to begin the study of organic chemistry we must revisit ...consider the different ways to draw...
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To begin the study of organic chemistry we must revisit the different types
of organic compound you met during the Higher course. It is essential you are
able to recognise functional groups; name the compounds and draw their
structures.
R
or
Remember that aldehydes and ketones are
called CARBONYL compounds.
Ethers and nitriles are not part of the Higher course. They are included here
to complete the set of compounds required for Advanced Higher.
There are many types of formulae used by chemists, including molecular,
structural, and skeletal. The most universal formulae used worldwide is
skeletal, which is used as it is the quickest and most effective way of drawing
chemical structures.
Consider the different ways to draw propan-2-ol.
The molecular formula of a compound simply shows the number of each type of
atom in the compound. No structural information is provided.
Writing a molecular formula tells us nothing about the structure of the
compound - the molecular formula C3H8O has three isomers -
Without further information we would not be able to tell which of these
isomers the molecular formula C3H8O was referring to.
Drawing out full structural formulae also has its drawbacks.
Consider the molecule amoxicillin, one of the most commonly used antibiotics in
the penicillin family.
While this is not a particularly
large molecule it would be very
Full structure
Propan-2-ol Propan-1-ol Ethyl methyl ether
time consuming to draw out its
full structural formula.
To deal with this problem, organic chemists developed the system of skeletal
structures which are quick to draw and show the basic structure and the
functional groups present in a molecule.
The skeletal structure of amoxicillin is
shown opposite.
There are some basic rules for drawing
skeletal structures.
1. Carbon chains are drawn in a zig-zag format where each corner or
end-point represents a carbon atom – no hydrogen atoms are shown.
The structures shown above all have SIX carbon atoms.
2. Double bonds are shown with two lines
and triple bonds with three lines.
The structures shown each have FOUR carbon atoms.
3. All heteroatoms (atoms other than carbon or hydrogen) must be shown.
Any hydrogen atom attached to a heteroatom must be shown.
The full structure and the corresponding skeletal structure of butan-2-ol are
shown below.
This table shows the names, full structures and skeletal structures for a
variety of organic molecules.
Name Full structure Skeletal structure
hex-3-ene
ethanal
propanoic acid
propanone
2-chlorobutane
diethylether
ethylethanoate
1. Name the following molecules.
2. Write the molecular formula and draw the full structural formula for each of the
following compounds.
3. Draw skeletal structural formula for vitamin A and vitamin C.
a b
c d
a b c d
e f g h
4. The skeletal structures for the nine isomers of heptane are shown.
State the systematic names for all nine isomers.
5. Consider the following pairs of compounds, and determine whether they are isomers,
the same molecule or different compounds which are not isomers.
6. Benzene has a ring of delocalised electrons and can be represented in a variety of
ways
Write the molecular formula of these aromatic molecules.
6. Drawing skeletal structural formulae can help identify the type of reaction occurring
or the functional groups in a molecule
a. Identify the type of reaction in each of the following
(i) (ii)
(iii)
(iv)
b. Atenolol and enalapril are drugs used in treatment of heart disease.
Label and name all the functional groups in these two drug molecules.
a b
From a chemical point of view, the most fundamental data about a compound
are: which elements are present and how much of each are in the compound?
This information leads to the empirical formula for the material.
The empirical formula shows the simplest whole number ratio of the different
atoms in a compound.
The alkenes C2H4, C3H6, C4H8 all have the same empirical formula - CH2.
A hundred years ago, chemists didn’t have much in the way of instruments that
could analyze substances. They had a balance and a Bunsen burner. So the
natural thing to do was to weigh something first, then burn it and then weigh
the ashes.
The percentage of carbon, hydrogen, nitrogen and sulphur in a compound can
be found by burning a known mass of the material in an oxygen atmosphere and
measuring the mass of carbon dioxide, water, nitrogen and sulphur dioxide
produced.
Using this apparatus, a known
mass of the compound is
burned. Any hydrogen turns to
water which is trapped in the
first canister. The mass of
water is found and from this
the mass of hydrogen is
calculated.
Any carbon in the compound is converted to
carbon dioxide which is trapped by the
second canister. The mass of the carbon
dioxide is found and from this the mass of
carbon in the compound is calculated.
Other elements are analysed in a similar
manner.
modern combustion analyser.
1. Complete combustion of 1.75 g of an organic compound containing only carbon,
hydrogen and oxygen gave the following results.
Mass of carbon dioxide produced = 3.51 g
Mass of water produced = 1.43 g
The relative molecular mass of the compound was found to be 88 g.
