Nolan's Things

Question 1

How many atoms do orbitals exist for?

Answer 1

One - Hydrogen. All other atoms are called ‘Hydrogen Like’, due to the complexity of the schrodinger equation, it cannot be solved for elemements with more than one electron, i.e. H, He⁺, Li ²⁺. There is no exact solution for other elements, so their orbitals are only an approximation

Question 2

How do you calculate the effective nuclear charge?

Answer 2

Z* = Z-S where S = the sheilding constant

Question 3

What are the rules for cacluating the sheilding constant for an ns valence electron?

Answer 3

1. Any electrons to the right of the valence electron (VE) are unimportant and can be ignored. 2. All other electrons in the nsnp group sheild the valence electron to the extent of 0.35 3. Electrongs in the n-1 shield to the exxtent of 0.85 4. Electrons in n-2 or lower shield to the extent 1.00 Where the values are normalized to 1, and hense unitless

Question 4

How o you order and group the electrons using Slater’s Rule to calculate the shielding constant?

Answer 4

In order, rather than energetically, i.e. (1s)(2s2p)(3s3p)(3d)(4s4p)(4d)(4f)(5s5p)(5d)(5f) S and P orbitals are similar enough to group together for this model.

Question 5

How would you cacluate S for a nf valence electron?

Answer 5

1. 1. Any electrons to the right of the valence electron (VE) are unimportant and can be ignored. 2. All other electrons in the nsnp group sheild the valence electron to the extent of 0.35 3. Electrons in n-1 or lower shield to the extent 1.00 Where the values are normalized to 1, and hense unitless

Question 6

Why is there a differece between the calcuation of d or f shell valence electrons and s and p shell valence electrons?

Answer 6

Due to the diffuseness of the d/f orbitals. D and f orbitals always sit above s and p which are much more penetrating orbitals, and have a much higher sheilding effect, when compared to the d or f orbitals.

Question 7

Why can the effective neuclear charge of not be used to caclulate ionization energy?

Answer 7

It is a QUALITATIVE measurment, and can only be used to compare electrons. It assumes that all electrons are sheilded equally by electrons lying above them in the order, and it does not take more idffuse orbitals into account.

Question 8

What is the effective neuclear charge of the valence s electrongs in ⁴⁸Cd?

Answer 8

(1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)10(5s)2 (28 * 1) + (18 * 0.85) + (1 * 0.35) = 43.65

Question 9

What is the effective neuclear charge of the valence f electron in ⁶¹Pm

Answer 9

Figure this one out loser.

Question 10

What is the reason behind the Lanthanide Contraction?

Answer 10

The Lanthanide contraction is due to the increase of mass at the core of the atom, but due to the diffuse nature of f-orbitals, there is a DRAMATIC shrinkage of the atomic radii across the period.

Question 11

What causes Relativistic effects?

Answer 11

Relativistic Effects are corrections that effet electrons differently, depending on the electron speed relative to the speed of light. They are more prominent in heavy atoms where the large ionic radii means that the far away electrons can travel almost at the speed of light.

Question 12

What are relativistic effects? Give one example

Answer 12

Relativistic effects in chemistry can be considered to be perturbations, or small corrections, to the non-relativistic theory of chemistry, which is developed from the solutions of the Schrödinger equation. These corrections affect the electrons differently depending on the electron speed relative to the speed of light.

An example is the yellow colour of cold, compared with the silvery colour of the metals around it.

Question 13

What is the general stable oxidation state of the Lanthanides?

Answer 13

+3

Question 14

What is the most popular stable oxidation state of Yb?

Answer 14

The electronic configuration is 6s² 5d¹ 4f¹³

It’s most popular oxidation state is +2.

As it has a f13 subshell, which is almost full, the d electron is promoted to provide the stability of the full f-shell, and the atom will then happily lose it’s s electrons.

Question 15

What is the most poular oxidation state of Ce?

Answer 15

The electronic configuration is 6s² 5d¹ 4f¹

It’s most popular oxidation state is +4, as this provides it with a nobel gas like electronic configuration.

Question 16

What is the electronic configuration of Sm, and hense it’s most stable oxidation state?

Answer 16

The electronic configuration is 6s² 5d¹ 4f⁵

It’s most popular oxidation state is +2. It’s d electron is promoted easily into the f shell, giving it 6 electrons in it’s f-shell, which is almost a half filled shell and hence is more stable than the resulting 3+ state

Question 17

Name the only radioactive Lanthanide?

Answer 17

Promethium

Question 18

What are the Reactivity trends across the Lanthanides?

Answer 18

Lathanides are very lewis acidic. They are starved of electrons and go after lone pairs. They have very high co-ordination numbers, and can fit 8/12 ligands arouund them in complexes, so long as they fit. They are also oxophilic and halogenphilic.

Question 19

What are some of the problems with NMR associated with paramagnetic materials?

Answer 19

Paramagnetism, and interaction with the magnetic fields results in huge phasing problems, as well as ‘fold overs’ where the peaks on the graph are shown much lower than they should be, as they went off the scale. Lu and La are the exceptions to the Ln problem with NMR as they are not paramagnetic and hense have no problems.

Question 20

What are some of the uses of Lanthanides?

Answer 20

The metals are brittle and hard, and can be used and oil drilling and added to metal alloys, espicially steel. Zeolites are used in petroleum cracking.

Question 21

1. Can you figure out the electronic config of f-electrons?
2. Explan the stable oxidations states?
3. Discuss general reactivity and properties of the lanthanides?

Answer 21

Yes?

Question 22

What is the starting and end point of alkene polymerization using a transition metal catalyst?

Answer 22

Metal-Hydride bond starting point, with b-hydride eminination as the chain ending step

Question 23

What is a pro and a con of using a lanthanide based catalyst for alkene polymerization?

Answer 23

Lanthanides do not require a co-catalyst, however they are incredible oxygen sensitive.

Question 24

Draw the Alkene chain formation steps

Answer 24

Question 25

Draw Alkene chain termination

Answer 25

Question 26

What causes chain termination of Alkenes when using a transition metal or lanthanide catalyst?

Answer 26

Due to limited space around the metal centre, it becomes more and more difficult tod eliver ethylene to the reaction site. If the ethylene is not delivered fast eough then the metal centre will react with any nearby electron density, which usually is a hydrogen atom.

Question 27

What is the most stable Metal-Alkane bond?

Answer 27

M-Me. This is the only R group which does not have the possibility of undergoing b-hydride elmination.

Question 28

Which lanthanides are the best for alkene polymerization and why?

Answer 28

Late transition metal catalyized hydrogenation of alkenes differes to the early version due to a difference in size. The large the ion, the more binding sites available, and the ionic character increases, making better, longer chains.