INTRODUCTION

A molecule is formed if it is more stable and has lower energy than the individual atoms. Normally only electrons in the outermost shell of an atom are involved in bond formation and in this process each atom attains a stable electronic configuration of inert gas. Atoms may attain stable electronic configuration in three different ways by loosing or gaining electrons by sharing electrons. The attractive forces which hold various constituents (atoms, ions etc) together in different chemical species are called chemical bonds. Elements may be divided into three classes.

$\bullet$Electropositive elements, whose atoms give up one or more electrons easily, they have low ionization potentials.

$\bullet$Electronegative elements, which can gain electrons. They have higher value of electronegativity.

$\bullet$Elements which have little tendency to loose or gain electrons.

Three different types of bond may be formed depending on electropositive or electronegative character of atoms involved.

Electropositive element + Electronegative element = Ionic bond (electrovalent bond)

Electronegative element + Electronegative element = Covalent bond

or less electro positive + Electronegative element = Covalent bond

Electropositive + Electropositive element = Metallic bond.

ELECTROVALENCY

This type of valency involves transfer of electrons from one atom to another, whereby each atom may attain octet in their outermost shell. The resulting ions that are formed by gain or loss of electrons are held together by electrostatic force of attraction due to opposite nature of their charges. The reaction between potassium and chlorine to form potassium chloride is an example of this type of valency.

Here potassium has one electron excess of it's octet and chlorine has one deficit of octet. So potassium donates it's electron to chlorine forming an ionic bond.

Here the oxygen accepts two electrons from calcium atom. It may be noted that ionic bond is not a true bond as there is no proper overlap of orbitals.

Criteria for Ionic Bond:

One of the species must have electrons in excess of octet while the other should be deficit of octet. Does this mean that all substance having surplus electron and species having deficient electron would form ionic bond? The answer is obviously no. Now you should ask why? The reasoning is that in an ionic bond one of the species is cation and the other is anion. To form a cation from a neutral atom energy must be supplied to remove the electron and that energy is called ionization energy. Now it is obvious that lower the ionization energy of the element the easier it is to remove the electron. To form the anion, an electron adds up to a neutral atom and in this process energy is released. This process is called electron affinity.

So for an ionic bond one of the species must have low ionization energy and the other should have high electron affinity. Low ionization energy is mainly exhibited by the alkali and alkaline earth metals and high electron affinity by the halogen and chalcogens. Therefore this group of elements are predominant in the field of ionic bonding.

Energy Change During the Formation of Ionic Bond

The formation of ionic bond can be consider to proceed in three steps

(a) Formation of gaseous cations

$A(g)+I.E.\to A^{+}(g)+e^{-}$

The energy required for this step is called ionization energy (I.E)

(b) Formation of gaseous anions

$X(g)+e^{-}\to X^{-}(g)+E.A$

The energy released from this step is called electron affinity (E.A.)

(c) Packing of ions of opposite charges to form ionic solid

$A^{+}(g)+X^{-}(g)\to AX(s)+energy$

The energy released in this step is called lattice energy.

Now for stable ionic bonding the total energy released should be more than the energy required.

From the above discussion we can develop the factors which favour formation of ionic bond and also determine its strength. These factors have been discussed below :

(a) Ionization energy: In the formation of ionic bond a metal atom loses electron to form cation. This process required energy equal to the ionization energy. Lesser the value of ionization energy, greater is the tendency of the atom to form cation. For example, alkali metals form cations quite easily because of the low values of ionization energies.

(b) Electron affinity: Electron affinity is the energy released when gaseous atom accepts electron to form a negative ion. Thus, the value of electron affinity gives the tendency of an atom to form anion. Now greater the value of electron affinity more is the tendency of an atom to form anion. For example, halogens having highest electron affinities within their respective periods to form ionic compounds with metals very easily.

(c) Lattice energy: Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three dimensionally in a definite geometric pattern to form ionic crystal.

Since the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied with the release of energy referred to as lattice energy. Lattice energy may be defined as the amount of energy released when one mole of ionic solid is formed by the close packing of gaseous ion.

In short, the conditions for the stable ionic bonding are:

(a) I.E. of cation forming atom should be low:

(b) E.A. of anion – forming atom should be high;

(c) Lattice energy should be high.

Born Haber Cycle

Determination of lattice energy

The direct calculation of lattice enthalpy is quite difficult because the required data is often not available. Therefore lattice enthalpy is determined indirectly by the use of the Born – Haber cycle. The cycle uses ionization enthalpies, electron gain enthalpies and other data for the calculation of lattice enthalpies. The procedure is based on the Hess's law, which states that the enthalpy of a reaction is the same, whether it takes place in a single step or in more than one step. In order to understand it let us consider the energy changes during the formation of sodium chloride from metallic sodium and chlorine gas. The net energy change during the process is represented by $\Delta$H_f.

Example1. Calculate the lattice enthalpy of$MgB{r}_2$. Given that

Enthalpy of formation of $MgB{r}_2$ = -524 kJ $mo{l}^{-1}$

Some of first & second ionization enthalpy (IE₁ + IE₂ ) = 148 kJ mol^-1

Sublimation energy of Mg = +2187 kJ$mo{l}^{-1}$

Vaporization energy of $B{r}_2$(I) = +31kJ $mo{l}^{-1}$

Dissociation energy of $B{r}_2$(g) = +193kJ$mo{l}^{-1}$]

Electron gain enthalpy of Br(g) = -331 kJ $mo{l}^{-1}$.

Solution: $\Delta {H^{\circ }}_f=S+I.E+\Delta H_{vap}+D+2\times E.A.+U$

or $U=\Delta {H^{\circ }}_f-\left(S+I.E+\Delta H_{vap}+D+2\times E.A.\right)$

or U = -524 - [2187 - 148 + 31 + 193 + 2 ´ (-331)]

= $-524-1897$= $-2421kJmo{l}^{-1}$

Characteristics of ionic compounds:

The following are some of the general properties shown by these compounds

(i) Crystalline nature: These compounds are usually crystalline in nature with constituent units as ions. Force of attraction between the ions is non-directional and extends in all directions. Each ion is surrounded by a number of oppositely charged ions and this number is called co-ordination number. Hence they form three dimensional solid aggregates. Since electrostatic forces of attraction act in all directions, therefore, the ionic compounds do not posses directional characteristic and hence do not show stereoisomerism.

(ii) Due to strong electrostatic attraction between these ions, the ionic compounds have high melting and boiling points.

(iii) In solid state the ions are strongly attracted and hence are not free to move. Therefore, in solid state, ionic compounds do not conduct electricity. However, in fused state or in aqueous solution, the ions are free to move and hence conduct electricity.

(iv) Solubility: Ionic compounds are fairly soluble in polar solvents and insoluble in non-polar solvents. This is because the polar solvents have high values of dielectric constant which defined as the capacity of the solvent to weaken the force of attraction between the electrical charges immersed in that solvent. This is why water, having high value of dielectric constant, is one of the best solvents.

The solubility in polar solvents like water can also be explained by the dipole nature of water where the oxygen of water is the negative and hydrogen being positive, water molecules pull the ions of the ionic compound from the crystal lattice. These ions are then surrounded by water dipoles with the oppositely charged ends directed towards them. These solvated ions lead an independent existence and are thus dissolved in water. The electrovalent compound dissolves in the solvent if the value of the salvation energy is higher than the lattice energy of that compounds.

$AB+Latticeenergy\to A^{+}+B^{-}$

These ions are surrounded by solvent molecules. This process is exothermic and is called solvation.

$A^{+}+x\left(solv.\right)\to {\left[A{\left(solv.\right)}_x\right]}^{+}+energy$

$B^{-}+y\left(solv.\right)\to {\left[B{\left(solv.\right)}_y\right]}^{-}+energy$

The value of solvation energy depend on the relative size of the ions. Smaller the ions more is the solvation. The non-polar solvents do not solvate ions and thus do not release energy due to which they do not dissolve ionic compounds.

(v) Ionic reactions: Ionic compound furnish ions in solutions. Chemical reactions are due to the presence of these ions. For example

$N{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$

$BaC{l}_2\to B{a}^{2+}+2C{l}^{-}$

COVALENCY

This type of valency involves sharing of electrons between the concerned atoms to attain the octet configuration with the sharing pair being contributed by both species equally. The atoms are then held by this common pair of electrons acting as a bond, known as covalent bond. If two atoms share more than one pair then multiple bonds are formed. Some examples of covalent bonds are

CO-ORDINATE COVALENCY

A covalent bond results from the sharing of pair of electrons between two atoms where each atom contributes one electron to the bond. It is also possible to have an electron pair bond where both electrons originate from one atom and none from the other. Such bonds are called coordinate bond or dative bonds. Since in coordinate bonds two electrons are shared by two atoms, they differ from normal covalent-bond only in the way they are formed and once formed they are identical to normal covalent –bond.