Use the results to calculate the empirical formula and the molecular formula of the
organic compound.
Step 1 – calculate masses of C,H and O in the compound
The mass of any element in a compound is found using the formula
Mass of C = {C/CO2} x 3.51 = {12/44} x 3.51 = 0.957 g
Mass of H = {H/H2O} x 1.43 = {2/18} x 1.43 = 0.159 g
In general if oxygen is present in a compound its mass can be determined by
subtracting the total mass of the other elements from the mass of the
original sample.
Mass of O = mass of sample – {mass of C + mass of H}
= 1.75 – {0.957+0.159} = 0.634 g
Step 2 – determine the empirical formula
This is found by determining the number of moles of each element and converting the values
to a whole number mole ratio.
Element C H O
Mass(g) 0.957 0.159 0.634
Moles 0.957/12 = 0.0798 0.159/1 = 0.159 0.634/16 = 0.0396
Mole ratio 0.0798/0.0396 = 2 0.159/0.0396= 4 0.0396/0.0396 = 1
The empirical formula is C2H4O
Step 3 – determine the molecular formula
The molecular formula can be found using the formula........
Multiplying the empirical formula by the
constant will produce the molecular formula.
Constant = 88/{GFM C2H4O} = 88/44 = 2
Molecular formula = 2 x C2H4O = C4H8O2
2. A hydrocarbon has a composition by mass of
88.25% carbon and 11.75% hydrogen.
Calculate the empirical formula of the compound.
3. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen.
Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg
of H2O. The molar mass of salicylic acid is 138. What is its molecular formula?
Mass of C = {C/CO2} x 97.1 = {12/44} x 97.1 = 26.48 mg
Mass of H = {H/H2O} x 17.0 = {2/18} x 17.0 = 1.889 mg
Mass of O = 43.5 – {26.48+1.889} = 15.131 mg
Element C H
Mass(%) 88.25 11.75
Moles 88.25/12 = 7.354 11.75/1 = 11.75
Mole ratio 7.354/7.354 = 1 11.75/7.354= 1.6
Element C H O
Mass(g) 26.46 1.889 15.131
Moles 26.46/12 = 2.205 1.889/1 = 1.889 15.131/16 = 0.946
Mole ratio 2.205/0.946 = 2.33 1.889/0.946= 2 0.946/0.946 = 1
In this example the % mass is given.
These numbers can be used directly to
find the empirical formula.
note.... the scaling is often the
tough part, here we have an
initial ratio of 1 : 1.6, so you
need to find the first multiple of
1.6 that is a whole number
(or very close to it).
Avoid the temptation to round
the first two decimal places.
The empirical formula is C5H8
Empirical formula mass, C7H6O3,is 138 138/138 = 1 molecular formula is C7H6O3
Multiply by 3 to give empirical formulaC7H6O3
1. Dianabol is one of the anabolic steroids that has been used by some athletes to
increase the size and strength of their muscles. The molecular formula of dianabol,
which consists of carbon, hydrogen, and oxygen, can be determined using the data
from two different experiments.
The apparatus shown below was used for the first experiment.
A 14.765 g sample of dianabol was burned, and 43.257 g CO2 and 12.395 g H2O
formed in the absorption chambers.
a. (i) What practical steps need to be undertaken to obtain the values for the masses
of CO2 and H2O formed?
(ii) Calculate the empirical formula of dianabol.
b. In the second experiment, the molecular mass of dianabol is found to be 300.
What is the molecular formula for dianabol?
2. Caffeine, a compound found in coffee, tea, and cola drinks that has a marked
stimulatory effect on mammals. The combustion analysis of caffeine shows that it
contains 49.5% carbon, 5.2 % hydrogen, 28.9 % nitrogen, and 16.5 % oxygen by mass,
and its experimentally determined molecular mass is 195.
Use this information to calculate the molecular formula of caffeine.
3. An antibiotic compound containing the elements C,H,N,S and O was completely burned
in oxygen. Combustion of 0.3442 g of the antibiotic gave 0.528 g of carbon dioxide,
0.144 g of water, 0.128 g of sulfur dioxide and 0.056 g of nitrogen.
a. Calculate the mass of carbon, hydrogen, sulfur and oxygen in the compound.
b. Calculate the empirical formula of the compound.
c. The molecular mass of the compound is 172.1.
Calculate the molecular formula of the compound.
Instrumental analysis is a field of analytical chemistry that investigates
compounds using scientific instruments.
