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If enough energy is supplied to an electron of a particular atom, it can be promoted to successively higher energy levels. If the supplied energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus.
The amount of energy required to remove the most loosely bound electron, that is outermost electron from an isolated gaseous atom of an element in its lowest energy state, that is ground state to produce cation is known as Ionization Potential or Ionization Energy of that element.
M(g) + Ionisation EnergyM+(g) + e
It is generally represented as I or IP and it is measured in electron volt (eV) or kilo calories (kcal) per gram atom.
One electron volt (eV) being the energy gained by an electron falling through a potential difference of one volt.
1 eV = (Charge of an electron) × (1 volt)
= (1.6 × 10-19 Coulomb) × (1 Volt)
= 1.6 × 10-19 Joule
= 1.6 × 10-12 erg
Calculate the ionization potential of Hydrogen atom in eV.
We know that, the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization potential of the hydrogen atom.
Thus the Ionization Potential of the Hydrogen atom, 
E H = 2 π 2 m e 4 ( 1 1 2 - 0 )
EH  = 2.179 × 10-11 erg
= 2.179 × 10-18 Joule
= (2.179 × 10-18)/(1.6 × 10-19) eV
= 13.6 eV
Calculate the second ionization potential of Helium (given ionization potential of Hydrogen = 13.6 eV)
The ground state electronic configuration of helium is 1S2. The second ionization potential means removal of the second electron from the 1S orbital against the nuclear charge of +2.
So we have:
IPHe = (2π2mZ2e4/h2)[(1/n12) - (1/n22)]
= Z2 × IPH
∴ Second Ionization Potential of Helium,
= 22 × 13.6
= 54.4 eV

Successive Ionization Potentials:

  1. The electrons are removed in stages one by one from an atom. The energy required to remove the first electron from a gaseous atom is called its first Ionization potential.
    M (g) + IP1 M+ (g) + e
  2. The energy required to remove the second electron from the cation is called second Ionization potential.
    M+ (g) + IP2 M+2 (g) + e
  3. Similarly we have third, fourth Ionisation potentials.
    M+2 (g) + IP3 M+3 (g) + e
    M+3 (g) + IP4 M+4 (g) + e
The values show that these increase in order:
IP1IP2 IP3IP4
The successive increase in their values is due to the fact that it is relatively difficult to remove an electron from a cation having higher positive charge than from a cation having lower positive charge or from neutral atom.

Factor Affecting the Magnitude of Ionisation Potential and its Periodic Variation:

The following factors influence the magnitude of the Ionization Potentials:
  • The distance of the electron from the nucleus:

The greater the distance of an element from the positive charge nucleus of an element, the weaker will be the attraction and hence the lower the ionization potential of that element.
If an atom is raised to an exited state by promoting one electron to a higher energy level, then that electron is further from the nucleus and it is observed that this excited electron is more easily detached than when the electron is in its ground state.
Thus on moving from top to bottom in a group the ionization potential of the elements decreases with the increase of their nucleus electron separation.
1A IIA IIIA IVA VA VIA VIIA 0
H
13.5
He
24.5
Li
5.4
Be
9.3
B
8.3
C
11.2
N
14.5
O
13.6
F
17.4
Ne
21.6
Na
5.1
Mg
7.6
Al
6.0
Si
8.1
P
11.0
S
10.4
Cl
13.0
Ar
15.8
K
4.3
Ca
6.1
Ga
6.0
Ge
7.8
As
9.8
Se
9.8
Br
11.8
Kr
14.0
Rb
4.2
Sr
5.7
In
5.7
Sn
7.3
Sb
8.6
Te
9.0
I
10.4
Xe
12.1
Cs
3.9
Ba
5.2
Tl
6.1
Pb
7.4
Bi
7.2
Po
8.4
At Rn
10.7

  • The Charge on the Nucleus that is, Atomic Number:
The higher the charge on the nucleus the more difficult it is to remove an electron and hence the higher the value of ionization potential.
Thus the value of ionisation potential generally increases in moving from left to right in a period, since the nuclear charge of the elements (atomic number) also increases in the same direction.
Li Be B C N O F Ne
+3 +4 +5 +6 +7 +8 +9 +10
5.4 9.3 8.3 11.2 14.5 13.6 14.4 21.6
With increasing atomic number electrons are added to orbitals of the same principal quantum number. These electrons add little to the size of the atom, the increasing nuclear charge bring about a contraction in size. In effect therefore ionisation potential steadily increases along a period.
  • Completely - filled and half - filled orbitals:

According to the Hund's rule atoms having half - filled or completely filled orbitals are comparatively more stable and hence move energy is needed to remove an electron from such atom.
The ionization potential of such atoms is therefore relatively higher than expected normally from their position in the periodic table.
A few regulation that are seen in the increasing vale of ionization potential along a period can be explain on the basis of the concept of the half - filled and completely filled orbitals.
Be and N in the second period and Mg and P in the third period have slightly higher vale of ionization potentials than those normally expected.

