Question

answer the questions below about the highlighted atom in this lewis structure:

in how many sigma bonds does the highlighted atom participate?

in how many pi bonds does the highlighted atom participate?

what is the orbital hybridization of the highlighted atom?

Explanation:

Part 1: Sigma Bonds

Step1: Identify bonds of highlighted C

The highlighted C (carbon) is bonded to one H (single bond) and one C (triple bond? Wait, no, the structure: the highlighted C is in \( \text{C} \equiv \text{C} \)? Wait, no, the Lewis structure is \( \text{H}-\text{C}-\text{C} \equiv \text{C}-\text{H} \)? Wait, no, looking at the structure: the middle is \( \text{C}-\text{C} \equiv \text{C} \)? Wait, the highlighted C is the rightmost in the triple bond? Wait, the structure: let's parse the bonds. The highlighted C (red) is connected to a C (left, with triple bond? Wait, no, the bonds: single bonds are sigma, double has 1 sigma 1 pi, triple has 1 sigma 2 pi. The highlighted C: it has a single bond to H (sigma), and a triple bond to the left C? Wait, no, the structure is: H-C (with H above, below) - C (with H above, below) ≡ C - H. Wait, the highlighted C is the right C in the triple bond? Wait, no, the bonds: the highlighted C (red) is bonded to H (single bond, sigma) and to the left C (triple bond: 1 sigma, 2 pi). Wait, no, let's count: for the highlighted C (let's call it C3), it has:

• Bond to H: single bond (1 sigma)
• Bond to C2 (left C): triple bond? Wait, no, the structure is H - C1 - C2 ≡ C3 - H, with C1 having H above, below; C2 having H above, below. Wait, no, the drawing: H | H - C - C ≡ C - H | | H H. So C1: bonded to H (left), H (top), H (bottom), and C2 (right, single bond? Wait, no, the bonds: C1 is connected to H (left, single), H (top, single), H (bottom, single), and C2 (right, single? Then C2 is connected to C3 (right, triple bond? Wait, no, the middle is C2 - C3 with a triple bond? Wait, no, the structure is:

C1: bonded to H (left), H (top), H (bottom), C2 (right, single bond: sigma)

C2: bonded to C1 (left, single: sigma), H (top), H (bottom), and C3 (right, triple bond: 1 sigma, 2 pi)

C3: bonded to C2 (left, triple bond: 1 sigma, 2 pi) and H (right, single bond: sigma)

Wait, no, the highlighted C is C3? Wait, the problem says "highlighted atom" is the red C, which is in \( \text{C} \equiv \text{C} \) (wait, no, the structure shows \( \text{C} \equiv \text{C} \)? Wait, the Lewis structure: H - C (with H above, below) - C (with H above, below) ≡ C - H. So the bonds:

• C3 (highlighted) has:

• Bond to H (right): single bond (sigma, 1)

• Bond to C2 (left): triple bond (1 sigma, 2 pi)

So total sigma bonds: 1 (from H) + 1 (from C2 triple bond's sigma) = 2? Wait, no, wait: in a triple bond, there is 1 sigma and 2 pi. So the bond between C2 and C3: 1 sigma, 2 pi. Then C3's bonds:

• C3 - H: single bond (sigma, 1)

• C3 - C2: triple bond (sigma, 1; pi, 2)

So total sigma bonds: 1 + 1 = 2? Wait, no, wait: let's re-express. For any carbon, the number of sigma bonds is equal to the number of atoms it's bonded to (since each single bond is sigma, double has 1 sigma, triple has 1 sigma). Wait, the highlighted C: how many atoms is it bonded to? It's bonded to H (1 atom) and C2 (1 atom). So two atoms, so two sigma bonds? Wait, no: single bond to H (1 sigma), triple bond to C2 (1 sigma). So total sigma bonds: 2.

Step2: Confirm sigma bond count

Each single bond is 1 sigma, double has 1 sigma 1 pi, triple has 1 sigma 2 pi. So for the highlighted C:

• Bond to H: single (1 sigma)

• Bond to C (left): triple (1 sigma)

Total sigma bonds: \( 1 + 1 = 2 \).

Step1: Identify pi bonds in highlighted C's bonds

The highlighted C is in a triple bond with the left C. A triple bond has 2 pi bonds (since triple bond = 1 sigma + 2 pi). The other bond (to H) is single (no pi). So pi bonds: 2.

Step2: Confirm pi bond count

Triple bond contributes 2 pi bonds. So the highlighted C participates in 2 pi bonds.

Step1: Determine steric number (SN)

Steric number (SN) = number of sigma bonds + number of lone pairs. The highlighted C has:

• Sigma bonds: 2 (to H and to C)

• Lone pairs: 0 (carbon in organic compounds, bonded to 2 atoms, no lone pairs)

SN = 2 + 0 = 2.

Step2: Determine hybridization from SN

• SN = 2: sp hybridization (linear geometry, 2 hybrid orbitals, used for sigma bonds; remaining 2 p orbitals for pi bonds, which matches triple bond: 1 sigma (sp-sp), 2 pi (p-p)).

So hybridization is sp.

Answer:

(Sigma Bonds): 2

Part 2: Pi Bonds