Chem 105 chapter 6 homework

Apply molecular orbital theory to determine the bond order of F2 . 3 0 1.5 0.5 2 2.5 1

1

How many hybrid orbitals do we use to describe each molecule? N2O5 Express your answer as an integer.

10 The interior nitrogens are each bonded to three atoms, so each is sp2 hybridized (three hybrid orbitals), and the interior oxygen is bonded to two atoms and has two lone pairs, so the interior oxygen is sp3 hybridized (four hybrid orbitals). So, the total number of hybrid orbitals is 2(3)+4=10 .

How many hybrid orbitals do we use to describe each molecule? C2H5NO (4 C−H bonds and one O−H bond) Express your answer as an integer.

14 In this molecule, the left carbon is bonded to four atoms and has no lone pairs so the carbon on the left is sp3 hybridized (four hybrid orbitals). The right carbon is bonded to three atoms and has no lone pairs so the carbon on the right is sp2 hybridized (three hybrid orbitals). The nitrogen is bonded to two atoms and has one lone pair so the nitrogen is sp2 hybridized (three hybrid orbitals). The oxygen is bonded to two atoms and has two lone pairs so the oxygen is sp3 hybridized (four hybrid orbitals). So, the total number of hybrid orbitals is 4+3+3+4=14 .

The structure of hypoxanthine, a naturally occurring purine derivative How many sigma bonds are present in hypoxanthine?

15 σ bond(s) A single bond always corresponds to a sigma ( σ ) bond, which is a pair of orbitals overlapping end to end. A double bond always corresponds to one σ bond and one pi ( π ) bond. The structure shows 11 single bonds and 4 double bonds. Therefore, there are 15 σ bonds in the molecule.

The structure of hypoxanthine, a naturally occurring purine derivative How many pi bonds are present in hypoxanthine?

4 πbond(s) Only one sigma ( σ ) bond forms between any two atoms, as it is a result of end-to-end overlap of orbitals. Additional bonds must be pi ( π ) bonds and require side-by-side overlap of p orbitals. A double bond always contains one σ bond and one π bond. The structure shows four double bonds; thus, there are four π bonds in the compound.

How many hybrid orbitals do we use to describe each molecule? HOCN (with no formal charges) Express your answer as an integer.

6 In this molecule there are two interior atoms, carbon and oxygen. The carbon is bonded to two atoms and has no lone pairs so the carbon is sp hybridized (two hybrid orbitals). The oxygen is bonded to two atoms and has two lone pairs so the oxygen is sp3 hybridized (four hybrid orbitals). So, the total number of hybrid orbitals is 2+4=6 .

Choose the set of diagrams in which the electrons involved in bonding are circled.

All the unpaired electrons of the P atom and the unpaired electron of each H atom contribute to the formation of the PH3 molecule.

Consider the structure of the amino acid alanine. Indicate the hybridization about each interior atom.

Assign the hybridization scheme from the electron geometry of interior atoms (determined in the hint for this part). The carbon bonded to three H atoms and one C atom has a tetrahedral electron geometry and, therefore, sp3 hybridization. The carbon bonded to two C atoms, one H atom, and one N atom has a tetrahedral electron geometry and, therefore, sp3 hybridization. The nitrogen bonded to two H atoms and one C atom has a tetrahedral electron geometry and, therefore, sp3 hybridization. The carbon bonded to two O atoms and one C atom has a trigonal planar electron geometry and, therefore, sp2 hybridization. The oxygen bonded to one H atom and one C atom has a tetrahedral electron geometry and, therefore, sp3 hybridization.

Complete the sentences to describe the bonding and geometry of C3H4 , using a valence bond approach.

Because of the perpendicular orientation of the π bonds formed between C1 and C2 and between C2 and C3, the C3H4 molecule is not planar. In fact, the plane containing C1 with two hydrogen atoms attached and the plane containing C3 with two hydrogen atoms attached are also perpendicular to each other. The molecule can thus be thought of as a two-bladed propeller.

According to Lewis theory, which species is most stable? CN+ CN CN−

CN− According to the Lewis structure, CN− would be the most stable as it has the octet satisfied for both carbon and nitrogen with a strong carbon-nitrogen triple bond.

