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Study construction of chemical bond formation in chemistry

The fundamental questions in chemistry ever since the beginning to study the nature of forces responsible for the construction of bond to form a molecule.

After nearly a century of confusion, Kekul, Van't Hoff , Le Bel, Lewis, and London, etc in the nineteenth century given the concept of the construction of chemical bonds in chemistry.
It was easily realized that the number of atoms or groups of atoms combines to forming the ions or molecular species.

Every element has a saturation capacity for the construction of bonds. The valency is commonly used for the saturation capacity of an element.

Definition of chemical bond

Chemical bond defines as the force holding together two atoms or groups of atoms forming an aggregate of ions or molecular species such that there occur lowering of energy.

A chemical bond represents the forces between atom or ions in a molecule. There will be many chemical compounds whose properties would indicate intermediate types.

Different types of bond in the chemical compound

To study different types of force holding together on atoms or ions to form a molecule in our definition the chemical compounds form mainly three types of bond.
  1. Ionic bond or electrovalent bond or electrostatic bond.
  2. Covalent bond.
  3. Metallic bond.
The ionic bond is electrostatic forces that bind together oppositely charged ions forms by the transfer of electron or electrons from an electropositive metal to an electronegative non-metal atom.
A covalent bond defined as a force holding together atoms through the sharing of electrons.
Types of chemical bonding
Chemical bonding

Construction of bond in the ionic compound

Ionic compounds are constructed by the transference of electron or electrons from one atom to another.
Elements that have the tendency to lose one or more electrons are called electropositive and elements that have a tendency to gain electrons are called electronegative.

Crystallographic studies show that there is no discrete sodium chloride molecule in the crystal of sodium chloride. each of sodium ion surrounded by six chlorine atom or vice versa. The two oppositely charged ions are held together by the electrostatic force of attraction.

Each sodium atom loses one electron to form uni positive sodium ion and neutral chlorine atom gains this electron to form uni negative ion. The two ions constructed a close-packed type ionic compound.
Na → Na⁺ + e
Cl + e → Cl⁻

Na⁺ + Cl⁻ → NaCl

Study single, double, and triple covalent bond

Covent bond is formed by the sharing of electrons. Lewis represents the structure of the covalent bond of hydrogen and hydrogen fluoride.

A single covalent bond is formed between two hydrogen atom and hydrogen and a fluorine atom. This type of bond formed by the sharing of only one electron between the bonded atom.

H⋅ + ⋅H → H⋅ ⋅H or H-H
H⋅ + ⋅F → H⋅ ⋅F or H-F

Double and triple covalent bonds formed when the atoms bonded together share two or three electrons. This type of bond formed in oxygen and nitrogen molecules.

It is not essential for covalent bonding, the sharing of electron equally to the partner. For the compound formation between boron trifluoride and ammonia both the electrons come from ammonia. Such type of bond is called a coordinate covalent bond. here ammonia as a donner and boron trifluoride is an acceptor.
NH₃ + BF₃ → BF₃←NH₃

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Bonding in metals - metallic bond

Metals are a good conductor of electricity and thermal energy and formed crystalline solid with high coordination numbers of 12 or 14.

Since all the atoms of metal, crystal are identical and they can not bound by an ionic bond. Two different types of atom formed this type of bond.

Covalent bonding also not possible for the formation of metallic crystals due to much weak van der Waals forces acting between covalent atoms.

Metallic crystal may be a collection of positive atomic cores immersed in a fluid of mobile electrons or sea of mobile electrons.

The force that binds a metal ion to the mobile electrons or sea of electrons is a known metallic bond. This concept is called the electrons sea model.
This model can explain the conductivity and heat conduction in the metal.
  • Under the influence of the applied electric field, it is possible to move the electrons through the crystal lattice. Thereby metals are the conductor of electricity.
  • Heat conduction by the consequence of freedom of motion in electrons. The higher energy transfers some energy to mobile electrons, which transfer one atomic core of the metal atom to another atomic core at another distinct core.
Most of the properties of the metal can explain this type of bonding but the heat capacity of metals difficult to explain by these types of bonding.

What is environmental chemistry?

Environmental chemistry is the branch of science which deals with the chemical changes in the environment. It includes our surroundings such as air, water, soil, forest, and sunlight. Broadly speaking the environment is three main components.
  1. Abiotic components.
  2. Biotic components.
  3. Energy components.

Abiotic, biotic, and energy components

The abiotic or non-living component or factor of ecosystems is any non-living factor which used or present in an ecosystem. Sunlight, oxygen, nitrogen, temperature, pH, and water is the most common abiotic factors present in our ecosystem. These nonliving components present in the lithosphere or atmosphere.

The biotic or living components are plants and animals including the man in our ecosystem. The most common abiotic factors are temperature, air currents, and minerals.

Energy components are globally leading end-to-end hydrocarbon production data management and accounting software, tracking hydrocarbons from production and transport to sales for revenue generation. These include solar energy,  geochemical, thermochemical, hydroelectric and nuclear energy.

Product of life cycle for our environment

Life originally appeared on our earth in an overall reducing atmosphere. In 1953 Miller and Urey showed that lighting discharges through a mixture of methane, water and ammonia, and hydrogen produce a significant amount of amino acids essential product of life cycle.

Abelson and other latter showed that solar ultraviolet ration radiation acts on carbon monoxide, carbon dioxide, nitrogen and a small amount of hydrogen cyanide. Hydrogen cyanide subsequently gives rise to glycine, adenine, and other biological significant product of the life cycle.
These compounds undergo a series of chemical complex evaluations over millions of years to produce living organisms.

Photosynthesis organisms appeared approximately 2.5 billion years ago when the atmosphere becomes gradually reach become dioxygen and sustain the aerobic of life. The diverse biological process of today has largely developed with this form of life.
Environmental chemistry product of life cycle
Environmental chemistry

What is an ecosystem in our environment

The coexistence animal, plants and microorganisms or biotic component living organisms and abiotic component non-living material or factors which may be inorganic carbon dioxide,  nitrogen, sulfur phosphorus or organic protein and carbohydrates and climate factors like temperature and humidity is called an ecosystem.
  • In short, an ecosystem is the coexistence of the population of all species of living and nonliving components that live together in a particular area.
  • An ecosystem is the living organism's presence in his physical environment.
  • Ecosystems are commonly different types and size namely marine, aquatic, or terrestrial. The marine ecosystem is present in sea and ocean to connect living organisms in the sea and non-living organisms in a seawater environment.
  • The law of conservation of energy follows in every ecosystem by conservation of matter into energy and energy into matter.
Physical chemistry articles for school and college students to study online
  1. What is Chemical equilibrium?
  2. Law of mass action
  3. What is chemical kinetics?
  4. Study properties of gases online
  5. What is the heat capacity of gases?

Environmental protection from pollution

The main aim of the environmental chemistry studies the course of environmental pollution and remedial measure necessity to free it from this environmental pollution.

A pollutant is a substance, chemical or factor which has the potential to adversely affect the natural characteristic of environmental, natural resources and assets.

To save our environment from this pollution we need to recycle household wastage, which included sewage and garbage, much toxic industrial wastage.

In 1990 the concept of green chemistry was introduced for protecting our environment from toxic elements used in or daily life. The production of substances of daily use by chemical reactions which neither employ toxic chemicals nor release toxic elements in the atmosphere.

The effort has been made by use chemicals from the various chemical reactions in the presence of ultraviolet light, sound waves, microwaves, and enzymes harmful to our environment.

What are the isotopes of radioactive elements?

Isotopes of an element are atoms of the same element with different atomic weights. Isotopes posts identical chemical properties because of their identical electronic structure. Isotopes of a particular element have the same number of protons but varying the number of neutrons inside the nucleus of an atom.
  1. When an Alpha particle emits from within the nucleus the mother element loss two units of atomic number and four units of mass number.
    If a radioactive element with mass number M and atomic number Z ejected an alpha particle the newborn element has mass number = (M - 4) and atomic number = (Z - 2).
  2. When a beta particle emits from the nucleus, the daughter element nucleus has an atomic number one unit greater than that of the mother element nucleus.
    If a radioactive element with mass number M and atomic number Z ejected a beta particle the newborn element has mass number the same and atomic number = (Z + 1).
₈₈Ra²²⁶ → ₈₈₋₂Rn²²²⁻⁴ + ₂He⁴(ɑ)
₈₈Ra²²⁶ → ₈₈Rn²²² + ₂He⁴(ɑ)
₉₀Th²³⁴ → ₉₁Pa²³⁴ + ₋₁e⁰(β)

Isotopes of lead and uranium

Uranium is the first discovered radioactive elements. Uranium in the 6th group or actinides of the periodic table with atomic number 92 and mass number 238. Uranium undergoes successive disintegration till the daughter elements become stable, non-radioactive isotopes of lead.
The mother element along with all the daughter elements down to the stable isotope of lead is called a radioactive disintegration series.

