11 Chemical Bonding II: Valence Bond and Molecular Orbital Theories

• Bond formation can be described in terms of the overlap of atomic orbitals.

• Use appropriate hybridization schemes to describe bonding in molecule.

• Explain how bonding and antibonding affect the bond order of a diatomic molecule.

• Delocalized p electrons can be described with the use of molecular orbital theory.

• The maps show the negative charge density caused by the p molecular orbitals of benzene.

• The Lewis theory is useful in our discussion of chemical bonding.
• The creation of a set of orbitals is a second approach.

• The purpose of this chapter is not to try to master theories of bonding.
• We want to find out how these theories give models that give deeper insights into the nature of chemical bonding than do Lewis structures alone.

• The hydrogen molecule can be used to discuss bonding theories.

• There are two H atoms that are very far apart.

• When the atoms are very far apart, the approach action energy is equal to zero.
• The types of forces that draw the atoms closer together are represented by the dashed intermediate distances and attractive interactions.
• Blue and atoms are pushed apart by repulsive interactions at small distances.

• The energy reaches its lowest value on the solid black line.
• In the internuclear distance, r, two H atoms combine into a H2 molecule through a covalent bond.

• The molecule vibrates, as the nucleus continuously moves back and forth.

• The observed properties of a given mol ecule should be explained by a theory of covalent bonding.
• There are many ways to understand bonding.
• Different methods have different strengths and weaknesses.
• The Lewis theory is easy to apply and can be written quickly.
• It is possible to propose shapes that are in agreement with the results of the experiment.

• The Lewis theory has problems with odd-electron species and situations in which it is not possible to represent a molecule through a single structure.

• The theories of chemical bonding are all based on quantum theory.

• The sharing of one or more electron pairs was associated with a chemical bond in previous chapters.
• It is tempting to describe the stability of a chemical bond in terms of simple electrostatics in which a negatively charged electron pair between a pair of positively charged nuclei serves as the "glue" that holds the nuclei together.
• The description is incomplete.
• The structure and stability of an atom can't be explained without using quantum mechanical principles.
• The stability of a molecule and the description of its bonds require principles from quantum theory.
• A quantum mechanical concept is what a covalent bond is.

• The description of a covalent bond between two atoms, in which an electron from each atom is moved from a point close to its own nucleus to a point between the two nuclei, is purely classical.

• Each electron is close to a nucleus, but the two electrons are far apart.
• The electron-electron repulsion is relatively low because the electrons are far apart.
• The movement of the electrons to a point between the two nuclei reduces the attraction of the electron-nucleus.
• The movement of electrons from near their respective nuclei to the internuclear region involves an increase in potential energy.
• It is difficult to rationalize how the sharing of electrons has a stabilizing effect if you focus only on the electrostatic interactions.

• Explaining the stability of a chemical bond requires us to consider the redistribution of electron density from the nucleus to the internuclear region.

• The net force of the electron on the nuclei tends to draw the nucleus together and less so in regions where the net force draws the nucleus apart.

• Figure 11-3(a) exaggerates how much of the electron density is in the internuclear region.

• The answer is no.
• The contraction of the electron density peak toward the individual atomic nuclei contributes more to the energy lowering of the system than the transfer of electron density into the internuclear region.
• The contraction leads to a decrease in potential energy because each electron is closer to one of the nuclei than it is to the isolated H atom.

• The attractive forces between the electron and each nucleus are represented by the gray arrows.

• The forces are parallel to the axis.

• The black line is the electron density in H2.
• The two nonbonded H atoms are represented by a red line.
• The blue line shows the difference between the electron densities in H2 and nonbonded atoms.
• Bond formation involves a redistribution of electron density into the internuclear region and a contraction of electron density toward the nucleus.
• Bond formation involves a transfer of electron density between the two nuclei.

• The main point of the preceding discussion is that, ultimately, it is a large decrease in potential energy caused by the contraction of electron density to regions closer to the nucleus that contributes most to the stability of a chemical bond.
• The relative importance of other factors, such as the electron-electron repulsion and the nuclear-nuclear repulsion, can be used to justify this assertion.

• Let's look at the total energy of the electrons.
• There are two different effects on the total energy of the electrons.
• The idea that the volume of space available to an electron decreases as it increases can be developed.

• When an electron moves within a larger region than it does in the vicinity of a single nucleus, the energy of the electron will be lowered.
• The net transfer of electron density into the internuclear region is quite small and is offset by a net contraction of the electron density toward the individual nuclei.

• The increase in energy is caused by the net contraction of electron density.
• The formation of a chemical bond involves a small increase in the energy of the electrons.

• The increase in the electron-nuclei attractions is caused by the net contraction of the electron density.

• The contribution of electron-electron repulsion is smaller than that of the electron-nuclear attractions.
• The transfer of electron density into the internuclear region causes the nuclear-nuclear contribution to be smaller than that of electron-nuclear attractions.

• Bond formation involves a small increase in the energy of the electrons and a small decrease in the potential energy of the electrons.
• Increased electron-nuclear attractions bring about a decrease in potential energy.

