## 2019

### 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 = (2Ï€²me⁴/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
Question
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 = (2Ï€²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
= Z² × IPH
∴ Second Ionization potential of Helium, = 22 × 13.6
= 54.4 eV

### Ionization energy trend

 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
Question
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 along a period
Question
Arrange the following with the increasing order of ionization energy Li, Be, B, C, N, O, F, Ne.
Liã„‘Bã„‘Beã„‘Cã„‘Oã„‘N〈Fã„‘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.

### Crystalline solids and amorphous solids

 Crystalline solids

### What is solid-state 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: crystals and amorphous substances.

### Crystalline solid definition

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.

#### Examples of 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 solids

In the crystalline solid 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. Crystalline solids are 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

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 solids

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.
CO₂, CCl₄, Ar, and most of the organic compounds are examples of these types of crystals.
This class further classified into three category
1. Non-polar molecular
The constituent particles of these types of crystalline solids are non-directional atom(H, He, etc.) or non-polar molecules(H₂, O₂, Cl₂, CO₂, CH₄, etc.). And the force operating between constituent particles(atoms or molecules) is a weak London force of attraction.
2. Polar molecular
The constituent particle of this type of crystalline solids is polar molecules(SO₂, NH₃, etc.)  and the force operating between constituent particles is the dipole-dipole attraction force.
3. H-bonded molecular
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, Î²-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 Ï€-bonds to one neighbor.
The resonance between structures having an alternative mode of Ï€ bonding makes all C-C bonds equal, 114.5 pm equal, consistent with a bond order of 1.33.
The Ï€ electrons are responsible for the electrical conductivity of graphite. Successive layers of C-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.
An extremely rare Lonsdalete allotrope found in certain meteorites, the tetrahedral units are stacked to form a hexagonal wurtzite types lattice.

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

Van't Hoff equation proposed equilibrium constant Kp of a reaction is constant at a given temperature. Equilibrium constant values can be changed with the change of temperature.
The quantitative relation, relation, known as Van't Hoff equation is derived by using the Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation is,
Î”G⁰ = Î”H⁰ + T[d(Î”G⁰)/dT]p
Zero superscripts are indicating stranded values

or,- (Î”H⁰/T² )= -(Î”G⁰/T² )+(1/T)[d(Î”G⁰)/dT]p
or, - (Î”H⁰/T² ) = [d/dT(Î”G⁰/T)]p

Van't Hoff isotherm is,

- RT lnKp = Î”G⁰

or, - R lnKp = Î”G⁰/T

Differentiating with respect to T at constant P,
- R [dlnKp/dT]p = [d/dT(Î”G⁰/T)]p

Comparing the above two equation we have,

dlnKp/dT = Î”H⁰/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(Î”H⁰), the faster the equilibrium constant (Kp) changes with temperature (T).
Separating the variables and integrating,
∫ dlnKp = (Î”H⁰/R)∫ (dT/T²)

Assuming that Î”H⁰ is independent of temperature.
or, lnKp = - (Î”H⁰/R)(1/T)+C
where C is integrating constant

The integration constant can be evaluated and identified as Î”S⁰/R using the relation,
Î”G⁰ = Î”H⁰ - TÎ”S⁰

Thus the Van't hoff Equation is,

lnKp = - (Î”H⁰/RT) + (Î”S⁰/R)

Problem
The heat of a reaction 2A + B ⇆ 2C, Î”G⁰(500 K) = 2 KJ mol⁻¹ find the Kp at 500 K for the reaction A + ½B ⇆ C.
Solution
Î”G⁰(500 K) for the reaction,
A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.
The relation is Î”G⁰ = - 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 and Van't Hoff equation

• For exothermic reaction, Î”H⁰ = (-) ve.
Examples are the formation of ammonia from hydrogen and nitrogen.
 N₂ + H₂ ⇆ 2NH₃ Î”H⁰ = (-) ve
• For endothermic reaction, Î”H⁰ = (+) ve.
Examples are the dissociation of hydrogen iodide into hydrogen and Iodine.
 HI ⇆ H₂ + I₂ Î”H⁰ = (+) ve
• For the reaction, Î”H⁰ = 0.
lnKp is independent of T. Provided Î”S⁰ does not change much with T.
 Van't Hoff equation

#### Enthalpy change of reaction from Van't Hoff equation

For the ideal system, H is not a function of P and standard enthalpy change (Î”H⁰) = enthalpy change (Î”H) and Van't Hoff equation is,
dlnKp/dt = Î”H/RT²

The integrated equation is,
lnKp = (Î”H/RT) + Î”S/R
However, S of an ideal gas depends strongly on P, so Î”S and Î”G per mole of reaction in the mixture differ quite substantially from Î”S⁰ and Î”G⁰.
If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(Kp₂/Kp₁) = (Î”H/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 P remains constant during the change of temperature.
Problem
Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.
Solution
Van't Hoff reaction isotherm is,
Î”G = - RT lnKa + RT lnQa.
But when the reaction attains an equilibrium,
Qa = Ka

∴ Î”G = 0

#### Assumptions from Van't Hoff equation

1. The reacting system is assumed to behave ideally.
2. Î”H is taken independent of temperature for a small range of temperature change.
Due to the assumption involved Î”H and Î”U do not produce the precise value of these reactions. As Î”G is independent of pressure for ideal system Î”H and Î”U also independent of pressure.
Î”G⁰ = - RT lnKp
or, [dlnKp/dT]T = 0
Problem
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?
Solution
Again Kp = Kc (RT)Î”Æ”
or, lnKp = lnKc + Î”Æ” lnR + Î”Æ” lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + Î”Æ”/T.
But dlnKp/dT = Î”H⁰/RT²
Hence, dlnKc/dT = Î”H⁰/RT² - Î”Æ”/T
or, dlnKc/dT = (Î”H⁰ - Î”Æ”RT)/RT²

∴ dlnKc/dT = Î”U⁰/RT²
Î”U⁰ 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) = (Î”U⁰/R){(T₂ - T₁)/T₁T₂)}
For ideal system Î”U⁰ = Î”U. Since the reaction occurs at constant volume in equilibrium is called Van't Hoff reaction isochore.

### Heat of a reaction from Van't Hoff equation

Van't Hoff equations give a quantitative expression of the Le-Chatelier Principle.
lnKp = - (Î”H/R)(1/T) + C
It is evident that for endothermic reaction (Î”H〉0), an increase of T increases the value of any of the reactions.
But for exothermic reaction (Î”Hã„‘ 0), with rising in temperature, a is decreased. This change of Kp also provides the calculation of the quantitative change of equilibrium 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 equilibrium, the system always reacts in a direction to reduce the applied stress.
Thus when the temperature is increased, the system at equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.
More Study chemistry

### Zero order kinetics

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

### 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

Question
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.
Questions
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
Question
If the rate of the reaction is equal to the rate constant. What is the order of the reaction?
Zero-order reaction.
Question
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.
Question
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
Question
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

According to Lewis's concept, 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 Adduct

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.

#### 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.

F⁻ã„‘Cl⁻ã„‘Br⁻ã„‘I⁻

Thus F - is the hardest and I - is the softest base.
 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

#### 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.

Question
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.

#### 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 metal 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

Question
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.

Question
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 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
 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.

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

[Pd] 4f⁷ 5S² 5P⁶ 5d¹ 6S²

Lutetium
[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.
Question

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.

Question

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
 5f-block elements

### What are the atomic orbitals?

A wave function represents an electron is the product of two parts, a radial part, and an angular part.
The square of the radial part of the wave function indicates the probability of finding the electron at any distance r from the nucleus.
The square of the angular part of the wave function gives the probability of finding in a particular direction from the nucleus.
The radial dependence and angular dependence of wave function taken together, tell us that a three-dimensional standing electron wave (orbital) can be a picture to have size, shape, and orientation of an orbital.
In order to describe the size, shape, orientation of an orbital four quantum numbers are necessary. These quantum numbers are designed as,
1. Principal quantum number
2. Azimuthal quantum number
3. Magnetic quantum number
4. Spin quantum number

#### Principal quantum number of an atom

The principal quantum number(n) is of primary importance in determining the size and hence the energy of an electron.
Hydrogen, the energy is fixed by the value of the principal quantum number(n). In other multi-electron atoms, the energy of each electron depends on the value of the principal quantum number of the electron.
As the value of the principal quantum number(n) increases the radius (the and electron separation increases that are the size of the orbital increases).
The energy also raised, principal quantum numbers(n) is always an integer and can assume the value,1,2,3,4.... but not zero.