Radioisotopes of uranium - 235 and uranium - 238

Uranium - 238 disintegrate ultimately to an isotope of lead. The entire route involves eight alpha and six beta emissions to form the isotopes of lead with atomic number 82 and mass number 206.
₉₂U²³⁸ → ₈₂Pb²⁰⁶ + 8 ɑ + 6 β
Radioisotopes of lead - uranium -238 disintegration series
Radioisotopes of lead - uranium
The mass number of all the above disintegration products are given by (4n +2) where n = 59 for Uranium - 238. This disintegration series is known as (4n + 2) series.

Uranium - 235 or (4n+3) series starts with uranium - 235 and ends with the stable isotope of lead - 207. Seven alpha and four beta emissions in the entire route of the overall process,
₉₂U²³⁵ → ₈₂Pb²⁰⁷ + 7 ɑ + 4 β
Isotopes of lead-207 and uranium-235 in 4n + 3 series
Isotopes of lead-207 and uranium-235

Isotopes of thorium- 232 in radioactive disintegration

Thorium - 232 undergoes successive disintegration till the daughter elements become stable, non-radioactive isotopes of lead with mass number 208.
The entire route involves six alpha and four beta emission called the thorium- 232 disintegration series or 4n series.
₉₀Th²³² → ₈₂Pb²⁰⁸ + 6 ɑ + 4 β
Isotopes of thorium-232 and lead-208 in 4n + 3 series
Isotopes of thorium 232 and lead 208
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Periodic table and group displacement law

In 1913 Soddy proposed the law for the position of radioisotopes in the periodic table when the knowledge of atomic structure was still incomplete. extensive chemical studies on the disintegration products of different series preceded the announcement of this law.

When an alpha particle is emitted in a radioactive disintegration step, the product is displaced two places to the left in the periodic table but the emission of a beta particle results in a displacement of the product to one place to the right.

Soddy observed that more than one product belonging to the same group on the periodic table. He further established that the product elements belonging to the same group had identical chemical properties though their radioactivity's were different. Soddy coined the term isotope for such elements occupying the same place on the periodic table.

₉₀Th²³² → ₈₈Ra²²⁸ + ɑ → ₈₉Ac²²⁸ + β → ₉₀Th²²⁸ + β
₉₂U²³⁸ → ₉₀Th²³⁴ + ɑ → ₈₉Ac²³⁴ + β → ₉₂U²³⁴ + β


A study of the displacement law further reveals that when a parent radio element emits one alpha and two beta particles successively the product isotope occupies the same group in the periodic table with the parent element.

Uses of radioactive isotopes

Isotopes have been used in a variety filed like medicine, biology, agriculture, trace analysis, and many other fields. The uses of isotopes broadly classified under the heads,

Medicinal uses of isotopes

  1. Radioactive iodine-131 is used in the diagnosis and treatment of thyroid gland disorder. After drinking a solution of sodium iodide containing sodium-131. The radioactive iodine moves preferentially to the thyroid gland. The radiation or beta emission destroys the malignant cells without affecting the rest of the body.
  2. Cobalt-60 is a good gamma-ray emitter. Cobalt-60 used to inhibit the growth of malignant tissue in the treatment of cancer.
  3. For abnormality of the circulation of blood, a small amount of a sodium chloride solution labeled with sodium-24 a beta emitter used. Sodium chloride solution injected into a vein of the patient.

Study online ionization energy in inorganic chemistry

The study of ionization energy is an important article in inorganic chemistry for school and college-level courses.

The electrons are raised to higher energy levels by absorption of energy from external sources. If energy supplied to electrons sufficient, electrons go completely out of the influence of the nucleus of an atom.

The amount of energy required to remove the most loosely bound electron or the outermost electron from an isolated gaseous atom of an element in its lowest energy state or ground state to produce a cation is known ionization energy.

M (g) + Ionization energy → M⁺ (g) + e

The process of ionization is an endothermic process since energy is supplied during ionization. Ionization energy generally represented I or IE and measured in electron volt or kilocalories per gram atom.

What is one electron volt?

One electron volt is the energy consumption by an electron falling through a potential difference of one volt. Electron volt simply represented eV.

∴ 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

Removal of an electron from hydrogen energy levels

The energy required for removing an electron from energy levels of the hydrogen atom called the ionization energy. Simply the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization energy of the hydrogen atom.

The ionization energy of the hydrogen atom
= (2π²me⁴/h²)[(1/n₁²) - (1/n₂²)]
where n₁ = 1 and n₂ = ∞.

∴ 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 for removal second and third electron

  1. The electrons are removed in stages one by one from an atom. The amount of energy required to remove the first electron from a gaseous atom called its first ionization energy.
  2. M (g) + IE₁ → M⁺ (g) + e
  3. The energy required to remove the second electron from a cation called second Ionization energy.
    M⁺ (g) + IE₂ → M⁺² (g) + e
  4. Similarly, we have third, fourth ionization.
    M⁺² (g) + IE₃ → M⁺³ (g) + e
    M⁺ (g) + IE₄ → M⁺⁴ (g) + e
Question
How to calculate the second ionization energy of helium atom if ionization energy of hydrogen is13.6 eV?

Answer
The ground state electronic configuration of helium 1S². The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.

∴ IEHe = (2π²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
= Z² × IEH.

∴ Second Ionization energy of helium = 2² × 13.6
= 54.4 eV.

Ionization energy trend in the periodic table

Trend of Ionization energy and periodic table
Ionization energy and the periodic table
The trend of ionization energy in periodic table influenced by the following factors
  • Atomic radius.
  • Change on the nucleus or atomic number.
  • Completely-filled and half-filled orbitals.
  • Shielding effect of the inner electrons.
  • Overall charge on the ionizing species.

The atomic radius of an element

Greater the atomic radius or the distance of an electron from the positive charge nucleus of an element, the weaker will be the attraction. Hence the energy required to remove the electron lower.

An atom raised to an excited state by promoting one electron to a higher energy level than the excited electron was more easily detached because of the distance between the electron and nucleus increases.

The Atomic radius decreases from left to right along a period of the periodic table because of the increasing charge on the nucleus of an atom. Thus when we move left to right along with period normally ionization energy increases.

When we moving from top to bottom in a group the ionization energy of the elements decreases with the increasing size of the atom.

Charge on the nucleus or atomic number

With the increasing atomic number change on the nucleus increases and more difficult to remove an electron from an atom. Hence grater would be the value of ionization.

Normally the value of ionization increases in moving from left to right in a period since with the increasing atomic number the change on the nucleus also increases.

The increase in the magnitude of ionization due to the increase in the electrostatic attraction between the outermost electrons and the nucleus of an atom. Thus it becomes more difficult to remove an electron.

Completely-filled and half-filled orbital of the atom

According to Hund's rule, an atom having half-filled or completely filled orbital comparatively more stable and hence more energy consumption to remove an electron from such atom.

The ionization of such an atom is therefore relatively difficult than expected normally from their position in the periodic table.
Few exceptions in the value of ionization energy in the periodic table can be explained on the basis of the half-filled and completely filled orbitals.

The ionization energy of group-15 elements is higher than the group-16 elements and group-2 elements are higher than the group-3 elements in the periodic table.
Boron and nitrogen in the second period and magnesium and phosphorus in the third period have a slightly higher value of ionization energy than those normally expected.

Nitrogen and phosphorus in group-15 elements with atomic number 7 and 15 have the electronic configuration

1S² 2S² 2P³
1S² 2S² 2P⁶ 3S² 3P³.

Removal of an electron from half-filled 2P and 3P suborbital of nitrogen and phosphorus required more energy.

Removal of an electron from the group-2 element of beryllium and magnesium with completely-filled S-subshell required more energy.
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Shielding effect of the inner electrons

The study of electrostatic attraction between the electrons and nucleus shows that an outer electron attracted by the nucleus and repelled by the electrons of the inner shell.
The combined effect of this attractive and repulsive force acting on the outer electron experiences less attraction from the nucleus. This is known as the shielding effect.
Thus a larger number of electrons in the inner shell, lesser the attractive force for holding outer electron.

The radial distribution functions of the S, P, d subshell show that for the same principal quantum number the S-subshell most shielding than the p-subshell and least shielding the d - orbital.

∴ Shielding efficiency: S〉P〉d.

As we move down a group, the number of inner-shells increases and hence the ionization tends to decreases.

Group-2 elements
Be〉Mg〉Ca〉Sr〉Ba.

Ionization trend in the second and third period

Ionization energy trend and atomic number
Ionization energy trend
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 more energy required for removing an electron from the atom. With the increase atomic number electrostatic attraction between the outermost electrons and the nucleus of an atom increases. Thus removal of an electron from an atom 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.

