LECTURE 1
ATOMIC STRUCTURE & BONDING
Objectives
After reading this lesson you will be able to;
•
Explain the structure of an atom.
•
Define valence electron and its importance in chemical bonding
•
Outline the four types of bonds and cite some examples i.e., ionic, covalent, metallic & van
der Waals.
ATOM
➢ The basic building block of all matter is called an atom.
➢ Atoms are a collection of various subatomic particles containing negatively charged
electrons, positively charged protons, and neutral particles called neutrons.
Electronic Configuration
➢ The arrangements of electrons in orbitals, subshells, and shells are called electronic
configurations.
➢ The electronic configuration of an atom describes the number of electrons that an atom
possesses, and the orbitals in which these electrons are placed.
Valence Electrons
➢ Valence electrons are the electrons in the outermost shell, or energy level, of an atom.
➢ They participate in the formation of a chemical bond.
Chemical Bond
➢ A force of attraction that holds two atoms together.
➢ Involves the valence electrons.
➢ When two atoms of the same or different elements approach each other, the energy of
the combination of the atoms becomes less than the sum of the energies of the two
separate atoms at a large distance.
➢ We say that the two atoms have combined or a bond is formed between the two. The
bond is called a chemical bond.
➢ Thus, a chemical bond may be visualized as an effect that leads to a decrease in energy.
➢ The combination of atoms leads to the formation of a molecule that has distinct
properties different from that of the constituent atoms.
➢ There are four major types of chemical bonds;
•
Ionic
•
Covalent
•
Metallic
•
Van der Waals
Ionic Bond
➢ Occurs from a transfer or loss/gain of electrons.
➢ Usually form between atoms of metals and atoms of non-metals.
➢ Here, the metal loses electrons to become a positively charged cation, whereas the
nonmetal accepts those electrons to become a negatively charged anion.
➢ NaCl, sodium chloride, or table salt, is the “classic” example of an ionic compound.
➢ Sodium is a metal, and chlorine is a non-metal. It has ionic bonds and a crystalline
structure.
➢ In solution, it separates into ions.
Figure 1: The ionic bond of sodium chloride.
Figure 2: The ionic bond of Sodium Sulphide.
Properties
➢ Ionic compounds have high melting points.
➢ Ionic compounds are hard and brittle.
➢ Ionic compounds dissociate into ions when dissolved in water.
➢ Solutions of ionic compounds and melted ionic compounds conduct electricity, but solid
materials do not.
➢ An ionic compound can be identified by its chemical formula: metal + nonmetal or
polyatomic ions.
Covalent Bond
➢ A force that bonds two atoms together by a sharing of electrons
➢ Each pair of shared electrons creates a bond
➢ Usually occurs between atoms of non-metals
Examples
- A hydrogen molecule, H2, has two hydrogen atoms linked by a covalent bond.
- A hydrogen atom has 1 electron in its only energy level thus unstable.
- Therefore, two hydrogen atoms combine by each contributing an electron.
-Other examples include chlorine molecules, oxygen, carbon (IV) oxide, etc.
Figure 3: Examples of covalent bond
Properties
➢ They are non-electrolytes (don’t conduct electric current) when in molten or solution
form.
➢ They have usually low melting and boiling points.
➢ They are often insoluble in water.
➢ They are usually soluble in organic solvents like benzene and propanone.
➢ Most of them are gaseous at room temperature or volatile liquids.
Metallic Bond
➢ The metallic elements have electropositive atoms that donate their valence electrons to
form a “sea” of electrons surrounding the atoms.
➢ The valence electrons move freely within the electron sea and become associated with
several atom cores.
➢ The positively charged ion cores are held together by mutual attraction to the electrons,
thus producing a strong metallic bond.
➢ Metals show good ductility since the metallic bonds are non-directional.
➢ There are other important reasons related to a microstructure that can explain why metals
exhibit lower strengths and higher ductility than what we may anticipate from their
bonding.
➢ Ductility refers to the ability of a material to be stretched or bent permanently without
breaking.
Example:
➢ The metal atoms Na, Cu, Ag, Fe, etc., are bound to each other in their crystals by metallic
bonds.
Properties
➢ High melting points.
➢ Good conductors of electricity.
➢ Good conductors of heat.
➢ High density.
➢ Malleable.
➢ Ductile.
Van der Waals Bond
➢ A secondary bond developed between atoms and molecules as a result of interactions
between dipoles that are induced or permanent.
➢ It’s a weak attractive force between atoms or nonpolar molecules caused by a temporary
change in dipole moment arising from a brief shift of orbital electrons to one side of one
atom or molecule, creating a similar shift in adjacent atoms or molecules.
➢ The various types of Van der Waals forces include; Keesom interactions, Debye forces,
and London dispersion forces.
➢ London dispersion forces arise due to the interactions between an instantaneous dipole
and an atom/molecule and are also known as instantaneous dipole–induced dipole forces.
➢ These forces are believed to be the weakest of all Van der Waals forces.
Properties of Van der Waals Forces
➢ Van der Waals forces are additive in nature, they are made up of several individual
interactions.
➢ These forces cannot be saturated.
➢ No directional characteristic can be attributed to these forces.
➢ They are not dependent on temperature (except dipole-dipole interactions)
➢ Van der Waals forces are short-range forces. Their magnitude is high when the
atoms/molecules in question are close to each other.
LECTURE 2
MOLE CONCEPT FORMULAE AND CHEMICAL REACTIONS
Objectives
By the end of this lecture the learners should be able to;
•
Define the terms mole, molar solution, and standard solution.
•
Write balanced chemical equations
•
Explain oxidation/reduction reactions.
•
Apply redox reactions to corrosion/reduction of ores.
•
Explain the role of water in ionization
Molecular and Formula Mass
➢ The molecular mass of a molecule is its average mass as calculated by adding the atomic
weights of the atoms in the molecular formula.
➢ The formula mass of a molecule is the sum average of the atomic weights of the atoms
in its empirical formula.
➢ For example, the empirical formula of glucose, which consists of carbon, hydrogen, and
oxygen in the ratio of 1:2:1 is CH2O (empirical formulae are also called simplest
formulae).
➢ The molecular formula of a substance is always an integral multiple of its empirical
formula (i.e., molecular formula = Xn where X is the empirical formula and n is an
integer). For example, the molecular formula of glucose is C6H12O6 which is 6 × its
empirical formula.
Difference between formula and molecular mass
Steps in Calculating Molecular Mass
i.
Determine the number of atoms of the elements present in the molecule.
ii.
Find the atomic mass of each atom using the periodic table of elements. Multiply the
atomic mass by the number of atoms of the elements present in the molecule.
iii.
Get the sum of all the products.
Steps in Calculating Formula Mass
i.
Determine the number of atoms of the elements present in the molecule.
ii.
Find the atomic mass of each atom using the periodic table of elements. Multiply the
atomic mass by the number of atoms of the elements present in the molecule.
iii.
Get the sum of all the products.
MOLE
➢ A mole is the quantity of a substance that contains the same number of particles.
➢ In counting the number of atoms, we use a constant number called Avogadro’s number
which is equivalent to a MOLE.
➢ Understanding the basic concept of a mole is the key to relating mass, mole, and number of
particles in elements, compounds, and chemical reactions.
➢ This chemical calculation used in chemistry is known as stoichiometry.
➢ One mole is the amount of substance that contains as many particles as there are present in
12 grams of Carbon-12 atom.
➢ The actual number of atoms in 12 g of Carbon-12 isotope is equal to 6.02X1023 particles
called Avogadro’s number.
Calculating the Number of Mole Given the Mass or Vice Versa
Mathematically, the number of moles is determined using the formula;
𝑀𝑜𝑙𝑒 =
𝑀𝑎𝑠𝑠 (𝑖𝑛 𝑔)
𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Let us calculate the number of moles of oxygen (O2) in 24.0 g of O2. From the definition of a
mole, we know that 1 mole of O2 = 32.0 g.
Solution:
Molar Solution
➢ A molar solution is defined as an aqueous solution that contains 1 mole (gram-molecular
weight) of a compound dissolved in 1 liter of a solution.
➢ In other words, the solution has a concentration of 1 mol/L or molarity of 1 (1M).
➢ Molar solutions are also useful in predicting corrosion rates. For example, steel corrosion
in a 1M hydrochloric acid solution can be assessed using weight loss and other
electrochemical techniques.