Spectroscopy is a technique that uses the interaction of energy with a sample
to perform an analysis. Nowadays with modern computer technology
spectroscopy can identify almost any organic molecule.
In our study of instrumental analysis you will be expected to interpret the
graphs , called spectra, produced by three different types of instrument
Mass spectroscopy
This technique provides information about MOLECULAR MASSES and the
structure of organic molecules.
Infra - red (IR) spectroscopy
This technique provides information about the FUNCTIONAL GROUPS
in organic molecules.
Proton nuclear magnetic resonance (NMR) spectroscopy
This technique uses radio-waves to provide information about the number of
HYDROGEN ATOMS and their environments in an organic molecule.
For all these different techniques you should be able to interpret simple
spectra (usually with the help of the data booklet) and identify the species
responsible for the various peaks present.
Given a simple organic molecule you should be able to suggest what its various
spectra might look like.
Mass spectroscopy
Points 1. Used to determine molecular mass and structure.
2. Molecules are changed into POSITIVE IONS - mostly +1
3. Ions are deflected into separate ion paths by a magnetic field
according to their mass/charge ratio (m/z).
4. As molecules ionise
they tend to break
into smaller ion
fragments .
Questions on mass spectrometry are likely to involve; identifying the formula for
ion fragments (including the molecular ion); deducing the molecular mass;
deducing s structure for the compound.
Remember to include the positive charge in the formula for any ion fragment.
1. The mass spectrum of methyl butanoate,
C5H10O2, is shown opposite.
a. How can the mass spectrum be used to confirm
the molecular mass of methyl butanoate?
b. Write the formula for the molecular ion of
methyl butanoate.
c. Draw the full structural formula for methyl butanoate.
d. Draw the skeletal formula for methyl butanoaote.
e. Write a possible formula for the ions which cause the peaks at m/z 71 and m/z 59.
f. Butyl methanoate is an isomer of methyl butanoate. Suggest an m/z value for a peak
which would appear on the spectrum for butyl methanoate which does not appear on
the spectrum for methyl butanoate.
2. The mass spectra of propanal and propanone are shown below.
a. For compound A, which group of atoms could be lost when the ion of m/z 43 forms f
a. Write the formula for the molecular ion of propanal.
b. Explain which spectrum, A or B, is the spectrum of propanone and which is the
spectrum of propanal.
Your answer must include reference to the ion fragments formed.
m/z
102
10 20 30 40 50 60 70 80 90 100 110
59
43
29 87
71
3. Compound L is a sweet smelling liquid containing the elements carbon, hydrogen and
oxygen. When 0.4080g of the liquid is completely burned in oxygen, 1.0560g of
carbon dioxide and 0.2160g of water were the only products.
The mass spectrum of compound L is shown below.
a. Calculate the empirical formula of compound L.
b. Calculate the molecular formula of compound L.
c. Write the formula for the molecular ion of compound L.
4. Fragmentation of the molecular ion of methylbutanone, (CH3)2CHCOCH3, gives rise to
dominant peaks at m/z = 71 and m/z = 43.
a. Draw a skeletal formula for methyl butanone.
b. Write a balanced equation to show how fragmentation of the molecular ion gives rise to
the peak at m/z = 71
c. Two fragmented ions are responsible for the peak at m/z = 43.
Give the formula for both these ion fragments.
5. The mass spectrum of a straight chain alkane is shown below.
a. State the name of the alkane.
b. Draw the skeletal formula for the
alkane.
c. Write the formula for the ion
fragments at m/z = 57 and m/z = 29.
6. The mass spectrum of methylene chloride, CH2Cl2, is shown below.
a. Draw a diagram of the structure of methylene chloride which clearly shows
the shape of the molecule.
b. Write the formula for the ion fragment responsible for the peak at m/z = 49.
c. The molecular ion, M+, is found at m/z = 84 indicating the molecular mass of
methylene chloride is 84.
(i) Write the formula for the molecular ion, M+.
(ii) Suggest what might be responsible for the peaks at any of the values higher
than 84. These peaks have values of 85, 86, 87 and 88.
7. Consider the mass spectrum shown below.
Use your knowledge of chemistry to discuss information displayed in the diagram.
m/z
When you sit in front of a fire, which gives
out infra-red (IR) radiation, you feel warm. This is because the IR radiation
absorbed by molecules in your skin is of a suitable frequency to increase their
vibrational energy. The greater a molecule's vibrational energy, the hotter it is.