It is represented as $\left[\to \right]$

Atom/ion/molecule donating electron pair is called Donor or Lewis base. Atom / ion / molecule accepting electron pair is called Acceptor or Lewis acid $\left[\to \right]$ points donor to acceptor

$N{H_4}^{+},N{H}_3$ has three (N – H) bond & one lone pair on N – atom. In $N{H_4}^{+}$ formation this lone pair is donated to H⁺ (having no electron)

$N{H}_3+H^{+}\to N{H_4}^{+}$

Lewis base Lewis acid

Properties of the coordinate compounds are intermediates of ionic and covalent compounds.

Maximum Covalency

Elements which have vacant d-orbital can expand their octet by transferring electrons, which arise after unpairing, to these vacant d-orbital e.g. in sulphur.

In excited state sulphur has six unpaired electrons and shows a valency of six e.g. in SF₆. Thus an element can show a maximum covalency equal to its group number e.g. chlorine shows maximum covalency of seven.

Example 2. Which of the following statement is/are not true for $\sigma$-bond.

1. It is formed by the overlapping of s - s or s - p orbitals

2. It is weaker than pi bond

3. It is formed when p-bond exists already.

4. It is resulted from partial overlapping of orbitals.

(A) 1, 2, 3, 4 (B) 2, 3 and 4

(C) 2 and 4 (D) 1, 2 and 4

Solution: (B)

Example 3. The types of bond present in ZnSO₄.7H₂O are only

(A) Electrovalent and covalent

(B) Electrovalent and co-ordinate

(C) Electrovalent, Covalent and co- ordinate

(D) None of these

Solution: (C)

RESONANCE

There may be many molecules and ions for which it is not possible to draw a single Lewis structure. For example we can write two electronic structures of O₃.

In (A) the oxygen - oxygen bond on the left is a double bond and the oxygen-oxygen bond on the right is a single bond. In B the situation is just opposite. Experiment shows however, that the two bonds are identical. Therefore neither structure A nor B can be correct.

One of the bonding pairs in ozone is spread over the region of all the three atom rather than associated with particular oxygen-oxygen bond. This delocalised bonding is a type of bonding in which bonding pair of electrons is spread over a number of atoms rather than localised between two.

Structures (A) and (B) are called resonating or canonical structures and C is the resonance hybrid. This phenomenon is called resonance, a situation in which more than one plausible structure can be written for a species and in which the true structure cannot be written at all.

Some other examples

(ii) Carbon-oxygen bond lengths in carboxylate ion are equal due to resonance.

(iii) Benzene

(iv) Vinyl Chloride

Difference in the energies of the canonical forms and resonance hybrid is called resonance stabilization energy and provides stability to species.

Rules for writing resonating structures

$\bullet$Only electrons (not atoms) may be shifted and they may be shifted only to adjacent atoms or bond positions.

$\bullet$The number of unpaired electrons should be same in all the canonical form.

$\bullet$The positive charge should reside as far as possible on less electronegative atom and negative charge on more electronegative atom.

$\bullet$Like charge should not reside on adjacent atom

$\bullet$The larger the number of the resonating structures greater the stability of species.

$\bullet$Greater number of covalency add to the stability of the molecule.

Example 4. Out of the following resonating structures for CO₂ molecule, which are important for describing the bonding in the molecule and why?

Solution: Out of the structures listed above, the structure (III) is wrong since the number of electron pairs on oxygen atoms are not permissible. Similarly, the structures (II) has very little contribution towards the hybrid because one of the oxygen atoms (electronegative) is show to have positive charge. Carbon dioxide is best represented by structures (I) and (IV).

FACTORS GOVERNING POLARIZATION AND POLARISABILITY (FAJAN'S RULE)

$\bullet$Cation Size: Smaller is the cation more is the value of charge density ($\varphi$) and hence more its polarising power. As a result more covalent character will develop.Let us take the example of the chlorides of the alkaline earth metals. As we go down from Be to Ba the cation size increases and the value of $\varphi$ decreases which indicates that BaCl₂ is less covalent i.e. more ionic. This is well reflected in their melting points. Melting points of

BeCl₂ = 405°C and BaCl₂ = 960°C.

