Electrons in Molecules Flashcards

1
Q

What are the M-shell orbitals

A

they have n=3
so are the 3s, 3p, and 3d orbitals

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2
Q

the number of radial nodes is equal to

A

n - l - 1
for the m shell l = 0, 1, 2 (hence 3s has two nodes, 3p has one node and 3d has non)

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3
Q

What is a radial node and what is a nodal plane

A

Radial node is a sphere where electrons will not be found
Nodal plane is a plane in the x, y or z direction where electrons are not found

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4
Q

How is the number of nodal planes worked out
e.g. work out the number of nodal planes for 3d orbital

A

no of nodal planes = n - 1
e.g. 3d = 3-1 = 2 nodal planes

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5
Q

As well as hydrogen, ions like He⁺, Li²⁺, Be³⁺, are also ‘one-electron atoms’
What is the difference however

A

Their nuclear charge (z) is larger
As z increases the electrons are more strongly attracted to the nucleus, so the orbital energies become more negative. The orbitals become less diffuse (shown by the radial wavefunction)

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6
Q

Electrons possess an intrinsic property called spin: this is describe by which quantum number

A

This is described by the spin quantum number ms
this always takes the value ±1/2 for electrons

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7
Q

If there is no electric or magnetic field, the two spin states…

A

… have the same energy (degenerate)

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8
Q

To completely describe an electron in the hydrogen atom…

A

… four quantum number must be given: n, l, ml, ms

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9
Q

True of false:
You can use the Schrodiger equation analytically for wavefunction, for multi-electron atoms

A

False
The reason for this is due to the electrostatic repulsion between pairs of electrons, which contributes to the potential energy of the attraction between the electrons and the nucleus

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10
Q

To overcome not being able to use the Schrodiger equation for multi-electron atoms we use the orbital approximation
This is?

A

In the orbital approximation each electron of a multi-electron atom occupies a one-electron orbital (this is specificed using electronic configuration)

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11
Q

The Pauli exclusion principle states…

A

… no two electrons in a multi-electron atom can have all four quantum numbers (n, l, ml, ms) the same
This means only two electrons with ‘paired’ spins (+1/2 and -1/2) can occupy each orbital

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12
Q

Explain the radial distributiom function of Lithium

A

The one-electron radial distribution functions for lithium show that the extra L-shell electron tend to be found outside the space occupied by the 1s electrons
The 1s electrons are said to shield the L-shell electrons from the full attraction of the nucleus
The 2s electrons are said to penetrate the nucleus more than the 2p electrons as the shielding of the shielding of the 2s orbitals by the 1s orbitals is less then the 2p (this lowers the energy of the 2s orbital, hence why it fills first)

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13
Q

The extent of shielding and penetration can be quantified by

A

The extent of shielding and penetration can be quantified by considering that the outer shell electrons experience a reduced effective nuclear charge (Zeff)

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14
Q

What is the Aufbau principle

A

States that the electrons in a multi-electron atom occupy the orbitals in the order
1s<2s<2p<3s<3p<4s<3d…
So as to achieve the lowest overall energy with each orbital occupied by two electrons

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15
Q

How do you convert energy in Jules into wavelength in nm
(1cm⁻¹ = 1.986x10⁻²³)
E.g. convert 5.89x10⁻¹⁷ into nm

A

1cm = 1x10⁻⁷
1.986x10⁻²³ / 1x10⁻⁷ = 1.986x10⁻¹⁶
1.986x10⁻¹⁶ / 5.89x10⁻¹⁷ = 3.372 nm

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16
Q

The solar spectrum contains emission lies due to electronic transition in the one-electron is C⁵⁺
What is the wavelength of the line corresponding to the n=2 to n=1 transition for this ion
Rh = 2.18x10⁻¹⁸ J

A

z = 6
-(6²)(2.18x10⁻¹⁸)(1/2² - 1/1²) = 5.89x10⁻¹⁷
5.89x10⁻¹⁷ J = 3.372 nm

17
Q

Ionisation energy is

A

The energy change for the reaction
A(g) → A⁺(g) + e⁻
This can be use using Koopmans theory to work out orbital energy and hence strodingers equation

