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CCEA Advanced Subsidiary Chemistry Notes for Module 1

1.1             Atomic Structure

 

Electrons, protons and neutrons as the constituent particles of the atom. Their location in the atom, their relative masses and charges. Atomic number, mass number and isotopes.

 

When you have finished this section you should be able to:

  • Describe the properties of protons, neutrons and electrons in terms of their relative charge and relative mass ;
  • Understand the importance of these particles in the structure of the atom ;
  • Define the terms atomic number, Z and mass number, A ;
  • Use values for atomic number and mass number to calculate the number of protons and neutrons in the nucleus ;
  • Explain the existence of isotopes
  • Use isotopic symbols to describe the composition of the nucleus.

All atoms are electrically neutral. The number of electrons in the shells is the same as the number of protons in the nucleus. The mass is made up almost entirely from the masses of the protons and neutrons. The masses of the proton and neutron are virtually identical.

 

Properties of sub-atomic particles

 

ParticleRelative MassRelative Charge
Proton (p)1+1
Neutron (n)10
Electron (e)0  (1/1837)-1

 

Evidence for particles

Electrons

J.J. Thompson (1897) experiments with cathode ray tubes.  He discovers a beam of rays emitted from the cathode when an electric current passes through a gas at low pressure.  The rays are deflected by electric and magnetic fields and consist of a stream of electrons.

 

Protons

Electric discharges through gases at low pressure produces a stream of particles from the anode.  Work with electric and magnetic fields shows them to be positively charged. Hydrogen gives the lightest particles, which are assumed to be protons.

Neutrons

Chadwick (1932) bombards beryllium with a-particles producing fast moving particles, which are not affected by electric or magnetic fields.  The particles are neutrons.

Relative sizes of atoms and nuclei

Atoms consist of a very small, dense nucleus, around which the electrons circulate in a comparatively large volume. The volume of the nucleus is about 10-44 m3 and it is composed of two different types of particles, protons and neutrons. These are known collectively as nucleons.

An atom is small, but its nucleus is smaller still. The radius of an atom is of the order of 10-10 m, but the radius of a nucleus is of the order of 10-15 m.

 

Do the next exercises will help you to appreciate the difference in size.

 

Exercise 1

Suppose a football, diameter 22 cm, is scaled up so that it becomes as big as the earth, diameter 13000 km.

Calculate whether an atom of diameter 0.32 nm (3.2 x 10-10 m) will become as big as:

A          a pin head, diameter 1mm

B          a 1p coin, diameter 1.9 cm

C          a football, diameter 22cm

D          a weather balloon, diameter 1.8 m

 

Exercise 2

If the nucleus of an atom were scaled up to the size of a pin head (say 1 mm diameter), how big would the atom be?

 

 

Since the mass of an atom is concentrated in its nucleus, the nucleus must be extremely dense. Estimate how dense it is by doing the next exercise.

 

 

Exercise 3

For atoms of elements at the beginning of the Periodic Table the volume of the nucleus, VN, is given by:

VN  =  1.73 x (relative atomic mass) x 10-45 m3

Use this expression to calculate the density of the sodium nucleus:

(a)    in kg m-3

(b)    in tonnes cm-3

(Remember that 1 mol of sodium atoms weighs 23.0 g and contains 6.02 x 1023 atoms)

 

Exercise 4

Calculate:

(a)    the volume occupied by a sodium atom (radius 1.86 x 10-10 m)

(b)    the fraction of the volume occupied by the nucleus.

(Hint: assume that both the atom and its nucleus are spheres with volume given by 4πr3/3 )

 

Most of your body is empty space too. If all the spaces between the nuclei were squeezed out, you would be only half as big as a flea, although you would weigh the same.

With all this empty space why does any object appear solid? The electrons in an atom move very rapidly around the nucleus. The electrons effectively form a shield around the nucleus, marking the limits of the atom’s volume and making it seem solid.

 

 

 

Atomic number, Mass number and Isotopes

Atomic number and mass number give us important information about an atom and are particularly useful in distinguishing one isotope of an element from another.

 

Atomic number

The atomic number  (Z) of an atom is the number of protons in the nucleus.

 

 

Mass number

The mass number (A) is the total number of particles in the nucleus.

 

 

Isotopes

Isotopes are atoms of an element that have different numbers of neutrons in the nucleus i.e. they have the same atomic number but different mass numbers.

e.g. 3517Cl and 3717 Cl.

 

 

Exercise 5

The table shows the mass number and number of neutrons in the nucleus, for four atoms, W, X, Y and Z.

 

W         X          Y          Z

Mass number                 36        39        40        40

Neutrons in nucleus      18         20        21         22

 

  1. a) Write down the atomic numbers of the four atoms.
  2. b) Which of the four atoms are isotopes of the same element?
  3. c) Use your Periodic Table to write isotopic symbols (e.g. 2713Al) for the four atoms.

 

 

Relative atomic mass, relative isotopic mass and relative molecular mass. The carbon-12 standard. The use of the mass spectrometer to obtain accurate atomic masses. (Details of the workings of the mass spectrometer are not required). Deduction of Relative Molecular Mass from a molecular ion peak. (Limited to ions with single charges).

 

When you have finished this section you should be able to:

  • Calculate the masses of coins relative to a chosen standard ;
  • Express masses in a variety of units ;
  • Define the terms relative atomic mass (Ar), relative isotopic mass and relative molecular mass (Mr) in terms of carbon-12 ;

 

 

Relative atomic mass

Atoms are so small that their masses, expressed in grams, are difficult to work with.  Some examples are listed in Table 1 below

 

Table 1

ElementAverage mass of atom g
H1.67355 x 10-24
He6.64605 x 10-24
Li1.15217 x 10-23
C1.99436 x 10-23
O2.65659 x 10-23
Na3.81730 x 10-23
Ar6.63310 x 10-23
U3.95233 x 10-22

 

The mass of an atom expressed as relative atomic mass (Ar) is much more manageable.