This is explained on the basis of extra stability of the completely - filled 2S - orbital in Be (2S2) and 3S - orbital in Mg (3S2) and of half - filled 2P - orbital in N (2S22P3) and 3P - orbital in P (3S23P3).
  • The Screening effect of the lower lying inner electrons:

Electrons provide a screening effect on the nucleus. The outermost electrons are shielded from the nucleus by the inner electrons.
The radial distribution functions of the S, P, d orbitals shows that for the same principal quantum number the S - orbital is the most penetrating., next is P - orbital and least penetrating is the d - orbital. Screening efficiency falls off in the order,
 SPd.

As we move down a group, the number of inner - shells increases and hence the ionisation potential tends to decreases.
Elements of II A Group:
Be〉Mg〉Ca〉Sr〉Ba
  • Overall Charge on the Ionizing Species:

An increasing in the overall charge on the ionizing species (M+, M+2, M+3, etc) will enormously influence the ionization potential since electron withdrawal from a positive charged species is more difficult than from a neutral atom.
Ionization energy trend up and down
First ionization energy chart.
The first ionization potentials of the elements very with their positions in the periodic table. In each of the table the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.
In first transition series electron filling up processes begins in the 3d level below a filled 4S level. During ionization process will a 4S electron or a 3d electron be lost first ? Explain with reference to Chromium.
Electronic Configuration of Chromium,
1S2 2S2 2P6 3S2 3p6 3d5 4S1 
Screening constant (σ) for 4S electron is,
σ = (2 × 1.0) + (8 × 1.0) + (8 × 0.85) + (5 × 0.85)
= 21.05
∴ Effective nuclear charge,
= (24 - 21.05)
= 2.95
Screening constant (σ) for 3d electron,
σ = (2 × 1.0) + (8 × 1.0) + (8 × 1) + (4 × 0.35)
= 19.4.
∴ Effective nuclear charge
= (24 - 19.4)
= 4.60
Hence a 3d electron is more tightly held than a 4S electron. So during ionization the 4S electron lost first.
Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
LiBBeCONFNe

Electron Affinity of an element is defined as the amount of energy released when an electron is added to a neutral gaseous atom in its lowest energy state that is, ground state to convert it into uni-negative gaseous ion. 
In simple words, the electron affinity of an atom is defined as the energy liberated when a gaseous atom captures an electron.
A(g) + ElectronA- (g) + Electron Affinity
Evidently electron affinity is equal in magnitude to the ionisation energy of the species formed. The definition of electron affinity as the exothermicity of the electron capture reaction would lead to the negative sign according to our usual thermodynamic convention. But electron affinity is normally describes with positive sign. 
Negative electron affinities are known. These negative values indicate that the species concerned does not like to have any more electron.

Measurement of Electron Affinity:

Electron affinities are difficult to obtain. These are often obtained from indirect measurements, by analysis of Born-Haber energy cycles in which one step is electron capture or by the direct study of electron capture from heated filaments.
Calculate the electron affinity of chlorine from the Born -  Haber cycle,  given the following date :  Lattice Energy = - 774 kJ mol-1 Ionization Potential of Na = 495 kJ mol-1 Heat of Sublimation of Na = 108 kJ mol-1 Energy for Bond Dissociation of chlorine (Cl2) = 240 kJ mol-1 and  Heat of Formation of NaCl = 410 kJ mol-1.
Born - Haber Cycle for formation of NaCl (S) is:
Electron affinity trend and exceptions
Born - Haber Cycle.
From the above Born - Haber cycle we can written as:
-UNaCl - INa + ECl - SNa - 1/2DCl - ΔHf = 0 
or, ECl = UNaCl + INa + SNa + 1/2DCl + ΔHf 
ECl = -774 + 495 + 108 + 120 + 410 
= 359 kJ mol-1
Factors Influencing the magnitude of Electron Affinity:
The magnitude of Electron Affinity (EA) is influenced by following factors such as,
  1. Atomic Size:
    Larger the atomic radius lesser the tendency of the atom to attract the additional electron towards itself and lesser is the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom.
    Thus Electron Affinity values decreases with increases atomic atomic radius.
  2. Effective Nuclear Charge:
    Higher the magnitude of effective nuclear charge (Z) towards the periphery of an atom greater is the tendency of the atom to attract the additional electron electron towards itself. 
    Again greater the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom.
    As a result higher the energy released when extra electron is added to form an anion. Thus the magnitude of Electron Affinity (EA) of atom increases with increasing Z value.
  3. Electronic Configuration:
    The magnitude of electron affinity depends on the electronic configuration of the atom. The elements having,
    nS2, nS2nP6, nS2nP6
    valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration.
  • Electron Affinity of Be and Mg are -0.60 eV and - 0.30 eV respectively.
These elements posses filled S-Orbital in their valence shell which is stable hence Be and Mg do not prefer to except the additional electron in order to form anion. In such cases energy is required to form anion and consequently the electron affinity of Be and Mg posses negative sign with low magnitude.
From the above discussion it is clear that elements of Group - IIA posses low electron affinity vales with negative sign due to stable nS2 valence shell electronic configuration.
Atom Electron Affinity Atom Electron Affinity
H 0.747 S 2.07
F 3.45 Se 1.70
Cl 3.61 Li 0.54
Br 3.36 Na 0.74
I 3.06 K 0.70
O 1.47 Be -0.60
C 1.25 Mg -0.30
Si 1.63 B 0.20
N -0.10 Al 0.60
P 0.70 Zn -0.90
Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.
Atomic number and the electronic distribution of lithium and beryllium are:
Li 3 1S22S1
Be 4 1S22S2
Lithium has an incompletely filled 2S sub-shell while beryllium has the sub shell filled. Thus lithium can receive an electron in its 2S sub-shell but for beryllium a still higher energy 2P level has to be made of. 
A filled shell or sub-shell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.
  • The elements of Group - VA (group - 15) having nS2 nP3 valence shell configuration also posses low electron affinity values:
Why electron affinity of Nitrogen is - 0.10 eV while that of Phosphorus is + 0.70 eV ?
Atomic number and the electronic distribution of Nitrogen and Phosphorus are:
N 7 1S22S22P3
P 15 1S22S22P63S23P3
Due to smaller size of Nitrogen atom when extra electron is added to the stable half - filled 2P orbital some amount of energy is required and hence electron affinity of nitrogen is negative. 
On the other hand, due to bigger size of P atom in compare to nitrogen less amount of energy released when the extra electron is added to the stable half - filled 3P orbitals and thus electron affinity of Phosphorus is expressed with positive sign.
  • Electron Affinity data on noble gases are not available.
Complete lack of evidence for stable noble gas anion is sufficient prof of their low electron affinity. An electron, even if accepted by a noble gas atom, would have to go into an orbital of next quantum shell and presumably the nuclear charge not high enough to hold an electron placed so far out.

Periodic Variations:

In a Group:
In a group when move from top to bottom the size of atom generally increases with increasing atomic number and consequently the magnitude of electron affinity decreases in the same direction.
There are some exceptions to this general rule as is evident from the following examples:
  1. Although the elements of the second period of the periodic table are relatively smaller in size than those of the third period elements, but the electron affinity values of elements of second period is smaller than the electron affinity values of third period elements.This unexpected behavior is explained by saying that the much smaller sizes of the second period elements give a very much higher value of charge densities for the respective negative ions. A high vales of electron density is opposes by the inter - electronic repulsion forces.
  2. Electron Affinity of fluorine is lower than that of Chlorine. The lower values of electron affinity for Florine due to the electron-electron repulsion in relatively compact 2P-Orbital of Florine atom.
    Thus the electron affinity values of halogens are:
    F Cl Br I
    - 3.6 eV - 3.8 eV - 3.5 eV - 3.2 eV
Explain why electron affinity of chlorine is more than fluorine?
The halogen possess large electron affinity values indicating their strong tendency to form anions. This is easily understandable because their electronic configurations are only one electron short of the next noble gas element.
The electron affinity of chlorine is greater than that of F. This is probably due to very small size of F atom which results in a very high charge density and consequently strong repulsion between existing valence shell electrons and the entering electron (new coming electron). On the other hand Cl being bigger size , charge density is small and thus such repulsion is not strong enough. Hence the electron affinity Cl is greater than that of F.
In a Period:
In a period, when we move from left to right Z value towards the new coming electron gradually increases with the increasing atomic number and hence the electron affinity gradually increases in the same direction.
A plot of electron affinities of elements up-to chlorine against atomic number shown as,
Electron Affinity and its Variation along a Period and a Group with Exception
Electron Affinities as Functions of Atomic Number
Why electron gain enthalpy of Be and Mg are almost zero?
Be and Mg have their electron affinity values equal to almost zero. Since Be and Mg have completely filled S orbitals.
Be 4 1S22S2
Mg 12 1S22S22P63S2
The additional electron will be entering the 2P-orbital in the case of Be and 3P-orbitals in the case of Mg which are of considerably higher energy than the 2S- Orbitals respectively.
Why electron affinity of noble gases are zero?
Inert gases in which the nS and nP orbitals are completely filled (nS2 nP6 configuration) the incoming electron must be go into an electron shall having the larger values for the principal quantum number, n. Thus inert gas have their electron affinity values equal to zero.

The emission spectrum of hydrogen atom was determined experimentally and it was possible to relate the frequencies of different spectral lines by simple equation,
ν = R[1/n12 - 1/n22]
Where n1 and n2 are integers and R is a constant, called Rydberg constant after the name of the discoverer. The existence of discrete spectral emission could hardly be explained by any model at Rydberg's time.

Energy of an Electron in Bohr Orbits:

The Energy associated with the permitted orbits is given by ,
En = - 1/2(e2/r), where r = n2h2/4π2me2
Thus, En = - (2π2me4/n2h2)
Again E1 = - (2π2me4/h2)
En = E1/n2
Thus the values of E2, E3, E4, E5 etc. in terms of E1(- 13.6 eV) are given as,
E2 = -13.6/22 = - 3.4 eV
E3 = -13.6/32 = - 1.51 eV
E4 = -13.6/42 = - 0.85 eV
E5 = -13.6/52 = - 0.54 eV
As n increases the energy becomes less negative and hence the system becomes less stable.

Explanation of the Rydberg Equation:

The explanation of Rydberg equation is now simple. The most stable orbit must be the one with the lowest energy (ground state).
When n = 1 the energy of an electron is lowest and it the lowest energy state and the states with n 〉 1 are the higher energy exited states.
Provided the right amount of energy is supplied the electron may jump from n = 1 to another higher level. Conversely energy will be released in the form of light of definite frequency when the exited electron returns to its ground state.
Since energy and frequency of the emitted light are connected by Plank Relation
E = hν 
or, ν = E/h
The energy corresponding to a particular line in the emission spectrum of hydrogen atom is the energy difference between the initial state 1 and the final state 2.
So that, E= E2 - E1
= - (2π2me4/n12h2) - [- (2π2me4/n22h2)]
= (2π2me4/h2)[1/n12 - 1/n22]
Then ν corresponding to the energy E is given by,
ν = E/h = (2π2me4/h3)[1/n12 - 1/n22]
ν = R[1/n12 - 1/n22]
Where R is the Rydberg Constant.
This relation tells us that as we go to higher and still higher orbits the energy gap between two immediate orbits continues to decreases.
  • Value of the Rydberg Constant:
Using the value of 9.108 × 10-28 gm for the mass of an electron, the Bohr Theory Predicts the following value of Rydberg Constant.
R = (2π2me4/h3)
= {2 × (3.1416)2 × (9.108 × 10-28 gm) × (4.8 × 10-10 esu)4}/(6.627 × 10-27 erg sec)3 
= 3.2898 × 1015 cycles/sec
Evaluate R in terms of wave number () instead of frequency. Wave number and frequency are connected with,
λν = c and ν = c/λ = c
Thus, = ν/c
where c is the velocity of light.
Recall the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second. Wave number () on the other hand stands for the number of waves connected in unit length that is per centimeter (cm-1) or per metre (m-1).
Thus, R = 3.2898 × 1015 sec-1/2.9979 × 1010 cm sec-1
= 109737 cm-1
= 10973700 m-1
The experimental values of R is 109677 cm-1 (10967700 m-1) showing a remarkable agreement between experiment and Bohr's Theory.

Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom:

Putting n = 1, n = 2, n = 3, etc in Rydberg Equation we get the energies of the different stationary states for the hydrogen electron. The transition energies can be calculated from the Rydberg Equation. The experimental spectra of hydrogen atom also exhibited several series of lines .
Energy Level Diagram for the Hydrogen Spectrum
Hydrogen atom contains only one electron but its spectrum gives a large number of lines - Explain.
This is because any given sample of hydrogen contains almost infinite number of atoms.
Under the normal conditions the electron of the each hydrogen atom remains in the ground state near the nucleus that is n = 1 (K - Shell). 
When energy is supplied to this sample of hydrogen gas by passing an electron discharge through the gas in the discharge tube, individual atom absorbed different amounts of energy. Some of atoms absorbed such energy to shift their electron to third energy level (n = 3) while some others may absorbed large amount of energy to shift their electron to the Fourth (n = 4), Fifth (n = 5), Sixth (n = 6) and Seventh (n = 7) energy level.
The electrons in the higher energy level are relatively unstable and hence drop back to the lower energy level with the emission of energy in the form of line spectrum containing various lines in of particular frequency and wave length.
  • Lyman Series:
Transition to the ground state (n = 1) from n = 2, n = 3, n = 4, etc exited states constitute the Lyman Series in the ultraviolet region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = R[1/n12 - 1/n22]
For Lyman Series, we have n1 = 1 and n2 = 2, 3, 4 ....
Thus, (i) When n2 = 2,
1/λ = 109677{1 - (1/4)}
or, 1/λ = {(109677 × 3)/4} cm-1
λ = {4/(109677 × 3)}
= 1215 × 10-8 cm
= 1215 Å
Thus, (i) When n2 = ∞,
1/λ = 109677{1 - (1/∞2)}
or, 1/λ = 109677 cm-1
λ = (1/(109677)
= 912 × 10-8 cm
= 912 Å
  • Balmer Series:
Thus the transition occurs to the ground state (n = 1) from n =2.
Thus the transition occurs to the ground state (n = 1) from n =2.
Transition to the n = 2 level from n = 3, n = 4, n = 5, etc exited states constitute the Balmer Series in the visible region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/22) - (1/n22)]
  • Pschen Series:
Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Pschen Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/32) - (1/n22)]
  • Brackett Series:
Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Brackett Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/42) - (1/n22)]
  • Pfund Series:
Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/52) - (1/n22)]
Series of Lines n1 n2 Spectral Region Wavelength
Lyman Series 1 2,3,4..etc Ultraviolet ㄑ4000 Å
Balmer Series 2 3,4,5..etc Visible 4000 Å to 7000 Å 
Paschen Series 3 4,5,6..etc Near Infrared 〉7000 Å
Brackett Series 4 5,6,7..etc Far Infrared 〉7000 Å
Pfund Series 5 6,7,8..etc Far Infrared 〉7000 Å
The wave number of the experimental spectra of hydrogen atom in Lyman Series is 82200 cm-1. Show that the transition occur to the Ground state (n =1) from n = 2.
(R = 109600 cm-1)
We know that, Wave number,
= R[1/n12 - 1/n22]
For Lyman series n = 1 and 
Wave Number () = 82200 cm-1
Thus, 82200 = 109600(1/12 - 1/n22)
or, 1/n22 = 1 - (822/1096)
= 274/1096 
=1/4
∴ n2 = 2
Which of the Balmer lines fall in the visible region of the spectrum wave length 4000 to 7000 Å ? (Given  R = 109737 cm⁻¹)
The Balmer Series of the hydrogen spectrum comprise the transition from n 〉2 levels to the n = 2 level.
4000 Å = 4000 × 10-8 cm = 4 × 10-5 cm and 7000 Å = 7 × 10-5 cm
Therefore wave number,
= 1/λ = (1/4) × 105 to (1/7) × 105 cm⁻¹
Transition energy of the Balmer lines,
= 1/λ = 109677[(1/22) - (1/n22)]
Taking () = (1/4) × 105 cm-1
We have,
(1/4) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/4 ⋍ 1.1{(1/4) - (1/n22)}
or, n2 ⋍ 44
So n2 = 7 ( nearest whole number).
Taking () = (1/7) × 105 cm-1
We have,
 (1/7) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/7 ⋍ 1.1 {(1/4) - (1/n22)}
or, n2 ⋍ 3
So n2 = 3 ( nearest whole number).
So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.
Calculate the wave length of Hɑ and Hβ of the Balmer Series.
We know that, for the Balmer Series,
= 1/λ = 109677[(1/22) - (1/n22)]
Again for Hɑ line n2 = 3 and for Hβ line n2 = 4
Thus the wave length for Hɑ - line,
1/λ = 109677 {(1/22) - (1/32)}
= 109677 × (5/36)
λ = 36/(5 × 109677)
= 6.564 × 10-5 cm
= 6564 Å
Again the wave length for Hβ - line,
1/λ = 109677 {(1/22) - (1/42)}
= 109677 × (3/16)
λ = 16/(3 × 109677)
= 4.863 × 10-5 cm
= 4863 Å

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