• We will deepen our understanding of chemical bonding by using a quantum mechanical perspective and describing bond formation in terms of orbitals.

• The region of high electron probability in a H atom is described in Chapter 8 through the mathematical function called a 1s orbital.
• The two orbitals are not the same.
• When the two atoms are close enough that their atomic orbitals overlap, a bond can be formed.
• Bond formation can be imagined through a redistribution of orbitals.

• There is an increase in electron probability density between the two positively charged nuclei.
• The system's energy is lowered.

• Sometimes the overlap of half-filled orbitals is used to create a covalent bond in the method.
• The region of orbital overlap and both nuclei are described by an electron probability density.

• The diagram shows the formation of hydrogen-to-sulfur bonds in hydrogen sulfide.
• The line joining the centers of the H and S atoms has a maximum overlap between the 1s and 3p orbitals of the H and S atoms.
• This is in agreement with the observed angle.

• There are schematic representations throughout the rest of the book.

• It is not possible to choose which lobe is positive.
• The other side of the brain is always negative once this choice is made.

• The Valence Bond Method is used to describe a phosphine molecule.

• Four steps are used when applying the method.
• The first thing we do is identify the orbitals of the central atom.
• We sketch the orbitals.
• We sketch the orbital overlap by bonding atoms to the central atom.
• We describe the structure.

• The overlap involves the central atom's orbitals.

• The 3p orbitals are half-filled.

• The H atoms are in the same plane.

• The bond angles are 90 degrees.

• The H bond angle is 90 degrees, and the sured bond angles are 93 to 94 degrees.
• They are in agreement.

• The expected geometry of nitrogen triiodide is illustrated in this example.

• Only bonding orbitals are used in this exam.

• We are quickly disappointed if we try to extend Section 11-2 to a greater number of molecules.
• Our descriptions of geometry based on the simple overlap of atomic orbitals do not correspond to observed measurements.

• The highly reactive molecule CH2 is only observed under specially designed circumstances.

• Under normal laboratory conditions methane, CH4 is the simplest hydrocarbon observed.
• The molecule is stable and has a formula that is consistent with the Lewis theory.
• Imagine that one of the 2s electrons in a ground-state C atom absorbs energy and is promoted to the empty 2p orbital.

• H bonds are based on the C atom's 2p orbitals.
• The fourth bond would be directed to the fourth H atom.

• The model bond angles are the same as predicted by the CH4 VSEPR theory.
• The molecule has a tron configuration that does a poor job of explaining the bond angles in CH4.

• The problem is not with the theory but with the way the situation has been.

• The assumption worked well for H2S and PH3 but we have no reason to expect the same results in all cases.

• The new orbitals are more appropriate for bonding.
• The C atom's orbitals are imagined to be transformed.

• They are simply added or subtracted because they are a linear combination of an isolated C atom of the C atom in CH4 orbitals.

• The diagram shows the s orbital in blue, suggesting that the wave function has only positive values.

• In discussions of bonding, the outermost portions of the atomic orbitals contribute the most to the overlaps.

• Four different combinations of CH4 are needed to represent one of the new orbitals.
• The four carbon sions represent the combinations.

• There are more atomic orbitals than there are hybrid bond angles.

• 3 orbitals on the C atom are used to form a bond with a H atom.

• Below is a list of the orbitals of carbon.

• The bonding in the CH molecule was described by the 3 hybridization scheme.

• It is important to emphasize the following points before using the concept of hybridization.

• There is a symbol that identifies the numbers and types of orbital involved in the hybrid scheme.

• An after-the-fact rationalization of the experimentally observed shape of a molecule is the objective of a hybridization scheme.
• It is not a physical phenomenon.
• We can't observe the distribution of electron density between pure and hybrid orbitals.

• It's not appropriate to describe bond formation in terms of hybrid orbitals.
• In organic chemistry, the concept of hybridization is used a lot.

• The central atoms in H2O and NH3 need an sp3 hybridization scheme.
• The bond angles in water and NH3 are similar to these angles.
• Bonding in NH3 can be described in terms of the following diagram.

• The three half-filled sp3 orbitals are involved in bond formation.

• The observed bond angles do not conform to the expected H bond angle.

• H bonds to open up to hybrid orbitals.

• There is a single theory that is consistent with all the available evidence.

• The sp2 hybridization scheme is related to the trigonal-planar electron group geometry.

• Bonding in a linear arrangement of atoms is described by the sp hybrid orbitals.
• When this scheme is used to describe bonds formed by atoms other than beryllium, there are two un hybridized p orbitals that are oriented along the y and Z axis.

• Themolecular are left un hybridized.
• The scheme is further outlined in theory, and the diagrams of the valence-shell are shown here.

• CONCEPT ASSESSMENT can be used to form bonds.

• The text states that orbital mixing is a mathematical process of changing pure atomic orbitals for isolated atoms into new atomic orbitals for bonding atoms.
• A hybrid atomic orbital is a result of a mathematical combination of wave functions describing two or more atomic orbitals.

• The linear combinations of the s and p are depicted.
• The three-dimensional forms of the sp hybrid are shown above the maps.

• We have previously said that hybridization is not a real phenomenon, but an after-the-fact rationalization of an experiment.

• There is no better example of this point than the issue of the sp3d and sp3d2 hybrid orbitals.

• In the previous section, we used either the experimental geometry or the geometry predicted by the VSEPR theory to help us decide on the appropriate hybridization scheme for the central atom.
• We used the concept of hybridization in Chapter 10.

• The known geometries of CH4, H2O, and NH3 are accounted for by the sp3d and sp3d2 hybrid orbitals.

• The same number of hybrid orbitals should be produced by the same scheme as there is by the central atom.

• An sp3 hybridization scheme for the central atom predicts that four hybrid orbitals are distributed in a tetrahedral fashion.
• Depending on how many hybrid orbitals are involved in orbital overlap and how many contain lone-pair electrons, the structures are either trigonal-pyramidal or angular.

• Some important applications to organic chemistry will be considered in the next section.

• Predict the shape of the XeF4 molecule with a hybridization scheme consistent with this prediction.

• A plausible Lewis structure is what you should write.

• The Xe atom's valence shell must be expanded to accommodate 12 electrons in Chemical Bonding II: Valence Bond and Orbital Lewis structure.

• The central atom has an electron-group geometry.
• Bond pairs and lone pairs are the electron groups.

• The four pairs of bond electrons are directed to the corners of a square, and the lone pairs of electrons are found above and below the plane of the Xe and F atoms.

• Pick a scheme that matches the prediction.
• There are only a few pairs of electrons above and below the plane of the Xe and F atoms.

• The number of electron pairs in the geometry of the electron-group determines how many orbitals are used in the scheme.
• The shape of and bonding in a molecule can be described with a combination of VSEPR and hybridization theory.
• The sp3d2 hybridization scheme for xenon is not the best way to describe the bonding in XeF.

• There is a central atom in the ion Cl2F+.

• The central atom in the ion BrF + 4 is the subject of a possible hybridization scheme.

• There isn't a correct method for describing structures.
• The experimental evidence from which the structure is established is the only correct information.

• You may be able to rationalize the evidence by using one method or another once the evidence is in hand.
• The valence bond method seems to do a better job of explaining the observed 92deg bond angle than the VSEPR theory.

• The method used by the VSEPR theory suggests a 109.5 degree angle to the electron-group geometry.
• The predicted bond angle is less than geometry because of the data that was changed to accommodate lone-pair-lone-pair and lone approximate molecular pair-bond-pair repulsions.

• Unless you have specific information to suggest otherwise, describing a molecule in bonding based on shape is a good bet.
• It's important to remember that both geometry.

• There are two different types of overlap when multiple bonds are described.
• Specific examples of the carbon-to-carbon double bond in ethylene, C2H4 and the carbon-to-carbon triple bond in acetylene will be used in our discussion.

• There is a carbon-to-carbon double bond in the Lewis structure.

• There is a molecule called ethylene.
• The bond angles are close.
• The CH2 groups are coplanar.

• sp2 is the scheme that produces a set of hybrid orbitals.

• The line joining the two atoms overlaps.
• Above and below the plane of the carbon and hydrogen atoms, there is a region of high electron density.

• A p bond is formed by side-to-side fashion.
• The phase of the p orbitals is retained.

• An alternative definition for s and p bonds is based on the number of nodal planes that are parallel to the bond axis.

• CH2 p bond in a double bond and group out of the plane of the other would reduce overlap of two in a triple bond.

• The C2H4 molecule is planar and the double bond is rigid.

• The s bond involves more overlap than the p bond.

• The bond angles are 120 degrees.

• One of the carbon-to-carbon bonds is a s bond.

• One p bond is formed by a pair of p orbitals, which are oriented along an axis in the plane of the page.
• The lower part of the diagram shows the formation of the other p bond.
• One of the p orbitals has a transparent surface to distinguish it from the other p bond.

• The number of electron groups around a central atom affects the number of atomic orbitals.
• The carbon atom in formaldehyde contains three electron groups, which means that three atomic orbitals are hybridized to form three sp2 orbitals.

• The central atom is C, and the terminal atoms are H and O.
• There are 12 valence electrons.
• This structure requires a carbon-to-oxygen double bond.

• The central C atom has an electron-group geometry.
• The central C atom is the basis of the s-bond framework.
• The distribution of three electron groups suggests a trigonal-planar molecule.

• The scheme that conforms to the electron-group geometry is identified.
• sp2 hybrid orbitals are associated with a trigonal-planar orientation.

• The orbital set sp2 + p is produced by the hybridized C atom.

• Two sp2 hybrid orbitals are used to form bonds with H atoms.

• A s bond with oxygen is formed by the remaining sp2 hybrid orbital.

• The central and terminal atoms have bonding orbitals.

• The bonds are used to bond two H atoms and one O atom.
• We look at the number of bonds and lone pairs around the atom to determine the electron-group geometry and bond angles.

• There is a plausible bonding scheme for dimethyl ether.

• It is difficult to draw three-dimensional sketches to show overlaps.
• Straight lines are used to draw bonds between atoms.
• They are labeled s or p, and the orbitals that overlap are indicated.

• A Lewis structure is the first step in describing a bonding scheme.
• A description of the species obtained by experiment is sometimes the starting point.

• A formula with bond angles is given.
• The scheme should be consistent with this structure.

• The angle of the figure is close to the angle of the book.
• An sp3 hybridization scheme is employed by the O atom on the Bonding and structure of right.

• The lone pair electrons are not shown in the original structure.
• The observed bond angles can be used to infer the presence of lone pairs.
• The sp3 hybridization at the oxygen atom would have been deduced using a Lewis structure and VSEPR theory.

• An industrial solvent is acetonitrile.
• The scheme should be consistent with its structure.

• The molecule diazine has a formula.

• The combination of Lewis structures, VSEPR theory, and the valence bond method makes for a potent description of bonds.
• Most of the time, they are satisfactory.
• Sometimes a greater understanding of structures and properties is needed by chemists.
• Oxygen is paramagnetic, H + 2 is a stable species, and the electronic spectrum of molecule is not explained by any of these methods.
• A different method of describing chemical bonding is needed to address these questions.

• We will only give an overview and focus on the application of the theory to the diatomic molecule.
• We will start our overview by comparing the conceptual model we introduced in Chapter 8 for multielectron atoms with the molecular orbital theory.

• In a similar way, we imagine that each molecule has a set of orbitals and can be built up by placing electrons into them.
• Unlike atomic orbitals, which are centered on a single nucleus,molecular orbitals are defined by all the nuclei.

• Let's look at the results of a calculation for F in more detail.

• The equation for the F molecule is used to solve the total wave function.

• The idea is that because a molecule is composed of atoms, the AOs of those atoms can be used as the basis of a method for describing how each electron in a molecule interacts with all nuclei simultaneously.
• The atomic orbitals can be used as a basis for describing a molecule.
• The premise of the LCAO method is that each MO can be represented as a linear combination of all AOs.

• The following ideas are incorporated into the LCAO method.

• When the atoms combine to form a molecule, the AOs are replaced by a set of molecular orbitals.
• This conceptualization applies what we know about electrons in atoms and extends that description to Molecules.
• Deducing the appropriate combinations of AOs is a difficult aspect of applying the approach.
• The appropriate combinations are simple for diatomic molecules.

• The total number of AOs is the same as the total number of MOs.

• The ten AOs can be combined to form ten MOs that are appropriate for describing electrons in the F molecule.

• The classification is dependent on the presence or absence of nodal planes.
• An antibonding orbital is labeled with an Nodal planes asterisk and is characterized by the alignment of a nodal plane between the two nuclei.

• The subscript on s or p is a reference to the atomic orbitals that are combined in the LCAO method to form a particular molecule.

• The formation of bonding and antibonding from 1s and 2p orbitals is illustrated.

• Bonding, antibonding, or nonbonding can be classified as MOs.
• The electron density is the square of the wave function that represents a MO in the internuclear region.
• Most MOs are nonbonding.

• We used a conceptual model for multielectron atoms.
• There is only one difference here, we are filling MOs, not AOs.
• The rules we used for atoms, such as the Pauli exclusion principle, also apply to the molecule.

• Chemical Bonding II: Valence Bond and Molecular Orbital Theories Orbital is two and if there are two electrons in one MO, they must have opposite spins.

• The nucleus has a cylindrical shape.

• The electron density is zero at the center of atom A, which is why the orbital is described as antibonding.

• There are points between the nuclei.

• s1s* is the abbreviation for B from the subtraction of two 1s orbital.

• A stable species has more electrons in bonding orbitals than in antibonding orbitals.
• The bond order is 1.

• The regions with the highest density of electron density are between the two nuclei.
• The internuclear region has low electron density and the regions outside the two nuclei have high electron density.
• There is a defining feature of an antibonding orbital.

• The first period elements, H and He, are some of the Diatomic Molecules of the First-Period Elements.

• There is a single electron in this species.
• It enters the s1s orbital.

• There are two electrons in this molecule.
• The bond order is 2.
• The bond in H2 is described as single covalent by Lewis theory.

• There are three electrons in this ion.
• There are two electrons in the s1s and one in the s*1s.

• Two electrons are in the s1s and two are in the s*1s.
• The bond order is 2.
• He2 is not a stable species.

• The bond energies of H are estimated.

• The strength of a bond is determined by its order.
• We double the strength if we double the bond order.

• The bond order is equivalent to a single bond.

• The bond order is 1 We should 2.

• The bond energy of Li2 is 106 kJ.

• We had to combine only 1s orbitals for diatomic molecule and H and He.
• We must work with both 2s and 2p orbitals in the second period.

• The 2s atomic orbitals are the same as the 1s atomic orbitals, except they are at a higher energy.
• There is a different situation for combining 2p atomic orbitals.
• s2p and s*2p are s-type molecular orbitals.
• The phase of the 2p orbitals must be taken into account when forming the bonding and antibonding combinations.
• Adding the two wave functions increases the phase electron density in the internuclear region and produces a s2p orbital.
• When to create a bond.

• The other two pairs must combine in a parallel or side-to-side fashion.
• The extra electron density is not found along the internuclear axis, so it is produced in a less direct way.
• This is a characteristic of antibonding character, as a node is formed between the nuclei in the nodal plane.

• The atomic orbital energy levels are related to the energy-level diagram for the molecular orbitals.
• The relationship between the 2s and 2p atomic orbitals is the same as the one between the 2s and 2p molecular orbitals.
• The end-to-end overlap of 2p orbitals should be more extensive than side-to-side overlap, which should result in a lower energy.

• Predicting that don't match experimental results is what we will do if we use this assumption.

• The phases of the orbitals are depicted in the different colors depicted in these figures.

• The electron density between the nuclei contributes to a chemical bond.
• All antibonding orbitals have a nodal plane that is parallel to the internuclear axis.

• The axis is the internuclear axis.

• The electron density in the same region between the nuclei is produced by the 2s and 2p orbitals.
• The two s orbitals are very similar in shape and energy, and they mix to form modified s orbitals.
• The original s2s and s2p are not present in the modified s orbitals.
• The modified s2s go down in energy and the modified s2p goes up in energy.
• The modified s2p is pushed up in energy above the p2p orbitals.

• The p2p orbitals are at a lower energy than s2p because the energy difference between the 2s and 2p orbitals is smaller.

• The modified s2s and s2p orbitals are described above.

• We start with the s1s and s*1s orbitals and assign electrons to them.
• In order to increase energy, we add electrons to the second principal shell.
• There are some properties listed in the figure.

• The two atoms are different and we can assign electrons to them.

• The assignment of eight electrons to the following diagram is consistent with the observation that the 2 is diamagnetic.

• The plained features of the O2 molecule help us understand some of the previously unex designation.

• The paramagnetism of O2 is explained by this.

• The bond order is two because there are eight and four antibonding electrons.

• Writing a Molecular Orbital Occupancy Diagram and Determining Bond Order Represent bonding in O + 2 with a Molecular Orbital Occupancy diagram and determine the bond order in this ion is an example.

• The O + 2 ion has 11 electrons.
• The ideas on page 488 allow us to assign these to the available molecular orbitals.
• We can remove one electron from the diagram.

• 2.5 is the bond order.

• Because the 2s-2p separation in oxygen is large, the diatomic molecule uses the molecular orbital occupancy diagram without 2s-2p mixing.

• 2 are 112, 121, 128, and 149 pm.

• P bonds are always accompanied by a s bond.

• We can extend the ideas we developed for Homonuclear diatomic species to give us an idea of the bonding in Heteronuclear diatomic species.
• The atomic number increases the energy of a given orbital.

• The relative energy ordering of the orbitals of each type is indicated by the symbol s or p preceded by an integer.
• The lowest energy is labeled 1s, the next lowest 2 and so on.

• The 1s and 2s orbitals are not shown.

• We must first classify the occupied molecular orbitals as bonding, antibonding, or nonbonding before we can cal culate the bond order.
• The 2p orbitals are antibonding.
• It is difficult to classify the s orbitals.

• The tentative classification of the s orbitals is supported by both quantum mechanical calculations and experimental data.

• The order for the free radical, NO, which has just one more electron than CO, can be used to predict the bond orbital structure of CO.

• 2.5 becomes the bond order.
• The bond energy in NO is predicted to be less than in CO.

• Determine the bond order for the cyanide ion and write the ground state electron configuration.

• The number of valence electrons is determined by the number of atoms in the molecule.

With the assumption that 4s and 5s are nonbonding, the bond order is 12 + 42>2

• The isoelectronic molecule CO has a bond order of 3.
• The Lewis structure has a triple bond.

• Determine the bond order and write the electron configuration.

• Determine the bond order by writing the electron configuration.

• The 2s and 3s orbitals are weakly bonding and weakly antibonding, respectively.

• In Section 11-4, we discussed delocalized electrons, such as those in C2H.

• In the fragrant aromas section, we will combine ideas from the bonding theories we have studied so far to consider bonding in benzene and a few other polyatomic molecules.

• We will describe the bonding in the molecule by using two different theories.

• The application of molecular orbital theory to polyatomic mole cules is not easy.
• The combinations of atomic orbitals contributing to the various molecular orbitals are not easily deduced.

• We will use the molecular orbitals to help us understand the bonding in each case.

• Friedrich Kekule advanced the first good proposal for the structure of benzene in 1865.
• The C6H6 molecule consists of a flat, hexagonal ring of six carbon atoms joined by alternating single and double covalent bonds.

• Two other C atoms are joined to one H atom.
• Kekule suggested that the sin first determined the X-ray gle and double bonds, and then that the carbon-to-carbon bonds are all alike.

• This view is suggested by a figure.

• We can learn a lot more about bonding in the benzene flat.

• The six remaining 2p orbitals are used to build the delocalized p bonds.

• The true structure contributes equally to the two resonance structures.
• Each corner of the hexagonal structure has a carbon atom and a hydrogen atom.

• We don't need to think about the delocalized p or the resonance hybrid for benzene.
• Localized bonds in benzene are not the p bonds.

• Three of the p molecular orbitals are bonding.
• The next two p-bonding molecular orbitals have the same energy.
• The fact that the two orbitals are higher in energy than the one with no nodes should not come as a surprise, as we have already seen.
• The next pair of orbitals, which are antibonding p orbitals, have two nodes and the final one has three.

• The three bonding orbitals fill with six electrons and the three antibonding orbitals are empty.
• The bond order for the six electrons is 16 - 2 - 3.

• Three of the six p molecular orbitals are filled with an electron pair.
• There are no antibonding molecular orbitals at higher energy.

• Each pair of C atoms have a half-bond between them.

• The concept of resonance was introduced in Section 10-5.

• In points 1 and 2, 2 hybrid orbitals are discussed.

• The bond angle is calculated using the predicted trigonal-planar electron-group geometry and assignments of electrons.
• The central O atom has a discussion of the hybridization scheme chosen for them.

• The O atom uses the orbital set sp2 + p.

• Four of these are bonding electrons and ten are lone-pair electrons.

• There is no contribution to the wave function from the central atom.

• The p molecular orbitals are shown on the right.

• The second is antibonding.
• A nonbonding molecular orbital has the same energy as the atomic orbitals from which it is formed, and it doesn't detract from bond formation.

• The four remaining electrons are assigned to the p orbitals.
• Two go into the bonding and two into the nonbonding.
• The antibonding orbital is empty.

• The bond order is 12 - 2.

• This is the same as averaging the two Lewis structures.
• Section 10-5 describes the O bond length suggested by this method.

• The NO 3 ion has two bonds in the p system.

• We first look at the electrons associated with the s bond framework.
• Reasoning out the number of each type ofbonding, antibonding, nonbonding in the p system is the key to solving the problem.
• The final step is to assign electrons in the p system.

• The NO 3 ion has a total number of electrons.
• The NO 13 * 6 + 5 + 12 = 24 3 ion and the corresponding s bond framework are shown below.
• The N atom has three atoms and is hybridized.

• The number of electrons in the p system is 24 and the number of electrons in the s bond framework is 18.
• There are a lot of nonbonding molecular orbitals in the system.

• The diagram shows how the six p electrons are assigned to the p orbitals.

• The bond order of each nitrogen-to-oxygen bond is 11.

• Represent chemical bonding in the molecule SO3 by using a combination of local and delocalized orbitals.

• Represent chemical bonding in the ion NO 2 by using a combination of local and delocalized orbitals.

• There are four 2p orbitals on the p-bonding molecule, one on the N atom and two on the three oxygen atoms.
• The antibonding p orbital is at the highest energy.
• We expect to get four molecular orbitals since we are combining four 2p orbitals.
• Each of the remaining two molecular orbitals will have a single point of contact.

• The NO 3 is for the nitrogen atom.
• It must be nonbonding with respect to nitrogen.

• The p-bonding molecular orbital has all the p orbitals in phase, while the p-antibonding orbital has all the p orbitals out of phase.
• There is a nitrogen atom at each of the nonbonding orbitals.

• We will see how it is used to explain the colors of plants.

• The contiguous p system is a common feature of these molecules.

• Antibonding bonds can be found in carotene.

• It takes very little energy to get an electron from the HOMO to the LUMO.
• The colors we see are caused by the absorption of the light's electrons across the energy gap between the HOMO and the LUMO.

• The ideas introduced in this chapter can be used to describe bonding in metals.
• You can find the Appendix to Chapter 11 Bonding in Metals on the website.

• We presented a wide range of views of chemical bonding in Chapters 10 and 11.
• Each of these models has deficiencies, and uncritical use can lead to incorrect conclusions.

• The more fundamental view of molecular shapes is still an issue.
• We might wonder how the theories are related.
• The significance of electron density calculations will be stressed in this section.

• Let's see if it is possible to describe the bonding in SF6 while keeping the octet rule.

• The electronic structure is a combination of resonance structures and Valence Bond.
• In the case of SF6, four electron pairs bond six F atoms to a central S atom.
• When resonance structures are written in this way, the assumption is that the bond lines represent covalent bonds.
• The molecule SF6 requires a total of 15 structures of the type shown.
• The collective charge of -2 on the six F atoms is implied by this description.

• The suggested charges on the S and F atoms can be compared with those obtained from a quantum-mechanical calculation.
• There is a charge of +3.17 on sulfur and -0.53 on each fluorine.
• To describe bonding through hyperconjugation that is in better agreement with the quantummechanical calculation, we would have to use additional resonance structures with higher charges and fewer covalent bonds.
• Adoption of this large number of structures is not justified because it is cumbersome.

• Lewis did not consider exceptions such as PCl5 and SF6 to be of great significance when he created his rule of eight.
• The rule of two is more important than the rule of eight because the density of the electron pair is the most important factor in understanding bonding.
• A special type of bonding is not the reason for bonding in the first place.

• The bonds in these molecules are similar to the bonds in other molecules.