 

Exercise 6

Collect as many British coins of each kind (1p, 2p, 5p, 10p) as you can.  Weigh a group of each kind to the nearest 0.01 g and calculate the average mass of each of the denominations to the nearest 0.001 g.

Enter your results in Table 2.

 

Table 2

CoinNumber of coinsTotal mass gAverage mass g
1p
2p
5p
10p

 

You can calculate the relative mass of each of the coins using

 

Relative mass of coin =    average mass of coin

mass of standard

 

 

Fill in column 1 of Table 3

 

Exercise 7

Table 3

 

Coin

1 Average mass gRelative mass
2 Mass OCU3 Mass CCU4 Mass SCU
1p
2p
5p
10p

 

(a)       Define the unit of mass, the OCU (for one-penny coin unit).

Let one OCU equal the average mass of a one penny coin.

1.00 OCU =                                g

 

(b)      Calculate the relative mass of each type of coin on the OCU scale using

Relative mass of coin =   average mass of coin

mass of OCU

Fill in column 2

 

Exercise 8

(a)                Define a second unit of mass, the SCU (for silver coin unit) .

Using the data in Table 1 calculate

 

Average mass of 5p coin =                                  g

1.000    SCU = average mass of 5p coin

1.000    SCU =                            g

 

(c)       Calculate the relative mass of each type of coin using the expression:

Relative mass of coin =       average mass of coin

Mass of SCU

 

Record the values in column 4 of Table 1.

 

Exercise 9

According to the Royal Mint, the mass of a newly minted 2p coin is 7.128 g.  We define a second unit, the CCU (copper coin unit), as one half the mass of a newly minted 2p coin.

1.000 CCU = ½ x 7.128 g

1.000 CCU = 3.564 g

Using the defined value for the CCU, calculate the relative masses for column 3.

 

The relative atomic mass scale

You have now completed a series of exercises using coins to illustrate how relative mass changes as the choice of standard changes.  Now you will do a similar exercise using masses of atoms instead of coins, where you calculate the relative atomic masses on the hydrogen, oxygen and carbon-12 scales

 

Exercise 10

Use the values in Table 1 to calculate atomic masses relative to

(a)       hydrogen,

(b)      oxygen,

(c)       carbon-12

in a similar way to that in which you calculated relative masses of coins.

 

Complete Table 4.

Some values are included as a check.

(The mass of an atom of carbon-12 = 1.99252 x 10-23 g).

 

Table 4

 

Element

Relative atomic mass (Ar)
H scaleO scale12C scale
H1.000001.00790
He4.00276
Li6.88459
C12.01110
O16.0000
Na22.8096
Ar39.9496
U238.030

 

In most of you’re A level work, you use relative atomic masses expressed to three significant figures (e.g. He = 4.00, O = 16.0, U = 238).  To this degree of precision, the oxygen scale and the carbon-12 scale can be regarded as the same, but you should not use the hydrogen scale as it differs so much from the others.

 

 

 

RELATIVE ATOMIC MASS (Ar)

Atoms are so light that their actual masses are not used.  Instead, each atom is compared to a standard atom.  The atom chosen as standard is the most common isotope of carbon, carbon-12.

The carbon-12 atom has a mass of exactly 12.0000 units and all other atoms are given a mass relative to the carbon-12 standard.

For example, a magnesium atom is twice as heavy as a carbon-12 atom.

 

The relative atomic mass of an element

 

relative atomic mass of an element =          mass of one atom of the element                  

(1/12) x mass of one atom of carbon-12

 

 

 

RELATIVE MOLECULAR MASS (Mr)

The Relative Molecular Mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of a compound.

For example: find the Relative Molecular Mass of sulphuric acid.

 

Formula H 2SO4

2 atoms of H  =    2.00

1 atom of S    =   32.0

4 atoms of O  =  64.0

TOTAL    =  98.0

 

R.M.M. sulphuric acid is 98.

 

RELATIVE FORMULA MASS

The Relative Formula Mass of an ionic compound equals the sum of the Relative Atomic Masses of all the atoms in a formula unit of the compound.

 

For example: find the Relative Formula Mass of magnesium chloride.

 

Formula MgCl2

 

1 atom of Mg                     = 24.0

2 atoms of Cl                     = 71.0

TOTAL                              = 95.0

 

R.F.M. magnesium chloride is 95.0.

 

 

Find the Relative Formula Mass of hydrated copper sulphate crystals,  CuSO4.5H2O.

1 atom of copper Cu                  =   64.0

1 atom of sulphur S                   =   32.0

10 atoms of hydrogen H            =   10.0

9 atoms of oxygen O                 = 144

TOTAL                                     = 250

R.F.M. hydrated copper sulphate is 250.
The Mass Spectrometer

 

When you have finished this section you should be able to:

  • Understand the principles of a simple mass spectrometer, limited to ionisation, acceleration, deflection and detection ;
  • Identify peaks on a simple mass spectrum and use them to calculate the relative abundance and masses of ions ;
  • Calculate the relative atomic mass of an element from (a) a mass spectrum, (b) percentage abundance data ;
  • Sketch a mass spectrum, given relevant data.

 

The mass spectrometer is used to obtain accurate atomic masses by measuring the mass and relative abundance of the isotopes of an atom.

 

  1. The sample is vaporised – atoms must be in a gaseous state.
  2. Positive ions are formed. Atoms are bombarded by electrons and positive ions are formed .

X(g)                         X+(g)  +  e-

  1. The positive ions are accelerated by an electric field. The slits restrict the ions to a narrow beam.
  2. A magnetic field deflects the ions depending on their mass/charge ratio. Ions with high m/z ratio are deflected less than those with low m/e ratio.
  3. Ions with the correct m/z ratio pass through the slit and arrive at the detector.
  4. The charge received at the detector is amplified and turned into a sizeable electric current.
  5. The electric current operates a pen recorder, which traces a peak on the recording.
  6. If the magnetic field is kept constant while the accelerating electric field varies continually, ions of different m/z ratio are deflected one after the other into the detector and a trace is obtained.

 

Interpreting Mass Spectra

E g. the mass spectrum of rubidium Rb.

 

 

 

 

Height of peak

 

 

 

 

 

 

 

 

m/z           85              87

 

Note:

  1. The height of each peak is proportional to the amount of each isotope present (i.e. it’s relative abundance).

 

  1. The m/z ratio for each peak is found from the accelerating voltage for each peak. Many ions have a +1 charge so that the m/z ratio is numerically equal to mass m of the ion.

 

Exercise 11
Refer to the diagram of the mass spectrum of rubidium previously to answer this question.

 

(a)    Describe the two isotopes of rubidium using isotopic symbols.

 

(b)    What information can you get from the heights of the peaks on the mass spectrum?

 

Calculating the relative atomic mass of an element

 

  1. Measure the height of each peak.

 

85 Rb = 5.82 cm

87 Rb = 2.25 cm

 

Therefore the ratio 85 Rb : 87 Rb is

5.82 : 2.25

 

  1. Calculate the percentage relative abundance

% abundance =       amount of isotope  x  100

total amount of all isotopes

 

85 Rb =          5.82         x 100 = 72.1 %

(5.82 + 2.25)

% 87 Rb = =       2.25         x 100 = 27.9 %

(5.82 + 2.25)

 

  1. Calculate the Ar

Ar (Rb) = (72.1 x 85)   +   (27.9 x 87)      =  85.6

100

 

 

Exercise 12

Use the mass spectrum shown below to calculate:

(a)    the percentage of each isotope present in a sample of naturally occurring lithium;

(b)    the relative atomic mass of lithium.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3       4        5       6       7        8

mass/charge ratio

 

Exercise 13

The mass spectrum of neon consists of three lines corresponding to mass/charge ratios of 20, 21 and 22 with relative intensities of 0.910; 0.0026; 0.088 respectively.

Calculate the relative atomic mass of neon.

Exercise 14

The percentage abundance of the stable isotopes of chromium are:

5024Cr – 4.31%; 5224Cr – 83.76%; 5324Cr – 9.55%; 5424Cr – 2.38%.

 

(a)    Sketch the mass spectrum that would be obtained from naturally occurring chromium.

(b)    Calculate the relative atomic mass of chromium, correct to three significant figures.

(c)    Label each peak on the mass spectrum using isotopic symbols.

 

 

 

 

 

 

The next exercise involves a molecular element.

 

Exercise 15
The element chlorine has isotopes of mass number 35 and 37 in the approximate proportion 3:1.

 

 

30            40                50              60              70               80

mass/charge ratio

 

Interpret the mass spectrum of gaseous chlorine shown above indicating the formula (including mass number) and charges of the ions responsible for each peak.

 

 

 

Exercise 16

Calculate the relative atomic mass of potassium, which consists of 93.0% 39K and 7.0% 41K.

 

 

 

 

Additional Exercises

(a) Chlorine consists of isotopes of relative masses 34.97 and 36.96 with natural abundances of 75.77% and 24.23% respectively.

Calculate the mean relative atomic mass of naturally-occurring chlorine.

(b)          Calculate the relative atomic mass of natural lithium which consists of 7.4% of 6Li (relative atomic mass 6.02) and 92.6% of 7Li (relative atomic mass 7.02).

 

(c) Copper (atomic number 29) has two isotopes, the first of relative atomic mass 62.9 and abundance 65%, the second of relative atomic mass 64.9 and abundance 35%.

Calculate the mean relative atomic mass of naturally-occurring copper.

 

(d)                                  Using mass spectrometry, the element gallium has been found to consist of 60.4 per cent of an isotope of atomic mass 68.93 and 39.6 per cent of an isotope of atomic mass 70.92.

Calculate, to three significant figures, the relative atomic mass of gallium.

 

 

 

Mass Spectra of Molecules

Molecules produce more complex mass spectra than atomic spectra.  The simplest ion produced is the parent molecule with one electron removed.

 

M (g)                                        M+ (g)  +  e-

 

M+ is referred to as the molecular ion.

 

It is possible for the molecular ion to break apart to give fragment ions.

M+ (g)                                       Y+ (g)  +  Z+ (g) etc.

The relative molecular mass of a molecule can be determined from the molecular ion peak.

 

e.g. The mass spectrum of copper (II) nitrate

 

 

 

molecular ions

 

 

peak

 

 

height

 

 

 

 

 

 

 

 

63   65          79   81               125   127                  187 189

mass/ charge  (m/z)

 

m/zIon
6363 Cu+
65
7963 CuO+
81
12563 CuNO3+
127
18763 Cu(NO3)2+
189

 

Exercise 17

A sample of water containing 1H, 2H and 16O was analysed in a mass spectrometer. The trace showed peaks at mass numbers 1, 2, 3, 4, 17, 18, 19 and 20.

Suggest ions which are responsible for these peaks.

 

 

The main use of spectrometry today is in identifying and analysing organic molecules which split up to give a great many different ions.

A simple introduction to the spectroscopic evidence for discrete energy levels in an isolated hydrogen atom. The equation E=hv .

 

When you have finished this section you should be able to:

  • Describe the difference between a continuous spectrum, a line emission spectrum and an absorption spectrum.
  • Explain the emission of light by atoms in terms of ground states and excited states.
  • State the relationship between wavelength and frequency (v=c/λ).
  • State the relationship between energy and frequency of radiation (E=hv).
  • Describe the emission spectrum of atomic hydrogen and explain how it is produced.
  • Sketch at least one series of lines from the hydrogen spectrum.

 

Emission Spectroscopy
When white light is passed through a prism on to a screen a spectrum of coloured light is observed. This appears as broad bands of colour merging into one another with no sharp boundaries. White light consists of a continuous range of wavelengths, which we distinguish as colour. This is called a continuous spectrum. A continuous spectrum shows a continuous range of wavelengths.