Atoms in a molecule are always vibrating, and
the frequency of this vibration depends on the
atoms mass and the length or strength of its
bonds. The vibrational frequency of atoms lies
in the IR region of the electromagnetic
spectrum and so they can absorb IR radiation.
IR radiation causes parts of a molecule to
vibrate, and the wavelengths absorbed will
depend on the type of chemical bond and the
group of atoms at the end of these bonds.
Obviously it is difficult to show molecules
vibrating on this paper.
The diagrams show some of the vibrations
that are possible. Molecules can bend, twist,
rotate and stretch. If the molecule is
subjected to energy of the correct
frequency (i.e. IR frequencies) then certain
bonds will absorb this energy – this is what an
IR spectrometer detects and displays.
The approximate frequency range for
infra-red radiation is from 1x1013 – 4x10 14 Hz.
This equates to wavenumber values of around
400 cm-1 to 4000 cm-1. Wavenumbers are used
on the x-axis of IR spectra, while the Y-axis is
usually % transmission.
The higher the wavenumber the higher the energy of the radiation.
Infra-red (IR) Spectroscopy
IR spectra are often used to
identify the functional groups
in an organic molecule.
Functional groups are seen at
characteristic frequencies
(wavenumbers) no matter
which compound they are in.
The spectrum above is for butanal and so it has
the C=O functional group at 1750 cm-1 and the
C-H stretch at 2900 cm-1.
For this spectrum, the fingerprint region will be unique to butanal.
This IR correlation table is found in the data booklet.
This is the IR spectrum of ethanol. Important features are the C-O stretches
and the hydrogen bonded O-H stretch due to the hydroxyl functional group.
Notice that the O-H stretch is a
broad peak unlike the C-O
stretches which are relatively
sharp. This broad OH peak is very
characteristic of alcohols as the
H bonds cause more interaction between molecules, which in turn, causes more
possible vibrations and these will have a variety of frequencies and so the peak
appears over a wider wavenumber range.
Detailed analysis of IR spectra is a high order skill as many peaks for
different groups can occur at very similar wavenumbers.
For this course you only need to distinguish between the major functional
groups and correctly identify their presence on a given IR spectrum.
The examples on the next page highlight this.
1. By categorising appropriate peaks on each spectra, identify which spectra could be an
alkane, which could be an ester, which could be an alcohol and which a carboxylic acid.
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2. The infrared spectra shown below are those of the four compounds, A, B, C and D.
a. Write skeletal formula for A, B, C and D.
b. Match A,B, C and D to their appropriate spectra.
c. Calculate the energy, in kJ mol-1, of the vibration associated with a wavenumber of
3000 cm-1.
3. Spectrum of an organic compound A are shown below
a. Compound A has the empirical formula C2H4O.
Deduce the molecular formula of compound A.
b. The absorption peak at 1745 cm−1 in the infra-red spectrum can be used to help
identify A.
(i) Which bond is responsible for this absorption?
(ii) Which type of compound is A?
c. Draw a possible skeletal formula for compound A.
This type of spectroscopy works due to the fact that a spinning hydrogen
nucleus behaves like a tiny magnet. Normally the direction of the spinning
nuclei will be random but if the nuclei placed in a magnetic field the nuclei will
line up in the direction of the applied magnetic field, rather like how a compass
needle lines up with the Earth’s magnetic field
If energy of the correct frequency is now applied, the tiny magnet can flip
over so that it aligns against the magnetic field and will therefore move to a
higher energy state.
The energy required for this transition is in the radio frequency region of the
electromagnetic spectrum, typically from 60 mega hertz, MHz to 1000
megahertz, MHz.
When the energy is removed the nuclei
“flip back” to their lower energy state
and release the small amount of energy
they absorbed. It is this emitted energy
that a nuclear magnetic resonance
spectrometer measures.
This is a picture of one of the world’s strongest NMR spetrometers.
Strength 21.1 Tesla
Type Superconducting
Cost $16 million
Weight 36,287 kg (40
tons)
Height 4.7 meters (15
feet, 6 inches)
Operating
temperature
-271.45 ° C
(-456.61 ° F)
Length of
superconducting
cable
153 km (95
miles)
Proton nuclear magnetic resonance (NMR) spectroscopy
The spectrum produced in NMR gives detailed information about the
hydrogen atoms present in organic molecules.
1. Hydrogen atom environment
2. Number of atoms in each environment.
What is meant by hydrogen environment?
