Spectroscopy & UV Flashcards

(40 cards)

1
Q

What is spectroscopy?

A

The study of interaction of matter and electromagnetic radiation

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2
Q

What is electromagnetic radiation (EMR)?

A

The radiant energy that displays the properties of both particles and waves

There are different types of EM waves that make up the electromagnetic spectrum (EMS)

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3
Q

Example of EMR

A

Visible light

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4
Q

What range is visible light?

A

400 to 700 nm

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5
Q

What are gamma rays?

A

Emitted from the nuclei of some radioactive elements and because of their high energy, can severely damage biological organisms

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6
Q

What are X-rays?

A

Less harmful than gamma except in high doses

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7
Q

What is UV light?

A

Responsible for sunburn

Repeated exposure leads to skin cancer by damaging DNA molecules in skin cells

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8
Q

What is infrared radiation?

A

Felt as heat

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9
Q

What are radio waves?

A

Have the lowest energy (frequency)

E.g. Radio, TV

Also used in NMR and in magnetic resonance imaging (MRI)

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10
Q

Nature & properties of electromagnetic radiation

A
  • Has 2 components = an electric field and a magnetic field which are in planes at right angles to each other
  • A given point in space experiences a periodic disturbance in electric and magnetic fields as the wave from radiation passes by
  • EM waves can travel through a vacuum
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11
Q

What is frequency (v) in EMR?

A

The number of crests of the wave that passes a point per second (number of cycles)

Unit is s-1 also known as Hz

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12
Q

What is wavelength (λ) in EMR?

A

Distance between 2 adjacent crests

Unit is in metres but nm can be used

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13
Q

What is velocity (c) in EMR?

A

In a vacuum, EM waves travel with the same velocity

This is ALWAYS 2.998 x 10^8 ms-1

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14
Q

What equation relates frequency, velocity and wavelength?

A

C = λ x v

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15
Q

What is a particle of EMR called?

A

A photon

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16
Q

How is the energy (E) of a photon calculated?

A

E = h x v

OR

E = h x c/ λ

h is planks constant = 6.626 x 10-34 J s

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17
Q

Key things about energy of a photon

A

Directly proportional to frequency (v)

Inversely proportional to wavelength (λ)

Units is Joules

18
Q

What is the wavenumber?

A

1/λ —> Ṽ

Units is m-1

19
Q

Describe UV spectroscopy

A
  • UV provides radiation
  • Monochromator selects a single wavelength from the wide range provided by the light source
  • i0 is incident light = light going into the sample
  • i is the light not absorbed by the sample
  • The cell (cuvette) can be glass or plastic for visible light but must be quartz for UV light (glass and plastic absorb UV)
20
Q

What factors contribute the amount of absorbance?

A
  1. The thicker the sample, the more absorption takes place
  2. More absorption seen in more concentrated samples (more molecules present to absorb light)
  3. Some molecules are better than others at absorbing a particular wavelength = an inherent property of the molecule
21
Q

What does absorbance (A) of a sample depend on?

A
  1. Concentration (c) of the absorbing species in mol L-1
  2. Length of the light path (l) through the cell in cm
  3. Molar absorption coefficient (E) in L mol-1 cm-1

E relates to the inherent property of a molecule to be better at absorbing particular wavelengths than others

22
Q

What is the Beer-Lambert law equation?

A

A = E x c x l

23
Q

How can the beer-lambert equation be used in measuring incident and transmitted light?

A

A = E x c x l —> A = log10(I0/I0)

24
Q

How can transmission be calculated using absorption data?

A

Transmission (T) = I/I0 SO A = -log10(T)

25
How can we calculate transmitted light (I)?
I = I0 x 10^-Ecl
26
Basic premise of molecular orbital theory
When atoms get close enough to bond, their atomic orbitals reshape themselves and become a set of molecular orbitals that no longer belong to any one atom but instead are orbitals for the entire molecule
27
6 basic rules of molecular orbital theory
1. Number of molecular orbitals is equal to the number of atomic orbitals 2. If 2 AOs combine then, of the 2 molecular orbitals creates, one will be a bonding orbital (lower energy) and one will be an antibonding orbital (higher energy) 3. Electrons enter the lowest orbital available 4. Maximum number of electrons in an orbital is 2 5. Hund’s Rule —> electrons are alone before they pair up 6. Orbitals on an atom that can’t combine with any on the other become non-bonding orbitals
28
Molecular orbitals formation
- A sigma bond is by s-s overlap and p-p overlap (head to head) - A bond formed by 2p orbitals overlapping side to side is a pi bond - Non-bonding orbitals are higher in energy than pi
29
MOs in UV spectroscopy
- Antibonding orbitals (π* and sigma*) are the excited states - Bonding (pi and sigma) & non-bonding orbitals are ground states - Visible light is 400nm (violet) to 750nm (red) - UV is 200-400nm - Energy of UV is usually insufficient to promote electrons from sigma bonds but is in the region covered by pi (π) electrons and non-bonding (n) electrons - 2 types of transitions seen = n to π* and π to π*
30
What is the normal configuration (state) of a molecule?
Ground state All electrons are in the available orbitals with the lowest energy
31
What does HOMO stand for?
Highest occupied molecular orbital Normally a non-bonding or π orbital
32
What does LUMO stand for?
Lowest unoccupied molecular orbital
33
Describe transitions between states
- a change between energy levels is called a transition - the lowest energy transition moves an electron between the HOMO and LUMO orbitals & generally, the excited states will also be vibrationally excited - excited states only last for very short periods of time (1-10 nanoseconds) as the high energy state is energetically unstable - the extra energy is lost through a relaxation process such as emission of light or heat
34
Key information about MOs in UV
Only compounds with π electrons or non-bonding electrons produce UV spectra n to π* is less energy than π to π* Therefore, the lower the energy the longer the λ (wavelength)
35
What is λmax?
The wavelength corresponding to the maximum of the absorption band = top of the peak
36
Why does the n to π* transition have a smaller peak?
n to π* transitions result from the fact that the lone pair in electrons are concentrated in a different region of space from π electrons This make the n to π* transition less probable than the π to π*
37
Transition intensities
- n to π* transitions often occur at lower energy (longer wavelength) than π to π* as they are less probable - A given photon of n to π* light must encounter many more molecules of light before it is absorbed (compared to π to π*) - π to π* absorptions are generally much more intense (strong absorption) than n to π* absorptions (weak) - Difference in intensity is 2-3 orders of magnitude (i.e. 100-1000 times)
38
Effects of conjugation on wavelength
1. Conjugation decreases the energy gap between HOMO and LUMO 2. Hence, less energy is required for electronic transitions 3. Therefore, transitions occur at longer wavelengths 4. If a compound has enough double bonds it will absorb visible light and the compound will be coloured
39
Effect of conjugation on transitions
Because of the more extensive π-systems of conjugated double bonds (molecules has more than one double bond), both the n to π* and π to π* transitions occur at a longer wavelength
40
What is a chromophore?
Molecule or part of a molecule that can be excited by absorption E.g. Chlorphyll a —> absorbs blue and red light so appears green from reflected light