 

  i.r.    red      orange      yellow      green      blue      indigo       violet    u.v.
visible range

 

Coloured light emitted from sources such as sodium street lamps, neon lights, discharge tubes consist of a limited number of coloured lines on a black background. This type of spectrum is called an atomic emission spectrum or a line spectrum.

An emission spectrum shows a limited number of wavelengths, which appear as distinct lines. In the visible range, the lines appear in different colours.

Hydrogen spectrum in the visible region

red                                             blue                          indigo         violet

 

 

 

 

Increasing frequency

All substances give emission spectra when they are excited in some way either by heat or the passage of an electric current.

An absorption spectrum is seen when white light is passed through a substance. Black lines can be seen where light of some wavelengths has been absorbed.

 

When viewed through a spectrometer, the emission spectrum of hydrogen is seen to be a series of lines.

 

 

NOTE:

  • The atomic spectra consist of discrete (i.e. separate) lines.
  • The atoms only absorb or emit light of certain frequencies.
  • The lines get closer together as frequency increases (or wavelength decreases) until they converge to form a continuum.

 

Explanation

To explain the above observations Niels Bohr in 1913 put forward his picture of the atom. As the emission spectrum consists of lines, the electrons must only emit radiation of a particular wavelength or frequency. He suggested that the electrons in an atom can exist only in certain definite energy states, not an infinite number of values: their energy is quantised. An electron in an orbit distant from the nucleus has a higher energy than one in an orbit close to the nucleus.

When an electron in an atom absorbs energy it becomes excited and is promoted to a higher energy level, moving to an orbit further away from the nucleus. The electron is unstable in this higher energy state and falls back to a lower energy state, emitting radiation. Since the energy levels themselves can only have certain fixed values then the difference in energy can only have certain values and therefore fixed frequencies (i.e. certain colours).

 

 

Orbit of energy E2

 

 

E2

 

 

ΔE

 

 

 

E1

 

 

The small amount of energy absorbed or emitted when an electron undergoes a transition between two energy levels is called a quantum and the frequency of the radiation is given by:

 

ΔE  = E2 - E1  =  hv

Where h = Planck’s constant (6.63 x 10-34 Js)

v = frequency of radiation

 

[The symbol Δ (Greek ‘delta’ ) is used to refer to the difference between two values of a physical quantity – in this case energy]

 

Bohr assigned quantum numbers to the orbits. The orbit of lowest energy (nearest the nucleus) was given the quantum number 1 (n = 1). An electron in this orbit is in its ground state. The next energy level has quantum number 2 and so on.

 

 

The convergence of lines interpreted as convergence of energy levels leading to a value for the ionisation energy. Origin of flame spectra and an experimental study of flame coloration by the chlorides of Li, Na, K, Ca, Sr and Ba.

 

When you have finished this section you should be able to:

  • Explain what is meant by the convergence limit of a series of lines in the hydrogen emission spectrum.
  • Calculate a value for the ionisation energy of hydrogen by obtaining the convergence limit graphically.

 

Convergence Limits and Ionisation Energy

You will have noticed that the lines in the visible region of the hydrogen emission spectrum (the Lyman series) get closer and closer together at higher frequencies. It follows therefore that the electron energy levels must also get closer together and eventually come together.

E∞                                                                                      n= ∞

E6                                                                                                            n=6

E5                                                                                                            n=5

E4                                                                                                            n=4

 

 

E3                                                                                                            n=3

 

 

 

 

E2                                                                                                            n=2

 

 

 

We can make use of this fact to calculate the ionisation energy for hydrogen.

The next exercise suggests graphical methods in which you use the frequencies of radiation emitted in the Lyman series to calculate a value for the ionisation energy of hydrogen.

 


Exercise 18

The table below shows the frequency of lines in the Lyman series in the emission spectrum of hydrogen. Complete the table and then use the information to.

obtain the first ionisation energy of hydrogen by one of the methods below

 

Energy level, n, of excited electronFrequency, v

/1015 s-1

Δv

/1015 s-1

1/n2
22.4660.457
32.9230.160
43.083
53.157
63.197
73.221
83.237
93.248

 

 

(a) Plot a graph of Δv (vertical axis) against v (horizontal axis). (You can use either the higher or the lower value of v in each pair as long as you are consistent. Better still, plot two lines using both values in each pair.)

Extrapolate the curve to Δv = 0 and estimate a value of v at this point.

 

OR

 

Plot a graph of v (vertical axis) against l/n2 (horizontal axis). Extrapolate the

line to 1/n2 = 0 (i.e. n = ∞ ) and estimate a value of v at this point.

 

(b) Use Planck's constant (6.63 x 10-34 Js) to calculate a value for the ionisation energy from this frequency.  Insert the value of v at Δv = 0 into the equation ΔE = hv.

Hint: Remember that ionisation energy refers to one mole of electrons.

 

(c) Compare the value you have calculated with the value given in a data book for the first ionisation energy of hydrogen.

 

 

 

Origin of Flame Spectra

When certain metal salts are heated strongly in a Bunsen flame characteristic colours are produced.

 

Metal
Flame colour
Sodium NaIntense golden yellow
Potassium KLilac
Barium BaPale green
Strontium SrRed
Calcium CaOrange/red
Lithium LiRed

 

 

 

 

Atom in ‘excited’ state

 

 

 

 

 

energy                                                               hv     energy given out

absorbed                                                                   as light

 

energy

 

 

 

atom in               atom returns

ground state       to ground state

 

 

At room temperature nearly all the atoms are in the ground state i.e. the electrons occupy the obits of lowest energy. In a flame, or other energy source, electrons move to orbits of higher energy. The resulting excited statesare not stable and each excited electron soon falls to a lower energy state. In the change a definite amount of energy called a quantum, leaves each atom. The energy appears as radiation of a particular frequency, which may be visible and coloured. Since the energy levels in the atom or ion have fixed values, the emission spectrum consists of a set of lines at frequencies (or wavelengths) characteristic of the element. The emission spectra you have been looking at consist of a series of coloured lines, each line corresponding to a particular energy drop, from higher energy levels to a lower one. The greater the number of electrons making a particular transition the more intense the corresponding spectral line.

 

 

 

 

 

 

 

 

The evidence, from graphs of first ionisation energies of elements up to krypton and from successive ionisation energies of an element, for the existence of the main energy levels and s, p and d orbitals. Shapes of s and p orbitals.

When you have finished this section, you should be able to:

  • Define first ionisation energy.
  • Write an equation to show the first ionisation energy of an atom.
  • Write equations representing second third and subsequent ionisation energies of an atom.
  • Deduce the electron arrangement of an element from a graph of log10 ionisation energy against number of electron removed.
  • Use a graph of successive ionisation energies against number of electrons removed to provide evidence for the existence of sub-shells.
  • Draw the shapes of an s-orbital and a p-orbital.
  • Describe briefly what is meant by the wave-particle duality of an electron.
  • Give a simple, non-mathematical description of an orbital in terms of probabilities.

 

Ionisation Energy

Normally an electron in a hydrogen atom occupies the lowest possible energy level and is said to be in the ground state. If a photon has sufficiently high energy the electron may be completely freed from the influence of the nucleus and become an ion.

M                                 M+  +  e-

Once this happens the electron has kinetic energy and since this is not quantised light of all frequencies can be absorbed.

 

n=6

 

 

 

n=3

 

 

n=2

 

 

 

n=1

 

 

 

 

v

 

The convergence of lines in the visible spectrum (the Balmer series) corresponds to a transition from the n=2 to the n=∞ i.e. the energy level where the electron has just escaped from the atom. The convergence frequency can be used to find the ionisation energy of the atom.

The first ionisation energy of an element is the energy required to remove one electron from each of one mole of atoms in the gas phase to form one mole of gaseous ions.

M (g)                            M+ (g)  +  e-

 

Successive Ionisation Energies

Evidence for the arrangement of electrons in shells of different energies is provided by values of successive ionisation energies for elements i.e. the removal of one electron after another.

The important point about ionisation energies after the first one is that an electron is removed from a positively charged ion each time.

 

Exercise 19

(a) Write equations to show the first, second and third ionisation energies of aluminium.

(b) Would you expect the values of these ionisation energies to increase or decrease, in the order :1st, 2nd, 3rd?

(c) Explain your answer to (b).

 

 

Exercise 20

(a) For the element calcium, plot log10 (I.E.) against the number of the electron (one, two, three etc.) removed.

 

Number of electron removedIonisation energy (I.E.) /kJ mol-1Log10 (I.E.) / kJ mol-1
15902.77
211453.06
349123.69
464743.81
581453.91
6104964.02
7123204.09
8142074.15
9181924.26
10203854.31
11570484.76
12633334.80
13700524.85
14787924.90
15863674.94
16940004.97
171049005.02
181116005.05
194947905.69
205277595.72

 

(b) What information does this graph give about the electron configuration of the calcium atom?

(c) Why were you asked to plot log10 (I.E.) and not just ionisation energy?

(d) Explain why the ionisation energy increases when successive electrons are removed from a given shell.

 

 

 

 

This exercise fits the Bohr planetary model of the atom. The large jumps in the value of the ionisation energy indicate shells of different energies.

The electrons fall into four groups. The higher the ionisation energy, the more difficult it is to remove the electron and therefore the closer it is to the nucleus. From the graph the electronic configuration of calcium is 2.8.8.2.

 

 

2                                      n=4

8                                       n=3

8                                       n=2

 

 

 

 

 

2                                       n=1

 

Generally the maximum number of electrons in each shell is 2n2, where n is the principal quantum number.

 

 

 

First Ionisation Energies of Successive Elements

If we take a closer look at the ionisation energies of electrons in a given shell we find that the Bohr picture of the atom is oversimplified. There is evidence for a structure within each shell.

Exercise 21

Plot a graph of first ionisation energy against atomic number for the first twenty elements (up to calcium).

Label the vertical axis: ‘1st ionisation energy / kJ mol-1’ and extend the scale from zero to 2500 (in intervals of 500).

Label the horizontal axis: ‘Atomic number’ and extend the scale from zero to 20.

Label each point with the symbol for the element and join each point to the next with a straight line.

ElementI.E. kJ mol-1
H1318
He2378
Li526
Be905
B807
C1092
N1408
O1320
F1687
Ne2086
Na502
Mg744
Al584
Si792
P1018
S1005
Cl1251
Ar1526
K425
Ca596

From the graph points between Li - Ne and Na – Ar represent the filling up of a shell with electrons.

Na        Mg        Al         Si         P          S          Cl         Ar

2.8.1     2.8.2     2.8.3     2.8.4.    2.8.5     2.8.6     2.8.7     2.8.8

 

Energy Levels and Sub-shells
Each shell can be further divided into subsections called sub-shells containing a group of two and a group of six electrons. Detailed studies of ionisation energies and spectral lines have led to the conclusion that the energy levels are split into sub-shells as shown below.

 

 

Quantum shellSub-shellsNumber of electronsNumber of orbitals
n = 1      K shellOne sub-shell

1s sub-shell

 

2

1
n = 2      L shellTwo sub-shells

2s sub-shell

2p sub-shell

 

2              8

6

 

1

3

n =3       M shellThree sub-shells

3s sub-shell

3p sub-shell

3d sub-shell

 

2

6            18

10

 

1

3

5

n=4         N shellFour sub-shells

4s sub-shell

4p sub-shell

4d sub-shell

4f sub-shell

 

2

6             32

10

14

 

1

3

5

7

 

K 2.8.8.1

4p

n=4                  1

3d

1                           4s               ­

6                          3p        ­¯  ­¯  ­¯

n = 3                 8

2                          3s                 ­¯

 

 

 

6                          2p        ­¯  ­¯  ­¯

n= 2                  8

2                           2s                ­¯

 

 

 

 

 

 

n=1                   2                                              2                           1s                ­¯

 

 

 

We can see from this exercise that the Bohr model of the atom has its shortcomings. This does not mean that the planetary model is completely useless but we will now consider the question of models in general before going on to a more sophisticated picture of the atom.

