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### Ionization Potential

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 Energy → M+(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,
${E}_{H}= 2{\mathrm{Ï€}}^{2}m{e}^{4}\left(\frac{1}{{1}^{2}}- 0\right)$
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 = (2Ï€2mZ2e4/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 H13.5 He24.5 Li5.4 Be9.3 B8.3 C11.2 N14.5 O13.6 F17.4 Ne21.6 Na5.1 Mg7.6 Al6.0 Si8.1 P11.0 S10.4 Cl13.0 Ar15.8 K4.3 Ca6.1 Ga6.0 Ge7.8 As9.8 Se9.8 Br11.8 Kr14.0 Rb4.2 Sr5.7 In5.7 Sn7.3 Sb8.6 Te9.0 I10.4 Xe12.1 Cs3.9 Ba5.2 Tl6.1 Pb7.4 Bi7.2 Po8.4 At Rn10.7

• 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.
 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 (Ïƒ) for 4S electron is,
Ïƒ = (2 × 1.0) + (8 × 1.0) + (8 × 0.85) + (5 × 0.85)
= 21.05
∴ Effective nuclear charge,
= (24 - 21.05)
= 2.95
Screening constant (Ïƒ) for 3d electron,
Ïƒ = (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.
 Liã„‘Bã„‘Beã„‘Cã„‘Oã„‘N〈Fã„‘Ne

### Electron Affinity

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) + Electron→A- (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:
 Born - Haber Cycle.
From the above Born - Haber cycle we can written as:
 -UNaCl - INa + ECl - SNa - 1/2DCl - Î”Hf = 0 or, ECl = UNaCl + INa + SNa + 1/2DCl + Î”Hf ∴ 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, nS2, nP6, nS2, nP6
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 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.

### Law of Mass Action

The Low of Mass Action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864. The basis of their formulation is the observation of huge deposit of Sodium Carbonate on Egyptian take shore. Large amount of NaCl in take water and CaCO3 on the take shore made the reverse reaction possible.
 CaCl2(aq) + Na2CO3(aq) ↓ CaCO3(s) + 2NaCl(aq)
The foreword reaction occurs spontaneously in the laboratory.
They States the Low as,
The rate of chemical reaction at a constant temperature, is directly proportional to the active mass of the reactants.
The active mass is thermodynamic quantity. We assumed active mass as,
1. Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally.
2. Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low.
3. For pure solid and pure liquid, active mass is assumed to be unity since their mass does not effect the rate of reaction.

### Application of Low of Mass Action to the Chemical Reaction:

Let us Consider a reaction,
 A + B ⇆ C + D
Let the reacting system contains reactants only and CA and CB are their Concentrations in molar units.
According to the mass action low, the rate of the foreword reaction,
 Rf ∝ CA × CB ∴ Rf = Kf × CA × CB
Though the rate equation is an experimental quantity and powers of concentration terms are determined experimentally, here we consider a special type of rate equation where the stoichometric coefficients are raised in the of the conc. term(this is true for the elementary or one step reaction).
As the reaction proceeds in the foreword direction, conc. of A and B decreases and Rf also decreases. When the products are getting accumulated in the system, backwards reaction also starts and the rate of the backward reaction,
 Rb ∝ CC × CD ∴ Rb = Kb × CC × CD
Here Kf and Kb are the rate constants of the foreword and backward reaction and they do not depends on Conc. at a given temperature.
As the reactions proceeds in the foreword reactions Rf is decreasing but Rb is increasing. A state is then attain when they are equal. This state is called the chemical equilibrium. There will be no further change in the composition of the system.
Thus at equilibrium,
Rf = Rb
or, Kf × CA × CB = Kb × CC × CD
 ∴ Kf/Kb = (CC × CD)/(CA × CB)
CA, CB, CC and CD are the equilibrium concentration of A, B, C and D.
Again, Kf/Kb = Kc, called concentration equilibrium constant of the reaction.
At a given temperature for a reaction, Kc is constant does not depend on the concentration of the reacting components.
 Kc = (CC × CD)/(CA × CB)