Consider the molecules shown below...............
a. Propan-1-ol has FOUR different b. Propan-2-ol has THREE
hydrogen environments. hydrogen environments.
c. Benzene has ONE hydrogen d. Ethanal has TWO
environment. hydrogen environments.
e. But-2-ene has TWO hydrogen f. 1,4-dimethyl benzene has TWO
environments. hydrogen environments.
C C C O
H
H
H H
H
H
H
HC C C
H
H
H H H
HO
H
H
C C C C
H
H
H H H H
H
H
Consider the NMR low resolution NMR spectrum for ethanol.
Reading the spectrum
1. There are three peaks indicating there are three types of hydrogen
environment.
2. The relative areas under the peaks are proportional to the number of
hydrogen atoms giving rise to each signal – in this case 1:2:3
To make this easier to evaluate most modern NMR spectrometers
integrate and record the area of each signal. The lines A,B and C on this
spectrum are called INTEGRATION lines.
Measuring the height of these lines will also give the 1:2:3 ratio.
Tetramethylsilane
{TMS}
This NMR spectrum of an ester has an
integration signal ratio of 23:67
which cancels down to 1:3. This ratio
shows the 3 H atoms in one
environment and the 9 H atoms in the
second environment.
3. One molecule can contain many hydrogen environments. Each environment
will release a different frequency of energy when it drops down from its
excited state to line up with the magnetic field. Each different hydrogen
environment will appear in a different position in the NMR spectrum. This
is called Chemical Shift which has the symbol delta, δ, and is measured in
parts per million. (parts per million of what I hear you ask – we do not need
to go into this for AH chem.)
4. The peak at zero ppm is for the compound tetramethylsilane(TMS). This
substance is used as a reference chemical against which all other chemical
shifts are compared. This substance is used because it is inert, can be
easily removed from the test sample and it has a signal frequency lower
than almost all other compounds.
5. There is a table of chemical shift
values on page 15 of the data
booklet and this should always be
referred to when attempting NMR
questions. The table shown here is a
simplified version of the data
booklet table.
You will be expected to be able to
assign peaks to individual spectra,
identify compounds form there
spectra and to sketch simple
spectra for given compounds
Functional Group Chemical
Shift
Alkane 0.8-1.2
1.6
Benzyl 2.3
Carbonyl 2.2
Amine 2.3
Alcohol 3.3
Alkyl Halide 3.6
Alkene 4.5-6.0
Benzene 6.0-9.0
Alcohol 0.5-4.5
Very Broad
Carbox. acid 9.0-15.0
R CH3
CHC CH3
CH3
R C CH3
O
R N CH3
HO CH3
H3C Cl
H2C CH2
H
R OH
R
O
OH
1. How many different hydrogen environments do each of the following molecules have?
a. b. c.
d. ` e. f.
2. Use the data booklet to find the approximate chemical shift values for the hydrogen
atoms in the molecules in Q1 and then draw the NMR spectrum for each of them. You
should draw a vertical line for each signal and remember to have them at the
appropriate height ratios
Here is the spectrum for but-2-ene as an example.
a.
b.
c.
d.
e.
f.
3. An NMR spectrum was recorded for each of these straight-chain hydrocarbons:
CH4, C2H6, C3H8 and C4H10.
a. (i) Which compound(s) would only show one peak ?
(ii) How many peaks would you expect for both the structural isomers of C4H10 ?
b. A compound of formula C5H12 has only one peak on its NMR spectrum.
Draw the structural formula of this compound.
4. The approximate values and integrals for the NMR spectrum for one of the isomers
of C4H10O is given in the table below.
a. Draw the skeletal formula of this isomer.
b. Using the information in the table draw the NMR spectrum of this isomer.
c. Draw the isomer of this molecule which is a tertiary alcohol.
5. Phenylethanal is one of the aldehyde molecules responsible for the smell of chocolate.
The skeletal formula for phenylethanal is shown opposite.
3-methylbutanal and 3-methylbutanal are another two
aldehydes which contribute to the smell of chocolate.
a. Draw a skeletal formula for both 3-methylbutanal and 2-methylbutanal
b. The low resolution proton NMR spectrum shown is for one of these three aldehydes.
Explain which of the three aldehydes
would give this proton NMR spectrum.
Isomer Group Approximate
value Integral
CH3CH
2CH
2OCH
3 CH
3 0.8–1.3 3
CH2 1.2–1.8 2
CH2O 3.5–4.0 2
OCH3 3.0–4.0 3
All the NMR spectra we have looked at so far have been low resolution. With
high resolution NMR the spectra are more complicated but provide a clearer
picture of molecular structure. In this sense NMR instruments are like digital
cameras and HDTVs: more megapixels means better resolution which means
more information and clearer pictures (and much higher price tags!)
