Ochem lecture 3: sp3, sp2, sp and implications

Context of in-class discussion

• Free to talk during clicker questions; students may discuss with roommates or hot mates as part of the activity.

• Instructor noted permission to walk around and engage with others if needed.

• Some students tried to correct slides or images mid-discussion (e.g., confusion about which picture was shown).

• A student attempt to list or count atoms/electrons/nodes appeared garbled (e.g., sequences like 56891011121111, 131416171920, etc.), with an explicit word “atoms” followed by later phrases like “six six, I believe. 34. 56. Yep.” — unclear transcription for this portion; no clear numerical pattern established.

Key topic introduced: carbon hybridization and bond count

• The discussion centers on how many substituents (attachments) a carbon atom has, and how that relates to hybridization states $sp^3$, $sp^2$, and $sp$.

• Mentioned sequence: four $\rightarrow$ three $\rightarrow$ two $\rightarrow$ one, linked to hybridization levels ($sp^3$, $sp^2$, $sp$) and the idea of tetrahedral geometry.

• The student notes that double bonds affect the number of attachments considered as sigma bonds:

  • A carbon with a double bond still must have four bonds total, but one of those bonds is a pi bond, not a sigma bond.

  • Therefore, when a carbon has a double bond, it typically has three sigma bonds (to two hydrogens and to the other carbon, in many cases), with the fourth valence used in the pi bond.

Hybridization recap and relationships

• General idea: carbon (and other second-row elements) can hybridize to form sigma bonds that determine molecular geometry.

• Hybridization states and typical geometries:

  • $sp^3$: four sigma bonds, tetrahedral arrangement $\sim109.5^\text{o}$

  • $sp^2$: three sigma bonds, trigonal planar arrangement $\sim120^\text{o}$

  • $sp$: two sigma bonds, linear arrangement $\sim180^\text{o}$

• Important distinction: the total valence of carbon is four, but the presence of pi bonds (from double or triple bonds) uses p orbitals and does not contribute to the count of sigma bonds.

Detailed mapping of hybridization to bond count

• $sp^3$ (tetrahedral)

  • Sigma bonds: 4

  • Pi bonds: 0

  • Typical example: methane, $CH_4$

  • Electron-domain geometry corresponds to four regions of electron density around carbon.

• $sp^2$ (trigonal planar with a remaining p orbital for pi bonding)

  • Sigma bonds: 3

  • Pi bonds: 1 (from the remaining p orbital overlapping with another p orbital)

  • Typical examples: ethene, $CH<em>2=CH</em>2$; in each carbon, there are three sigma bonds (two C–H or C–H and C–C) and one pi bond in the C=C bond.

  • With a double bond, one of the carbon’s valence sites is used as a pi bond; thus the carbon has three sigma bonds, not four.

• $sp$ (linear, two sigma bonds and two pi bonds in a triple bond)

  • Sigma bonds: 2

  • Pi bonds: 2 (as in a triple bond consists of one sigma and two pi bonds)

  • Typical example: acetylene, $HC\equiv CH$ (each carbon has one H and one sigma bond to the other carbon; the triple bond includes two pi bonds and one sigma bond between carbons)

Explanation of the double bond effect on attachment count

• When a carbon forms a double bond, one of its valence interactions is a pi bond.

• Pi bonds do not count toward the sigma bond count.

• Carbons involved in double bonds typically have fewer sigma-bonding partners:

  • They usually have three sigma bonds.

  • Example: In $CH<em>2=CH</em>2$, each carbon has two C–H sigma bonds and one C–C sigma bond.

  • The remaining valence for that carbon is used in the pi bond to the other carbon.

• This leads to a carbon in a C=C bond being described as $sp^2$ with three sigma bonds and one pi bond.

Numerical and symbolic references (LaTeX)

• Bond-count mapping across hybridizations:

  • $sp^3: \text{sigma bonds} = 4$, $\text{pi bonds} = 0$

  • $sp^2: \text{sigma bonds} = 3$, $\text{pi bonds} = 1$

  • $sp: \text{sigma bonds} = 2$, $\text{pi bonds} = 2$

• Geometric angles:

  • $\theta_{\text{tetrahedral}} \approx 109.5^\text{o}$

  • $\theta_{\text{trigonal planar}} \approx 120^\text{o}$

  • $\theta_{\text{linear}} = 180^\text{o}$

• Total valence of carbon is four: $n_{\text{valence}} = 4$, with the sum of sigma and pi bonds accounting for this valence when considering all bonds.

Concrete examples to anchor understanding

• Methane: $CH_4$

  • Carbon: $sp^3$, 4 sigma bonds, tetrahedral geometry.

• Ethene: $CH<em>2=CH</em>2$

  • Each carbon: $sp^2$, 3 sigma bonds (two C–H, one C–C), 1 pi bond in the C=C; molecule is planar around the double bond.

• Ethyne: $HC\equiv CH$

  • Each carbon: $sp$, 2 sigma bonds (one to H, one to the other C), 2 pi bonds in the C$\equiv$C triple bond; molecule is linear.

Practical implications and relevance

• Predicting geometry and reactivity:

  • Saturated hydrocarbons ($sp^3$) exhibit simple tetrahedral geometry.

  • Unsaturated hydrocarbons ($sp^2$ and $sp$) have planar or linear geometries.

  • These geometries influence physical properties and reaction mechanisms.

• Planarity and conjugation:

  • $sp^2$ centers enable pi-systems and the potential for conjugation (e.g., in benzene-like rings, though not explicitly covered).

  • $sp$ centers contribute to extended linear conjugation.

• Isomerism and reactivity:

  • The presence of double or triple bonds changes the number of attached substituents.

  • They impose restrictions on rotation, affecting stereochemistry and reaction pathways.

Real-world relevance and connections

• Hybridization concepts are fundamental to:

  • Organic synthesis

  • Materials science (e.g., polymers with specific geometries)

  • Biochemistry, where bond arrangements dictate molecular function.

• Understanding hybridizations helps explain:

  • Why certain hydrocarbons are gases, liquids, or solids at room temperature.

  • Why some hydrocarbons are more reactive in addition or elimination reactions.

Miscellaneous transcript notes and context

• The speaker noted casual classroom practices:

  • Discussion with roommates during clicker questions.

  • Correction about a slide/photo, illustrating real-time classroom dynamics and potential distractions.

• Unclear numerical sequences in the transcript (e.g., "56891011121111", "131416171920", etc.):

  • Followed by "atoms" and later musings ("six six, I believe. 34. 56. Yep.").

  • These appear to be garbled or unfinished student notes.

  • They do not provide meaningful content for the hybridization topic and should be treated as transcription noise.

Summary takeaways

• Carbon hybridization correlates with the number of sigma bonds and the presence of pi bonds:

  • $sp^3\rightarrow 4\text{ sigma bonds (tetrahedral)}$

  • $sp^2\rightarrow 3\text{ sigma bonds + 1 pi bond (trigonal planar)}$

  • $sp\rightarrow 2\text{ sigma bonds + 2 pi bonds (linear)}$

• Double bonds reduce the number of sigma attachments around the carbon by consuming one valence bond as a pi interaction, leading to three sigma bonds for the carbon involved in the double bond.

• Visualizing geometry via angles helps predict molecular shape and reactivity: $109.5^\text{o}, 120^\text{o}, 180^\text{o}$ for $sp^3$, $sp^2$, and $sp$, respectively.