Question 2:Drag the words into the correct boxesmolecular orbitals are more likely to place electrons between the nuclei.molecular orbitals are less likely to place electrons between the nuclei.bondingantibondingantibondingantibondingmolecular orbitals are higher in energy than the corresponding atomic orbitalsbondingin isolated atomsbondingantibondingmolecular orbitals are lower in energy than the corresponding atomic orbitals inisolated atoms.electrons inelectrons inmolecular orbitals destabilize the molecule.molecular orbitals stabilize the molecule.Checkmolecular orbitals are identified by an asterisk.Reuse<> EmbedH-P

1. "molecular orbitals are more likely to place electrons between the nuclei" - This matches the…

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Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter10: Molecular Structure And Bonding Theories

Section: Chapter Questions

Problem 10.123QE

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Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. liS ability to bond to iron in hemoglobin is the reason that CO is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: MCO a. On the basis of electronegativities, would you expect the carbon atom or the oxgen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. 4-54 and 4-55.) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.
Given the following molecules and/or ions, characterize them according to the formatshown in the table below. The central atom of the molecule/ion is shown in boldface.The central atom or molecules marked with an asterisk (*) do not obey the octet rule. H3O+ A =Bonding electronsB =Sigma-bondsC= Nonbonding electron pairsD =Hybrid orbitals havingnonbonding electron pairsE =Hybrid orbitals usedF= 3-D structure(NAME)G =Polar (P) or nonpolar (NP) MOLECULE OR IONS LEWIS STRUCTURE A B C D E F G
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Given the following molecules and/or ions, characterize them according to the formatshown in the table below. The central atom of the molecule/ion is shown in boldface.The central atom or molecules marked with an asterisk (*) do not obey the octet rule. PCl5* OF2 SF4* A=Bonding electronsB= Sigma-bondsC =Nonbonding electron pairsD= Hybrid orbitals havingnonbonding electron pairsE =Hybrid orbitals usedF= 3-D structure (NAME)G =Polar (P) or nonpolar (NP) MOLECULE OR ION LEWIS STRUCTURE A B C D E F G
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What hybridization is required for central atoms that have a tetrahedral arrangement ofelectron pairs? A trigonal planar arrangement of electron pairs? A linear arrangement ofelectron pairs? How many unhybridized p atomic orbitals are present when a centralatom exhibits tetrahedral geometry? Trigonal planar geometry? Linear geometry? Whatare the unhybridized p atomic orbitals used for?
Explain what is wrong with each molecular geometry and provide the correct molecular geometry, given the numbers of lone pairs and bonding groups on the central atom. Match the words in the left column to the appropriate blanks in the sentences on the right. a bent a trigonal bipyramidal an octahedral a trigonal planar a square planar a trigonal pyramidal a tetrahedral a seesaw a linear In structure (a), four pairs of electrons give cause lone pair-bonded pair repulsions and would have Reset electron geometry. The lone pair would molecular geometry. Help In structure (b), five pairs of electrons give electron geometry. The lone pair occupies an equatorial position to minimize lone pair-bonded pair repulsions, and the molecule would have molecular geometry. In structure (c), six pairs of electrons give electron geometry. The two lone pairs would occupy opposite positions to minimize lone pair-lone pair repulsions, and the molecule would have molecular geometry.
Match each description with the VSEPR molecular geometry. All descriptions are of the regions around the central atom. (ER = electron region; ERG = electron region geometry; MG = molecular geometry; LP = lone pair; BA = bonding area) %3D 1 BA between 2 atoms [Choose] 4 ER (as 2 LP + 2 BA) or 3 ER (as 1 LP + 2 BA) [Choose] 3 ER consisting of 0 LP and З ВА [Choose] 4 ER consisting of 1 LP and З ВА [Choose ] 4 ER consisting of 0 LP and [Choose] 4 BA
ONDING Counting sigma and pi bonds in a small molecule Answer the questions below about the highlighted atom in this Lewis structure: H H | | + H- | | H H •C-C-C=N: In how many sigma bonds does the highlighted atom participate? In how many pi bonds does the highlighted atom participate? What is the orbital hybridization of the highlighted atom? 0 0 0 X 1/5
2
Consider the Lewis Structure for the molecule, PBra. On your own, make a sketch of the VSEPR shape, and use that to answer the following questions: Br- -P-Br: :Br: Part A What is the electronic geometry of this molecule? O tetrahedral O trigonal pyramidal O bent O see-saw trigonal bipyramidal O t-shaped O trigonal planar O O O O OO

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