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Chapter 8: Q35E (page 450)
Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
No matter how the molecular orbitals are arranged, an odd number of electrons can never be coupled. It will remain paramagnetic indefinitely.
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Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
(a) \({\rm{H}}\) and \({{\rm{H}}_{\rm{2}}}\)
(b) \({\rm{N}}\)and \({{\rm{N}}_{\rm{2}}}\)
(c) \({\rm{O}}\)and \({{\rm{O}}_{\rm{2}}}\)
(d) \({\rm{C}}\)and \({{\rm{C}}_{\rm{2}}}\)
(e) \({\rm{B}}\)and \({{\rm{B}}_{\rm{2}}}\)
A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, \({{\rm{H}}_{\rm{3}}}{\rm{CCN}}\). It is present in paint strippers. (a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule. (b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form \({\rm{\sigma }}\) bonds. (c) Describe the atomic orbitals that form the \({\rm{\pi }}\) bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
True or false: Boron contains \({\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{1}}}\)valence electrons, so only one \({\rm{p}}\)- orbital is needed to form molecular orbitals.
Calculate the bond order for an ion with this configuration: \({\left( {{{\bf{\sigma }}_{{\bf{2s}}}}} \right)^{\bf{2}}}{\left( {{\bf{\sigma }}_{{\bf{2s}}}^{\bf{*}}} \right)^{\bf{2}}}{\left( {{{\bf{\sigma }}_{{\bf{2px}}}}} \right)^{\bf{2}}}{\left( {{{\bf{\pi }}_{{\bf{2py}}}}{\bf{,}}{{\bf{\pi }}_{{\bf{2pz}}}}} \right)^{\bf{4}}}{\left( {{\bf{\pi }}_{{\bf{2py}}}^{\bf{*}}{\bf{,\pi }}_{{\bf{2pz}}}^{\bf{*}}} \right)^{\bf{3}}}\).
Another acid in acid rain is nitric acid, HNO3, which is produced by the reaction of nitrogen dioxide, NO2, with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO2? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)
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