Lecture 2 - Orbital theory, electron configurations and hybridisation Flashcards

1
Q

The Bohr Model of the atom

A
  • electrons are arranged in shells/energy levels around a positively charged nucleus.
  • electrons jump between orbits by absorbing/ moving out or emitting/moving in electromagnetic radiation
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2
Q

Shells are given numbers. What are these called?

A

Principle quantum numbers.

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3
Q

What is the principle quantum number for the lowest energy shell?

A

N= 1

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4
Q

The higher the principle quantum number….

A

the more energy and the further from nucleus.

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5
Q

Shells are divided into…

A

subshells

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6
Q

How many subshells does the shell n=3 have?

A

3

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7
Q

name the four subshells

A

s, p, d and f

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8
Q

Subshells are made up of….

A

orbitals

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9
Q

What is an orbital?

A

An orbital is a region of space around the nucleus containing up to two electrons with opposite spins. They are region of space that you are likely to find an electron.

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10
Q

Electrons are not point particles. What are they? what is the implication of this on our idea of orbitals?

A
  • electrons have wave-like properties. They should be described as a wavefunction.
  • There is a possibility of finding an electron at any given location around the nucleus.
  • This is why the modern theory of the atom describes the positions of electrons around the nucleus in terms of probabilities- depending on its energy level, it exists only in certain regions around the nucleus. These are orbitals.
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11
Q
  • In the s subshell there is…
  • in the p subshell there is…
  • In the d subshell there is…
  • In the f subshell there is…
A
  • one s oribtal, total of 2 electrons
  • three p orbitals, tot. 6 electrons
  • 5 d orbitals, tot. 10 electrons
  • 7 f orbitals, tot. 14 electrons
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12
Q

How do orbitals normally fill? What is the exception?

A

Orbitals fill singly from the lowest to the highest energy. There is an anomaly with transition metals. 4s subshell is of a lower energy than the 3d subshell so the 4s subshell fills first.

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13
Q

Describe the shape of an S orbital, the effect on probability of finding electrons and the effect of the size of the shell

A
  • Sphere shaped. Highest probability of finding the electron is at the nucleus.
  • As space shells become bigger and further from the nucleus, there is a greater area in which you may find the electron.
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14
Q

Describe the shape of a p orbital

A

two ellipsoids with a point of tangency at the nucleus- in other words, shaped like dumbbells.

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15
Q

There are 3 p orbitals. How are they positioned?
Name the three

A

At right angles to each other.
2py, 2px, 2pz

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16
Q

With p orbitals, there is no node at the nucleus (dumbbell shaped). What are the effects of this?

A

No probability of finding an electron in a p orbital at the nucleus.

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17
Q

Explain the valence shell electron pair repulsion theory.

A
  • electrons are negatively charged and repel each other.
  • atoms with higher number of electrons in their valence shells (outer shells) will form fewer bonds and therefore have lone pairs of electrons.
  • lone pairs sit closer to the nucleus of an atom and so exert more repulsion on the other valence electrons.
  • lone pairs compress bond angles.
  • lone pairs repel more than bonding pairs.
    therefore, the greatest bond angles are between lone pairs of electrons.
  • the shape of molecules is therefore determined by the type and number of electron pairs surrounding the nucleus.
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18
Q

Draw a linear molecule.
Describe linear molecule

A
  • example CO2
  • two bonding pairs, no lone pairs
  • bond angle of 180
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19
Q
  • Draw a trigonal planar molecule
  • Describe.
A
  • BF3 is an example
  • 3 bonding pairs, no lone pairs
  • 120 bond angle
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20
Q
  • Draw a trigonal pyramidal molecule
  • Describe
A
  • 3 bonding pairs, one lone pair
  • 107 bond angle
21
Q
  • Draw a tetrahedral molecule
  • Describe
A
  • CH4/ methane is an example
  • 4 bonding pairs, no lone pairs.
  • 109.5
22
Q
  • Draw a non-linear/ v shaped molecule
  • Describe
A
  • H2O is an example
  • 2 lone pairs and 2 bonding pairs
  • 104.5 bond angle
23
Q
  • Draw an octahedral molecule
  • Describe
A
  • SF6
  • no lone pairs, 6 bonding pairs
  • bond angle 90
24
Q

What is the valency of an atom?
What is the valency of carbon?

A

The number of bonds an atom must make to complete the outer shell. Carbon has four outer electrons. To have a full outer shell carbon must form four covalent bonds. Therefore, it has a valency of four.

25
Q

Based on the one s and three p orbitals of carbon, the proposed structure of methane is not what it actually is. What is that reasonable, proposed structure?

A
26
Q

What is the actual shape of methane? Why does it have this shape?

A

Tetrahedral. This is the shape which minimises repulsion between bonds and males the structure most stable.

27
Q

You can’t get the tetrahedral shape of methane from ‘pure’ 2s and 2p orbitals. So, what did Linus Pauling propose?

A

None of the bonding orbitals are 100% p or 100% s. Instead, carbon has hybrid orbitals made up of partial s and partial p characters.

28
Q

NOTE: these examples refer to carbon as having one electron in 2s and three electrons in 2p, rather than having one 2s orbital containing 2 electrons and two p orbitals. We call this the ………… electron configuration of carbon. How do we achieve this? Draw it.

A
  • excited/promoted
  • achieve this through energy input.
29
Q

Sp^3 orbital hybridisation of carbon. Explain

A

Combines the s orbital and all three p orbitals. Some of the s character and some of the p character are hybridized and four new orbitals are formed- Sp^3 hybrid orbitals. You can think of Sp^3 hybrid orbitals as being made up of 25% of the 2s orbital and 75% of the 2p orbitals. The orbitals are now equal in energy.
This creates a tetrahedral shape.