18
Q

For one-electrons atoms ΔiE is related to

A

The energy of the outermost occupied orbital
ΔiE = -Eorbital
However because electron repulsion this is not exactly the case for a multi-electron atoms

19
Q

The molecular potential energy curve shows

A

A mimimum which corresponds to the equilibrium bond length Re
The depth of De of the minimum relative to D0 the energy of two infinitely separated atoms is closely related to the bond dissociation energy
Shows the addition potential energy associated when two atoms are in a molecule relative to the potential energy for two infinitely separated atoms which don’t interact

20
Q

An exact anlytical solution for the Schrodinger equation is not possible for multi-electron atoms. Instead, the Born-Oppenheimer approximation must be used, which assumes…

A

That the (heavy) nuclei are stationary which the (much lighter) electrons move in the resulting electrostatic potential
Using this appoximation the Schrodinger equation can be solved numerically for the electrons with a given internuclear seperation (bond length)
The calculation can be repeated for different separations so that the variation of the energy of the molecule with the bond length can be obtained (molecular potential energy curve)

21
Q

There are 2 contributions to the potential energy in H₂⁺

A
  • An electrostatic attraction between the negatively charge electron AND each of the two positively charged H nuclei
  • An electrostatic replusion between the two positively charged H nuclei
22
Q

A molecular orbital (Ψ) is

A

A single electron wavefunction on a molecule

23
Q

Molecular orbitals are formed through an approach called

A

Linear combination of atomic orbital (LCAO)

24
Q

Why does a bonding interaction occur

A
  • As atomic orbitals are single-electron wavefunctions, they have a wave character, so the atomic orbitals on one atom interact with those on another atom. If this interference occurs constructively, an in-phase combination occurs
  • The increased electron density between the nuclei hold the positive nuclei together - this means a bond is formed
  • The molecular orbital bonding arising is called a bonding orbital
25
Q

Why does an antibonding interaction occur

A
  • As an atomic orbital are single-electron wavefunctions, they have a wavecharacter, so the atomic orbitals of one atom interact with those on another atom
  • If this interference occurs destructively, an out-of-phase combination occurs
  • Forms an antibonding orbital
26
Q

The energies of two MOs relative to that of the Hydrogen atom can be calculated as a function of

A

The internuclear seperation

27
Q

Which line show on the graph denotes the bonding or antibonding orbital

A

The red line (E+) shows that if an electron occupies this MO it is energetically favourable for a bond to form - bonding orbital
The blue line (E-) shows there is no energy advantage for bond formation, and this orbital is known as an anti-bodnig MO

28
Q

The square of the wavefunction given the probability density of the electron
These graphs are formed, explain their shape

A

For the bonding MO, there is a build-up of electron density between the atoms as R decreases, so that the electron has an energetically favourable interaction with both nuclei
For the antibonding MO the electron density builds up outside the region between the two nuclei, pulling them apart

29
Q

Diatomic molecule like H₂⁺ have cylindrical symmetry, meaning?

A

They are unchanged by a rotation through an arbitrary angle about the internuclear axis

30
Q

What is inversion symmetry

A

The symmetry operation involves moving from an arbitrary point through the centre of the molecule and out an equal distance on the other side

31
Q

For homonuclear diatomic molecules the bonding MO and Antibonding MO has an identical value at intersion symmetry
What is the name for the symmetry for the bonding and antibonding MO

A

The bonding MO is said to posses gerade symmetry (german = ‘even’)
The antibonding MO has an identical value but opposite sign and is said to posses ungerade symmetry (german = ‘odd’)
Subscript g and u are used to denote this

32
Q

Why in the MO diagram for H₂⁺, is the antibonding MO raised more than the bonding MO is lowered

A

Because at the seperation value Re, the antibonding molecular orbital is raised in energy more than the bonding MO is lowered from the y axis