#### Azimuthal quantum number of an atom

The general geometric shapes of an electron wave or orbital are described by the azimuthal quantum number. This quantum number related to the electron in that state.
Thus, l = 0, 1, 2, 3.....(n-1)
Therefore an electron having principal quantum number n, assumed the values l is 0 to (n-1).

#### Magnetic quantum number of an atom

The magnetic quantum number associated with the orientation of the electron wave with respect to a given direction, usually that of a strong magnetic field. This quantum number hasn't an effect on the shape of orbital or on the energy of an electron.
For a given value of l, ml can have any integral value between +1 to -1.

∴ ml = + l, (l - 1), (l - 2), (l - 3) ..... 0 ..... - 1, - 2, - l

#### Spin quantum number of an atom

Besides the three quantum numbers, it also has a fourth quantum number namely spin quantum number (s), which was necessary to completely describe an electron in a particular shell.
The electron itself regarded as a small magnet. A beam of a hydrogen atom can be split into two beams by a strong magnetic field. This indicates that there are two kinds of the hydrogen atom in which can be differentiated on the basis of their behavior in a magnetic field.
It has been postulated that each electron spin around its axis like a lope and they behave like a magnet. A spinning electron can have only two possibilities.
The electron can either spin clockwise or counterclockwise. The two directions of spin represent as(↑↓).
This four quantum number s = 1/2 is independent of the other three quantum numbers.
Two directions of spin are represented as (↑↓) can have two possible ms values +1/2 and -1/2 depending on the direction of rotation of the electron about its axis.

#### How to find the quantum numbers of orbitals?

 Quantum energy levels of an atom
Question
What are the four quantum numbers of the 19th electron of chromium?
1S² 2S² 2P⁶ 3S² 3P⁶ 4S¹ 3d⁵
19th electron means 4S¹ electron. Thus for 4S electron n =4, and l =0 and for S¹ electron m = 0 and s = +½.
∴ The four quantum numbers of the 19th electron of chromium are

4, 0, 0, +½
Question
Write the correct set of four quantum numbers for the valence electron of rubidium.
The correct set of four quantum numbers for the valence electron of rubidium atom are,
5, 0, 0, +½
Question
How many electrons in an atom can have the following quantum numbers n=4 and l=1?
6 electrons in an atom can have the following quantum numbers n =4 and l =1.
Question
How many possible orbitals are there for n = 4?
Thus the number of possible orbitals when

n = 4 is [1(4S) + 3(4P) + 5(4d) + 7(4f)] = 16
Question
How many possible orbitals are there for n = 3, l =1, and ml =0 ?
Thus the number of orbitals is 1, 3S orbital.

### What are atomic orbitals? and S, P, d subshell

An orbital is a region in space where there is a high probability of finding an electron.
The S orbitals penetrate the nucleus most while the P and d subshell cannot penetrate the nucleus.
This means that S orbital electron can efficiently screen the nuclear charge from other electrons compared to the other P and d electrons.
The wave function of the electron in an atom is called orbital. The wave function is plotted against distance and space in three dimensional marked by a curve will give the shape of the orbitals.
The probability of finding an electron in space around the nucleus involves two aspects, radial probability, and angular probability.
It is not possible to represent completely in one diagram on paper the directional properties of electron orbitals. An angular probability distribution must be combined mentally to have an overall shape of the orbital.

#### Orbital electron distribution in s sublevel

The angular probability distribution is greater interest and importance, an S orbital electron has no angular dependence because the relevant wave function is independent of angles Î¸ and Î¦.
There is, therefore, an equal chance of discovering the electron in any direction of the nucleus.
 S subshell of an atom
With the nucleus at the origin of the Cartesian axes, the sphere of the radius(r) represents the probability of finding the electron in S orbital.
An S orbital electron has a spherically symmetrical probability distribution.

#### Shape of P subshell of an atom

The P orbitals are three orientations is represents as, Px, Py, and Pz. The orientations of the orbital plane correspond to ml = 1, 0, -1 is mutually at the right angles.
So the orbitals designated Px, Py and Pz are mutually perpendicular and they are concentrated along the respective coordinate axis X, Y and Z. Unlike the S orbitals, the angular part of the P wave function is dependent on Î¸ and Î¦.
 P subshell of an atom

#### Shape of d-subshell of an atom

These orbital arises when n =3 (M - shell), that is the orbitals start with the 3rd main energy level.
When l = 2(d - orbital),
m = -2, -1, 0, 2, 1
indicating that d - orbitals have five orientation, that is,

dxy, dxz, dyz, dx² - y² and dz²
All these five d - orbitals, in the absence of a magnetic field, are equivalent in energy and are, therefore, said to be five-fold degenerate.
 d subshell of an atom

### Concept of pH and pOH

 pH scale
The concept of pH and pOH plays a key role in an acid-base neutralization reaction. In an aqueous medium, it is usually represented as H₃O⁺.

A naked hydrogen ion has a vanishingly small size ( radius ~10⁻¹³ cm =10⁻¹⁵ Ã…) and therefore has a very high (charge/radius) ratio(~10⁵). It is expected to be the most effective in polarizing other ions or molecules according to Fajan's rules.

In HO₃⁺ there are assumed to coordinate bonds from water oxygen to the proton, thus giving the proton a helium electronic configuration.

### Evidence of pH concept

Perchloric acid (HClO₄) reacts vigorously with water and It gives a series of hydrates which are :
HClO₄ (112°C)

HClO₄, H₂O (+ 50°C)

HClO₄, 2H₂O (- 17.8°C)

HClO₄, 3H₂O (- 37°C )

HClO₄, 3.5H₂O (- 41.4°C)

Of these hydrates the most remarkable is the monohydrate, melting at the much higher temperature than the covalent anhydrous acid. It is very stable and can be heated to around 100⁰C without decomposition.

The monohydrate is about ten times vicious as the anhydrous acid. It has the same crystal lattice as the ionic ammonium perchlorate, showing that it is an ionic compound, [H₃O⁺][ClO₄].

### Water as an acid and as a base

We know that water dissociates weakly to H⁺ and OH⁻ ions. Regardless of what other ions are present in water, there will always be an equilibrium between H⁺ and OH⁻ ions.

H₂O ⇄ H⁺ + OH⁻

The proton, however, will be solvated and is usually written as [H₃O⁺]. For simplicity, we will write H⁺ only. The above equilibrium will have its own equilibrium constant:

K = ([H⁺] × [OH⁻])/[H₂O]
or, K × [H₂O] = [H⁺] × [OH⁻]

The square brackets indicate concentrations. Recognizing the fact that in any dilute aqueous solution, the concentration of water molecules (55.5 moles/liter) greatly exceeds that of any other ion, [H₂O] can be taken as a constant. Hence,

K×[H₂O] = Kw = [H⁺]×[OH⁻]

Where Kw is the dissociation constant of water (ionic product of water). The value of H⁺ in pure water has been determined as 10⁻⁷ M so that Kw becomes,

Kw = [H + ] × [OH - ]
= 10⁻⁷ × 10⁻⁷
= 1.0 × 10⁻¹⁴ M
 Ionic product of water

The above relation tells us that in aqueous solution the concentration of H⁺ and OH⁻ are inversely proportional to each other.

If H⁺ concentration increases 100 fold, that of OH⁻ has to decrease 100 fold to maintain Kw constant.

### Dissociation of water into H⁺ and OH⁻

Dissociation of water into H⁺ and OH⁻ ions are an endothermic reaction.

Endothermic reaction

The Reactions in which heat is absorbed by the system from the surroundings are known as endothermic reactions.
H₂O + 13.7 kcal → H⁺ + OH⁻

#### Le-Chatelier's Principle

If a system is in equilibrium, a change in any factors that determine the condition of equilibrium will cause the equilibrium to shift in such a way as to minimize the effect of this change.

Thus according to Le-Chatelier’s principle, increasing temperature will facilitate dissociation, thus giving higher values of Kw. The value of Kw at 20⁰ C, 25⁰ C, and 60⁰ C are 0.68×10⁻¹⁴, 1.00×10⁻¹⁴, and 9.55×10⁻¹⁴ respectively.

### Concept of pH

The dissociation of water, Kw, has such low value that expressing the concentrations of H⁺ and OH⁻ ions of a solution in terms of such low figures is not much convenient and meaningful.

Such expressions necessarily have to involve the negative power of the base 10. Sorensen proposed the use of a term known as pH, defined as:

pH = - log[H⁺] = log(1/[H⁺])

Thus for a solution having H⁺ concentration, 10⁻¹ M has a pH = 1And for a solution having H⁺ concentration, 10⁻¹⁴ M has a pH = 14.

For such solutions having H⁺ concentration in the range of 10⁻¹ M to 10⁻¹⁴ M is more convenient and meaningful to express the acidity in terms of pH rather than H⁺ concentrations.

The use of small fractions or negative exponents can thus be avoided. For monobasic acid molarity and normality are the same while they are different for poly-basic acid.

Thus 0.1 M H₂SO₄ is really 0.2 N H₂SO₄ and the pH of the solution is,

pH = -log[H⁺] = -log(0.2)
= 0.699

It follows from these relations that the lower the pH, the more acidic the solution is. If the acidity of a solution goes down 100 fold its pH goes up by the two units. For example, a solution of pH 1 has [H⁺] which is 100 times greater than that of pH 3.

### Calculation of pH

#### 0.002 M HCl solution

Since HCl is a strong electrolyte and completely dissociated.
Thus, [H⁺] = [HCl] = 0.002= 2×10⁻³ M
∴ pH = - log[H⁺]
= - log(2×10⁻³)
= (3 - log2)
= 2.7

#### 0.002 M H₂SO₄ solution

For H₂SO₄, [H⁺] = 2[H₂SO₄]
= 2 × 0.002 = 4 × 10⁻³ M
pH = -log[H⁺] = - log(4×10⁻³)
= (3 - log4)
= 2.4

#### 0.002 M acetic acid solution

For Acetic Acid [H⁺] = √(Ka× [CH₃COOH]
= √(2 × 10-5× 2 × 10⁻³)
= 2 × 10⁻⁴
Thus, pH = - log[H⁺]
= -log(2 × 10⁻⁴)
= (4 - log2)
= 3.7

### Concept of pOH

The corresponding expression for the hydroxide ion is,
pOH = - log[OH-]

If the acidity of a solution goes down 100 fold its pH goes up by two units. A solution of pH 1 has [H⁺] which is 100 times greater than that of pH 3. Taking the case of OH⁻ ions, the pOH will go down by two units (from 13 to 11).
The product of [H⁺] and [OH⁻] is 10⁻¹⁴ and that this has to remain constant.
Thus, [H⁺] [OH⁻] = 10⁻¹⁴
or, log[H⁺] [OH⁻] = log10⁻¹⁴
or, log[H⁺] + log[OH⁻] = -14
or, -log[H⁺] - log[OH⁻] = 14

pH + pOH = 14

We have from the definition,
pH = -log[H⁺] and pOH = -log [OH⁻]

### Acidic, basic and neutral solution

We can now proceed to differentiate between neutral, acidic or basic on the basis of relative concentrations of H⁺ and OH⁻ ions on the one hand, and on the basis of pH on the other.

#### pH and pOH of a neutral solution

A neutral solution is one in which the concentrations of H+ and OH- ions are equal.
Thus, [H⁺] = [OH⁻] = 10⁻⁷ M
In terms of pH, we have the following relations,
[H⁺] = 10⁻⁷ M
or, pH = 7

#### pH and pOH of acidic solution

[H⁺] 〉[OH⁻]
or, [H⁺] 〉 10⁻⁷ M
and [OH⁻] 〈 10⁻⁷ M
In terms of pH, we have the following relations,
[H⁺] 〉 10⁻⁷ M
or, pH 〈 7

#### pH and pOH of basic solution

[H⁺]〈 [OH⁻]
or, [OH⁻] 〉10⁻⁷M
and [H⁺]〈 10⁻⁷M
In terms of pH, we have the following relations,
[H⁺] 〈 10⁻⁷ M
or, pH 〉 7

A mathematical definition of pH provides a negative value when [H⁺] exceeds 1 M. However pH measurements of such concentrated solutions are avoided as these solutions are not likely to be dissociated fully.

The concentration of such strongly acid solutions is best expressed in terms of molarity than in terms of pH.

### Problems solutions

Problem

Calculate the [H⁺], [OH⁻] and pH of a solution obtained by dissolving 28 gm KOH to make 200 ml of a solution.

Solution
[OH⁻] = (28×1000)/(56×200)
= 2.5M
(Molecular Weight of KOH = 56 gm)
[H⁺] = (1 × 10⁻¹⁴)/[OH⁻]
= (1 × 10⁻¹⁴)/(2.5)
= 4 × 10⁻¹⁵
pH = - log[H⁺]
= - log4 × 10⁻¹⁵
= (15 - log4)
Problem

Calculate the [H⁺], [OH⁻] and pH of a solution prepared by diluting 20 ml of 0.1M HCl to one lit.

Solution
[H⁺] = (20 × 0.1)/1000 = 0.002 = 2×10⁻³
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
= (1×10⁻¹⁴)/(2×10⁻³)
= 0.5 × 10⁻¹¹
pH = - log[H⁺]
= - log(2 × 10⁻³)
= (3 - log2)
= 2.7
Problem

The pH of a solution is 4.5. Calculate the concentration of H⁺ ion.

Solution
We have from the definition,
pH = - log[H⁺] = 4.5
or, log[H⁺] = - 4.5
∴ [H⁺] = 3.16 × 10⁻⁵

### Le-Chatelier principle

Le-Chatelier principle predicts quantitatively the effect on the system at equilibrium when some of the variables such as temperature, pressure, and concentration of the equilibrium of a chemical reaction.

### Le-Chatelier principle definition

If a system at equilibrium is subjected to change, the system will react in such a way so as to oppose or reduce the change if this is possible that is the system tends to balance or counteract the effects of any imposed stress.

#### Effect of pressure

According to the Le-Chatelier principle with the increase of pressure, the reaction will shift in a direction where the no of moles is reduced thus the system will try to lower the pressure.
N₂ + 3 H₂ ⇆ 2 NH₃
The increase of pressure is to shift the reaction in a direction where the sum of the stoichiometric number of gaseous molecules is lowered thus lowering the pressure.
In other words, an increase in pressure shifts the equilibrium to the low volume side of the rection whereas a decrease of pressure shifts it to the high volume side.

#### Effect of temperature

According to the Le-Chatelier principle with the increase of temperature, the equilibrium will shift in the endothermic direction that is shifted to the high enthalpy side.
If the reaction proceeds from low enthalpy side to high enthalpy side heat is absorbed and it is for this reason this direction is known as endothermic direction.
N₂ + 3 H₂ ⇆ 2 NH₃ Î”H = -22 kcal
where Î”H = Î£Hproduct - Î£Hreactant
Thus the enthalpy of the reactants in the above reaction higher than that of the products. Thus with the increases in temperature backward reaction favors and thus the equilibrium shifted to the higher enthalpy side and the production of ammonia is decreased.
With the decrease of temperature, the equilibrium will shift in the exothermic direction that is shifted to the low enthalpy side.
If the reaction proceeds from high enthalpy side to low enthalpy side heat is released and it is for this reason this direction is known as exothermic direction.
Thus with the decreases in temperature forward reaction favors and thus the equilibrium shifted to the low enthalpy side and the production of ammonia is increased.
N₂ + O₂ ⇆ 2 NO Î”H = +44 kcal
Enthalpy of the reactants in the above reaction lower than that of the products. Thus with the increases in temperature forward reaction favors and thus the equilibrium shifted to the lower enthalpy side and the production of NO- is increasing.
Decreases in temperature for the above reaction backward reaction favors and the equilibrium shifted to the high enthalpy side and the production of NO is decreased.

Addition of inert gas(He, Ne, Ar, etc) at constant temperature by two way
Constant volume
The addition of inert gas at constant volume can not affect the equilibrium since the concentration of the total reacting components remain unchanged.
Constant pressure
When inert gas is added to the system at equilibrium at constant pressure the volume of the reacting system is increased and thereby total concentration is decreased.
According to the Le-Chatelier principle, the system will move in the direction in which no of moles is increases and thereby the concentration of the system is also increased.

