**Title: Understanding Lewis Structures - Bonding and Hybridization****Objective:** Analyze the Lewis structure and answer questions regarding the highlighted atom.**Structure:**The molecule depicted in the Lewis structure is as follows:- There is a chain of carbon atoms: three carbon (C) atoms are present.- The first carbon atom is bonded to three hydrogen (H) atoms.- The second carbon atom is bonded to two hydrogen (H) atoms and directly to the first and third carbon atoms.- The third carbon atom is highlighted in red and is triple-bonded to a nitrogen (N) atom, which has a lone pair of electrons.**Questions:**1. **In how many sigma bonds does the highlighted atom participate?** - ❒2. **In how many pi bonds does the highlighted atom participate?** - ❒3. **What is the orbital hybridization of the highlighted atom?** - ❒**Key Concepts:**- **Sigma (σ) Bonds**: Formed by the direct overlap of orbitals between two atoms.- **Pi (π) Bonds**: Formed by the sideways overlap of p orbitals above and below the plane of the atoms involved.- **Hybridization**: The concept of mixing atomic orbitals into new hybrid orbitals suitable for the pairing of electrons to form chemical bonds.*Note for Students:* Use the Lewis structure to identify the types of bonds and predict the hybridization.

Sigma bondsSigma bonds are formed by head to head overlapping of atomic orbitals. All single bonds are sigma bond.Pi bondsPi bonds are formed by the lateral overlapping of 2 atomic orbitals.Double bond (-C=C)- consist of 1 sigma bond and 1 pi bond.Triple bond (-C≡C-) consist of 1 sigma and 2 pi bonds.Hybridization:Hybridization of carbon (C) in hydrocarbons:• Alkane-C-C- in this case all bonds are single and the hybridization of the carbon is sp3.In sp3 hybridization all bonds are (σ) sigma bond.• Alkene-C=C- in this case double bond is there and those carbon atoms contains double bond between them are sp2 hybridized.In sp2 hybridization, we have 3 (σ) sigma bonds and 1 π bond• Alkyne-C≡C- in this case triple bond is there and those carbon atoms containing triple bond between them are sp hybridized.In sp hybridization, we have 2 (σ) sigma bonds and 2 π bond•Number of Sigma bonds the highlighted atom participated =2There are 2 Sigma bonds (-C-C- and -C-N) the highlighted atom participated.• Number of pi bonds the highlighted atom participated = 2Since, the highlighted atom bonded to another N atom by a triple bond that means there are 2 pi bonds the highlighted atom participated.• Orbital hybridization of the highlighted atom = sp-C≡N in this case triple bond is there and those carbon atoms contains triple bond between them are sp hybridized.In sp hybridization, we have 2 sigma(σ) bonds and 2 π bonds.• Number of Sigma bonds the highlighted atom participated = 2• Number of pi bonds the highlighted atom participated = 2• Orbital hybridization of the highlighted atom = sp

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Chemistry

Answer the questions below about the highlighted atom in this Lewis H H ││ HIC C-C=N: | | H H 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? 0 0 0

Answer the questions below about the highlighted atom in this Lewis H H ││ HIC C-C=N: | | H H 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? 0 0 0

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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