• We seem to think about bonding in terms of atomic orbitals.
• The concept of sp3 hybrid orbitals was introduced in order to describe bonding in the methane molecule.
• The geometry of a molecule can be determined by using a theory.

• The results of quantum-mechanical calculations have already been used to construct electrostatic potential maps.
• Let's use the results of the calculations to improve our understanding of the bonding in SF6 and similar molecules.

• In atomic units, the isodensity lines are shown in color in the order of 1 to 8.

• The Densities are truncated at 2 o'clock.
• The Bohr radius is adapted from Matta and Gillespie.

• First, consider a simpler molecule.
• The most striking feature of this diagram is that the electron density is very high at each nucleus; in order to show other features in the diagram, we truncated the very large maxima.
• There is a ridge of electron density between the sulfur and chlorine atoms.
• The ridge of electron density contributes to bond formation by transferring electron density from the atomic orbitals to the internuclear region.

• A con tour map is an alternative representation of the electron density distribution.

• The bond paths are connected by lines.
• The bond critical points are depicted in red Xs.

• Similar to a path along the valley floor between the "mountains" of electron density, this bifurcated line is a path tracing the minimum in the electron density.
• The type of bond connecting a pair of atoms in a molecule can be described with the electron density at the bond critical point.
• The higher the bond order is, the greater the electron density.

• We will consider the sulfuric acid molecule and the sulfate anion to decide whether or not to use expanded valence shells.

• We include less of the electron density distribution if we choose a surface with a higher electron density value.

• The values for the outer isodensity envelope are set in the three figures for both species.

• If there are bonds in the molecule that have more electron density at their bond critical points, then electron density will still appear in the three-dimensional representation.
• The stick from the ball-and-stick model is apparent, but between the sulfur and the two oxygen atoms that do not, the electron density between the sulfur and the two oxygen atoms has disappeared.
• The electron density between the sulfur atom and the nonprotonated oxygen does not disappear until the density of the calculated surface is increased to 0.33 Au.
• There is more electron density in the bonds between the sulfur atom and the nonprotonated oxygen atoms than in the bonds between the sulfur and oxygen.
• The double bond between the central S atom and the two terminal O atoms can be represented in a Lewis structure that reduces the formal charges seen in the octet Lewis structure of H2SO4.

• The bond critical density is 0.28 Au for the four bonds between the sulfur atom and the oxygen atoms.
• The bond between sulfur and the nonprotonated oxygen atoms is greater than the bond between sulfur and the protons.
• Maybe the Lewis structure that reduces formal charges is the best because of the higher bond critical point in the sulfate anion.
• The four sulfur-oxygen bonds have the same bond critical point, which can be used to write resonance structures.

• We have reached a point where using expanded shells seems to be the best way to minimize formal charges.
• A detailed analysis of the wavefunction of the sulfate anion suggests that the simple octet structure is the dominant form.

• The answer to several questions posed in this section may be found in the work of R. J. Gillespie.
• Lewis structures should be written as Lewis would have written them.

• The C bond dissociation enthalpy is very low.
• An analysis of the electron densities in SiF4 shows that they have a combination of very strong bonds.

• You should not be too concerned by the controversy over how to write Lewis structures in the chemical literature.
• Our approach to depicting the electronic structure of a molecule is based on the simplest Lewis structure and its use in determining the shape of a molecule through VSEPR theory.
• To understand experimental results, such as bond enthalpy values, we must analyze a computed electron density map for that molecule rather than rely on the Lewis structure.

• The structures of the molecule are studied.
• The method involves passing a high-energy photon through a sample of molecule.
• The focus on feature for Chapter 11 on the MasteringChemistry site is for a discussion of Photoelectron Spectroscopy.

• The simpler ideas of the Lewis model are not related to the electronic structure of molecule numbers.
• The transfer of atomic orbitals is used to generate them.

• The electron-group electron density that extends over several atoms in a mole geometry is predicted by the VSEPR theory.

• The concept of 11-4 Multiple Covalent Bonds--End-to-end overlap hyperconjugation was used in SF6.
• In terms of bond paths and critical points, side-to-side overlap is lyzed.

• A double bond consists of one s bond charge density point and Lewis structures.
• A triple bond consists of one s with the lowest formal charge and two p bonds.
• The bonding in these compounds is represented by the shape of way.
• Lewis structures without expanded are added as required to complete the bonding description, and a species determines the s-bond framework.

• Detonators for high explosives use hydrogen azide, HN3 and its salts as unstable substances.
• In air-bag safety systems in automobiles, NaN3 is used.
• The following data is from a reference source.
• The time is 114 pm.
• The Nc is 180 degrees.

• Write two structures to the resonance hybrid and describe a scheme for each structure.

• Data from Table 10 can be used to estimate the bond order.
• We can predict a likely geometric shape of the molecule by applying VSEPR theory to the Lewis structures.
• We can propose bonding schemes for the central atoms with this information.

• The average bond length for a single bond, double bond, and triple bond is 146 pm, 123 pm, and 112 pm, respectively.

• The Lewis structures have Na and Nb atoms as central atoms, the Nc and H atoms as terminal atoms, and bonds reflecting the observed bond lengths.