 Le-Chatelier principle

#### Effect of catalyst on equilibrium

Catalyst can speed up the reaction it does not affect the equilibrium of the reaction. A reversible reaction, the equilibrium state is one in which the forward and reverse reaction rates are equal.
The presence of a catalyst, speeds up both the forward and backward reaction, thereby allowing the system to reach equilibrium faster.
This is very important that the addition of a catalyst has no effect on the final equilibrium position of the reaction. Thus we can not increases the production of the product.
Catalysts can be lowering the transition state and the reaction proceeds faster rate. It can be lowering the energy of the transition state(rate-limiting step), catalysts reduce the required energy of activation to allow the reaction proceeds faster rate and reach equilibrium more rapidly.

#### Properties of the reacting system

The le-chatelier principle provides the reacting system with some special features.
1. For example, if the volume of the nonreactive system is decreased by a specific amount the pressure rises correspondingly.
In the reactive system, the equilibrium shifted to the low volume sides (if Î”V ≠ 0), so the pressure increases become less than in the non-reactive system.
The response of the system is moderate in the shift in the equilibrium position makes the reactive system higher compressibility than the non-reactive one.
2. Similarly, if the fixed quantity of the heat is supplied to the non-reacting system temperature is corresponding increases.
In a reactive system, such amount of heat supplied does not increase the temperature so much as the non-reacting system. Since the equilibrium is a shift to the higher enthalpy side and the temperature is less increased.
This shift of equilibrium makes the heat capacity much higher than the non-reactive system. This is useful since the reacting system is chosen as a heat storage medium.

Question
Why does vapour pressure of a liquid decrease with the addition of a nonvolatile solid solute?
The pure solve is the mole faction x1 = 1 but when the non-volatile solute is added to the solvent the mole fraction of the solvent is decreased from 1 that is x1 ã„‘1.
To reduce the effect according to the Le-Chatelier principle the solvent is less vapourised and the mole fraction of the solvent in a solution is thus improved. Thus there occurring a lowering of vapour pressure.
Question
N₂ + 3 H₂ ⇆ 2 NH₃ Î”H = -22 kcal what would most likely happen if increasing the pressure of the reaction?
According to the Le-Chatelier principle yield of NH₃ is increased if pressure is increasing.
Question
N₂ + 3 H₂ ⇆ 2 NH₃ Î”H = -22 kcal equilibrium shifted to forward direction when
(A) the concentration of NH3 increases
(B) pressure is decreasing
(C) the concentration of N2 and H2 decrease
(D) pressure increases and temperature decreases
(D) pressure increases and temperature decreases

### Bohr's model of an atom

According to Rutherford's atomic theory, the entire mass of an atom was concentrated in a tiny positively charged nucleus around which the extranuclear electrons are moving in constant motion in the orbit.
But a moving charged particle emits radiation, will then loss kinetic energy and eventually hit the nucleus.
To resolve the anomalous position Bohr's atomic theory was proposed which is based on several novel postulates.
 Energy of an electron
The following formulas will be widely used in solving the questions of Bohr's atomic theory.
Question
The energy of an electron in the first Bohr orbital of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?
The energy of an electron in nth orbit,
En = (E₁/n²) × Z²

For Li⁺², Z = 3 and in the excited state of Li⁺², n = 2

Putting the values on the above equation we have,

Energy of Li⁺² = (-13.6/4) × 3²
= - 30.6 eV
Question
What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Ã….

The radius of an electron in nth orbit,
rn = r₁ × n²

Second orbit n = 2

Thus the radius of the second orbit of a hydrogen atom,
r₂ = 0.529 × 2²
= 2.12 Ã…

### Quantum level of an atom

Size shape and orientation of an orbital, we need to know about the four quantum numbers and these quantum numbers are designed as, principal quantum number(n), azimuthal quantum number(l), magnetic quantum number(m) and spin quantum number(s).
The following rules will be widely used in solving the questions of quantum number orbitals.
Principal quantum number is denoted by n