According to MO theory, which species is most stable? CN+ CN CN−

CN− Both CN+ and CN have bond orders of 2.5. CN− has a bond order of 3.0. According to MO theory, the structure with the highest bond order is most stable; therefore, CN− is the most stable of the three according to MO theory.

Complete orbital diagrams (boxes with arrows in them) to represent the electron configuration of valence electrons of carbon before and after sp3 hybridization.

Carbon is a group 4A element and, therefore, has 4 valence electrons. Its electrons configuration is [He]2s22p2 . Before hybridization, the 2s orbital of carbon is lower in energy than the 2p orbital. Therefore, two electrons fill the 2s orbital first, and then the two remaining electrons singly occupy two 2p orbitals. The resulting orbital diagram for unhybridized carbon is [He]↿⇂↿↿ After hybridization, the number of valence electrons does not change, but all orbitals become equal in energy. Therefore, each orbital should be filled by one electron first. Since there are no remaining electrons to pair, the resulting diagram for a hybridized carbon atom is [He]↿↿↿↿ with four sp3 hybridized orbitals.

Determine the correct hybridization (from left to right) about each interior atom in CH≡CCH2Cl . 1st C sp ; 2nd C sp2 ; 3rd C sp2 1st C sp ; 2nd C sp3 ; 3rd C sp2 1st C sp2 ; 2nd C sp2 ; 3rd C sp3 1st C sp ; 2nd C sp ; 3rd C sp3

Draw the correct Lewis structure and count the number of electron groups on each atom; then refer to the table below to determine the correct hybridization for each atom. The first and second C atoms have two single bonds and therefore two electron groups. Two electron groups form sp hybrid orbitals. The third C atom have four single bonds and therefore four electron groups. Four electron groups form sp3 hybrid orbitals.

Use molecular orbital theory to determine which molecule is diamagnetic.

F2

Draw Lewis structure for CN

Note that the electrons are arranged so that the free radical resides on the carbon atom and all formal charges are neutral in the molecule.

Draw Lewis structure for CN−

Note that the only way to draw the structure to satisfy the octet on both atoms is to draw a carbon-nitrogen triple bond with a lone pair on each atom. Therefore, the negative formal charge must reside on the carbon atom.

Draw Lewis structure for CN+

Note that the positive formal charge resides on the less electronegative carbon atom.

Use molecular orbital theory to predict which species has the strongest bond. O2+ O2− O2 All bonds are equivalent according to molecular orbital theory.

O2+ The species with the highest bond order has the strongest bond. The bond order is the number of electrons in bonding molecular orbitals minus the number of electrons in antibonding molecular orbitals all divided by two. The electron configuration of the molecular orbitals is σ2s2σ∗2s2σ2p2π2p4π∗2p1 so there are eight electrons in bonding orbitals and three electrons in antibonding orbitals. The bond order is 8−3/2=2.5 .

Complete orbital diagrams (boxes with arrows in them) to represent the electron configurations, without hybridization, for all the atoms in PH3 .

Phosphorus (P) is a group 5A element and, therefore, has five valence electrons. Two electrons doubly occupy the 3s orbital. Then, according to the aufbau principle, three remaining electrons singly occupy each 3p orbital. The resulting orbital diagram for P is [Ne]↿⇂↿↿↿ Hydrogen (H) is a group 1A element and, therefore, has one valence electron in the 1s orbital. The resulting orbital diagram for H is ↿

What bond angle do you expect from the unhybridized orbitals? How well does valence bond theory agree with the experimentally measured bond angle of 93.3 ∘ ?

Since all three p orbitals are orthogonal, the angle between the P−H bonds should be 90 ∘ according to the valence bond theory. As you can see, the actual bond angle does not considerably differ from the one predicted by the valence bond theory.

Use molecular orbital theory to predict whether or not each of the following molecules or ions should exist in a relatively stable form.

Species with stable bonds are associated with a positive bond order (more electrons in bonding MOs than in antibonding MOs), whereas species with a negative or zero bond2865999 order do not form a chemical bond. Calculations of bond orders for each species are summarized in the hint for this part. H22− , Ne2 , and F22− have a bond order of 0 and, therefore, are unstable. He22+ has a bond order of 1; therefore, this ion is relatively stable.

According to this energy diagram, is H2O stable? Explain.

Stability of a molecule can be shown using the calculated bond order. If the bond order is positive, the molecule is stable. In other words, a stable bond requires more electrons in bonding MOs than in antibonding MOs. Electrons in nonbonding MOs don't take part in formation of bonds. The bond order is thus given by bond order=(number of e− in bonding MOs)−(number of e− in antibonding MOs)/2 A water molecule has four electrons in bonding MOs, four electrons in nonbonding MOs, and zero electrons in antibonding MOs. The bond order of H2O is therefore 2. With a bond order of 2, the molecule is stable.

The compound C3H4 has two double bonds. Draw its Lewis structure.

The Lewis structure of C3H4 contains three carbon atoms connected successively with double bonds, with two hydrogen atoms attached to each terminal carbon atom:

Choose a three-dimensional sketch of the molecule showing orbital overlap.

The bonds between the P atom and the H atoms are σ bonds. In a σ bond, one lobe of the p orbital overlaps with a sphere of the s orbital. Since three p orbitals are mutually perpendicular, each orbital of the H atoms should overlap with the orbitals of the P atom that are along different axes.

Consider the structure of the amino acid aspartic acid. Indicate the hybridization about each interior atom in the given structure.

The carbon atoms in positions 1 and 4, the carboxylic acid groups, have three electron groups around the atom, which results in a trigonal planar electron geometry and requires sp2 hybridization. The carbon atoms in positions 2 and 3, oxygen atoms of the OH groups, and the nitrogen atom all have four electron groups around the atom. Four electron groups corresponds to a tetrahedral electron geometry, which requires sp3 hybridization.

Choose the correct sketch of the lowest energy bonding molecular orbital.

The lowest energy bonding orbital of CO , σ2s , resembles the electron density distribution of a σ bond in valence bond theory. In second-period heteronuclear diatomic molecules, the molecular orbitals show a greater electron density at the more-electronegative element. The atomic orbitals of the more-electronegative elements contribute more to the bonding molecular orbital than those of the less-electronegative element. O is more electronegative than C , so there is more electron density at O than at C .

Draw an MO energy diagram for CO . (Use the energy ordering of O2 .)

The number of valence electrons in the orbital diagram of CO is equal to the sum of valence electrons in the C atom (four) and the O atom (six). To fill the orbital diagram with 10 electrons, start with the orbitals on the lowest energy level. A molecule or ion with 10 valence electrons has two paired electrons in the σ2s molecular orbital, two paired electrons in the σ∗2s orbital, two paired electrons in the σ2p orbital, and four paired electrons on both π2p orbitals.

Do the two theories agree? Yes No

Yes Both Lewis and MO theory predict that CN− will be the most stable molecule. Therefore, the two theories agree in this case.

Predict the bond order in CO . Express your answer as an integer or to one decimal place.

bond order =3

The structure of hypoxanthine, a naturally occurring purine derivative Insert the lone pairs in the molecule. What kinds of orbitals do the lone pairs occupy? only sp2 only sp3 sp2 and sp3 only sp

sp2 and sp3 The lone pairs on the N atoms in position 1 and position 3 occupy sp2 orbitals; the lone pairs on the N atoms in positions 2 and 4 occupy sp3 orbitals. You can determine the hybridization by counting the number of electron groups around the atom.

The results of a molecular orbital calculation for H2O are shown here. Examine each of the orbitals and classify them as bonding, antibonding, or nonbonding. Assign the correct number of electrons to the energy diagram.

Bonding orbtials are lower in energy than the atomic orbitals from which they are formed. Nonbonding orbitals remain localized to an atom, while antibonding orbitals are the result of destructive interference between atomic orbitals, and therefore are higher in energy than the atomic orbtials from which they are formed. The filling of molecular orbitals obeys the same rules as the filling of atomic orbitals: the principle of minimum energy, Pauli's principle, and Hund's rule. Water has eight electrons to place, six from oxygen and two from the two hydrogen atoms. When assigning the electrons of a molecule to MOs, fill the lowest energy MOs first with a maximum of two spin-paired electrons per orbital. When assigning electrons to two MOs of the same energy, follow Hund's rule: Fill the orbitals singly first, with parallel spins, before pairing.

Bromine can form compounds or ions with any number of fluorine atoms from one to five. Write the formulas of all five of these species. Express your answers as chemical formulas separated by commas. Enter your answers in order of increasing number of fluorine atoms.

BrF,BrF2−,BrF3,BrF4−,BrF5