➢ This information can then be used to perform calculations and evaluate steel corrosion in
different situations.
Standard Solution
➢ A standard solution is any chemical solution that has a precisely known concentration.
➢ To prepare a standard solution, dissolve a known mass of solute and dilute the solution to
a precise volume.
➢ Standard solution concentration is usually expressed in terms of molarity (M) or moles
per liter (mol/L).
➢ Not all substances are suitable solutes for standard solutions.
➢ In most cases, the reagent must be stable, pure, and preferably of high molecular weight.
Stoichiometry
➢ Stoichiometry is a section of chemistry that involves using relationships between
reactants and/or products in a chemical reaction to determine desired quantitative data.
➢ It is derived from two Greek words, stoikhein which means element and metron meaning
measure, so stoichiometry literally translated means the measure of elements.
➢ In order to use stoichiometry to run calculations about chemical reactions, it is important
to first understand the relationships that exist between products and reactants and why
they exist, which require understanding how to balance reactions.
Balancing Equations
➢ In chemistry, chemical reactions are frequently written as an equation, using chemical
symbols.
➢ The reactants are displayed on the left side of the equation and the products are shown on
the right, with the separation of either a single or double arrow that signifies the direction
of the reaction.
➢ The significance of single and double arrows is important when discussing solubility
constants, but we will not go into detail about it in this module.
➢ To balance an equation, it is necessary that there are the same number of atoms on the left
side of the equation as the right. One can do this by raising the coefficients.
Reactants to Products
➢ A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of
a chemical reaction.
➢ It includes the elements, molecules, or ions in the reactants and in the products as well as
their states, and the proportion for how much of each particle reacts or is formed relative to
one another, through the stoichiometric coefficient.
➢ The following equation demonstrates the typical format of a chemical equation:
2Na(s)+2HCl(aq)→2NaCl(aq)+H2(g)
➢ In the above equation, the elements present in the reaction are represented by their
chemical symbols.
➢ Based on the Law of Conservation of Mass, which states that matter is neither created nor
destroyed in a chemical reaction, every chemical reaction has the same elements in its
reactants and products, though the elements they are paired up with often change in a
reaction.
➢ In this reaction, sodium ( Na ), hydrogen ( H), and chloride ( Cl ) are the elements present
in both reactants, so based on the law of conservation of mass, they are also present on
the product side of the equations.
➢ Displaying each element is important when using the chemical equation to convert
between elements.
Examples
1. Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
Pb(OH)4 (aq)+H2SO4 (aq)→Pb(SO4)2 (aq)+H2O (l)
Soln:
Pb(OH)4 (aq)+2H2SO4 (aq)→Pb(SO4)2 (aq)+H2O (l)
2. Fe(s) + O2(g) → 2Fe2O3(s)
Soln.
4Fe(s) + 3O2(g) → 2Fe2O3(s)
3. Propane ( C3H8 ) burns in this reaction:
C3H8 (s)+5O2 (g)→4H2O (l)+3CO2 (g)
If 200 g of propane is burned, how many g of H2O is produced?
Soln:
Calculating the mass of H2O
mass H2O = 200 g C3H8 × (1 mol C3H8 / 44.096 g C3H8) × (4 mol H2O / 1 mol C3H8) ×
(18.015 g H2O / 1 mol H2O
mass H2O = 327 g
Oxidation & Reduction Reactions
Oxidation
➢ A process that involves the addition of oxygen or any electronegative element or the
removal of hydrogen or any electropositive element.
➢ According to electronic concept oxidation is defined as the process in which an atom or
ion loses one or more electrons.
Reduction
➢ A process that involves the addition of hydrogen or any electropositive element or the
removal of oxygen or any electronegative element.
➢ According to electronic concept reduction is defined as the process in which an atom or
ion gains one or more electrons.
➢ Oxidation–reduction or redox reactions are reactions that involve the transfer of electrons
between chemical species.