 

Models and their usefulness

You are probably more familiar with the term 'model' to describe a scaled-down version of an everyday object, such as a model car.  But model cars differ enormously in the accuracy and detail with which the original is reproduced.  At one extreme is the miniature replica, identical in every feature, including an engine, which burns fuel and moves the car along.  Less refined is the familiar 'Dinky toy' model, which is recognisable in many details as a particular make of car, but has no working parts.  Even less refined is the sort of model an architect might use for a new town centre plan - a solid, car-shaped block, to give an impression of the relative sizes of road and car.  Finally, at the other extreme, there is the model a motorist would use after a car accident - just a rectangle on a map to indicate where the vehicle was at the time of impact.

None of these models is strictly 'correct' but each has its own use in a given situation.  In a similar way, there are several models of the atom - from the solid sphere of John Dalton to the planetary model of Rutherford and Bohr with electrons as particles circling a central nucleus in fixed orbits.

A further model, the orbital model, treats electrons as waves.

As we move more deeply into the subject, we meet increasingly detailed and sophisticated descriptions of atomic structure.  But must we therefore discard earlier descriptions as 'wrong' and useless?  In fact, there is no need to do so.  For many purposes, such as explaining the states of matter, there is no point in using anything more complicated than the simple 'billiard ball' picture; in other cases, the planetary model is all that is needed.

Scientific theories are rather like models: they can be simple or elaborate, depending on the job they have to do.  It is usually more sensible to ask, not whether a model is 'right' but whether it is useful.  This is an important underlying theme in chemistry (and in science generally).

Before considering the orbital model of the atom, we will spend a short time on the nature of electrons.

 

 

Wave-particle duality

Until now, it has been convenient to think of electrons as minute, almost mass-less particles, but there is evidence to suggest that they also behave as waves. Electron beams can behave like beams of light.  For example, they can be diffracted, and diffraction is a property of waves.

Clearly then, we need more than one model for the electron.  To explain some properties, we regard electrons as particles; to explain others we regard them as waves.  In other words, they appear to have a dual nature.  This phenomenon is known as 'wave-particle duality'.

 

It was by treating electrons as waves and applying to them mathematical methods, known as wave mechanics, that scientists came up with pictures of electron distribution.  We go on to consider this model, where electrons do not occupy fixed orbits as in the planetary model, but orbitals, which describe how their charges are spread out in small regions of space.

 

 

Modern Atomic Theory

Based on the diffraction of an electron beam de Broglie (1924) suggested that electrons may be thought of as behaving as waves. Developing this idea Schrödinger devised a branch of mathematics known as wave mechanics which is the basis of modern atomic theory.

Instead of a precise path for an electron as in the Bohr model of the atom, Schrödinger’s wave mechanics defines a volume of space in which there is a high probability of finding an electron – referred to as an atomic orbital.

In modern atomic theory the mathematical solution of the Schrödinger wave equation determines the orbital and this in turn gives information about energy levels available to an atom. The solution involves 4 quantum numbers with permitted values shown below.

  1. Principal quantum number n

n = 1, 2, 3, 4,                                  ∞

  1. Subsidiary quantum number l

l = 0, 1, 2, 3,                                   (n-1)

  1. Magnetic quantum number m

m = (-l)                                  -1, 0, 1, 2,                                    (+l)

  1. Spin quantum number s

s = ± ½

 

The four numbers are all necessary to precisely calculate the energy levels of an electron and they can also be used as labels for the electrons (generally only the n and l quantum numbers are used).

 

l valueCode letter
0s
1p
2d
3f

e.g. a 3s electron denotes one with n=3 and l=0.

The number of electrons occupying an orbital is indicated by a superscript i.e. 3s1.

 

The shapes of orbitals

The shapes of the s, p, d and f orbitals (from the initial letters of the words sharp, principal, diffuse and fundamental, referring to lines in the hydrogen spectrum) have been worked out using wave mechanics. You need to be familiar with the shapes of s- and p-orbitals. (The d-orbitals are shown just for the fun of it).

As a help in picturing the orbitals imagine you were able to take a large number of photographs of a hydrogen atom, containing one electron. By superimposing these photographs, you would get an impression of where the electron spends most of its time. The picture you would get would be something like the one below.

 

The picture is itself an over-simplification since it is restricted to two dimensions. The complete model is three-dimensional and spherical. Since even the two-dimensional picture is tedious to draw, we often use instead a boundary round the region where the probability of finding an electron is high - about 98% - as shown below.

 

                        The shape of s, p and d orbitals

Note :

  • The orbitals are 3-dimensional and not precisely defined. The charge density falls off sharply at a certain distance from the nucleus.
  • All s-orbitals are spherical 1s < 2s < 3s etc.
  • All p-orbitals are dumbell shaped in 3 directions in space 2p < 3p < 4p.

 

Exercise 22

(a) Explain why p-orbitals are labelled ’px’, ‘py’ and ‘pz’ .

(b)  How many electrons can be held in

(i)  an s-orbital

(ii)   a set of p-orbitals?

(c) use one word in each case to describe the shape of

(i) an s-orbital

(ii) a p-orbital.

 

 

Electronic structures of atoms and ions up to krypton in terms of main energy levels; the s, p and d notation and electrons-in-boxes notation using the building-up principle. Division of the Periodic Table into s, p and d blocks. spd notation: any question, including these on bonding, asking for an electronic arrangement/configuration/structure should use spd (even if this not specifically stated) and statements, showing an electron arrangement as 2.8.7 will not be credited.

When you have finished this section, you should be able to :

  • State the maximum number of electrons that an orbital can hold.
  • Use the aufbau principle to work out the order in which the orbitals are filled in a given element.
  • Write down the electron configuration of any element or ion up to krypton using

(a) the electrons-in boxes method

(b) s, p, d, f notation.

 

Electronic configurations of many electron atoms

The above theory applies to hydrogen, the simplest element which has a single electron in the lowest energy orbit, the 1s. The theory can also be applied to other atoms if we imagine going through the elements in order of increasing atomic number, adding an extra electron each time.

