
Concept explainers
(a)
Interpretation:
The electron geometries of all nonhydrogen atoms in the four listed species are to be determined.
Concept introduction:
Electron geometry around an atom is determined on the basis of the Valence Shell Electron Pair Repulsion (VSEPR) Theory. Electron geometry describes the orientation of the electron groups in an atom’s valence shell. An electron group is a lone pair or a bond between the two atoms. The bond, whether a single, double or triple, counts as just one electron group.
Since all electrons have the same charge, the electron groups repel each other. They try to move (orient themselves) as far away from each other as possible in order to minimize these repulsions. This results in a linear geometry (
The number of electron groups and geometry is determined on the basis of the Lewis structure of the molecule/ion.
(b)
Interpretation:
The hybridization of all nonhydrogen atoms in the given four species is to be determined.
Concept introduction:
The concept of hybridization of atomic orbitals is used in Valence Bond (VB) Theory to account for the electron and molecular geometry around an atom. A hybrid orbital is a combination of one or more atomic orbitals from the valence shell of an atom. It typically involves an s orbital and a number of p orbitals from the valence shell, resulting in the same total number of hybrid orbitals of the same energy and shape. In heavy atoms, those from Group 3 onward, the valence shell d orbital may also be involved if the atom has an expanded octet. The orientation of these orbitals is same as the electron geometry of the atom. The number of hybrid orbitals required is the same as the number of electron groups. If the number of electron groups is two, two hybrid orbitals are needed. These are formed by a combination of the s and one p orbital, giving

Trending nowThis is a popular solution!

Chapter 3 Solutions
Organic Chemistry: Principles And Mechanisms
- V Biological Macromolecules Drawing the Haworth projection of an aldose from its Fischer projection Draw a Haworth projection of a common cyclic form of this monosaccharide: H C=O HO H HO H H OH CH₂OH Explanation Check Click and drag to start drawing a structure. Xarrow_forwardComplete the mechanismarrow_forwardComplete the mechanismarrow_forward
- 8 00 6 = 10 10 Decide whether each of the molecules in the table below is stable, in the exact form in which it is drawn, at pH = 11. If you decide at least one molecule is not stable, then redraw one of the unstable molecules in its stable form below the table. (If more than unstable, you can pick any of them to redraw.) Check OH stable HO stable Ounstable unstable O OH stable unstable OH 80 F6 F5 stable Ounstable X Save For Later Sub 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy C ཀྭ་ A F7 매 F8 F9 4 F10arrow_forwardJust try completing it and it should be straightforward according to the professor and TAs.arrow_forwardThe grading is not on correctness, so if you can just get to the correct answers without perfectionism that would be great. They care about the steps and reasoning and that you did something. I asked for an extension, but was denied the extension.arrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning

