Priyam Study Centre

A Page about Chemistry and Related Topics.

Mar 18, 2019

Van der Waals Equation

In 1873, Van der Waals modified the Ideal Gas Equation,
PiVi = RT
    By incorporating the size effect and inter molecular attraction effect of the real gases. These above two effects are discussing under the Volume Correction and Pressure Correction of the Ideal Gas Equation.

Volume Correction:

    Molecules are assumed to be a hard rigid spheres and in a real gas, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of 1 mole gas molecules,
Vi = (V - b)
    where V is the molar volume of the gas and b is the volume correction factor.
    Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses.
    Let us take r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
What is the van der waals equation?
Encounter of the gas molecules.
    Then it can be shown that the volume correction term or effective volume of one mole gas molecules,
    b = 4 × N0 (4/3) π r3
    When two molecules encounter each other the distance between the centers of the two molecules would be σ. They can not approach beyond this distance. Thus the sphere of radius σ will occupy a space unavailable for a pair of molecules.
    This excluded volume = 4/3 π σ3 for a pair of molecules.
    Thus a single molecule,
    = 1/2 × 4/3 π σ3
    The effective volume for Avogadro number of molecules (present in 1 mole gas),
b = N0 × 2/3 × π σ3
or, b = N0 × 2/3 × π (2r)3
or, b = 4 N0 × 4/3 × π (r)3
Thus, b = N0 × 2/3 × π (2r)3
∴ b = 2/3 × N0 × π σ3
    Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule. The equation becomes,
    Pi (V - b) = RT

Pressure Correction:

    Pressure of the gas is developed due to the wall collision of the gas molecules. But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular attraction as in ideal gas Pi.
    Thus, Pi 〉P
    or, Pi = P + Pa
    Where Pa is the pressure correction term originating from attractive forces. Higher the inter molecular attraction in a gas, greater is the magnitude of Pa.
    Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision. Both contributions are diminished by the attractive forces and the strength with which these operate is roughly proportional to the density of the molecules.
    Therefore the average pressure exerted by the molecules is decreased by Pa is proportional to the square of the density.
    Pa ∝ 1/V2
    Since, density ∝ 1/V
    ∴ Pa = a/V2
    Where a = constant for the gas that measures the attractive force between the molecule.
    Thus, Pi = P + (a/V2)
    Using the two corrections we have Van der Waals equation for 1 mole Real Gas is,
(P + a/V2 )(V - b) = RT
    To convert the equation for n moles volume has to change as it is the only extensive property in the equation.
    Let V be the volume of n moles gas hence, V = v/n. Putting this term in the Van der Waals equation, we have,
what is the Van der Waals equation?
Van der Waals equation for n Moles Real Gas
    Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 270C. Given, a = 1.4 atm lit2 mol-2 and b=0.04 lit mol-1 using Van der Waals equation.
    Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.
    P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:

    (a)For the real gas a=0 (that is no inter molecular attraction )
    but b≠0 (size considers):
    We have Van der Waals Equation,
    (P+ an2/V2)(V - nb) = nRT but a=0
    Hence, P = nRT/(V - nb) 〉Pi
    Since, Pi = nRT/V only.
    (b)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):
    It means that the molecular size effect (repulsive interaction) creates higher pressure then that observed by the ideal gas where molecules have no volume.
    We have Van der Waals Equation,
    (P+ an2/V2)(V - nb) = nRT but b=0(no size).
    Hence, P = (nRT/V - an2/V2)ㄑPi
    Since, Pi = nRT/V only.
    Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of Van der Waals Constant a and b:

    From the Van der Waals Equation,
    (P+ an2/V2)(V - nb) = nRT
    where, Pa = an2/V2
    Where Pa is the pressure correction term, called often the internal pressure of the gas.
    Then, a = Pa × (V2/n2)
Thus the unit of a = atm lit2 mol-2
    Again, nb = Unit of volume (say liter)
Hence, the unit of b = lit mol-1
    Find the Units and Dimensions of Van dar Walls constant 'a' and 'b' from Van der walls Equation.
    In SI system,unit of 'a' = N m4 mol-2
    In CGS system,unit of 'a' =dyne cm4 mol-2
    In SI system,unit of 'b' = m3 mol-1
    In CGS system,unit of 'b' = cm3 mol-1

Significance of Van der Waals constants 'a' and 'b':

    a’ term originates from the inter molecular attraction and Pa = an2/V2. Thus, ‘a’ is a measure of internal pressure of the gas and so the attractive force between the molecules. Higher the value of ‘a’ grater is the inter molecular attraction and more easily the gas could be liquefied.
    Thus, a of CO2 = 3.95 atm lit2 mol-2 while, a of H2 = 0.22 atm lit2 mol-2.
    Another constant, ‘b’ measures the molecular size and also a measure of repulsive forces. The value of b can also be utilized to calculate the molecular diameter(σ). The grater the value of ‘b’ larger the size of the gas molecule.
    Thus, b of CO2 = 0.04 lit mol-1 while, b of H2 = 0.02 lit mol-1.

For a Van der Waals gas the expression for Boyle's temperature:

    Mathematical condition for Boyle’s temperature is,
    TB = [d(PV)/dP]T When P→0
    From the Van der Waals Equation for 1 mole gas,
    (P + a/V2)(V - b) = RT
    or, P = RT/(V - b) - a/V2
    Hence, PV = {RTV/(V - b)} - a/V
    Thus, TB = [d(PV)/dP]T
    = [RT/(V-b)-{RTV/(V-b)2}+a/V2][(dV/dP)]T
    = [{RT(V - b) - RTV
    }/(V - b)2 + a/V2] [(dV/dP)]T
    =[{- RTb/(V - b)2} + a/V2] [(dV/dP)]T
    But when T = TB,
    [d(PV)/dP]T = 0 and [(dV/dP)] ≠ 0
    Thus, We have, RTBb/(V - b)2 = a/V2
    or, TB = (a/Rb) {(V - b)/V}2
    Since P → 0, V is large.
    Thus, (V - b)/V ≃ 1
Hence,TB = a/Rb
    Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit2 mol-2 and b = 0.04 lit mol-1.
    TB = 427 K
    Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.
  1. When, a = 0, Van der Waals Equation, P (V - b) = RT or, PV = RT + Pb Differentiating with respect of pressure at constant temperature, [d(PV)/dP]T = b, but b ≠ 0 hence [d(PV)/dP]T ≠ 0. The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]T ≠ 0.
  2. Again when a = b = 0, Van der Walls equation becomes PV = RT; or, [d(PV)/dP]T = 0.
    That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.

Explanation of Amagat’s Curves:

Amagat’s Curves.
Van der Waals equation for 1 mole real gas,
(P + a/V2)(V - b) = RT
or, PV - Pb + (a/V) + (a/V2) = RT
Neglecting the small term (a/V2), we get,
PV = RT + Pb - (a/V)
Using ideal gas equation for small term,

a/V = aP/RT and taking Z = (PV/RT),
We have, Z = 1 + (1/RT){b - (a/RT} P
This Shows that, Z = (T,P)
    This equation can be uses to explain Amagat curve qualitatively at low pressure and moderate pressure region.
(a) For CO2 ‘a’ is very high since the gas is easily liquefied.
    Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve .
    That is slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with increase of Pressure and it is also found in the curve of carbon dioxide.
(b) For H2 ‘a’ is very small and It is not easily liquefied.
    So, a/RT〈 b and slope of Z vs P curve for H2 becomes (+) ve and the value of Z increases with pressure.
(i) When T〈 TB
or, T〈 a/Rb
or, b〈 a/RT
and {b−(a/RT)} = (-)ve.
    That is the value of Z decreases with increases with P at the moderate pressure region of the curve of CO2 , Z〈 1 and more compressible.
(ii) When T = TB = a/Rb
or, b = a/RT
hence, {b - (a/RT)} = 0
    That is Z = 1 the gas shows ideal behavior. The size effect compensates the effect due to inter molecular attraction of the gas. Z runs parallel to P axis up to certain range of pressure at low pressure region.
(iii) When T 〉TB
or, T 〉a/Rb
or, b 〉a/RT
hence {b - (a/RT} = (+)ve.
    That is value of Z increases with increase of P when T 〉TB.
    The size effect dominates over the effect due to inter molecular attraction. For hydrogen and helium, 0°C is grater then their TB values and Z vs P slope becomes (+)ve.
    At very low pressure P→0 and high temperature,the volume is very large and both the size effect and attraction effect becomes negligible that is Pb and (a/RT) P are negligible and Z=1 that is the gas obeys ideal behavior.

Mar 17, 2019

Critical Constants of a Gas

A gas can be liquefied by lowering temperature and increasing of pressure. But influence of temperature is more important. Most of gases are liquefied at ordinary pressure by suitable lowering of the temperature. But a gas can not be liquefied unless its temperature is below of a certain value depending upon the nature of the gas whatever high the pressure may be applied.
This temperature of the gas is called its Critical Temperature (Tc). A gas can only be liquefied when the temperature is kept below Tc of the gas.

Definitions of Critical Constants:

    Critical Temperature (Tc):
    Critical Temperature (Tc) is the maximum temperature at which a gas can be liquefied, that is the temperature above which a liquid cannot exist.
    Critical Pressure (Pc):
    Critical Pressure (Pc) is the maximum pressure required to cause liquefaction at the temperature (Tc).
    Critical Volume (Vc):
    Critical Volume (Vc) is the volume occupied by one mole of a gas at critical temperate (Tc) and Critical Pressure (Pc).

Andrews Isotherms:

    In 1869, Thomas Andrews carried out an experiment in which P - V relations of carbon dioxide gas were measured at different temperatures.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Andrews Isotherms.

Following are observed from this graph:

  1. At high temperature, such as T4, the isotherms look like those of an Ideal Gas.
  2. At low temperatures, the curves have altogether different appearances. Consider, for examples, a typical curve abcd. As the pressures increases, the volume of the gas decreases (curve a to b).
  3. At point b liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure P.
  4. At the point C, liquefaction is complete and thus the cd is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that ab represents the gaseous state, bc, liquid and vapour in equilibrium, and cd shows the liquid state only.
  5. At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.
  6. At temperatures Tc the horizontal portion is reduced to a mere point. At temperatures higher then Tc there is no indication of qualification at all.
Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.

Continuity of State:

    It appears from the Amagat curve at T there is discontinuity or break during the transformation of gas to liquid. But it is not so.
    The continuity of the states from the gas to liquid can be explains from the following isotherm pqrs at T.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Continuity of State.
    The gas at A is heated to B at constant volume (V) along AB. Then the gas is gradually cooled at constant P along BC, the volume is reduced considerably. The gas is again cooled at constant V until the point D is reached.
    No where in the process liquid would appear. At D, the system is highly compressed gas. But from the curve, this point is the representation of the liquid state.
    Hence there is hardly a distinction between the liquid state and the gaseous state. There are no line of separation between the two phases. This is known as the principle of continuity of state.

Critical Phenomena and Van der Walls Equation:

(P + a/V2)(V - b) = RT
V3 - (b + RT/P)V2 + (a/P)V - ab/P = 0
    This equation has three roots in V for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Critical Phenomena and Van der Walls Equation.

The main characteristics of the above isotherm are as follows:

  1. At higher temperature and higher volume region, the isotherms looks much like the isotherms for a Ideal gas.
  2. At the temperature lower than Tc the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V1, V2 and V3 at pressure P1.
    The section AB and ED of the Van der walls curve at T1 can be realized experimentally. ED represents Supersaturated or Super cooled vapour and AB represents super heated liquid. Both these states are meta-stable.
    These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium.
  3. The section BCD of the Van der Walls isotherms cannot be realized experimentally. In this region the slope of the P - V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease of pressure. The line BCD represents the meta- stable state.

Expression of Critical Constants in Terms of Van der Waals Constants:

    Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point.
    Thus the mathematical condition of critical point is,
    (dP/dV)T = 0 and (d2P/dV2)T = 0
    Van der Waals equation for 1 mole gas is,
    (P + a/V2)(V - b) = RT
    or, P = RT/(V - b) - a/V2
    Differentiating Van der Waals equation with respect to V at constant T,
    We get Slope,
    (dP/dV)T = - {RT/(V - b)2} + 2a/V3
    And Curvature,
    (d2P/dV2)T = {2RT/(V - b)3} - 6a/V4
    Hence at the critical point,
    - {RTc/(Vc - b)2} + 2a/Vc3 = 0
    or, RTc/(Vc - b)2 = 2a/Vc3
    and {2RTc/(Vc - b)3} - 6a/Vc4 = 0
    or, 2RTc/(Vc -b)3 = 6a/Vc4
    Thus, (Vc - b)/2 = Vc/3
∴ Vc = 3b
    Putting the value of Vc = 3b in RTc/(Vc - b)2 = 2a/Vc3.
    We have, RTc/4b2 = 2a/27b3
∴ Tc= 8a/27Rb
    Again the Van der Walls equation at the critical state,
    Pc = RTc/(Vc - b) - a/Vc2
    Putting the value of Vc and Tc,
∴ Pc = a/27b2
    The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm3mol-1. Calculate the value of Van der Waals constants a and b.
    We have Tc = 647 K,
    Pc = 22.09 Mpa = 22.09 × 103 kPa
    and Vc = 0.0566 dm3 mol-1
    Thus, b = Vc/3 = (0.0566 dm3 mol-1)/3
    ∴ b = 0.0189 dm3 mol-1
    a = 3 Pc Vc2
    = 3 (22.09 × 103 kPa)(0.0566 dm3 mol-1)2
    ∴ a = 213.3 kPa mol-2

Compressibility Factor at the Critical State (Zc):

    The Critical coefficient is defined as,
    Thus the value of Critical Coefficient,
    = {R × (8a/27Rb)}/{(a/27b2) × 3b}
    = 8/3
    = 2.66
    Thus the value of Compressibility factor at the critical state (ZC)
    = PCVC/RTC
    = 3/8
    = 0.375

Van der Waals constants in terms of critical constants:

    Van der Waals constants can be determined from critical constants Tc and Pc of the gas. Vc in the expression is avoided due to difficulty in its determination.
    We have, b = Vc/3
    but, PcVc/RTCc = 3/8
    or, Vc = (3/8) × (RTc/Pc)
∴ b = (1/8)(RTc/Pc)
    Again, a = Pc × 27b2
    = 3 × Pc × (3b)2
    = 3 PcVc2
    = 3Pc × (3RTc/8Pc)2
∴ a = (27/64)(R2Tc2/Pc)
    Calculate Van der Waals constants for Ethylene. (TC = 280.8 K and PC = 50 atm). 
    a = 0.057 lit mol-1 and b = 4.47 lit2 atm mol-2
    Argon has (TC = - 122°C, PC = 48 atm). What is the radius of the Argon atom?
    1.47 × 10-8 cm
Critical Constant Values of some Commonly used Substance
Gas TC (0K) PC (atm.) VC (c.c.)
Oxygen 154.28 49.7 74.4
Hydrogen 33.2 12.8 69.7
Nitrogen 125.97 33.5 90.0
Ammonia 405.5 111.3 72.0
304.15 72.9 94.2
Helium 5.2 2.25 61.55
Methane 190.3 45.6 98.8

Mar 8, 2019

Slater's Rules

    Slater proposed some set of empirical rules to calculate the screening constant (σ) of various electrons present in different orbitals of an atom or an ion. Once we get the value of screening constant it is easy enough to find the effective nuclear charge (Z*).

Screening Effect:

    The Valence electron in a multi - electron atom is attracted by the nucleus and repelled by the electrons of inner-shells.
Slater's rule practice problems
Screening Effect
    The combined effect of this attractive and repulsive force acting on the valence electron is that the Valence - electron experiences less attraction from the nucleus. This is known as the screening effect.

Slater's Rules to Calculate the Screening Constant (σ):

