (bromo-propyne), we analyze the molecular geometry and bonding around each central atom. The first carbon (C1) is bonded to one hydrogen and triple-bonded to a second carbon (C2). In this arrangement, C1 has two regions of electron density: one sigma bond with hydrogen and one sigma bond as part of the triple bond. Triple bonds contain one sigma and two pi bonds, but hybridization only depends on sigma bonds or lone pairs—so C1 is sp hybridized.
The second carbon (C2) is in the middle of the molecule, triple-bonded to C1 and single-bonded to the third carbon (C3). It also has two regions of electron density—thus, it is also sp hybridized. This gives a linear geometry around both C1 and C2.
The third carbon (C3) is single-bonded to C2 and bromine (Br). Additionally, bromine has three lone pairs, but we’re focusing on the central atom C3. Carbon prefers a total of four bonds, so if C3 is saturated (i.e., tetravalent), it will also have hydrogens to satisfy its valency (if not explicitly shown). Therefore, it has four sigma bonds in total and no lone pairs, meaning it is sp³ hybridized, with tetrahedral geometry.
Thus, the hybridization states are:
- C1: sp
- C2: sp
- C3: sp³
This matches answer E) sp / sp / sp³.
