Concept explainers
(a)
Interpretation:
Whether the presence of one or more lone pairs on the central atom is desired or not in diagram (a) has to be indicated.
Concept Introduction:
Valence Shell Electron Pair Repulsion model predicts shape by inclusion of bond angles and most distant arrangement of atoms that leads to minimum repulsion. For the molecules that have no lone pairs around the central atom the bonded-atom unshared -pair arrangement is decided by the table as follows:
In order to determine the shape the steps to be followed are indicated as follows:
- 1. Lewis structure of molecule should be written.
- 2. The type electron arrangement around the central atom should be identified around the central atom. This essentially refers to determination of bond pairs and unshared or lone pairs around central atoms.
- 3. Then bonded-atom unshared -pair arrangement that can maximize the distance of electron pairs about central atom determines the shape.
(b)
Interpretation:
Whether the presence of one or more lone pairs on the central atom is desired or not in diagram (b) has to be indicated.
Concept Introduction:
Refer to part (a).
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
- Consider an iodine tetrafluoride cation (IF4+): What is the elemental symbol of its central atom? How many valence electrons does this molecule contain? How many lone electron pairs surround the central atom? How many regions of electron density surround the central atom? For remaining items in this question, place the appropriate letter in the box: What is the electron geometry of this molecule? A = linear B = trigonal planar C = tetrahedral D = trigonal bipyramidal E = octahedral What is the molecular shape of this molecule? A = tetrahedral B = trigonal pyramid C = bent or angular D = T-shape E = Sawhorse or seesaw Is this a polar molecule? A = Yes B = No What is the hybridization of the central atom? A = sp B = sp2 C = sp3 D = sp3d E = sp3d2arrow_forwardExcept for nitrogen, the elements of Group 5A(15) all form pentafluorides and most form pentachlorides. The chlorine atoms of PCI5 can be replaced with fluorine atoms one at a time to give, successively, PCI4F, PCI3F2, .. PF5. (a) Given the relative sizes of F and CI, would you expect the first two F substitutions to be at axial or equatorial positions? O axial equatorial This answer has not been graded yet. (b) Which of the five fluorine-containing molecules have no dipole moment? (Select all that apply.) PCIF4 PCI3F2 PCI2F3 PF5 PCI4Farrow_forwardConsider a hypothetical molecule with the formula X4Y2Z. The atoms X, Y, and Z have electronegativities of 2.5, 3.0, and 3.5, respectively. The molecule adopts a tetrahedral geometry with X atoms at the corners, Y atoms in the middle of the edges, and Z atoms at the center of the molecule. Explain the polarity of this molecule and the factors contributing to it.arrow_forward
- Please don't provide handwriting solutionarrow_forward1) Why is it important to understand the molecular geometry or shape of molecules and ions in chemistry? H2Se CCl4 CH2Cl2 HCN NH3Cl+ NH2- CH3COCH3 CH3COOH 2) Among the eight molecules and ions studied in this experiment, (a) which molecule(s) or ion(s) are polar, and which ones are non-polar? (b) What is the difference between polar and non-polar molecules or ions? (c) What two factors dictate the polarity of the molecule or ion?arrow_forwardThere are two molecules with the formula C2H6O. Draw electron-dot structures for both. (The connection of atoms is different in the two structures.)arrow_forward
- Depending on the protein under study, the bond angle of a drug molecule can become critical to successfully deactivating a viral protein. For this reason, chemists are frequently concerned with the 3D shape of their molecules and their bond angles. The oxygen atom shown in the structure has a tetrahedral electronic geometry, meaning we would predict that it would have 109.5° bond angles. However, the actual structure, the bond angles are smaller than 109.5°. Explain why this compression occurs. Make sure to discuss what’s happening around/what groups are present around the oxygen atom.arrow_forwardThe hydrocarbon cyclobutane, C4H8, is represented above. At high temperatures, cyclobutane quickly decomposes into ethene, C2H4. (see attached image) (a) Draw a Lewis electron-dot diagram of the ethene molecule in the following box, and estimate the value of the H−C−H bond angle in ethene.arrow_forwardDoes your molecular exhibit resonance? If so, show all possible forms. Does your molecule have any isomers? If so, show Lewis structures of them as well. for CH3CH2OHarrow_forward
- Answer the questions in the table below about the shape of the sulfur tetrabromide (SBr_(4)) molecule. How many electron groups are around the central sulfur atom? Note: one "electron group" means one lone pair, one single bond, one double bond, or one triple bond. What phrase best describes the arrangement of these electron groups around the central sulfur atom? (You may need to use the scrollbar to see all the choices.arrow_forwardIodine trichloride, ICl₃, is a bright yellow solid and an oxidizing agent. Based on your Lewis structure for ICl₃, how many electron domains are on the central atom?arrow_forward0=c=0; Lewis electron-dot diagrams for CO2 and SO2 are given above. The molecular geometry and polarity of the two substances are the same because the molecular formulas are similar the same because C and S have similar electronegativity values different because the lone pair of electrons on the S atom make it the negative end of a dipole D different because S has a greater number of electron domains (regions of electron density) surrounding it than C hasarrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning