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aa The f-elements > Meet the Lanthanides 1 > Flashcards

Meet the Lanthanides 1 Flashcards

1
Q

What are the 4f elements called

A
  1. Lanthanides
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2
Q

What are the 5f elements called

A
  1. Actinides
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3
Q

What is an application of yttrium

A
  1. Cancer treatment drugs
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4
Q

What is an application of cerium

A
  1. Catalytic converters
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5
Q

What are two applications of praseodymium

A
  1. Super-strong magnets
  2. Lasers
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6
Q

What is an application of neodymium

A
  1. extremely strong permanent magnets
  2. lasers
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7
Q

What are two applications of europium

A
  1. color tv screens
  2. fluorescent glass
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8
Q

What is an application of gadolinium

A
  1. NMR imaging
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9
Q

What is an application of terbium

A
  1. TV sets
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10
Q

What are two applications of holmium

A

1 lasers
2. high strength magnets

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11
Q

What is an application of thulium

A
  1. High temp superconductor
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12
Q

What is an application of ytterbium

A
  1. lasers
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13
Q

What is one of the biggest applications of the f-elements

A
  1. catalysts- especially solid-state
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14
Q

What caused the difficulties in separating the lanthanides and obtaining the elements in pure form

A
  1. The pronounced chemical similarities between the elements
  2. Due to the 4f orbitals
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15
Q

How many 4f orbitals are there for any atom

A
  1. 7
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16
Q

What are the 2 sets of 4f orbitals and why does this occur

A
  1. Two solutions to the schrodinger equation so depends on symmetry
  2. General set- low symmetry situations
  3. Cubic set- higher symmetry cubic environment
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17
Q

What are the 7 4f orbitals in the general set

A
  1. z^3
  2. xz^2
  3. yz^2
  4. x(x^2-3y^2)
  5. y(3x^2-y^2)
  6. z(x^2-y^2)
  7. xyz
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18
Q

What are the 7 4f orbitals in the cubic set

A
  1. xyz - in general
  2. z(x^2-y^2) - in general
  3. z^3 -in general
  4. x^3
  5. y^3
  6. x(z^2-y^2)
  7. y(z^2-x^2)
19
Q

Describe briefly the orbitals

A
  1. X^3, Y^3 and z^3 are on the axis orbitals
  2. x(z^2-y^2), y(z^2-x^2), z(x^2-y^2), xyz each have 8 lobes
  3. x(z^2-y^2), y(z^2-x^2), z(x^2-y^2) are related to each other by 90 degree rotations about the x,y and z-axis respectively
20
Q

Do 4f-orbitals possess nodes

A
  1. They possess conical and planar nodes
  2. NONE possess radial nodes
21
Q

What are the quantum numbers of the 4f orbitals

A
  1. Principal quantum number n=4 (shell)
  2. secondary quantum number l=3 (types of orbital possible)
  3. magnetic quantum numbers = -3, -2, -1, 0, +1, +2, +3. (orbital orientations for each l value)
22
Q

Describe radial extension of 4f orbitals and draw radial distribution function

A
  1. 4f orbitals have only limited radial extension - extremely contracted- no radial nodes
  2. Inefficiently shielded from Zeff
23
Q

What is Zeff

A
  1. the effective nuclear charge is the actual amount of positive charge experienced by an electron in a multi-electron atom.
24
Q

What is the result of the 4f orbitals having limited radial extension

A
  1. Electrons accommodated behave in a core-like manner
  2. thus are relatively uninvolved in bonding and are also unaffected by the ligand environment to any great degree.
  3. They don’t react well with environment around them
25
Q

Write the electronic configuration for all 4f elements

A
  1. La [Xe]5d16s2
  2. Ce [Xe]4f15d16s2
  3. Pr [Xe]4f36s2
  4. Pm [Xe]4f56s2
  5. Sm [Xe]4f66s2
  6. Eu [Xe]4f76s2
  7. Gd [Xe]4f75d16s2
  8. Tb [Xe]4f96s2
  9. Dy [Xe]4f106s2
  10. Ho [Xe]4f116s2
  11. Er [Xe]4f126s2
  12. Tm [Xe]4f136s2
  13. Yb [Xe]4f146s2
  14. Lu [Xe]4f145d16s2
26
Q

Which are the 4 elements which don’t follow the pattern in expected electronic configuration

A
  1. Ce
  2. Pr
  3. Gd
  4. Lu (just because filled up f though)
27
Q

Why does Ce have the electronic configuration: [Xe]4f15d16s2

A
  1. At the beginning of the series the 5d orbitals are very similar energy to 4f orbitals.
  2. Can get lower energy conformation by putting one e- in both orbitals.
  3. If both in 4f orbital they would repel each other as close
  4. That repulsion leads to promotion as orbitals are very very close in energy at beginning of series
  5. As the atomic number (Z) increases the 4f orbitals rapidly contract and are stabilized relative to the 5d orbitals.
  6. This is because the 4f set are only poorly shielded by the core electrons. so cheaper to put 3 electrons in close together f orbitals than higher energy 5d orbital as 4f orbital energy been pulled down.
  7. Thus the ground state of La is [Xe]5d16s2 and that of Ce is [Xe]4f15d16s2 not [Xe]4f26s2.
28
Q

Why does Gd have electronic configuration: [Xe]4f75d16s2

A
  1. At Gd the 4f orbitals are half filled and the next electron is added to a 5d orbital giving [Xe]4f75d16s2
  2. As half-filled shell gives extra stabilisation
  3. Same as Cu and Cr
  4. To do with exchange energy
29
Q

What is exchange energy

A
  1. effect that arises due to parallel electrons being indistinguishable and interchangeable and leads to stabilisation of configurations that contain parallel electrons.
  2. The more pairs of parallel electrons present in an atom the greater the exchange energy
  3. Why half-filled shells are relatively stable
30
Q

What is pairing energy

A
  1. The pairing energy P is the energy penalty for putting two electrons in the same orbital, resulting from the electrostatic repulsion between electrons
31
Q

What is the favoured oxidation state of all lanthanides

A
  1. +3 oxidation state
  2. Always most stable
  3. In each case the sum of the first three ionization energies are less than the energy associated with the process
32
Q

Why is it rare to get +4 oxidation state

A
  1. In most cases the extra energy to remove the fourth electron cannot be compensated for by bond formation.
33
Q

Why is +3 OS the most stable

A
  1. As electrons are removed from the neutral atoms all of the valence (4f, 5d, 6s) orbitals are stabilized.
  2. As successive electrons are removed from the neutral Ln the stabilizing effect on the orbitals occurs in the order of their principal quantum number 4f>5d>6s.
  3. Easy to lose 1st 2 e- form s block
  4. 3rd e- in the f-orbital is high enough in energy to remove
  5. Once three electrons have been removed, the extra stabilization of the 4f orbitals is so large that the 4f electrons are so tightly held as to be inaccessible.
  6. (A further effect of the core-like character of the 4f orbitals.)
34
Q

Which Ln elements can form stable 2+ ions

A
  1. Eu2+ (4f7),
  2. Yb2+ (4f14)
  3. Sm2+ (4f6).
35
Q

Why can certain Ln elements form stable 2+ ions

A
  1. stability is achieved at half-filled (Eu2+) and completely filled 4f shells
  2. This is also responsible for higher third ionisation energy of Eu and Yb and very low values for subsequent elements.
36
Q

What is chemical evidence for the Ln2+ ions

A
  1. observation that like the alkaline earths Ca and Ba, europium and ytterbium dissolve to form blue solutions containing Ln2+ ions and solvated electrons in liquid ammonia.
37
Q

What is a consequence of the +2 OS

A
  1. Sm and Eu make good reducing agents
  2. They still want to give e- away but can be made- want to form +3 OS
38
Q

What element can form the +4 OS

A
  1. Ce [Xe]4f0
  2. Chemically accessible due to higher energy of the 4f orbitals at the start of series which are not sufficiently stable to prevent the loss of the fourth electron.
39
Q

What is consequence of +4 OS

A
  1. Ce4+ is good oxidising agent as not stable enough to prevent loss of the fourth electron
40
Q

Describe the trend in radii of Ln3+ ions and Ln metal as you go across the group

A
  1. Both the metallic and ionic radii show a gradual reduction with increasing atomic number.
  2. This is with the exception of the sharp discontinuities for the metallic radii of Eu and Yb which are some 0.2 Å higher than would be expected by extrapolation.
41
Q

What is the cause reduction of radii as atomic number increases

A
  1. The reduction in the Ln metal and Ln3+ radii is referred to as the lanthanide contraction - ALWAYS in exam
  2. Arises from the poor ability of the 4f electrons to screen the other valence electrons from the increasing nuclear charge. i.e the effective nuclear charge increases across the series.
42
Q

Why are there sharp discontinuities for the metallic radii of Eu and Yb

A
  1. This can be rationalised in terms of a structure Ln2+ (e-)2 for Eu and Yb rather than the Ln3+ (e-)3 adopted by the other lanthanides with the result that the other electrons are not pulled so near to the nucleus.
43
Q

What is a consequence of the lanthanide contraction

A
  1. Consequences for the metals of the subsequent second and third transition series.
  2. The reduction in radius caused by the poor screening ability of the 4f electrons results in the ions of metals from the same groups having very similar radii and chemistries (e.g. Zr4+ and Hf4+ or Mo3+ and W3+).

aa The f-elements (11 decks)

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