Concept explainers
Piperine, the active ingredient in black pepper, has this Lewis structure.
- (a) Give the values for the indicated bond angles.
- (b) What is the hybridization of the nitrogen?
- (c) What is the hybridization of the oxygens?
(a)
Interpretation:
The indicated bond angles has to be given.
Concept Introduction:
Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.
Explanation of Solution
The given structure of Piperine is given below:
.
The bond angles to be determined are given below:
.
Consider the
Consider the
Consider the
(b)
Interpretation:
The hybridization for nitrogen atom in the molecule has to be designated.
Concept Introduction:
Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.
Geometry of different types of molecule with respect to the hybridizations are mentioned below,
Explanation of Solution
The given structure of Piperine is shown below:
.
Consider the nitrogen atom. There will be four electron regions around nitrogen atom. It is a type of
Consider the right carbon. There will be three electron regions in the molecule and hence the electron-region geometry will be triangular planar. For a molecule having tetrahedral geometry, the hybridization will be
(c)
Interpretation:
The hybridization for oxygen atoms in the molecule has to be designated.
Concept Introduction:
Refer to (b).
Explanation of Solution
The given structure of Piperine is shown below:
.
Consider the nitrogen atom. There will be four electron regions around nitrogen atom. It is a type of
Consider the oxygen atoms with single bonds. There will be four electron regions around both oxygen atoms. They are a type of
Consider the double bonded oxygen atom. There will be three electron regions around oxygen atom. It is a type of
Want to see more full solutions like this?
Chapter 7 Solutions
Chemistry: The Molecular Science
- (a) Write a single Lewis structure for SO3 , and determine the hybridization at the S atom. (b) Are there other equivalent Lewis structures for the molecule? (c) Would you expect SO3 to exhibit delocalized π bonding?arrow_forwardA useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H3CCN. It is present in paint strippers.(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.(c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.arrow_forwardButadiene, C4H6, is a planar molecule that has the followingcarbon–carbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle C—C bond length in butadiene (1.48 Å) is a little shorter than the average C—C single bond length (1.54 Å). Does this imply that the middle C—C bond in butadiene is weaker or stronger than the average C—C single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle C—C bond.arrow_forward
- Organic compounds (a) A simple molecule with alternating single and double bonds is buta-1,3-diene. (i) Draw its skeletal structure. (ii) Write its molecular formula and empirical formula. (iii) How many carbon atoms have sp2 hybridization in buta-1,3-diene?arrow_forward(a) Describe the hybridization of the central atom of a molecule with a see-saw shape. (b) Describe the hybridization of the central atom of a molecule with a trigonal planar shape. (c) Describe the hybridization of the central atom of a molecule with a trigonal bipyramidal shape.arrow_forwardIn ozone, O3, the two oxygen atoms on the ends of the molecule are equivalent to one another. (a)What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the π electrons? (d) How many electrons are delocalized in the π system of ozone?arrow_forward
- (a) Find the angle u between adjacent nearest-neighbor bonds in the silicon lattice. Recall that each silicon atom is bonded to four of its nearest neighbors.The four neighbors form a regular tetrahedron— a pyramid whose sides and base are equilateral triangles. (b) Find the bond length, given that the atoms at the corners of the tetrahedron are 388 pm apart.arrow_forward4. (a) Draw the shape of the atomic valence orbitals formed by the overlaping of two fluoride 2p atomic orbitals. (b) Draw the molecular orbital diagrams for F2 and F2*. Identify their bond order and magnetic properties. (c) An unstable nucleus exhibit radioactivity. (i) Explain how the number of protons and neutrons in a radioactive nucleus can be used to predict its probable mode decay. (ii) Illustrate your answer in (i) with a schematic graph.arrow_forwardDescribe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds. (a) H3PO4, phosphoric acid, used in cola soft drinks (b) NH4NO3, ammonium nitrate, a fertilizer and explosive (c) S2Cl2, disulfur dichloride, used in vulcanizing rubber (d) K4[O3POPO3], potassium pyrophosphate, an ingredient in some toothpastesarrow_forward
- Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.(a) H3PO4, phosphoric acid, used in cola soft drinks(b) NH4NO3, ammonium nitrate, a fertilizer and explosive(c) S2Cl2, disulfur dichloride, used in vulcanizing rubber(d) K4[O3POPO3], potassium pyrophosphate, an ingredient in some toothpastesarrow_forwardIn hydrogen isocyanide molecules (HCN), both carbon (C: 1s^2 2s^2 2p^2 ) and nitrogen (N: 1s^2 2s^2 2p^3 ) atoms undergo sp hybridization. (a) Use Orbital Hybridization theory to determine the molecular shape of HCN molecules. (b) Explain how the C atom binds to the N atom in HCN molecules.arrow_forward(a) Triazine, C3 H3 N3, is like benzene except that in triazineevery other C¬H group is replaced by a nitrogen atom.Draw the Lewis structure(s) for the triazine molecule. (b) Estimatethe carbon–nitrogen bond distances in the ring.arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY