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What is ionization energy?

    Ionization energy is the minimum energy required to remove an electron completely from the gaseous atomic species. If the source energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus of an atom.
    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 to produce cation is known as Ionization energy.
    M(g) + Ionization Energy → M⁺(g) + e
    It is generally represented as I or IP and it is measured in electron volt (eV) or kilocalories (kcal) per gram atom. One electron volt (eV) being the energy consumption 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 eV = 1.6 × 10⁻¹² erg

Energy consumption for removal of an electron

    We know that the energy corresponding to the transition from n = ∞ to n = 1 gives the energy consumption or ionization potential of the hydrogen atom.
Thus the ionization potential of the hydrogen atom,
EH = (2π²me⁴/h²)[(1/n₁²) - (1/n₂²)]
∴ EH = 2.179 × 10⁻¹¹ erg
= 2.179 × 10⁻¹⁸ Joule
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV
∴ EH = 13.6 eV

Energy consumption for removal second and third electron

  1. The electrons are removed in stages one by one from an atom. The energy consumption to remove the first electron from a gaseous atom is called its first Ionization potential.
    M (g) + IP₁ → M⁺ (g) + e
  2. The energy consumption 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 Ionization potentials.
    M⁺² (g) + IP₃ → M⁺³ (g) + e
    M⁺ (g) + IP₄ → M⁺⁴ (g) + e
    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 the 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

Ionization energy trend

Ionization potential trend in periodic table
Ionization potential trend

Ionization energy trend for atomic radius

    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 excited 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.

Ionization energy trend for the nuclear charge of an atom

    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 ionization 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.
    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 brings about a contraction in size. In effect, therefore, ionization potential steadily increases along a period.

Ionization energy trend for filled and half-filled orbital

    According to Hund's rule atoms having half-filled or completely filled orbital are comparatively more stable and hence more energy consumption to remove an electron from such atom.
    The ionization of such atoms is therefore relatively difficult than expected normally from their position in the periodic table. A few regulations that are seen in the increasing value of ionization energy along a period can be explained 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 a slightly higher value 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 (2S² 2P³) and 3P - orbital in P (3S² 3P³).

Ionization energy trend for shielding effect

    Electrons provide a shielding effect on the nucleus of an atom. The outermost electrons are shielded from the nucleus by the inner electrons.
    The radial distribution functions of the S, P, d orbitals show 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. shielding efficiency falls off in the order, S〉P〉d.
    As we move down a group, the number of inner - shells increases and hence the ionization potential tends to decreases.
Elements of II A Group: Be〉Mg〉Ca〉Sr〉Ba
    In the first transition series electron filling up processes begins in the 3d level below a filled 4S level. During the ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to chromium.
    During ionization, the 4S electron lost first.

Ionization energy trend for overall charge

    An increase 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 positively charged species is more difficult than from a neutral atom.
    The first ionization of the elements varies with their positions in the periodic table. In each of the tables, the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.

Ionization energy trend for valence electrons in an atom

    If an atom has valence electronic configuration is nS² nP⁶, that is the atom attains stable noble gas configuration.
    The removal of an electron from such atoms needed greater energy thus the ionization potential of such atoms also increases.

Trend of ionization energy along a period

Ionization energy trend in periodic table
Ionization energy trend along a period
    Arrange the following with the increasing order of ionization energy Li, Be, B, C, N, O, F, Ne.
    The greater the charge on the nucleus of an atom the more energy consumed for removing an electron from the atom.
    More energy consumed means increasing the ionization potential of an atom. With the increase in nuclear charge the electrostatic attraction between the outermost electrons and the nucleus increases. Thus removal of an electron from an atom is more difficult.
    The values of ionization energy generally increase in moving left to right in a period since the nuclear charge of an element also increases in the same direction.

different types of crystalline solids with examples
Types of crystalline solids

    Study crystalline solid in chemistry

    Solids are characterized by their definite shape and also their considerable mechanical strength and rigidity. The rigidity due to the absence of translatory motion of the structural units (atoms, ions, etc) of the solids. Here we study crystalline solids and amorphous solids.
    These properties are due to the existence of very strong forces of attraction amongst the molecules or ions. It is because of these strong forces that the structural units (atoms, ions, etc) of the solids do not possess any translatory motion but can only have the vibrational motion about their mean position.
    Liquids can be obtained by heating up to or beyond their melting points. In solids, molecules do not possess any translatory energy but posses only vibrational energy. The forces of attraction amongst them are very strong.
    The effect of heating is to impart sufficient energy to molecules so that they can overcome these strong forces of attraction. Thus solids are less compressible than liquids and denser than the liquid.
    Solids are generally classified into two broad categories: crystalline and amorphous substances.

Definition and examples of crystalline solids

    The solids which posses a definite structure, sharp melting point, and the constituents may be atoms, ions, molecules have order arrangement of the constituents extends in long-range order called crystalline solids.
    NaCl, KCl, Sugar, and Ice, quartz, etc are the examples of crystalline solids possess a sharp melting point. The pattern is such that having observed it in some small region of the crystal, it is possible to predict accurately the position of the particle in any region under observation.

Properties of crystalline

In the crystalline, the constituents may be atoms, ions, molecules.

  1. Crystalline solids are a sharp melting point, flat faces and sharp edges which is a well-developed form, are usually arranged symmetrically.
  2. Definite and the ordered arrangement of the constituents extends over a large distance in the crystal and called the long-range order.
  3. Crystalline solids (other than those belonging to the cubic class), on the other hand, are enantiotropic in nature. In this case, the magnitude of the property depends on the direction along which it is measured.

Amorphous solid definition and examples

    The solids which do not possess a definite structure, sharp melting point, and the constituents may be atoms ions, molecules do not have order arrangement of the constituents extends over a short-range the solids called amorphous solids.
    Amorphous solids such as glass, pitch, rubber, plastics, etc although possessing many characteristics of crystalline such as definite shape rigidity and hardness, do not have this ordered arrangement and melt gradually over a range of temperature.
    For this reason, they are not considered as solids but rather highly supercooled liquids.

Difference between crystalline and amorphous solids

  1. Crystalline solids possess definite structure, sharp melting point but amorphous solids that do not possess a definite structure, sharp melting point.
  2. Crystalline solids constituents(atoms, molecules) have order arrangement of the constituents extends over a long-range in solids but amorphous solids the constituents may be atoms, molecules do not have order arrangement.

Classification of crystalline

    On the basis of nature of force operating between constituent particles(atoms, ions, molecules) of matter, crystalline solids are classified into four categories,

Molecular crystalline solids

    Forces that hold the constituents of molecular crystals are of Van der Waals types. These are weaker forces because of which molecular crystals are soft and possess low melting points.
    Carbon dioxide, carbon tetrachloride, argon, and most of the organic compounds are examples of these types of crystals.
    This class further classified into three category
  1. Non-polar binding crystals
    The constituent particles of these types of crystalline solids are non-directional atom (H, He, etc.) or non-polar hydrogen (H₂), oxygen (O₂), chlorine (Cl₂), carbon dioxide (CO₂), methane (CH₄), etc.. The force operating between constituent particles (atoms or molecules) is a weak London force of attraction.
  2. Polar binding crystals
    The constituent particle of this type of crystalline solids is polar sulfur dioxide (SO₂), ammonia (NH₃), etc.  and the force operating between constituent particles is the dipole-dipole attraction force.
  3. Hydrogen-bonded crystals
    The constituent molecule of these types of crystalline solids is polar molecule and these molecules are bounded each other by hydrogen bonding. An example of this type of crystal is ice.

Ionic crystalline solids

    The forces involved here are of electrostatic forces of attraction. These are stronger than the non-directional type. Therefore ionic crystals strong and likely to be brittle.
    They have little electricity with high melting and boiling point and can not be bent. The melting point of the ionic crystal increases with the decreasing size of the constituent particles.
    In ionic crystals, some of the atoms may be held together by covalent bonds to form ions having a definite position and orientation in the crystal lattice. CaCO₃ is an example of these types of crystalline solids.

Covalent crystalline solids

    The forces involved here are chemical nature (covalent bonds) extended in three dimensions. They are strong and consequently, the crystals are strong and hard with high melting points. Diamond, graphite, silicon, etc. are examples of these types of crystalline solid.

Metallic crystalline solids

    Electrons are held loosely in these types of crystals. Therefore they are good conductors of electricity. Metallic crystalline solids can be bent and are also strong.
    Since the forces have non-directional characteristics the arrangement ao atoms frequently correspond to the closet packing of the sphere.

Isotopic forms of the carbon atom

    Carbon has several crystalline isotropic forms only two of them are common diamond and graphite. There are four other rare and poorly understood allotropes, β-graphite, Lonsdaleite or hexagonal diamond, Chaoite (a very rare mineral) and carbon VI.
    The last two forms appear to contain -C≡C-C≡C- and are closer to the diamond in their properties. 

Structure of graphite

    The various amorphous forms of carbon like carbon black, soot, etc. are all microcrystalline forms of graphite.
    Graphite consists of a layer structure in each layer the C-atoms are arranged in hexagonal planner arrangement with SP² hybridized with three sigma bonds to three neighbors and one π-bonds to one neighbor.
    The resonance between structures having an alternative mode of π bonding makes all C-C bonds equal, 114.5 pm equal, consistent with a bond order of 1.33. 
    The π electrons are responsible for the electrical conductivity of graphite. Successive layers of Carbon-atoms are held by weak van der Waals forces at the separation of 335pm and can easily slide over one another.

Structure of diamond

    In diamond, each SP³ hybridized carbon is tetrahedrally surrounded by four other carbon atoms with C-C bond distance 154 pm. These tetrahedral belong to the cubic unit cell.
    Natural diamond commonly contains traces of nitrogen or sometimes very rarely through traces of al in blue diamonds.

How much is Van't Hoff equation - effect on temperature?

    Van't Hoff equation proposed chemical equilibrium of a reaction is constant at a given temperature. Equilibrium values can be changed with the change of temperature.
    The quantitative relation, known as Van't Hoff equation is derived by using the Gibbs - Helmholtz equation.
    The Gibbs - Helmholtz equation is,
ΔG⁰ = ΔH⁰ + T[d(ΔG⁰)/dT]p
Zero superscripts are indicating stranded values.

or,- (ΔH⁰/T² )= -(ΔG⁰/T² )+(1/T)[d(ΔG⁰)/dT]p
or, - (ΔH⁰/T² ) = [d/dT(ΔG⁰/T)]p

Van't Hoff isotherm is,

- RT lnKp = ΔG⁰

or, - R lnKp = ΔG⁰/T

Differentiating with respect to temperature at constant pressure,
- R [dlnKp/dT]p = [d/dT(ΔG⁰/T)]p

Comparing the above two-equation,

dlnKp/dT = ΔH⁰/T²
    This is the differential form of Van't Hoff equation.

Integrated form of van't Hoff equation

    The greater the value of standard enthalpy of a reaction(ΔH⁰), the faster the equilibrium constant (Kp) changes with temperature (T).
Separating the variables and integrating,
∫ dlnKp = (ΔH⁰/R)∫ (dT/T²)

Assuming that ΔH⁰ is independent of temperature.
or, lnKp = - (ΔH⁰/R)(1/T)+C
where C is integrating constant.

The integration constant can be evaluated and identified as ΔS⁰/R using the relation,
ΔG⁰ = ΔH⁰ - TΔS⁰

Van't hoff Equation is,

lnKp = - (ΔH⁰/RT) + (ΔS⁰/R).

    The heat of a reaction 2A + B ⇆ 2C, ΔG⁰(500 K) = 2 KJ mol⁻¹ find the Kp at 500 K for the reaction A + ½B ⇆ C.
ΔG⁰(500 K) for the reaction,
A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.
The relation is ΔG⁰ = - RT lnKp
or, 1 = - (8.31 × 10-3 ) × (500) lnKP
Thus, lnKP = 1/(8.31 × 0.5)
or, lnKP = 0.2406.

∴ KP = 1.27.

Exothermic and endothermic reaction

  • Exothermic reaction, ΔH⁰ = (-) ve.
    Formation of ammonia from hydrogen and nitrogen.
N₂ + H₂ ⇆ 2NH₃ ΔH⁰ = (-) ve
  • Endothermic reaction, ΔH⁰ = (+) ve.
    Dissociation of hydrogen iodide into hydrogen and Iodine.
HI ⇆ H₂ + I₂ ΔH⁰ = (+) ve
  • For the reaction, ΔH⁰ = 0.
    lnKp is independent of temperature. Provided ΔS⁰ does not change much with temperature.
Van't Hoff equation and equilibrium point
Van't Hoff equation

Enthalpy change of reaction

    For the ideal system, H is not a function of P and standard enthalpy change (ΔH⁰) = enthalpy change (ΔH) and Van't Hoff equation is,
dlnKp/dt = ΔH/RT²

The integrated equation is,
lnKp = (ΔH/RT) + ΔS/R
    However, S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS⁰ and ΔG⁰.
    If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(Kp₂/Kp₁) = (ΔH/R){(T₂ - T₁)/T₁T₂)}
    Where Kp₁ and Kp₂ are the equilibrium constants of the reaction at two different temperatures T₁ and T₂ respectively.
    Thus the determination of Kp₁ and Kp₂ at two temperatures helps to calculate the value of change of enthalpy of the reaction.
    The above relation is called Van't Hoff reaction isobar since pressure remains constant during the change of temperature.
Van't Hoff reaction isotherm is,
ΔG = - RT lnKa + RT lnQa.
When the reaction attains equilibrium point,
Qa = Ka.

∴ ΔG = 0

Assumptions from Van't Hoff equation

  1. The reacting system is assumed to behave ideally.
  2. ΔH is taken independent of temperature for a small range of temperature change.
    Due to the assumption involved ΔH and ΔU do not produce the precise value of these reactions. As ΔG is independent of pressure for ideal system ΔH and ΔU also independent of pressure.
ΔG⁰ = - RT lnKp
or, [dlnKp/dT]T = 0.
    Use Gibbs - Helmholtz equation to derive the Van't Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?
Again Kp = Kc (RT)ΔƔ
or, lnKp = lnKc + ΔƔ lnR + ΔƔ lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + ΔƔ/T.
But dlnKp/dT = ΔH⁰/RT²
Hence, dlnKc/dT = ΔH⁰/RT² - ΔƔ/T
or, dlnKc/dT = (ΔH⁰ - ΔƔRT)/RT²

∴ dlnKc/dT = ΔU⁰/RT²
    ΔU⁰ is the standers heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.
    The integrated form of the equation at two temperature is,
ln(KC₂/KC₁) = (ΔU⁰/R){(T₂ - T₁)/T₁T₂)}
    For ideal system ΔU⁰ = ΔU. Since the reaction occurs at constant volume in the equilibrium point is called Van't Hoff reaction isochore.

Heat of a reaction and Le-Chatelier principal

    Van't Hoff equations give a quantitative expression of the Le-Chatelier principle.
lnKp = - (ΔH/R)(1/T) + C.
    It is evident that for endothermic reaction (ΔH〉0), an increase in temperature increases the value of any of the reactions.
    But for exothermic reaction (ΔHㄑ 0), with rising in temperature, a is decreased. This change of Kp also provides the calculation of the quantitative change of equilibrium point yield of products.
    This above statement is in accordance with the Le - Chatelier Principle. Whenever stress is placed on any system in a state of the equilibrium point, the system always reacts in a direction to reduce the applied stress.
    When the temperature is increased, the system at equilibrium point will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.

Zero-order kinetics, the rate of these reactions does not depend on the concentration of the reactants.

zero-order kinetics

    Let us take a reaction represented as
    A → Product
    Let the initial concentration of the reactant a and product is zero. After the time interval t, the concentration of the reactant is (a-x) and the concentration of the product is x. Thus x is decreased concentration in zero-order reaction.

Zero-order kinetics in terms of product

    Thus the mathematical equation of zero-order kinetics in terms of product,
dx/dt = k₀
Where k₀ is the rate constant of the zero-order reaction.
or, dx = k₀dt

Integrating the above reaction,
∫dx = k₀ ∫dt
or, x = k₀t + c
where c is the integration constant of the reaction.

When t = o, x is also zero thus, C = o Thus the above equation is,
x = k₀ t
    This is the relationship between decreases of concentration of the reactant(x) within time(t).

Zero-order kinetics in terms of reactant

Rate equation in terms of reactant,
-d[A]/dt = k₀ [A]⁰ = k₀
Where [A] is the concentration of the reactant at the time t.
or, - d[A] = k₀dt
Integrating the above equation,
We have - ∫d[A] = k₀ ∫ dt
or, - [A] = k₀t + c
where c is the integration constant of the reaction.
If initial at the time t = 0 concentration of the reactant [A]₀ Then from the above equation,
- [A]₀ = 0 + c
or, c = -[A]₀
    Putting the value on the above equation,
- [A] = kt - [A]₀
    This is another form of the rate equation in zero-order kinetics.

The half-life of zero-order kinetics

    The time required for half of the reaction to be completed is known as the half-life of the zero-order reaction. It means 50% of reactants disappear in that time interval.

Half-life in zero-order kinetics

    If in a chemical reaction initial concentration is [A]₀ and after t time interval the concentration of the reactant is [A].
    Then, [A]₀ - [A] = kt
    Thus when t = t½, that is the half-life of the reaction, the concentration of the reactant [A] = [A]₀/2. Putting the value on the above equation,
We have [A]₀ - [A]₀/2 = k t½
or, k t½ = [A]₀/2
t½ = [A]₀/2k
    Thus for the zero-order kinetics the half-life of the reaction proportional to its initial concentration.

Zero-order kinetics reactions examples

    The only heterogeneous catalyzed reactions may have zero-order kinetics.
Zero-order kinetics reaction
Zero-order kinetics reaction

Characteristics of zero-order kinetics

  1. The rate of the reaction is independent of concentration.
  2. Half-life is proportional to the initial concentration of the reactant.
  3. The rate of the reaction is always equal to the rate constant of the reaction at all concentrations.

Unit of the rate constant in zero-order kinetics

    The rate equation in terms of product for the nth-order reaction is,
d[A]/dt = k [A]n
or, k = (d[A]/dt) × (1/[A]n)
    Thus the unit of rate constant(k) = (unit of concentration)/{unit of time × (unit of concentration)n}
    = (unit of concentration)1-n/unit of time
    Thus if zero-order kinetics the concentration is expressed in lit mole⁻¹ and time in sec
Then the rate constant = (lit mol⁻¹)/sec
= mol lit⁻¹sec⁻¹

Questions and answers of zero-order kinetics

    The rate constant of a chemical reaction is 5 × 10⁻⁸ mol lit⁻¹sec⁻¹. What is the order of this reaction? How long does it take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹?
    The reaction is a zero-order reaction and 3.92 × 10⁵ sec takes to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹.
    The half-life of a zero-order reaction is x and the reaction is completed on t₁ time. What is the relation between x and t₁?
    2x = t₁
    If the rate of the reaction is equal to the rate constant. What is the order of the reaction?
    Zero-order reaction.
    For a reaction, N₂ + 3 H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴mol lit⁻¹sec⁻¹, What is the order and the value of - d[H₂]/dt of this reaction?
    Zero-order reaction and the value of d[H₂]/dt = 3 × 10⁻⁴ mol lit⁻¹sec⁻¹.
    For the reaction N₂O₅ → 2NO₂ + ½ O₂, the rate of disappearance of N₂O₅ is 6.25 × 10⁻³ mol lit⁻¹sec⁻¹, what is the rate of formation of NO₂ and O₂ respectively?
    1.25 × 10⁻² and 3.125 × 10⁻³ mol lit⁻¹sec⁻¹
    For the reaction H₂ + Cl₂ → 2HCl on sunlight and taking place on the water. What is the order of the reaction?
    This is a zero-order reaction.

Stability of soft and hard acids bases complexes

Soft and hard acids and bases(SHAB) principles are very helpful in making the stability of the complex A: B.
According to soft and hard acids and base 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 basic properties, 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 cations vary widely in their preferences to bases. This preference 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 nitrogen, oxygen or fluorine prefer to coordinate with the proton. Bases in which the donor atoms are phosphorus, sulfur, iodine, bromine, chlorine or carbon prefer to coordinate with mercury.

Neutralization of acids and bases

Lewis defined, an acid-base neutralization 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

The species A is called Lewis acid or a generalized acid and B is called Lewis base or a generalized base. Strong acid and a strong base B will form the stable complex A: B.

What are soft and hard bases?

The donor atoms of the second category are of low electronegativity, high polarizability, and are easy to oxidize. Such donors have been called ‘soft bases' since they are holding on to their valence electrons rather loosely.

The donor atoms in the first group have high electronegativity, low polarisability and hard to oxidize. Such donors have been named ‘hard bases' by Pearson since they hold on to their electrons strongly.

Classify the following as soft and hard 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 a soft base.
  2. The quadrivalent nickel has quite a high positive charge. Compared to the bivalent nickel its size will be much smaller. Its electronegativity will be very high and polarisability will 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 high polarizability.
  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 polarizability. Hence H⁺ is a hard acid.

Properties of soft and hard bases

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 the size of the donor atoms. Thus, among the halide ions, softness increases in the order.


Thus F - is the hardest and I - is the softest base.
Soft and hard acids and bases
Soft and hard bases

What are soft and hard 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 are 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 stability.

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 polarizability. In keeping with the naming of the bases, such acids are termed as 'hard acids'.

N≫P; O≫S; F〉Cl〉Br〉I

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 polarizability and low electronegative. Again in keeping with the naming of the bases, such acids are termed 'Soft acids'.

N≫P; O≫S; F〉Cl〉Br〉I
Soft and hard acids and bases
Soft and hard acids

SHAB principle for acids and bases

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.

The hard acid - hard base may interact with strong ionic forces. Hard acids have small acceptor atoms and positive charge while the hard bases have small-donor atoms but often with a negative charge. Hence a strong ionic interaction will lead to the hard acid-base combination.

On the 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 pairs of electrons in their valence shell.

Application of SHAB principle

The SHAB concept is extremely useful in elucidating many properties of chemical elements and will be often referred to at appropriate places.
  1. BF₃ and BH₃ the boron is trivalent but quite different in behavior is noted.
    The presence of hard fluoride ions in BF₃ makes it easy to add other hard bases.
    The presence of soft hydride ions BH₃ makes it easy to add other soft bases.
  2. [CoF₆]⁻³ is more stable than [CoI₆]⁻³. 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 ores can also be rationalized by applying the 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 sulfides (MgS, CaS, and Al₂S₃) since the anion CO₃⁻² and O⁻² are hard bases and S⁻² is a soft base. Soft acids such as Cu⁺, Ag⁺ and Hg⁺², on the other hand, occur in nature as sulfides.
  4. The borderline acids such as Ni⁺², Cu⁺², and Pb⁺² occur in nature both as carbonates and sulfides. 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.

Examples of application of SHAB principle

AgI2- is stable, but AgF2- does not exist. Explain.

We know that Ag⁺ is a soft acid, F⁻ is hard to base and I⁻ is the 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 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 to 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 explains why Hg(OH)₂ dissolved readily in acidic solution but HgS does not.

What are lanthanides and actinides?

The f block elements appear in two series characterized by the filling of 4f and 5f orbitals in the respective third inner principal quantum energy level from outermost.

The 4f series contains fourteen elements cerium to lutetium with the atomic number from 58 to 71 and are called Lanthanides as they appear after lanthanum.

The 5f series contains fourteen elements thorium to lawrencium with the atomic number from 90 to 103 and are called actinides as they appear after actinium.

4f-block or Lanthanides

The 4f block elements have been variously called rare earth, lanthanoids, and lanthanum. The lanthanide atoms and their trivalent ions have the following general electronic configuration.
Lanthanide atoms
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
where n has values 1 to 14
Lanthanide(M⁺³) ions
[Pd] 4fn 5S² 5P⁶
where n has values 1 to 14
f block elements electronic configuration
4f-block elements
In 4f-block inner transition elements with increasing atomic number electrons are added to the deep-seated 4f orbitals. The outer electronic configuration of 4f-elements is 6S² and inner orbitals contain f -electrons.

Cerium, Gadolinium, and Lutetium

Only Cerium, Gadolinium, and Lutetium contain one electron in 5d orbital and the electronic configuration of the following elements are
[Pd] 4f¹ 5S² 5P⁶ 5d¹ 6S²

[Pd] 4f⁷ 5S² 5P⁶ 5d¹ 6S²

[Pd] 4f¹⁴ 5S² 5P⁶ 5d¹ 6S²
Electrons of similar spin developed an exchange interaction which leads to the stabilization of the system. For the electrons of similar spin, repulsion is less by an amount called exchange energy.

The greater the number of electrons with parallel spins the greater is exchanged interaction and the greater is the stability. This the basis of Hund's rules of maximum spin municipality.

For the f subshell, maximum stability will result if there are seven electrons with parallel spins in the seven f orbitals, each orbitals having one when the f subshell is half-filled.

Thus Gadolinium atom contains one electron in 5d orbital. 4f and 5d are very close in terms of energy levels. In such a case, the half-filled orbital is slightly more stable than orbital with one additional electron by increasing exchange energy. Thus when we add the next electron in half-filled 4f-orbital it may land on 5d-orbital.

Ytterbium to Lutetium

In the Lutetium atom, the maximum capacity of electrons in 4f orbitals is 14. Thus when we come from Ytterbium to Lutetium the next electron land on 5d orbital.

Why is the +3 oxidation state so common and stable in lanthanides?


The nature of lanthanoids elements is such that three electrons are removed comparatively easy to give the normal trivalent state.

The ground state electronic configuration of the natural lanthanoids atoms is
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
The electronic configuration of trivalent ions with the reducing three electrons is
[Pd] 4fn 5S² 5P⁶
where n is 0 to 14 from Lanthanum to Lutetium.

As a consequence, the f electrons can not participate in the chemical reactions thus +3 oxidation state is common and stable in lanthanides.


Why does praseodymium possess electronic configuration 4f³ 6s² instead of the expected one 4f² 5d¹ 6s²?


This can be explained by (n+l) rules, the orbital which has a higher value of (n+l) is the higher energy orbitals.

4f-orbital, (n+l) = 4+3 = 7
5d-orbital, (n+l) = 5+7 =7

Thus for the above case which has the highest number of principal quantum number n is higher energy orbital. Thus 5d orbital is the higher energy orbitals.

Again electrons are fed into orbitals in order of increasing energy until all the electrons have been accommodated.

Thus praseodymium posses electronic configuration 4f³ 6s².

5f-block actinides atoms

The 5f-block elements from thorium to lawrencium from the second series of inner transition elements are called man-made elements or Actinides. The Actinides atoms and their trivalent ions have the following general electronic configuration.
Actinides atoms
[Rn] 4fn 5d1-2 6S²
where n has values 1 to 14
Actinides(M⁺³) ions
[Rn] 4fn
where n has values 1 to 14
f block elements electronic configuration
5f-block elements

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