Classical Idea of Oxidation and Reduction reactions:
Oxidation reactions involve:
1. Addition of oxygen:
C + O2 → CO2
(oxidation of carbon)
2. Addition of electronegative element:
Fe + S → FeS (oxidation of Iron)
3. Removal of hydrogen:
H2S + Br2 → 2 HBr + S (oxidation of sulphide)
4. Removal of electropositive elements:
2 KI + H2O2 → I2 + 2 KOH (oxidation of iodide)
➢ Oxidising agent is a substance which brings about oxidation. In the above examples O2,
S, Cl2, Br2, and H2O2 are oxidising agents.
Reduction reactions involve:
1. Addition of hydrogen:
N2 + 3 H2 → 2NH3
( reduction of nitrogen)
2. Addition of electropositive element:
SnCl2 + 2HgCl2 → SnCl4 + Hg2Cl2 ( reduction of mercuric chloride)
3. Removal of oxygen
ZnO + C → Zn + CO (reduction of zinc oxide)
4. Removal of electronegative element
2FeCl3 + H2 → 2FeCl2 + 2HCl (reduction of ferric chloride)
➢ A reducing agent is a substance which brings about reduction. In the above examples H2,
HgCl2 and C are Reducing agents.
Note: A substance, which undergoes oxidation, acts as a reducing agent while a substance,
which undergoes reduction, acts as an oxidizing agent.
Oxidation and Reduction in Terms of Electron Transfer
➢ This is the most commonly used definition of oxidation and reduction and the most
widely applicable.
➢ In this case, Oxidation is the loss of electrons and Reduction is the gain of electrons.
A very clever mnemonic to remember this concept is oil rig.
OIL
RIG
Oxidation is loss
Reduction is gain
➢ Oxidation and Reduction reactions are always interlinked.
➢ Because electrons are neither created nor destroyed in a chemical reaction, oxidation and
reduction always occur in pairs, it is impossible to have one without the other.
➢ In the below reaction Magnesium gets oxidized by losing two electrons to oxygen which
gets reduced by accepting two electrons from magnesium.
Common Redox Reactions
The three common redox reactions are discussed below:
1. Combustion reaction – It is a type of redox reaction which occurs between molecular
oxygen and compound to form oxygen-containing products.
2C8H18+25O2 → 16CO2(g)+18H2O
2. Disproportionation reaction – It is a type of redox reaction where a single reactant is
reduced and oxidized. It is also known as an auto-oxidation reaction.
3ClO−(aq) → ClO3−(aq)+2Cl−(aq)
3. Single replacement reaction – It is a type of redox reaction that involves two elements
switching places within a compound. It is also known as a single displacement reaction.
Zn(s)+2HCl(aq) → ZnCl2(aq)+H2(g)
Balancing Redox Reaction
Every chemical reaction must be balanced according to the “Law of conservation of mass”.
The chemical equations which involve oxidation and reduction can also be balanced with the
help of the following methods;
i.
Oxidation number method.
ii.
Ion electron method ( or half-reaction method)
i.
Oxidation number method:
The various steps involved in balancing a redox equation by oxidation method are discussed
here through an example.
Example: Balance the chemical equation by the oxidation number method
CuO + NH3 → Cu + N2 + H2O
Solution:
Step-1 : Write the oxidation number of each atom in the skeleton equation
Step-2 : Identify the atoms which undergo change in oxidation number.
Step-3 : Calculate the increase and decrease in oxidation number w.r.t reactant atoms.
Step-4 : Equate the increase and decrease in oxidation number on the reactant side.
Step-5 : Balance the number of Cu and N atoms on both sides of the equation.
Step-6 : Now balance H and O atoms by hit and trial method.
Note:(i) In the reactions taking place in acidic medium, balance the O atom by adding the
required number of H2O molecules to the side deficient in O atoms. Then balance the H
atoms by adding H+ to the side deficient in H atoms.
(ii) In the basic medium, first balance the number of negative charges by adding the required
number of OH– ions to the side deficient in the magnitude of the charges. Then add H2O
molecules on the other side in order to balance the OH– ions added.
Ion electron method ( or half reaction method):
It is based on the Principle that the electrons lost during oxidation half reaction in a particular
redox reactions is equal to the electrons gained in the reduction half reaction. The method is
called half reaction method. The balancing is completed in the following steps:
Example: Balance the chemical equation by ion-electron method:
Cr2O72- + Fe2+ + H+ → Cr3+ + Fe3+ + H2O
Step-1 : Write the oxidation number of each atom in the skeleton
equation:
Step-2 : Find out
the species involved in the oxidation and reduction half reactions:
Step-3 : Balancing oxidation half reaction:
As oxidation number increases 1, add one e– on the product side to balance change in O.N.