It can have integral values from 1 to ∞

The azimuthal quantum number denoted by l
For a given value of n

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

Total number of different values of l equal to n

The magnetic quantum number is denoted by ml
For a given value of l

ml = +l to -l

Total number of different values of ml equal to (2l + 1)
Question
Write the correct set of four quantum numbers for the valence electron of Rubidium.
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d¹⁰4P⁶5S¹
or, [Kr]³⁶ 5S¹
So the valence electron of rubidium means 5S¹ electron of a rubidium atom. Thus for 5S¹ electron, principal quantum number = 5, azimuthal quantum number = 0, magnetic quantum number = 0, and spin quantum number = +½.
So the correct set of four quantum numbers for the valence electron of Rubidium atom are,
5, 0, 0, +½
Question
What are the four quantum numbers of the 19th electron of Chromium atom?
The four quantum numbers of the 19th electron of chromium atom are 4, 0, 0, +½
Question
How many electrons in an atom can have the following quantum numbers n=4 and l=1?
n = 4 means principal quantum number 4 and l = 1 means the azimuthal quantum number is 1, that is P orbital.
Thus when n =1 and l = 0 we can describe 4P orbital. For P orbital magnetic quantum numbers = -1, 0, +1 and each magnetic quantum number has two spin quantum numbers.
Question
How many possible orbitals are there for n = 4?
n = 4 means principal quantum number 4.
Thus when n =4, l = 0, 1, 2, 3. That is 4S, 4P, 4d, and 4f orbitals. S orbital has one orientation of space, p orbitals have three orientation in space, d orbitals have five orientation in space, and f orbitals have seven orientation in space.
Thus the number of possible orbitals when n = 4 is (1 + 3 + 5 + 7) = 16
Question
How many possible orbitals are there for n = 3, l =1, and ml =0 ?
n = 3 means principal quantum number 3, l = 1 means azimuthal quantum number = 0, and ml = 0 means magnetic quantum number zero that is one orientation in space.
Thus the number of orbitals is 1, 3S orbital.
Question
An electron with the highest value of the magnetic quantum number as 3, what is its principal quantum number?
If the highest value of the magnetic quantum number is 3, then the highest value of the azimuthal quantum number is 3.
Thus the principal quantum number of this orbital is (l+1) = 3+1 = 4.
Question
What is the wavelength of the HÉ‘ line of the Balmer series of hydrogen?
The wavenumber of the Balmer lines, ⊽ = 1/Î» = 109677[(1/2²) - (1/n²)]
Again for HÉ‘ line n2 = 3
Thus the wavelength for HÉ‘ - line, 1/Î» = 109677 [(1/2²) - (1/3²)]
= 109677 × (5/36)
∴ Î» = 36/(5 × 109677)
= 6.564 × 10⁻⁵ cm
= 656.5 nm
Question
The lowest wavelength of the Lyman series of the hydrogen atom is x. What is the highest wavelength of the Paschen series of the He⁺² ion?
The highest wavenumber of the Lyman lines, ⊽max = 1/Î»min = R[(1/1²) - (1/∞²)]
or, 1/x = R
or, R = 1/x
The lowest wavenumber of the Paschen series of He⁺² ion is,
⊽min = 1/Î»max = R Z² [(1/3²) - (1/4²)]
or, Î»max = (9 × 16)/(4 × 7R)
∴ Î»max = 36x/7
Question
How many photons of light having the wavenumber a is necessary to provide 3 J of energy?
From the plank theory,
E = nhÎ½
Where n is the number of photons and E is the energy of the photon source.
3 = n h c ⊽ where ⊽ = wave number
∴ n = 3/hca
Question
What is the de-Broglie wavelength of an electron orbiting in the n = 6 state of a hydrogen atom? (r₀ = Bohr's radius)
We know that 2Ï€r = nÎ»

or, Î» = 2Ï€r/n
where r = 6² × r₀ =36 r₀
∴ Î» = (2Ï€ ×36 r₀)/6 = 12Ï€r₀
Question
Find out the number of unpaired electrons of an ion M+x(Z of M = 25) with the magnetic moment √15 B. M. and also find out the value of x?
We know that magnetic moment = √n(n+2)
Thus √15 = √n(n+2)
or, √3(3+2) = √n(n+2)
∴ n = 3
Again the electronic configuration of M is [Ar]¹⁸ 4S² 3d⁵
We find out that the number of unpaired electrons of M+x ion is 3. For this electronic configuration, we need to remove four electrons from its valence shell.
Electronic configuration of M is
[Ar]¹⁸ 3d³
Thus x = 4