Concept explainers
(a)
Interpretation:
Whether the hybridization of each carbon in the right-hand ring is the same or not is to be determined.
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as a single-electron group.
(b)
Interpretation:
Whether the hybridization of the oxygen atom in the center ring is the same as that present in the
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as a single-electron group.
(c)
Interpretation:
The number of carbon-oxygen
Concept introduction:
The sigma bond is the covalent bond that is formed by the axial overlap of orbitals. The pi bond is formed by the lateral overlap of orbitals. Both types of bonds can be bonding or anti-bonding interaction. Bonding interaction takes place when there is an overlap of orbitals in the same phase and antibonding interaction occurs when the orbital overlap in opposite phase. Sigma bond is stronger than the pi bond.
All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.
(d)
Interpretation:
Whether all the lone pairs on oxygen atoms occupy the same type of hybrid orbitals or not is to be determined.
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Want to see the full answer?
Check out a sample textbook solutionChapter 11 Solutions
MOLECULAR NATURE OF MATTER ALEKS ACCESS
- Show work with explanation needed. Don't give Ai generated solutionarrow_forward7. Calculate the following for a 1.50 M Ca(OH)2 solution. a. The concentration of hydroxide, [OH-] b. The concentration of hydronium, [H3O+] c. The pOH d. The pHarrow_forwardA first order reaction is 46.0% complete at the end of 59.0 minutes. What is the value of k? What is the half-life for this reaction? HOW DO WE GET THERE? The integrated rate law will be used to determine the value of k. In [A] [A]。 = = -kt What is the value of [A] [A]。 when the reaction is 46.0% complete?arrow_forward
- 3. Provide the missing compounds or reagents. 1. H,NNH КОН 4 EN MN. 1. HBUCK = 8 хно Panely prowseful kanti-chuprccant fad, winddively, can lead to the crading of deduc din-willed, tica, The that chemooices in redimi Грин. " like (for alongan Ridovi MN نيا . 2. Cl -BuO 1. NUH 2.A A -BuOK THE CF,00,H Ex 5)arrow_forward2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forward
- The following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forward
- The following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forwardSelect all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward3. Draw diagrams to represent the conjugation in these molecules. Draw two types of diagram: a. Show curly arrows linking at least two different ways of representing the molecule b. Indicate with dotted lines and partial charges (where necessary) the partial double bond (and charge) distribution H₂N* H₂N -NH2arrow_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