This process is known as the aufbau principle. (aufbau is a German word meaning to build up, pronounced ‘owf-bow’ – as in ‘bow-wow’)

There are three important rules which determine the order in which electrons occupy the vacant orbitals in an atom.

 

  1. Of the vacant orbitals available, the added electron always occupies the one with the lowest energy.
  2. Each orbital can hold a maximum of two electrons, which must have opposite spins and are said to be paired. This is the Pauli exclusion principle (no two electrons in an atom can have exactly the same values for the four quantum numbers).
  3. When electrons may occupy a set of orbitals of equal energy, the added electron will go into an empty orbital, keeping the spins the same, before spin-pairing occurs. (Hund’s rule).

Note:

  1. Electrons can be thought of as spinning on an axis like the earth. Unlike the earth, however, which spins in only one direction, an electron can spin in either of two directions.

The ↿ represents one direction of spin; the ⇂ arrow the other. Also in any one orbital when there are two electrons occupying it, they will spin in opposite directions. This is called spin-pairing and is shown as ↿⇂ .

 

As an example of how Hund’s rule operates, consider two electrons entering the

p-orbital. There seems to be three possible arrangements

 

px           py            pz                             px             py          pz                                         px           py          pz

↿⇂

 

But the first is the only one  which conforms to Hund’s rule.

 

The diagram below shows all of the orbitals of the first four shells , as well as some of the orbitals from the fifth to seventh shells. Each separate orbital is represented by a square box,     . The vertical axis represents energy.
Exercise 23

Use the diagram above to list the first ten sub-shells in order of increasing energy.

 

In an atom the orbitals are filled in order of increasing energy, starting from 1s.

An aid to remembering the order in which orbitals are filled is to write them down in columns as shown.

1s

2s         2p

3s         3p        3d

4s         4p        4d        4f

5s         5p        5d        5f

6s         6p        6d

7s         7p

The order of filling is then given by drawing diagonal lines through the symbols.

1s

2s         2p

3s         3p        3d

4s         4p        4d        4f

5s         5p        5d        5f

6s         6p        6d

7s         7p

 

The arrangement of electrons in orbitals

We start with hydrogen, the simplest element, which has one electron in the lowest energy orbital, the 1s. We can then build up the other elements in turn , imagining the addition of an extra electron each time.

1s

Hydrogen        ↿

 

1s

Helium            ↿⇂

 

2s

1s

Lithium             ↿⇂

 

 

Exercise 24

Draw in arrows to show the electron configurations of beryllium, boron, carbon and nitrogen.

 

Boxes-in-a-row

You can save time and space in writing down electron configurations by placing the orbital boxes side by side in a row.

1s

Hydrogen  ↿

1s

Helium  ↿⇂

1s                              2s

Lithium  ↿⇂ ↿

1s                              2s                                           2p

Carbon  ↿⇂  ↿⇂

1s                              2s                                           2p

Nitrogen  ↿⇂  ↿⇂

 

Exercise 25

Draw boxes-in-a-row to show the electron configurations of

(a)    oxygen, and

(b) fluorine.

 

 

Now we look at a further simplification in writing down electron configurations.

 

Using a noble gas 'core'

The 'boxes-in-a-row' method can become tedious, particularly if the atom contains many electrons.  In any case, we are most often concerned with the outermost electrons in an atom - the inner 'core' is not involved in chemical reactions.  Consequently, we can use a symbol for a noble gas to replace some of the arrows.  This is shown for two elements below.

2s                                                         2p

Boron                  [He]    ↿⇂

 

3s

Sodium                [Ne]    ↿

 

 

Exercise 26

Using boxes-in a-row, and noble gas cores, draw diagrams to show the electronic configurations of

(a) aluminium

(b) sulphur

(c) calcium

(d) rubidium

 

Another method of writing electron configurations does away with drawing boxes completely.

 

The s, p, d, f notation

In this method, the electron configuration of a hydrogen atom is represented as

1s1

 

 

 

 

Using this notation elements can be written as;

Carbon              1s22s22p2

Neon    1s22s22p6

Magnesium   1s22s22p63s2

Phosphorus   1s22s22p63s23p3

 

As previously noble gas cores can also be used.

Carbon              [He] 2s22p2

Phosphorus   [Ne] 3s23p3

 

Exercise 27

Using the s,p,d,f notation to write down the electron configurations for

(a) sulphur

(b) chlorine

(c) argon

(d) scandium

(e) cobalt

 

 

The electronic configurations of ions derived from atoms can be represented in a similar manner.

Li  1s22s1           Li+  1s2

F  1s22s22p5      F-   1s22s22p6

 

Exercise 28

Using the spdf notation, write the electron configuration of

(a)            (i)      The oxide ion , O2-.

(ii)        The magnesium ion, Mg2+.

(iii)       The bromide ion, Br-.

(b)   Which element in the Periodic Table has the same electron configuration as the ions (i) and (ii)?

 

Exercise 29

  1. The electronic energy levels of a certain element can be represented by1s22s22p63s23p1.

Sketch a graph showing the general form you would expect for the first five ionisation energies of the element.

 

  1. The electron energy levels of a certain element can be represented by 1s22s22p63s23p63d104s24p65s2.

(a)    What is the atomic number of the element?

(b)   In which Group of the Periodic Table should the element be?

(c)    The element forms an ionic bond when it reacts with oxygen. What will be the charge on the ion of the element?