Step-4: Balancing reduction half-reaction:
The decrease in oxidation number per Cr atom is 3 and the total decrease in O.N for two Cr
atoms is 6. Therefore, add 6e– on the reactant side. In order to balance O atoms add 7 H2O
molecules on the product side then balance H atoms by adding 14 H+ on reactant side.
Step-5 : Adding the two half reactions:
Role of water in Ionization
•
Water dissociates to form ions by transferring an H+ ion from one molecule acting as an
acid to another molecule acting as a base.
H2O(l)
+
acid
•
H2O(l)
H3O+(aq)
+ OH-(aq)
base
Acids react with water by donating an H+ ion to a neutral water molecule to form the
H3O+ ion.
HCl(g)
+
H2O(l)
acid
•
H3O+(aq)
+ Cl-(aq)
base
Bases react with water by accepting an H+ ion from a water molecule to form the
OH- ion.
NH3(aq)
+
H2O(l)
NH4+(aq)
+ OH-(aq)
base
acid
Water molecules can act as intermediates in acid-base reactions by gaining H+ ions from
•
the acid
HCl(g)
+
H2O(l)
H3O+(aq)
+ Cl-(aq)
and then losing these H+ ions to the base.
NH3(aq)
+
H3O+(aq)
NH4+(aq)
+ H2O
Application of Redox Reactions
i.
Corrosion
➢ Corrosion is the formation of compounds on the surface of a metal.’
➢ Corrosion of metal is a redox reaction in which occur when a metal loses electron and is
oxidised to form the metal ion.
➢ Metal is t corroded.
M → Mn+ + ne–
Rusting as a redox reaction
•
Rusting is the corrosion of iron. It is the most common corrosion of metal around.
•
For iron to rust, oxygen and water must be present.
•
Rusting is a redox reaction whereby oxygen acts as the oxidising agent and iron acts as
the reducing agent.
•
Figure shows the half-reactions of rusting.
•
The surface of iron in the middle of the water droplet serves as the anode, the electrode at
which oxidation occurs. The iron atoms here lose electrons to form iron(II) ions.
•
The electrons flow to the edge of the water droplet, where there is plenty of dissolved
oxygen. The iron surface there serves as the cathode, the electrode at
which reduction occurs. Oxygen gains the electrons and is reduced to hydroxide ions.
•
The iron(II) ions produced combine with the hydroxide ions to form iron(II) hydroxide.
•
Thus, the overall redox reaction is as follows:
•
The iron(II) hydroxide is then further oxidised by oxygen to form hydrated iron(III)
oxide, Fe2O3.xH2O) whereby x varies. The hydrates come in various shades of brown and
orange and together make up what is commonly known as rust.
•
In the presence of acids and salts, rusting occurs faster. These substances increase the
electrical conductivity of water, making water a better electrolyte.
For example:
(a) Iron structures such as bridges, fences and cars at coastal areas rust faster due to the
presence of salts in the coastal breeze.
(b) Iron structures in industrial areas rust quickly as these areas have air polluted with
acidic gases such as sulphur dioxide and nitrogen oxides.
ii.
Production of some important chemicals is also based on electrolysis, which in turn,
is based on redox reactions. Many chemicals, like caustic soda, chlorine, etc., are
produced using redox reactions.
iii.
Oxidation-reduction reactions also find their application in sanitising water and
bleaching materials.
iv.
The surfaces of many metals can be protected from corrosion by connecting them to
sacrificial anodes, which undergo corrosion instead. A common example of this
technique is the galvanisation of steel.
v.
The industrial production of cleaning products involves the oxidation process.
vi.
Nitric acid, a component of many fertilisers, is produced from the oxidation reaction
of ammonia.
vii.
Electroplating is a process that uses redox reactions to apply a thin coating of a
material on an object. Electroplating is used in the production of gold-plated
jewellery.
viii.
Many metals are separated from their ores with the help of redox reactions. One such
example is the smelting of metal sulphides in the presence of reducing agents.