### Equilibrium Constant:

Concentration Equilibrium Constant:
If we write the equation as,
Æ”1A1 + Æ”2A2 Æ”3A3 + Æ”4A4
Where Æ”1, Æ”2, Æ”3 and Æ”4 are stoichiometric coefficient.
 ∴ Kc = (C3Æ”3 × C4Æ”4)/(C1Æ”1 × C2Æ”2)
Where Kc is called concentration equilibrium constant of the reaction and C1C2C3 and C4 are the equilibrium concentration of A, B, C and D.
However the values of equilibrium constant of a chemical reaction depends on the mode of writing its Stoichiometric (balanced) equation.
Thus, for the reaction of formation of NH3 from N2 and H2, we can write the equation as,
N2 + 3 H2 2 NH3
The equilibrium constant can be written as,
 Kc = (CNH3)2/(CN2) (CH2)2
But if the equation written as,
1/2 N2 + 3/2 H2 NH3
 Then, K՛c = (CNH3)/(CN2)3/2(CH2)1/2
It is clear then Kc and c are not Equal in magnitude.
 Thus, Kc = (K՛c)1/2
The general rule is that if the equation is multiplied by n then the relation of Equilibrium concentration n will be raised to the power.
 That is, Kc = (K՛c)n
Pressure Equilibrium Constant:
When all the reactants and products are gases (that is, gas - phase reacting system), the expression of equilibrium constant for the equation,
Æ”1A1 + Æ”2A2 Æ”3A3 + Æ”4A4
Where Æ”1, Æ”2, Æ”3 and Æ”4 are stoichiometric coefficient.
 ∴ Kp = (P3Æ”3 × P4Æ”4)/(P1Æ”1 × P2Æ”2)
Where Kp is called Pressure equilibrium constant of the reaction and P1P2P3 and P4 are the equilibrium partial pressure of reacting components.
For the dissociation N2O4 2 NO2, obtain an expression for the fraction of original N2O4 dissociated at equilibrium in terms of Kp and total pressure.
The reaction is, N2O4 2 NO2
Let a mole of N2O4 is taken initially and x moles of N2O4 is dissociated at equilibrium then mole number of N2O4 and NO2 at equilibrium is (a - x) and 2x.
Total moles number at equilibrium = (a -x + 2x) = (a + x).
Equilibrium partial pressure of the components is {(a - x)/(a + x)}P and {2x/(a + x)}P where P = Total pressure of the reacting system.
The expression of the equilibrium constant,
Kp = (PNO2)2/PN2O4
= {2x/(a + x)}P/{(a - x)/(a + x)}P
Kp = (4x2P)/(a2 - x2)
The fraction of the original N2O4 dissociated at equilibrium É‘ = x/a.
Replacing (x/a) by É‘, we have,
 Kp = (4É‘2P)/(1 - É‘2)
Mole Fraction Equilibrium Constant:
If we write the equation as,
Æ”1A1 + Æ”2A2 Æ”3A3 + Æ”4A4
Where Æ”1, Æ”2, Æ”3 and Æ”4 are stoichiometric coefficient.
 ∴ Kx = (x3Æ”3 × x4Æ”4)/(x1Æ”1 × x2Æ”2)
Where Kx is called Mole fraction Equilibrium Constant of the reaction and x1x2x3 and x4 are the equilibrium mole fraction of reacting components.
 Equilibrium Constant.

### Relation Between Kp and Kc:

The interrelations of these Equilibrium constants are as follows,
Kp = (P3Æ”3 × P4Æ”4)/(P1Æ”1 × P2Æ”2)
The ideal gas Equation, PV = nRT, may be written as,
P = (n/V)RT = CRT
Where C is the concentration of gas expressed as amount per unit volume.
= {(C3RT)Æ”3 × (C4RT)Æ”4}/{(C1RT)Æ”1 × (C2RT)Æ”2}
 ∴ Kp = Kc(RT)Î”Æ” Î”Æ” = (Æ”3 + Æ”4) - (Æ”1 + Æ”2)
1. For the reaction in which total number of reactant molecules and of resultant molecules are same:
2. H2 (g) + I2 (g) 2HI
KP = (PHI)2/{(PH2)(PI2)}
= (CHIRT)2/(CH2RT) (CI2RT)
= [(CHI)2/{(CH2) (CI2)}] × [(RT)2/(RT)(RT)]
= Kc
Thus when (Æ”3 + Æ”4) = (Æ”1 + Æ”2),
Kp = Kc
3. For the reactions in which the number of molecules of reactants differ from that of the resultant:
4. 2SO2(g) + O2(g) SO3(g)
Here, Kp = Kc×(RT){1 - (2+1)}
= Kc×(RT)-2
Thus when, (Æ”3 + Æ”4) (Æ”1 + Æ”2),
Kp  Kc
 Reaction (Gaseous System) Relation between Kp and Kc H2 + Cl2 ⇆ 2HCl Kp = Kc CO + H2O ⇆  2H2 + CO2 Kp = Kc CO + NO2 ⇆ NO + CO2 Kp = Kc PCl5 ⇆ PCl3 + Cl2 Kp = Kc RT 2H2 + O2 ⇆ 2H2O Kp = Kc RT -1 2CO + O2 ⇆ 2CO2 Kp = Kc RT -1 N2 + 3H2 ⇆ 2NH3 Kp = Kc RT -2
Calculate the Kc value of the reaction N2 + 3H2 ⇆ 2NH3 at 4000C, Given Kp at the same temperature 1.64 × 10-4.
Kp = 0.5.
For Solution see,
At 1000C the vapour density of N2O4 is 25 at 1 atm. Show that Kp = 9.6.
N2O4 (g) 2NO2 (g)
Let 1 moles of N2O4 is taken initially (t = 0) and x mole of N2O4 has reacted at equilibrium.
So the mole number of each component is (1-x) and 2x and total moles at equilibrium,
(1-x+2x) = (1+x).
So total moles has increased from 1 to (1+x).
Let Volume is increases from V1 to V2
So, (1+x) = V2/V1
As density and hence vapour density is inversely proportional to volume so vapour density will decrease from d1 to d2.
Hence, (1+x) = V2/V1= d1/d2.
Molecular weight of N2O4 is 92 and vapour density,
= 92/2
= 46
Due to dissociation it is = 25.
∴ 1+x = 46/25
or, x = 0.84
Now partial pressure are,
PNO2 = {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
PN2O4 = {(1-x)/(1+x)}P
= 0.16/1.84
= 0.087
Kp = (PNO2)2/PN2O4
= (0.913)2/0.087
≃ 9.6

### Relation Between Kp and Kx:

 ∴ Kp = Kx(P)Î”Æ” Î”Æ” = (Æ”3 + Æ”4) - (Æ”1 + Æ”2)
Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.
Vant Hoff reaction isotherm is Î”G = - RT lnKa + RT lnQa
But when the reaction attain equilibrium, Qa = Ka
Thus, Î”G = 0.

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