54. Types of bonds, Ionic bonds.

Chemistry encompasses the study of elements and the various species (compounds/ions/complexes etc) which are formed from these elements. It is imperative that we understand how the atoms of elements are held together in molecules. Which forces hold these atoms together? Why do two or more atoms bond? What are the factors responsible for specific orientations of molecules? Let us begin our discussion on this very fundamental topic in chemistry – THE CHEMICAL BOND.

Types of bonds.

In Chemistry, there are primarily two types of bonding –

1)Primary bonding.
2)Secondary bonding.

Primary bonding	Secondary bonding
Sharing or transfer of electrons	NO sharing or transfer of electrons.
A strong force of attraction	A weak force of attraction
Short range	Long range
e.g.- Ionic,Covalent ,Co-ordinate covalent,Metallic bonds	e.g.- Hydrogen bonds, Vander Waal’s forces, and dipole interactions.

Primary bonds are formed by the transfer or sharing of electrons. These bonds are very strong. Secondary bonds are subtle forces of attraction and are not as strong as primary bonds. Though secondary bonds are weak they have significant effects on the molecules and their properties.

In this post, we start learning about the first type of primary bond – ‘the ionic bond’ in detail.

P1.THE IONIC BOND.

This bond is formed due to the electrostatic force of attraction between oppositely charged ions.

ELECTRO = electric charge i.e positive and negative charges
STATIC = stationary(which does not move).

electrostatic attraction = Coulombic attraction.

Charles_de_coulomb — Charles Augustin de Coulomb.

The electrostatic attraction was discovered by Charles Augustin de Coulomb ( a French Physicist) and thus this force is also called as Coulombic force of attraction.

Note that ‘Coulombic force’ is written with a capital ‘C’, as it denotes a scientist’s name.

Electrostatic/Coulombic force of attraction refers to the attraction between positive( +ve) and negative( -ve) charges on ions. e.g.- Na⁺, Mg⁺,(SO₄)^2-, Cl^– etc .

Like charges repel each other and opposite charges attract. This force of attraction between ions of opposite charge is the ionic bond.

How are ions formed?

Generally, except the hydrogen and helium atoms, all other atoms prefer to have 8 electrons in their valence/outermost shell. This phenomenon is referred to as ‘octet stability‘.

Why only 8 electrons? The s-, and p- orbitals can hold a maximum of 2, and 6 electrons respectively. When the octet is complete, the outermost s- and p- orbitals are filled with 8 electrons. This renders stability to the atom. It’s like a person feels satiated after a meal. The person’s tummy is full and he/she is fully satisfied.

Hydrogen and helium atoms have only 1s orbital in their atom. Thus, they just need only two electrons to gain stability. This is referred to as ‘duplet stability‘.

Positive ions–

The elements placed at the extreme left of the periodic table (metals), can lose valence electrons easily and attain an octet. Electrons are negatively charged, So, losing an electron means gaining a positive charge. Thus they show a greater tendency to form positive ions.
e.g.-Sodium Atom(Na)
Electronic configuration is (2,8,1) i.e it has one electron in its valence/outer shell. If it loses this electron, it will have eight electrons in its outer shell and so will have octet stability. So, it readily gives up this electron to form a cation.

Na → Na⁺ + e^–.

Negative ions-

The elements placed at the extreme right (except the zero group i.e. noble gases) have a tendency to accept electrons and form negative ions.
e.g.- The chlorine atom (Cl)

The electronic configuration is (2,8,7) i.e it has seven electrons in its valence/outer shell. If it gains one electron, it will have eight electrons in its outer shell and so will have octet stability. It readily accepts this electron from a sodium atom to form a chloride ion.

Cl+ e^–→ Cl ^–.

Attraction between positive and negative ions-

The oppositely charged ions attract each other and form an ionic bond. Ionic bonding involves the complete transfer of electrons. It is a chemical bond.

Species which looses electron/s → Cation (+ve ion) → electropositive elements form cations e.g. – metals like Na⁺, Mg²⁺,Ca²⁺,K⁺ etc.
Species that gains electron/s → Anion (-ve ion) → electronegative elements form anions. e.g.- non-metals like Cl^–,Br^–, CN^–, S^2-etc.

1 positive charge indicates the loss of one electron.
∴Na⁺→ one electron lost from sodium atom.
Ca²⁺→ two electrons lost from calcium atom.
Al³⁺→ three electrons lost from an aluminum atom.

Similarly, 1 negative charge indicates the gain of one electron from another species.
∴Cl^–→ one electron gained by chloride atom.
S²^–→ two electrons gained by a sulfur atom.

Let us study a few examples of molecules with ionic bonding –

Example 1 – Sodium chloride (NaCl molecule) – TABLE SALT.

Sodium Atom Na → electronic configuration is (2,8,1) i.e it has one electron in its valence/outer shell. If it loses this electron, it will have eight electrons in its outer shell and so will have octet stability. So, it readily gives up this electron to form a cation.

Na → Na⁺ + e^–.

Chlorine Atom Cl → electronic configuration is (2,8,7) i.e it has seven electrons in its valence/outer shell. If it gains one electron, it will have eight electrons in its outer shell and so will have octet stability. It readily accepts this electron from a sodium atom to form a chloride ion.

Cl+ e^–→ Cl ^–.

