![CHEMISTRY >CUSTOM<](https://www.bartleby.com/isbn_cover_images/9781309097182/9781309097182_largeCoverImage.gif)
(a)
Interpretation:
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in
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.
The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.
(a)
![Check Mark](/static/check-mark.png)
Answer to Problem 11.41P
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in
Explanation of Solution
The Lewis structure of
Iodine forms two single bonds with two fluorine atoms and three lone pairs are present on it so the hybridization of iodine in
The
The partial orbital diagram for an isolated
The partial orbital for hybridized
One s orbital, three p orbitals and one d orbital of central atom iodine combine to form five
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in
(b)
Interpretation:
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in
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.
The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.
(b)
![Check Mark](/static/check-mark.png)
Answer to Problem 11.41P
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in
Explanation of Solution
The Lewis structure of
Iodine forms three single bonds with three chlorine atoms and two lone pairs are present on it so five hybrid orbitals are required. The hybridization of
The atomic number of iodine is 53 so its electronic configuration is
The partial orbital diagram for an isolated
The partial orbital for hybridized
One s orbital, three p orbitals and one d orbital of central atom iodine combine to form five
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in
(c)
Interpretation:
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in
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.
The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.
(c)
![Check Mark](/static/check-mark.png)
Answer to Problem 11.41P
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom xenon in
Explanation of Solution
The Lewis structure of
Xenon forms four single bonds with four fluorine atoms and one double bond with oxygen and one lone pair is present on it so six hybrid orbitals are required. The hybridization of xenon in
The atomic number of xenon is 54 so its electronic configuration is
The partial orbital diagram for an isolated
The partial orbital for hybridized
One s orbital, three p orbitals and two d orbitals of central atom xenon combine to form six
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom xenon in
(d)
Interpretation:
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in
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.
The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.
(d)
![Check Mark](/static/check-mark.png)
Answer to Problem 11.41P
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom boron in
Explanation of Solution
The Lewis structure of
Boron forms one single bond with hydrogen and two single bonds with two fluorine atoms so three hybrid orbitals are required. The hybridization of boron in
The atomic number of boron is 5 so its electronic configuration is
The partial orbital diagram for an isolated
The partial orbital for hybridized
One s orbital and two p orbitals of central atom boron combine to form three
The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom boron in
Want to see more full solutions like this?
Chapter 11 Solutions
CHEMISTRY >CUSTOM<
- 93 = Volume 93 = 5.32× 10 3 -23 ст a √ 1073 5.32× 10 3 cm³arrow_forwardASP.....arrow_forwardQuestion 7 (10 points) Identify the carboxylic acid present in each of the following items and draw their structures: Food Vinegar Oranges Yogurt Sour Milk Pickles Acid Structure Paragraph ✓ BI UAE 0118 + v Task: 1. Identify the carboxylic acid 2. Provide Name 3. Draw structure 4. Take a picture of your table and insert Add a File Record Audio Record Video 11.arrow_forward
- Check the box under each structure in the table that is an enantiomer of the molecule shown below. If none of them are, check the none of the above box under the table. Molecule 1 Molecule 2 IZ IN Molecule 4 Molecule 5 ZI none of the above ☐ Molecule 3 Х IN www Molecule 6 NH Garrow_forwardHighlight each chiral center in the following molecule. If there are none, then check the box under the drawing area. There are no chiral centers. Cl Cl Highlightarrow_forwardA student proposes the following two-step synthesis of an ether from an alcohol A: 1. strong base A 2. R Is the student's proposed synthesis likely to work? If you said the proposed synthesis would work, enter the chemical formula or common abbreviation for an appropriate strong base to use in Step 1: If you said the synthesis would work, draw the structure of an alcohol A, and the structure of the additional reagent R needed in Step 2, in the drawing area below. If there's more than one reasonable choice for a good reaction yield, you can draw any of them. ☐ Click and drag to start drawing a structure. Yes No ロ→ロ 0|0 G Х D : ☐ பarrow_forward
- टे Predict the major products of this organic reaction. Be sure to use wedge and dash bonds when necessary, for example to distinguish between different major products. ☐ ☐ : ☐ + NaOH HO 2 Click and drag to start drawing a structure.arrow_forwardShown below are five NMR spectra for five different C6H10O2 compounds. For each spectrum, draw the structure of the compound, and assign the spectrum by labeling H's in your structure (or in a second drawing of the structure) with the chemical shifts of the corresponding signals (which can be estimated to nearest 0.1 ppm). IR information is also provided. As a reminder, a peak near 1700 cm-1 is consistent with the presence of a carbonyl (C=O), and a peak near 3300 cm-1 is consistent with the presence of an O–H. Extra information: For C6H10O2 , there must be either 2 double bonds, or 1 triple bond, or two rings to account for the unsaturation. There is no two rings for this problem. A strong band was observed in the IR at 1717 cm-1arrow_forwardPredict the major products of the organic reaction below. : ☐ + Х ك OH 1. NaH 2. CH₂Br Click and drag to start drawing a structure.arrow_forward
- NG NC 15Show all the steps you would use to synthesize the following products shown below using benzene and any organic reagent 4 carbons or less as your starting material in addition to any inorganic reagents that you have learned. NO 2 NC SO3H NO2 OHarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardShow work...don't give Ai generated solutionarrow_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
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305957404/9781305957404_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781259911156/9781259911156_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305577213/9781305577213_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780078021558/9780078021558_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079373/9781305079373_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781118431221/9781118431221_smallCoverImage.gif)