Due to the presence of a completely filled and half-filled orbital of beryllium and nitrogen, the ionization energy of beryllium and nitrogen slightly higher than the neighbor element boron and oxygen.
Ionization energy trend for the second period in the periodic table

LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe.

Overall charge on the ionizing species

An increase in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization since electron withdrawal from a positively charged species 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 energy.

Why we study the ionization energy of an element?

The study of the ionization energy of the element in a particular group of the periodic table is essential for the properties of the elements.

Lithium, sodium, potassium, rubidium, and cesium or alkali metal in the periodic table with a low value of ionization energy, point to the high reactivity of alkali metals for the formation of the ionic compound.

Online study crystalline solid in chemical science

All the branches of chemical science and all the science, technology depend on the structure of atom or molecules. Thus study crystalline solids is an important article for school or college level students who want to study this article in different books or websites.

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.

What is the definition of crystalline solid?

The definition of crystalline solids is the solid 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.

Sodium chloride, potassium chloride, sugar, and ice, quartz are examples of crystalline solids possess a sharp melting point.

The pattern of such crystal having observed in some small crystal region to predict accurately the position of the particle in any region under observation.

What are the properties of crystalline solid?

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 those belonging to the cubic class are enantiotropic in nature. The magnitude of the enantiotropic property depends on the direction along which is measured.

Amorphous solid or supercooled liquid

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 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 temperatures. 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 and sharp melting point but amorphous solids that do not possess a definite structure and 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.
Chemical science articles for school and college students to study online
  1. What is Chemical equilibrium?
  2. Law of mass action
  3. What is chemical kinetics?
  4. Study properties of gases online
  5. What is the heat capacity of gases?
  6. Hydrogen spectrum

Classification of crystalline on the basis of force

Study the basics of the nature of force operating between constituent particles or atoms, ions, molecules of matter, crystalline solids are classified into four categories.
Types of crystalline solids for study online
Types of crystalline solids

Molecular crystals, organic compounds

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. Hydrogen, helium atom or non-polar hydrogen, oxygen, chlorine, carbon dioxide, methane molecules are examples of these types of crystal. The force operating between constituent particles atoms or molecules is a weak London force of attraction.
  2. Polar binding crystals
    The polarity of bond shows in these types of crystalline solids. Sulfur dioxide and ammonia are examples where 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 ice.

Force of attraction between ionic crystal 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. Calcium carbonate is an example of these types of crystalline solids.

Covalent crystal or covalent bonding crystal

The forces involved here are chemical nature or covalent bonds extended in three dimensions. They are strong and consequently, the crystals are strong and hard with high melting points. Diamond, graphite, silicon 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 crystalline 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 (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 crystal

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 crystal

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 Van't Hoff equation - effect on temperature?

Van't Hoff equation proposed equilibrium of a chemical reaction is constant at a given temperature. The equilibrium constant, Kp values can be changed with the change of temperature.

This will be evident from the study of Kp values at different temperatures of a chemical reaction.
N₂ + O₂ ⇆ 2NO
TemperatureKp × 10⁴
2000° K4.08
2200° K11.00
2400° K25.10
2600° K50.30
The quantitative relation, known Van't Hoff equation connecting chemical equilibrium and temperature can be derived thermodynamically starting from Gibbs - Helmholtz equation.

The Gibbs - Helmholtz equation
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

- 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 equation is the differential form of Van't Hoff equation.

The 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 = integrating constant.

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

∴ Van't hoff equation

lnKp = - (ΔH⁰/RT) + (ΔS⁰/R).

Problem
The heat of a chemical reaction 2A + B ⇆ 2C, ΔG⁰ at 500 K = 2 KJ mol⁻¹.  Calculate  Kp at 500 K for the chemical reaction A + ½B ⇆ C.

Solution
ΔG⁰ at 500 K for A + ½B ⇆ C
= (2 kJ mol⁻¹)/2 = 1 kJ mol⁻¹.

ΔG⁰ = - RT lnKp.
∴ 1 = - 8.31 × 10⁻³ × 500 × lnKp
or, lnKp = 1/(8.31 × 0.5)
or, lnKp = 0.2406.

∴ Kp = 1.27.

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

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

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.

Problem
Show that the equilibrium point for any chemical reaction is given by ΔG = 0.

Solution
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.

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)ΔƔ
or, lnKp = lnKc + ΔƔ lnR + ΔƔ lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + ΔƔ/T.

dlnKp/dT = ΔH⁰/RT²
or, 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
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.

The 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.

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