The sodium and chloride ions have opposite charges on them and so they attract each other forming Na⁺Cl^–molecule.

The energy changes that take place during the formation of this ionic bond can be studied with the BORN – HABER CYCLE.

THE BORN – HABER CYCLE.

The formation of ions followed by the formation of ionic bonds between these oppositely charged ions is a favorable process. It is like a win-win situation for both the ions. The energetics of this bond formation also shows us how this process is energetically favored. To understand this we must first understand the following –

Less energy ⇒ More stability.

If the energy of a system decreases in a reaction, the reaction is energetically favored. To understand the energetics of any system, we need to know a few thermodynamic terms like internal energy, enthalpy, lattice energy, etc. Thermodynamics refers to the heat changes that take place during a reaction.

Thermo = relating to heat

1)Internal Energy (U) – All the energy of a system is called its internal energy. Suppose, we consider a system, enclosed in a box, then the internal energy is the sum of –

the kinetic energy of all the particles in that system (Energy associated with the movement of the particles) +
potential energy(Energy associated with the position of the particle) +
bond energy (energy in various bonds present in the system) etc.

However, internal energy changes with changes in temperature and pressure. e.g.- As the temperature increases, the molecules in the system start to move around more rapidly (as they get more heat energy). Thus, their kinetic energy increases and so does the internal energy of the system.

So when we study energy changes in a reaction, this parameter(internal energy) is not useful as in a reaction there are temperature changes(endothermic/exothermic reactions) and pressure changes involved. Thus, we use another parameter – ENTHALPY- while talking about energy changes in a reaction.

2)Enthalpy (H) – This term gives you the total heat content of the system. It considers the pressure and volume changes that occur during the reaction.

H = U + PV , where,
H ⇒ Enthalpy.
U ⇒ Internal energy.
P ⇒ Pressure.
V ⇒ Volume.

3) Dissociation energy(ΔH _dissociation)- The energy required to dissociate a compound is called dissociation energy. Dissociation of a compound is always an endothermic process and requires an input of energy.

4) Sublimation energy(ΔH_sublimation)– The energy required by a substance to change phase from solid to gas, is called sublimation energy.

5)Heat of formation( ΔH_formation)-The energy change during the formation of a compound from its elements is known as the heat of formation.

The stages of formation of an ionic compound MX are as follows –

1)The electropositive element M is converted from solid to its vapor phase, by providing the heat of sublimation to it.

M_solid + ΔH_sublimation → M_vapor

(2) In the vapor phase, the valence electron/s is/are lost to form a positive ion. Energy associated with this process is ionization energy.

M_vapor + I.E _(M) → M⁺ +e^–

(3) The electronegative element dissociates, and a negative ion is formed with the release of energy equivalent to the electron affinity (EA) of X.

X_(g) + e^– → X^– _(g) +EA_(X)

(5) The two ions combine to form the ionic solid MX. The heat of formation is released in this process. As heat is given out, the process is exothermic and this shows that the energy of the system decreases(as heat is given out), making it more stable.

M⁺ + X^–_(g) → MX_(s) + ΔH_formation

(6) Lattice enthalpy (U): The enthalpy change during the formation of one mole of sodium chloride from its constituent ions is called lattice energy of lattice enthalpy. It is an exothermic step. The gaseous ions form a solid.

M⁺(g) + X^–(g) → MX (s)

ΔH_sublimation, I.E, ΔH_{dissociation,}and EA, can be calculated experimentally.

However, WE NEED TO CALCULATE ΔH_formation.

How do we do that from the experimental data? We use Hess’s Law of Constant Heat Summation /Hess’s Law –

Regardless of the multiple stages or steps of a reaction, the total enthalpy(energy) change for the reaction is the sum of all changes.

ΔH_sublimation, I.E, ΔH_{dissociation,}and EA quantities give us energy changes at respective stages of the reaction and their values can be determined experimentally. Thus, we can calculate the total energy change by just adding them up.

∴ΔH_formation= ΔH_sublimation + I.E + ΔH_dissociation+ EA

e.g.– The formation of NaCl molecule –

Na_(s) → Na_(g) +107kJ mol ^-1

Na_(g) → Na⁺_(g) + e^–+495 kJ mol^-1

1⁄2Cl₂(g) → Cl_(g) +122 kJ mol ^-1

Cl_(g) + e^{– →} Cl –_(g) –349 kJ mol ^-1

Na⁺_(g) + Cl^–_(g) → NaCl _(s) U = -786 kj/mol

Na⁺_(g) + Cl^–_(g) → NaCl_(s)
Calculate ΔH_formation .

According to Hess’s law –

ΔH_formation= 107.3 +495.8+122 – 348.6 -786 = – 411 Kj/mol.

∴Na(s) + 1⁄2Cl₂₍g) → NaCl(s). ΔH_formation=– 411 kJ mol^-1.

Ionic compounds are generally solids at room temperature and have high melting points. This is due to the strong electrostatic force which binds the positive and negative ions together in a crystal lattice structure.

541

The melting point of NaCl is higher than potassium nitrate(KNO₃) 304^oC, as the anion (NO₃^–)is large compared to Cl^–ion, and the negative charge is dispersed over a larger volume. This weakens the attractive forces and lowers the melting point. Ammonium sulfate (NH₄SO₄), where both ions –(NH₄⁺) and (SO₄^–)are large, has the lowest melting point (235^oC).