30
Q

Explain the lobes in Sp3 hybrids

A
  • Sp3 hybrids have two lobes and are unsymmetrical about the nucleus. One of the two lobes is bigger than the other so can overlap more effectively with an orbital of another atom when it forms a bond.
31
Q

Draw the hybridization in methane- show the orbitals.

A
32
Q

Explain Sp3 hybridization in ethane

A

One Sp3 hybrid orbital from one carbon overlaps with that of the other carbon to form an Sp3-Sp3 sigma bond. The three remaining sp3 hybrid orbitals of each carbon atom will then overlap with 1s orbitals of three hydrogens to form the 6 C-H bonds in ethane.

33
Q

Ammonium ion is another example of what type of hybridization. Draw the molecule.

A

Sp3- tetrahedral.

34
Q

Explain the hybridization of nitrogen in ammonia. What is the molecular shape of ammonia?

A
  • The central nitrogen atom in ammonia is bonded to three hydrogens and has one lone pair. The ideal structure for four pairs of electrons is tetrahedral. So, hybridisation of the central nitrogen atom is Sp3.
  • But, because the lone pair sits closer to central atom and has a stronger repulsion than bonding pairs, the molecule is left with a piano stool arrangement of atoms about the central atom- this is called trigonal pyramidal.
35
Q

Describe hybridization of oxygen in water and it’s molecular shape.

A
  • Hybridisation of oxygen is still sp3.
  • But it has two lone pairs which repel more and reduce bond angle.
  • the molecule is therefore v shaped
36
Q

Sp^2 orbital hybridisation.

A

Combines the 2s orbital and two 2p orbitals. The third 2p orbital is left unchanged.
This creates a trigonal planar shape.

37
Q

Describe the orientation and bond angles of the Sp2 orbitals

A
  • Like sp3, the Sp2 hybrid orbitals are unsymmetrical about the nucleus and are strongly orientated in a specific direction so they can form strong bonds. The three sp2 orbitals lie in a plane of angles at 120 degrees to each other. The remaining p orbital is perpendicular to the Sp2 orbitals.
38
Q

Describe Sp2 orbital hybridization in ethene.

A
  • The two carbons with sp2 hybridization form a strong sigma bond in an sp2-sp2 head on overlap
  • The unhybridized p orbitals of each carbon will overlap sideways above and below the plane of the sigma bond to form a pi bond.
  • This results in the sharing of four electrons between the carbon atoms, to create a carbon-carbon double bond.
  • The two remaining Sp2 orbitals on each carbon atom will then form sigma bonds with the 1s orbitals of the hydrogen atoms, to form the 4 C-H bonds in ethene.
39
Q

Sp orbital hybridisation

A

Combines one 2s and one 2p orbital to form 2 hybrid sp orbitals. Two 2p orbitals remain unchanged. The two Sp orbitals are orientated 180 degrees apart.
This creates a linear shape.
The two remaining 2p orbitals are perpendicular.

40
Q

Describe sp hybridization in ethene CHCH

A
  • The two Sp hybridized carbons approach each other. One of the sp hybrid orbital from each carbon atom overlap head on with the other to form an Sp-Sp sigma bond.
  • At the same time, the Pz orbitals from each carbon overlap sideways to form a Pz-Pz pi bond. The two Py orbitals from each carbon overlap sideways to form a Py-Py pi bond.
  • 6 electrons are therefore now shared between two carbon atoms to form a carbon-carbon triple bond.
  • The remaining sp orbital on each carbon form a sigma bond with the 1s orbital in hydrogen.
41
Q

What is a sigma bond?

A

A sigma bond is a type of covalent bond. It is the strongest type of covalent bond.
Forms when atomic orbitals overlap head on.

42
Q

Sigma bonds can form between….

A

two s orbitals overlap (dihydrogen)
s and p orbitals overlap
s and sp orbitals overlap

43
Q

What are pi bonds?

A

A type of covalent bond formed from the overlap of two p orbital lobes on one atom with two orbital lobes on another. The sideways overlap of two p orbitals, above and below the plane of a sigma bond.
Pi bonds form with an existing sigma bond so they create double/triple bonds.

44
Q

In a pi bond there is a plane between molecular orbitals with no electron density. Why?

A

Due to the dumbbell shape of p orbital, in the node between lobes there is no electron density. So, when they overlap, there is a plane between molecular orbitals with no electron density.

45
Q

There is ………. around a sigma bond

A

rotation.

46
Q

pi bonds are fixed. What are the effects of this?

A

there is no free rotation. Leads to isomers.

47
Q

Give the shortcut for figuring out what hybridisation an atom has

A
48
Q

The exceptions for the hybridisation rule:

A
  1. Lone pairs adjacent to pi bonds/systems.
    - Lone pairs adjacent to pi bonds/systems tend to be in unhybridized p orbitals rather than in hybridised orbitals. In these cases, a hydrogen or oxygen that we might expect to be Sp3 hybridized is actually Sp2 hybridized.
  • You may wonder why because being in Sp3 is more stable than being in Sp2 because there is a greater electron pair repulsion in Sp2 (bigger bond angle) and so a rise in energy.
  • The answer= the lowering of energy from the overlap of the p orbital containing the lone pair of electrons, with the pi bond lowers the energy. It overcompensates for that energy gain from having sp2 hybridization rather than Sp3.
  • Also. Having lone pairs in unhybridized p orbitals allow for a better orbital overlap with adjacent pi systems (better overlap than if they were in an sp3 orbital). This is because the p orbital actually sits closer to the pi system/bond than the Sp3 orbital is.