Chemical bonding lecture 6

Question 1

What are triatomic nonhyrides? What shapes do they have?

Answer 1

They are molecules that are hydrogen deficient and are made from 3 atoms

They are linear if they have 16 or fewer valence electrons, and bent if they have 17 to 20 valence electrons.

Question 2

How does bonding occur in linear Triatomic nonhydrides?

Answer 2

With the linear structure, it is logical for the central atom to be sp hybridized with its px and py orbitals perpendicular to its two sp orbitals, the two atoms attached to the central atom don’t have to be hybridized, they each have a pz orbital that makes a sigma bond with the sp orbital of the central atom, as well as, a px and py orbital perpendicular to the pz orbital which are responsible for the pi bondings.

The 3 px orbitals combine to form 3 different MO:
1. A bonding orbital
2. A bonding orbital
3. A non-bonding orbital

the 3py orbitals combine similarly to the 3 px orbitals.

note: the anti-bonding orbital has the most energy, then the non-bonding orbital, then the bonding orbital.

Question 3

How does the bonding occur in non-linear triatomic molecules?

Answer 3

if the molecule lies in the x-y plane then the valance s, px, and py orbitals of the central atom can combine to form 3 sp^2 hybridized orbitals, two of them will partake in the sigma bond formation, one of them will have a lone pair, and the pz orbital will contribute to the pi bonding.

The outer atoms will have the p orbital points towards the central atom to form a sigma bond, the pz orbital will contribute to the pi bonding, and the other p and s will be AOs that do not partake in the bonding (in the molecular correlation diagram they will be represented by antibonding MOs localized on each respective atom (y3ani they will be at the same energy level) )

the 3 pz orbitals combine to form 3 different MOs which are:
1. A bonding orbital
2. A bonding orbital
3. A non-bonding orbital

Question 4

what is the bonding like in conjugated pi-electron systems?

Answer 4

all of the atoms have 3 sp^2 hybridized orbitals that form sigma bondings, and Pz orbital with one electron that contributes to the pi formation.

The Pz orbitals combine to form a set amount of MOs with the lowest energy one being fully bonding and the highest energy one being fully anti-bonding and the in-between are a mix of bonding and anti-bonding.

in the in-between orbitals, the bonding usually occurs between every other carbon-carbon( y3ani yk how a conjugate alternating pi bond system is) with the outer carbons usually having a bonding charter and the inner carbons having an anti-bonding character.

Such models are best described by delocalized MOs extending over the entire pi-electron system

note: the amount of MOs formed is the same as the amount of Pz orbitals.

Question 5

What is bonding like in benzene?

Answer 5

All of the carbons are in a cyclic chain with 3 orbitals of Sp^2 hybridization forming sigma bonds, and a Pz orbital having one electron contributing to the pi bonding.

There are 6 Pz orbitals in benzene (from each carbon) and they form 6 pi MOs characterized by their energies and the number of nodal planes (perpendicular to the molecular plane)

The lowest energy MO has no nodal planes (fully bonding), the next highest energy orbital has a degenerate pair and one nodal plane, the next highest energy orbital has a degenerate pair and two nodal planes, and the last orbital has 3 nodal planes(fully antibonding).

Benzene has 6 electrons between its Pz orbital, in which all are placed in the lowest 3 energy orbitals giving us equal electron distribution resulting in equal bond length between all C-C bonds.

Question 6

LCAO VS VB

Answer 6

LCAO provides way too much ionic charter where VB completely neglects ionic charter so for the most accurate approximation we need to use both LCAO and VB.

LCAO theory offers qualitative predictions for processes in diatomic and polyatomic molecules that involve promoting a molecule to an excited electronic state

Question 7

What is the lowest energy a geometric isomer has? How do we interconvert between cis-trans configurations? How does UV light contribute to the interconversion?

Answer 7

The isomer has the lowest energy when the Pz orbitals are parallel to each other (that also means when they have the most overlap).

In order to convert between the Cis and trans configurations rotations about the C=C bond need to be made, which means that the pi-bond needs to be broken then both the Pz orbitals are rotated by 180 degrees with respect to each other then reforming in the other configuration.

some isomers absorb UV light, which then causes the excitation of one of the electrons to go from the bonding to the anti-bonding pi orbital, effectively causing only one bond between the carbons meaning that the other Pz orbital can rotate freely to form the other configuration (ASK PROF)

Question 8

What is the shape of buckminsterfullerene and its bonding?

Answer 8

it a symmetric structure with 60 carbons arranged in a small net with 20 hexagonal faces and 12 pentognol faces.

All of the carbons have a SN of 3, so they all have 3 sp^2 orbitals forming sigma bonds, and a Pz orbtial contributing to the delocalized pi bond formation. The 60 Pz orbtials combine to form 60 MOs with the 60 electrons being found in the first 30 MOs

Question 9

What are fullerenes?

Answer 9

they are closed-cage carbon molecules that have an even number of atoms and properties like the buckminsterfullerene

Question 10

what is the total energy equation for exciting situations?

Answer 10

Etot = Etrans +Erot +Evib + Eel

Question 11

what are the two newly introduced quantum numbers? What do they represent?

Answer 11

J - Rotational energy

v(same as n) - vibrational energy

Question 12

what is a pure rotational transition?

Answer 12

energy changes involved in promoting a molecule from one rotational state to another without change in the vibrational quantum number.

Question 13

What absorption of light allows for the change in the rotational and vibrational states?

Answer 13

absorption of the microwave region allows for the change in the rotational state, and absorption of the infrared region allows for a change in the vibrational state as well as the rotational

Question 14

Why is Etrans considered to be constant?

Answer 14

Because a photon does not carry enough energy to substantially change the velocity of the center mass of the molecule

Question 15

what is a wave number?

Answer 15

it is a unit of energy and frequency that is often reported (where v = 1/lamda (cm^-1)). it is a convenient unit because it is proportional to energy

Question 16

What is the Boltzmann energy distribution? How can we compare the ratio of the number of molecules at different energy levels using the Boltzmann energy distribution?

Answer 16

It is an equation that represents (At thermal equilibrium) The fraction of the total molecules at energy level i with energy Ei

Ni/N = C gi e^(-Ei/kT)

Ni - the number of molecules with Ei
N - total number of molecules
gi - the number of distinct quantum states at Ei (degeneracy)
C- the normalization factor

the ratio of the number of molecules at different energy levels:

The normalization factor cancels and

Ni/Nj = gi/gj (e^(-(Ei -Ej)/Kt))

note: R is used instead of k if the units j pre mole is used

Question 17

What is the linear rigid rotor model? How does it represent the rotational motion?

Answer 17

it is a model in which the bond length is constant and the molecule rotates about an axis perpendicular to the intermolecular axis and passes through its center of mass.

Using this model the rotational motion is represented by a single moment of inertia, defined as:

I = u (Re)^2

Re - the equilibrium bond length

u - is the reduced mass (m1m2/(m1+m2))

Question 18

what is the quantization of the rotational angular momentum? The degeneracy of each level?

Answer 18

square of angular momentum:

J(J+1)(h/2pi)^2 J =0,1,2,3….

projection of the angular momentum on z-axis:

Mj(h/2pi) Mj = -J, -J +1,….. 0 ….., J-1, J

gj = 2J + 1

Question 19

The energy levels in the linear rigid rotor model given by?

Answer 19

Ej = h B J(J+1)

B = h/(8pi^2 I)

or

Ej = h c (B~) J(J+1)
B~ = h^2/(8 pi^2 c I)

note rearranging this expression can give use the bond length (Re)

Question 20

How does microwave absorption spectroscopy work? What are the requirements imposed on it?

Answer 20

When a given heteronuclear diatomic molecule with a dipole moment going from the partial negative charge to the partial negative charge rotates the dipole moment from the positive to the negative also rotates with it, this change in the dipole moment direction allows for the interaction of the oscillating electric field of the microwave light and absorption of a photon, these absorption are then represented on a spectrum.

This indicates that a molecule must have a permanent dipole to absorb or emit microwave radiation.

the two requirements:

1. energy is conserved
2. angular momentum is conserved by J = =-1 since a photon only carries 1 J ( if the photon is absorbed then +1, if a photon is emitted then -1)

Question 21

How can we calculate the energy of a photon during microwave spectroscopy?

Answer 21

Ephoton = 2 B~ (Ji+1)