 

(iii)             The electron configurations of the elements up to krypton are :

Element1s2s2p3s3p3d4s4p
H
He↿⇂
Li↿⇂
Be↿⇂↿⇂
B↿⇂↿⇂
C↿⇂↿⇂↿  ↿
N↿⇂↿⇂↿  ↿  ↿
O↿⇂↿⇂↿⇂ ↿  ↿
F↿⇂↿⇂↿⇂ ↿⇂ ↿
Ne↿⇂↿⇂↿⇂ ↿⇂ ↿⇂
Na↿⇂↿⇂↿⇂ ↿⇂ ↿⇂
Mg↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂
Al↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂
Si↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿  ↿
P↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿  ↿  ↿
S↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿  ↿
Cl↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿
Ar↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂
K↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂
Ca↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂
Sc↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂
Ti↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿  ↿↿⇂
V↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿  ↿  ↿↿⇂
Cr *↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿  ↿  ↿  ↿  ↿
Mn↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿  ↿  ↿  ↿  ↿↿⇂
Fe↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿  ↿  ↿  ↿↿⇂
Co↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂↿⇂ ↿⇂↿⇂ ↿⇂ ↿  ↿  ↿↿⇂
Ni↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂ ↿  ↿↿⇂
Cu *↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂
Zn↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂↿⇂
Ga↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂↿⇂
Ge↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂↿⇂↿  ↿
As↿⇂↿⇂↿⇂↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂↿⇂↿  ↿  ↿
Se↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂↿⇂↿⇂ ↿  ↿
Br↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿
Kr↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ ↿⇂

*Note :

  • In Cr the arrangement [Ar]3d54s1 with half filled 3d and 4s sub-shells is more stable than the alternative [Ar]3d44s2.
  • In Cu [Ar]3d104s1 with a completely filled 3d sub-shell and half filled 4s sub-shell is more stable than [Ar]3d94s2.

 

 

 

Electron structure and the Periodic Table

If we look at the Periodic Table as a whole we can see that it is divided into several major blocks. These correspond to the filling of the s, p, d and f sub-shells.

s-block                                                                             p-block

 

 

d-block

 

 

 

 

 

f-block

 

 

 

For the s- and p-block elements, the number of the last sub-shell to be occupied is the same as that of the period. For the d-block elements the sub-shell occupied is always one less than the period and for the f-block elements always two less than the period.

e.g. in the 6th period (n=6) going from caesium Cs to radon Rn, the 6s and 6p sub-shells are filled as well as the 5d and 4f sub-shells.

 

 

 

 

Ionisation energies of atoms broadly related to position in the Periodic Table; explanation in terms of nuclear charge, atomic radius, shielding and stability of filled and half-filled shells.

When you have finished this section, you should be able to:

  • Explain the term ‘periodic’ with reference to a graph of first ionisation energy against atomic number.
  • Explain the changes in first ionisation energy that take place across a Period and down a Group.

 

Ionisation energy and the Periodic Table

Ionisation energy gives a measure of the tendency of an atom to lose an electron.  It is always an endothermic process(ΔH +ve)

 

FIRST ionisation energy          X (g)                                X+(g) + e-   ΔH positive

SECOND ionisation energy       X+(g)                                X2-(g) + e-    energy required.

 

Its magnitude depends on the attraction of the nucleus for the electron removed.

The ionisation energy of an atom is influenced mainly by three factors:

 

(1)                The distance of the outermost electron from the nucleus:

As the distance increases the attraction of the positive nucleus for the negative electron will decrease and consequently, the I.E. decreases.

 

(2)               The size of the positive nuclear charge.

As the nuclear charge increases, its attraction for the outermost electron also increases and consequently, I.E. increases.

 

(3)               The shielding effect of the inner electrons.

The outermost electrons are repelled by all the other electrons in the atom besides being attracted by the positive nucleus.  The outermost electron is screened from the attraction of the nucleus by the repelling effect of the inner electrons.  In general this screening effect by inner electrons is more effective the closer these inner electrons are to the nucleus.  Thus

(a) electrons in shells of low principal quantum number (n=2, n=3)are more effective shields than electrons in shells of higher quantum number. (n=4, n=5) (b) electrons in the same shell exert a negligible shielding effect on each other.

 

 

General Trends In Ionisation Energy.

  1. Within a Group

First ionisation energy decreases down a Group due to the increasing size of atoms. The increasing number of electron shells and better shielding by inner shell electrons mean the outer electron becomes progressively further away from the nucleus and is less tightly held.

Na 2.8.1

K   2.8.8.1    etc.

 

  1. Across a Period

Ionisation energy generally increases across a Period because the atoms decrease in size from left to right. The trend is not entirely regular.

 

(a)     The electron is being removed from the same quantum shell but there is an increase in the nuclear charge, and therefore decrease in atomic radius, across a Period.

Element

Na        Mg        Al         Si         P           S          Cl        Ar

Electron structure

2.8.1    2.8.2      2.8.3     2.8.4     2.8.5     2.8.6     2.8.7     2.8.8

Nuclear  Charge

+11        +12       +13       +14       +15       +16       +17       +18

 

(b)     There are breaks in the 2nd and 3rd Period e.g. Be and N has a higher than expected I.E value.

Be 1s22s2

B ls22s22p1

All the sub-shells in beryllium are filled but the outer shell of boron contains only 1 electron.  In the same way as filled shells are associated with extra stability, there is also some extra energetic stability associated with half-filled sub-shells.  This means that the electronic structure of Be is more stable that expected and therefore its I.E. is greater than that of boron.  A similar situation arises with nitrogen which has a higher I.E. than oxygen

O  1s22s22p4

N  1s22s22p3

The half-filled 2p sub-shell in nitrogen with its evenly distributed charge is more stable than the 2p sub-shell in oxygen which contains four electrons.  This results in a higher first I.E. for nitrogen.

 

(c)        Ionisation energies increase from Scandium to zinc but only marginally compared to the increase across Period 3 from sodium to argon.

In crossing from Sc to Zn, the nuclear charge is increasing, but electrons are being added to an inner d sub-shell.  The inner d electrons shield the outer 4s electrons from the increasing nuclear charge much more effectively than outer shell electrons can shield each other.  Consequently the increase in I.E. from Sc to Zn is only slight.

 

In  general note that:

  • Noble gases are at the maximum of peaks due to the special stability of filled outer sub-shells.
  • Alkali metals are at the minimum (troughs) as only 1 electron has to be removed, leaving a stable noble gas core.