Rules for an electron in the nS, nP level:
  1. The first to do is to write out the Electronic Configuration of Elements of the atom or the ion in the following order and grouping.
  2. (1S) (2S, 2P) (3S, 3P) (3d) (4S, 4P) (4d) (4f) (5S, 5P) etc.
    It may be noted that so far as the screening effect is concerned the S and P electrons belonging to the same principal quantum shell have the same effect as advocated by Slater.
    Na atom:
    (1S)2(2S, 2P)8(3S)1
  3. An electron in a certain (nS, nP) level is screened only by electrons in the same level and by the electrons of lower energy level.Electrons lying above (nS, nP) level do not screen any (nS, nP) electron to any extent. Higher energy electrons have no screening effect on any lower energy electrons. Estimation of screening constant of the valence electron of the Sodium atom, (1S)2 (2S, 2P)8 (3S)1(The Valence electron will be excluded from our Calculation). Estimation of screening constant of the 2P electron of the Sodium atom, one 2P electron and 3S electrons will be excluded from our Calculation.
  4. Electrons of an (nS, nP) level shield a valence electrons in the same group by 0.35 each. This is also true for the electrons of the nd or nf that is for the electrons in the same group.
  5. Electrons belonging to one lower quantum shell, that is (n-1) shell shield the valence electrons by 0.85 each.
  6. Electrons belonging to (n-2) or still lower quantum shell shield the valence electron by 1.0 each. This means the screening effect is complete.
(a) For valence electron of Na atom:
Screening Constant (σ),
= (2 × 1) + (8 × 0.85) + (0 × 0.35) = 8.8Effective Nuclear Charge(Z),
= (11 - 8.8)

= 2.2
(b) For Na+ ion:
Electronic configuration of Na+ ion,
(1S)2(2S, 2P)8
Screening Constant (σ),
= (2 × 1) + (8 × 0.85) + (0 × 0.35)
= 8.8
Effective Nuclear Charge(Z),
= (11 - 8.8)
= 2.2
(c) For 2P electron of Na atom:
Screening Constant (σ),
= (2 × 0.85) + (7 × 0.35)
= 4.15
  • Evaluate the Z of Magnesium ion:
    Mg+2 : (1S)2(2S, 2P)8
    σ = (2 × 0.85) + (8 × 0.35)
    = 4.5
    Z = (12 - 4.5)
    = 7.5
  • Find out effective atomic number of Potassium ion:
    K+ : (1S)2 (2S, 2P)8 (3S, 3P)8
    σ = (2 × 1.0) + (8 × 0.85) + (8×0.35)
    = 11.6
    Z = (19 - 11.6)
    = 7.4
  • Estimate the σ and Z of the valence electron of the fluorine atom:
    F atom : (1S)2 (2S, 2P)7
    The valence electron has to be left out of our estimation.
    σ= (2×0.85)+(6×0.35)
    = 3.8
    Z= (9 - 3.8)
    = 5.2
  • Evaluate the Z of Fluoride ion:
    F - ion : (1S)2 (2S, 2P)8
    σ = (2×0.85)+(8×0.35)
    = 4.5
    Z⋆= (9 - 4.5)
    = 4.5

Rules for an electron in the nd, nf level:

    The above rules are quite well for estimating the screening constant of S and P orbitals. However when a d or f electron being shielded the (iv) and (v) rule replaced by a new rules for estimation of screening constant.
    The replaced rules are:
    All electrons below the nd or nf level contribute 1.0 each towards the screening constant.
  1. Estimate the screening constant for the outermost 4S electron of Vanadium:
    • Vanadium has atomic number 23 and the electronic configuration according to the Slater's Rule is:
      (1S)2 (2S, 2P)8 (3S, 3P)8 (3d)3 (4S)2
      We have consider only one electron of the two 4S electrons.
      Screening Constant (σ)
      = (2 ×1.0) + (8×1.0) + (8×0.85) + (3×0.85) + (1×0.35)
      = 19.7
  2. Screening Constant for a 3d electron of Vanadium:
      Electronic Configuration according to Slater's Rule is:
      (1S)2 (2S, 2P)8 (3S, 3P)8 (3d)3 (4S)2
      We have consider only two electron of the three 3d electrons.
      Screening Constant (σ)
      = (2 ×1.0) + (8×1.0) + (8×1.0) + (2 ×0.35)
      = 18.70
  3. Evaluate the effective nuclear charge of Vanadium(II) Ion:
      Vanadium(II) :
      (1S)2 (2S, 2P)8(3S, 3P)8(3d)3
      Screening Constant (σ):
      = (2 ×1.0) + (8×1.0+(8×1.0) + (3×0.35)= 18.70
      Z= (23 - 19.70)
      = 3.95
    In the first transition series electron filling up process begins in the 3d level below a filled 4S2 level. During ionisation process will a 4S electron or a 3d electron be lost first? Explain with reference to chromium.
    Chromium has atomic number 24 and the electronic configuration according to the Slater's Rule is:
    (1S)2 (2S, 2P)8 (3S, 3P)8 (3d)5 (4S)1
    ∴ Screening Constant (σ) for the 4S electron is:
    σ = (2×1.0) + (8×1.0) + (8×1.0) + (3×0.85) + (0×0.35)
    = 21.05
    Effective nuclear charge,
    Z = (24 - 21.05)
    = 2.95
    And the Screening Constant (σ) for the 3d electron is:
    σ = (2×1.0) + (8×1.0) + (8×1.0) + (4×0.35)
    = 19.4
    Effective nuclear charge,
    Z = (24 - 19.40)= 4.60
    Hence 3d electron is more tightly held than a 4S electron. So during ionisation the 4S electron will be lost.
    Calculate the effective nuclear charge of the hydrogen atom.
    Hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of single proton.
    Thus, σ= 0 and Z = 1.0 - 0 = 1.0
    Thus hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.
    Comment on the variation in effective nuclear charge for a 2P electron from carbon to oxygen.
    Electronic distribution according to the Slater's Rule is:
C (1S)2 (2S, 2P)4
N (1S)2 (2S, 2P)5
O (1S)2 (2S, 2P)6
    In carbon, 2P electron is screened by 1S2 2S2 2P1 electrons while in Nitrogen and Oxygen this is done by 1S2 2S2 2P2 and 1S2 2S2 2P3 electrons respectively.
    Thus Z for Nitrogen = Z for carbon + (1 nuclear charge) - Screening due to one 2P electron.
    = Z for Carbon + 1 - 0.35
    = Z for Carbon + 0.65
    and Z for Oxygen
    = Z for Nitrogen + 0.65
Thus effective nuclear charge will go up by the same amount from carbon to Nitrogen and then to Oxygen.

Mar 6, 2019

Soft and Hard Acids and Bases (SHAB) Principle

According to Lewis Concept an Acid-Base reaction involves an interaction of a vacant orbital of an acid (A) and a filled or unshared orbital of a base (B).
A + :B A : B 
Lewis Acid

Lewis Base

Adduct or Complex
The species A is called Lewis Acid or a generalized acid and B is called Lewis Base or a generalized base. A strong acid A and a strong base B, will form the stable complex A : B.
A concept known as Principle of Soft and Hard Acids and Bases(SHAB) Principle is very helpful in making a stability of the complex A : B.
According to this principle the complex A : B is most stable when A and B are either both soft or both hard.
The complex is least stable when one of the reactants (namely A and B) is very hard and the other one is very soft. 
In order to arrive at a comparative estimate of the donor properties of different bases, the preferences of a particular base to bind a proton H+ and methyl mercury (II) ion, [CH3HgB]+ was determined. Both the proton and methyl mercury cation can accommodate only one coordinate bond but the two cation vary widely in their preferences to bases. This preferences was estimated from the experimental determination of equilibrium constants for the exchange reactions:
BH+ + [CH3Hg(H2O)]+ [CH3HgB]+  + H3O+
The results indicate that bases in which the donor atom is N, O or F prefer to coordinate to the proton. Bases in which the donor atoms is P, S, I, Br, Cl or C prefer to coordinate to mercury.

Hard and Soft Bases:

The donor atoms in the first group have high Electronegativity, low Polaris ability and hard to oxidize. Such donors have been named ‘hard bases by Pearson, since they hold on to their electrons strongly. 
The donor atoms of second category are of low electronegativity, high polarisability, and are easy to oxidize. Such donors have been called ‘soft’ bases since they are hold on to their valence electrons rather loosely.
In simple terms hardness is associated with a tightly held electron shell with little tendency to polarise. On the other hand softness is associated with a loosely bound polarisable electron shell.
It will be seen that within a group of the periodic table softness of the Lewis bases increases with the increase in size of the donor atoms. 
Thus, among the halide ions softness increases in the order:
-Cl -Br -I -
Thus F - is the hardest and I - is the Softest base.
Definition of hard acid and soft acids and bases
Classification of Bases

Hard and Soft Acids:

After having gone through a classification of bases, a classification of Lewis acids is necessary. The preferences of a given Lewis acid towards ligands of different donor atoms is usually determined from the stability constant values of the respective complexes or from some other useful equilibrium constant measurements. When this is done, metal complexes with different donor atoms can be classified into two sets based on the sequences of their stabilities.
Hard acids have small acceptor atoms, are of high positive charge and do not contain unshared pair of electrons in their valence shell, although all these properties may not appear in one and the same acid. These properties lead to high electronegativity and low polarisability. In keeping with the naming of the bases, such acids are termed as 'Hard'Acids.
Hard Acids:

Soft acids have large acceptor atoms, are of low positive charge and contain ushered pairs of electrons in their valence shell. These properties lead to high polarisability and low electronegativity. Again in keeping with the naming of the bases, such acids are termed 'Soft' Acids
Soft Acids:
Difference between hard acid and soft acid and bases
Classification of Lewis Acids as Hard, Intermediate and Soft Acids
Classify the following as Hard and Soft Acids and Bases. 
(i) H- (ii) Ni+4 (iii) I+ (iv) H+
  1. The hydride ion has a negative charge and is far too large in size compared to the hydrogen atom. Its electronegativity is quite low and it will be highly polarisable by virtue of its large size. Hence it is Soft Base.
  2. Quadrivalent nickel has quite a high positive charge. Compared to bivalent nickel its size will be much smaller. Its electronegativity will be very high and polarisability will be low. Hence it is Hard Acid.
  3. Mono-positive iodine has a low positive charge and has a large size. It has a low electronegativity and a high polarisability. Hence it is a Soft Acids.
  4. H+ has the smallest size with a high positive charge density. It has no unshared pair of electrons in its valence shell. All these will give a high electronegativity and very low polarisability. Hence H+ is a Hard Acid.
Soft and Hard Acids and Bases (SHAB) Principle:
This principle also means that if there is a choice of reaction between an Acid and two Bases and two Acids and a Base,
A Hard Acid will prefer to combine with a Hard Base and a Soft Acid will prefer to combine with Soft Base and thus a more stable product will be obtained.
Hard Acid - Hard Base may interact by strong ionic forces. Hard Acids have small acceptor atoms and positive charge while the Hard Bases have small donor atoms but often with negative charge. Hence a strong ionic interaction will lead to Hard Acid - Base combination.
On other hand a Soft Acid - Soft Base combination mainly a covalent interaction. Soft Acids have large acceptor atoms, are of low positive charge and contain ushered pair of electrons in their valence shell.

Application of Soft and Hard Acids and Bases (SHAB) Principle:

  1. [CoF6]-3 is more stable than [CoI6]-3
  2. It will be seen that Co+3 is a hard acid
    F- is a hard base and I- is a soft base
    Hence [CoF6]-3 (Hard Acid + Hard Base) is more stable than [CoI6]-3 (Hard Acid + Soft Base).
  3. The existence of certain metal ores can also be rationalised by applying SHAB principle. Thus hard acids such as Mg+2, Ca+2 and Al+3 occur in nature as MgCO3, CaCO3 and Al2O3 and not as sulphides (MgS, CaS and Al2S3), since the anion CO3-2 and O-2 are hard bases and S-2 is a soft base.
  4. Soft acids such as Cu+, Ag+ and Hg+2 on the other hand occurs in nature as sulphides. The borderline acids such as Ni+2, Cu+2 and Pb+2 occur in nature both as carbonates and sulphides. 
    The combination of hard acids and hard bases occurs mainly through ionic bonding as in Mg(OH)2 and that of soft acids and soft bases occurs mainly by covalent bonding as in HgI2.
AgI2- is stable, but AgF2- does not exist. Explain.
We know that Ag+ is a soft acid, F- is hard base and I- is soft base.
Hence AgI2- (Soft Acid + Soft Base) is a stable complex and AgF2- (Soft Acid + Hard Base) does not exist.
Explains why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not?
In the case of Hg(OH)2 and HgS, Hg is a soft acid and OH- and S-2 is hard base and soft base respectively. Evidently HgS (Soft acid + Soft base) will be more stable than Hg(OH)2 (Soft Acid + Hard base). More stability of HgS than that of Hg(OH)₂ explain why Hg(OH)2 dissolved readily in Acidic Aqueous Solution but HgS does not.