Consider the molecule 1-bromopropane
This low resolution spectrum has three peaks
indicating three different hydrogen environments.
The ratio of the peaks is 2:2:3 which reflects the
number of each type of hydrogen.
The major difference seen in a high resolution spectrum is the phenomenon
known as multiplicity or coupling
This occurs because the magnetism of a hydrogen atom in one environment can
affect the magnetism of a hydrogen atom in a different environment on an
adjacent carbon atom.
When this happens the NMR signals splits into several signals forming new
signals which are termed doublets, triplets, quartets, etc
These split signals indicate the number of hydrogen atoms on carbon atoms
adjacent to the hydrogen atoms responsible for the signal.
This is the high resolution spectrum of 1-bromopropane
Analysis
Peaks There are three different signals so there are three
chemically different protons.
Splitting The signals include a triplet ( = 1.0)
sextet (= 1.8)
triplet ( = 3.4)
In general the individual signal seen on low resolution will split according to the
(n+1) rule where n = the number of hydrogen atoms on adjacent carbon atoms.
The signals due to the hydrogens attached to carbon ...
C1 triplet ( = 1.0) coupled to the two hydrogens on carbon C2 ( 2+1 = 3 )
C2 sextet ( = 1.8) coupled to five hydrogens on carbons C1 and C3( 5+1 = 6 )
C3 triplet ( = 3.4) coupled to the two hydrogens on carbon C2 ( 2+1 = 3 )
Integration The integration lines show that the ratio of protons is 2:2:3
The high resolution spectrum of pentan-2-one
Which peaks are due to hydrogens a,b,c and d?
Summary An NMR spectrum provides several types of information :-
• number of signal groups tells you ... the number of different
hydrogen environments
• chemical shift the general environment of the hydrogens
• multiplicity how many hydrogens are on adjacent atoms
• peak area the number of hydrogens in each environment
1. A compound with empirical formula C2H4O gives a molecular ion peak at m/z = 88.
The compound is neutral and has the following high resolution NMR spectrum.
Using all the information deduce the full structural formula for the compound.
2. The two spectra on the right were
obtained from a sample of propane.
a. Draw the full structural formula for
propane.
b. How many different types of hydrogen
atom do these spectra show.
c. Explain why the spectra appear different.
3. The high resolution proton NMR spectrum of butanone is shown below.
a. Name the compound responsible for the
peak at zero PPM
b. Draw a structure for butanone and
allocate the remaining peaks to the
correct hydrogen atoms in the molecule.
4. The high resolution spectrum of a straight chain ether with molecular formula C4H10O2
is shown below.
a. Deduce the name of the ether and draw its full structural formula.
b. Indicate on the spectrum which hydrogens in the ether give rise to each cluster of
signals.
A list of “learning outcomes” for the topic is shown below. When the topic is
complete you should review each learning outcome.
Your teacher will collect your completed notes, mark them,
and then decide if any revision work is necessary.
State that visible light is part of the electromagnetic spectrum with wavelengths between
400nm to 700 nm
State that blue light has higher energy than red light.
State that electromagnetic radiation travels in waves with a speed of
3x108 m s-1
Be able to use the relationship c = f where f is the frequency in Hertz and is
is the wavelength in metres.
State that an atomic emission spectrum is not continuous like the visible
spectrum but made of individual lines of a specific frequency.
Be able to explain the origin of lines in an emission spectrum
Need Help
Understand
Revise
Be able to use the formula E = L h c/1000 to calculate the energy, in kJ mol-1,
of a specified wavelength or frequency of light.
State the connection between the four quantum number n,l m and s and what
they indicate about atomic orbitals.
Be able to recognise s, p and d atomic orbitals from diagrams.
State the meaning of the term degenerate in relation to atomic orbitals.
Be able to write the electron configurations for atoms and ions of the first 36
elements in both spectroscopic and orbital box notation.
Be able to state the Aufbau principle, the Pauli exclusion principle and Hund’s
rule and to use the rules correctly to place electrons in atomic orbitals.
Be able to define and write appropriate equations for the first and second
ionisation energy.
Be able to explain ionisation energies in terms of the relative stablilities of
atomic orbitals.
I have discussed the learning outcomes with my teacher.
My work has been marked by my teacher.
Teacher Comments.
Date. __________________________________
Pupil Signature. __________________________
Teacher Signature. _______________________
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