• The formal charges on Na, Nb, and Nc are -1, + 1, and 0, respectively, in structure I.
• We can't say which structure is favored because the charges are so similar.
• The best way to make a decision is to compare the observed structure with the one suggested by the two schemes.

• The angle is expected to be close.

• The hybridization scheme in structure is preferred because of the Nb angle.
• On the right, we can see that Na is negatively charged as compared to Nb and Nc, in accord with our preferred structure.

• Melamine is a carbon-hydrogen-nitrogen compound used in the manufacture of adhesives, protective coating, and textile finishing.
• The molecule has a mass percent composition of 28.7% C, 4.80% H, and 66.64% N. The H atoms are outside the ring.

• Dimethylglyoxime is a carbon-hydrogen-nitrogen-oxygen compound with a mass of 116.12 U.
• In a separate experiment, the nitrogen in a 1.868 g sample of DMG is converted to NH31g2 and the NH3 is neutralized by passing it into 50.00 mL of 0.3600 M H2SO41aq2.
• After neutralization of the NH3 the excess H2SO41aq2 requires 18.63 liters of water.

• Lewis theory is linear and neither.

• C and N are using hybridized atomic orbitals.

• CO 2 is used to label each s and p bond.

• The geometric shape pictured in Table 10 is consistent with the model below.
• The acidic component of the juices.

• Malic acid is an organic acid that can be found in apples and other fruit.

• The structure of the molecule is shown here.

• sumbol root is a herb used as a stimulant.

• The preparation of the resins is done with dimethylolpropionic acid.

• In our discussion of bonding, we don't have an encoun bond method and molecular orbital theory describes a higher bond order.
• The energy bond can be used.

Is it correct to say that when a diatomic molecule loses it explains why the results are different?

• N21g2 has a high bond energy.

• The reason for your choice is the paramagnetism of B2.

Which of these ion has more atoms with the same electron configurations?

• The diagram for the molecule is called the molecular orbital diagram.

• The diagram is called the molecular orbital diagram.

• You would expect to find onance in which of the following.

• Lewis structures are consistent with the molec to-nitrogen bond.

• There is an overlap of sp hybrid orbitals in the compound.

• Potas bonding in N2 can be described by drawing orbital diagrams for the N atoms.

• Predict its geometric shape with the Explain.

• CH3NO3 is a rocket propellant.

• The molecule's skeleton is called CH3ONO2.
• NeF, NeF+, or N and three O atoms all lie in the same plane, but the NeF-.
• The NO3 group has energy-level diagrams in the same plane as the CH3 group.

• The bond angle species is 105deg.
• O is 125 degrees.

• The sketch of the molecule does not explain the metric shape.

• FONO2 is an oxidizer used as a rocket propellant.
• There is a reference source.

• The O2NOa plane has all the expected bond lengths and bond angles, but the label aF plane has no geometric structure.

• A solar cell that is 15% efficient in converting solar to three-dimensional sketch of the molecule and a plau electric energy produces an energy flow of 1.00 kW>m2 sible bonding scheme shows hybridization and when exposed to full sunlight.

• Draw a Lewis structure for the nitrite ion.

• If the power is produced at p bonding in this ion.

• The bond angles were marked by the hybridization values.

• A bonding scheme in the active center of many enzymes is proposed by Histidine, an essential amino acid.

• He2 isn't a stable molecule, but it does have an immune response.
• The structure of histidine supports the idea that a molecule can be formed.

• The bond length is 136 hours.

• NH2 tures can be written for this molecule, with the true *2 structure being a resonance hybrid of the two.
• Propose a bonding scheme for each structure.

• Pyridine is used in the synthesis of vitamins and drugs.
• The bond angles marked a and b are what the molecule can be thought of as.

• The anion I4 5 is V-shaped and has a 95 degree angle between the two arms of the C. There are many bonding and antibonding p-type schemes that are consistent with these observations.

• The isomers have one of the characteristics of antibonding.

• The bonding in these molecules can be described by using the bonding molecular orbitals.
• To determine which molecule has a nodal plane and a delocalized p system, you need to use molecular orbital theory.

• The F2Cl is bent.

• The central bond and single bond have different dou describe hybridization schemes.
• The atom has a different structure.

• HOON may be an intermediate of atmospheric reactions and a small contribution from an ion-pair structure HONO.

• The HOON is best represented by a combination of three resonance structures if you use the information to calculate the structure.

• You should convince yourself that the combinations use bond energies.
• The total enthalpy change required to break all the bonds in a Kekule structure of orbitals is determined by using bond energies from Table 10.3.

• 4H4O can be found in oat hulls, corn cobs, and other waste.

• The chemical behavior of the molecule suggests that it is a resonance hybrid of several con.

• Draw Lewis structures for the several contributing determine whether bond dipoles exist in a molecule and structures to the resonance hybrid.

• The s and p bonding in furan are involved in the cancellation and combine to produce.

• A bonding label can be created by combining a molecular orbital diagram for HF with combinations that do.

• Consider the Lewis structure of the molecule.