Chapter 8: Q9 E (page 446)
Why is the concept of hybridization required in valence bond theory?
Short Answer
Hybridization is necessary to determine the molecular geometry of complex.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 8: Q9 E (page 446)
Why is the concept of hybridization required in valence bond theory?
Hybridization is necessary to determine the molecular geometry of complex.
All the tools & learning materials you need for study success - in one app.
Get started for free
The bonding molecular orbital in \({{\rm{H}}_2}\) is lower in energy than an electron in \({\rm{1s}}\) orbital because the bonding molecular is more stable.
Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds: (a) \({\rm{ClNO}}\) (\({\rm{N}}\) is the central atom) (b) \({\rm{C}}{{\rm{S}}_{\rm{2}}}\) (c) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{CO}}\) (\({\rm{C}}\) is the central atom) (d) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{SO}}\) (\({\rm{S}}\) is the central atom) (e) \({\rm{S}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{S}}\) is the central atom) (f) \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{Xe}}\) is the central atom) (g) \({\rm{ClO}}{{\rm{F}}_{\rm{2}}}^{\rm{ + }}\) (\({\rm{Cl}}\) is the central atom).
Use valence bond theory to explain the bonding in\({{\rm{O}}_{\rm{2}}}\). Sketch the overlap of the atomic orbitals involved in the bonds in \({{\rm{O}}_{\rm{2}}}\).
Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
(a) \({\rm{N}}{{\rm{a}}_{\rm{2}}}^{{\rm{2 + }}}\)
(b) \({\rm{M}}{{\rm{g}}_{\rm{2}}}^{{\rm{2 + }}}\)
(c) \({\rm{A}}{{\rm{l}}_{\rm{2}}}^{{\rm{2 + }}}\)
(d) \({\rm{S}}{{\rm{i}}_{\rm{2}}}^{{\rm{2 + }}}\)
(e) \({\rm{P}}_{\rm{2}}^{{\rm{2 + }}}\)
(f) \({{\rm{S}}_{\rm{2}}}^{{\rm{2 + }}}\)
(g) \({{\rm{F}}_{\rm{2}}}^{{\rm{2 + }}}\)
(h) \({\rm{A}}{{\rm{r}}_{\rm{2}}}^{{\rm{2 + }}}\)
For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains \({\rm{77}}{\rm{.55\% }}\)Xe and \({\rm{22}}{\rm{.45\% \;F}}\)by mass.
(a) What is the formula of the compound?
(b) Write a Lewis structure for the compound.
(c) Predict the shape of the molecules of the compound.
(d) What hybridization is consistent with the shape you predicted?
What do you think about this solution?
We value your feedback to improve our textbook solutions.