$\bullet$Cationic Charge: More is the charge on the cation, the higher is the value of $\varphi$ and higher is the polarising power. This can be well illustrated by the example already given, NaBr and AlBr₃. Here the charge on Na is +1 while that on Al in +3, hence polarising power of Al is higher which in turn means a higher degree of covalency resulting in a lowering of melting point of AlBr₃ as compared to NaBr.

$\bullet$Noble Gas vs Pseudo Noble Gas Cation: A Pseudo noble gas cation consists of a noble gas core surrounded by electron cloud due to filled d-subshell. Since

d-electrons provide inadequate shielding from the nuclei charge due to relatively less penetration of orbitals into the inner electron core, the effective nuclear charge (ENC) is relatively larger than that of a noble gas cation of the same period. NaCl has got a melting point of 800°C while CuCl has got melting point of 425°C. The configuration of Cu⁺ = [Ar] 3d¹⁰ while that of Na⁺ = [Ne]. Due to presence of d electrons ENC is more and therefore Cl^– is more polarised in CuCl leading to a higher degree of covalency and lower melting point.

$\bullet$Anion Size: Larger is the anion, more is the polarisability and hence more covalent character is expected. An e.g. of this is CaF₂ and CaI₂, the former has melting

point of 1400°C and latter has 575°C. The larger size of I^– ion compared to F^– causes more polarization of the molecule leading to a lowering of covalency and increasing in melting point.

$\bullet$Anionic Charge: Larger is the anionic charge, the more is the polarisability. A well illustrated example is the much higher degree of covalency in magnesium nitride

(3Mg⁺⁺ 2N^3–) compared to magnesium fluoride (Mg⁺⁺ 2F^—). This is due to higher charge of nitride compare to fluoride. These five factors are collectively known as Fajan's Rule.

Example 5. The melting point of KCl is higher than that of AgCl though the crystal radii of Ag⁺ and K⁺ ions are almost same.

Solution: Now whenever any comparison is asked about the melting point of the compounds which are fully ionic from the electron transfer concept it means that the compound having lower melting point has got lesser amount of ionic character than the other one. To analyse such a question first find out the difference between the 2 given compounds. Here in both the compounds the anion is the same. So the deciding factor would be the cation. Now if the cation is different, then the answer should be from the variation of the cation. Now in the above example, the difference of the cation is their electronic configuration. K⁺ = [Ar]; Ag⁺ = [Kr] 4d¹⁰. This is now a comparison between a noble gas core and pseudo noble gas core, the analysis of which we have already done. So try to finish off this answer.

DIPOLE MOMENT

Difference in polarities of bonds is expressed on a numerical scale. The polarity of a molecule is indicated in terms of dipole moment$\left(\mu \right)$. To measure dipole moment, a sample of the substance is placed between two electrically charged plates. Polar molecules orient themselves in the electric field causing the measured voltage between the plates to change.

The dipole moment is defined as the product of the distance separating charges of equal magnitude and opposite sign, with the magnitude of the charge. The distance between the positive and negative centres called the bond length.

Thus,

$\mu =electricch\arg e\times bondlength$= $q\times d$

As q is in the order of ${10}^{-10}$ esu and d is in the order of ${10}^{-8}$ cm, $\mu$is the order of ${10}^{-18}esucm$. Dipole moment is measured in 'Debye' unit (D)

$1D={10}^{-18}esucm=3.33\times {10}^{-30}coulombmetre$

Note:

(i) Generally as electronegativity difference increase in diatomic molecules, polarity of bond between the atoms increases therefore value of dipole moment increases.

(ii) Dipole moment is a vector quantity

(iii) A symmetrical molecule is non- polar even though it contains polar bonds. For example, $C{O}_2,B{F}_3,CC{l}_{4}etc$because summation of all bond moments present in the molecules cancel each other.

(iv) Unsymmetrical non-linear polyatomic molecules have net value of dipole moment. For example, $H_{2}O,C{H}_{3}OH,N{H}_3$etc.

Calculation of Resultant Bond Moments

Let AB and AC are two polar bonds inclined at an angle $\theta$ their dipole moments are ${\mu }_1$ and${\mu }_2$.

Resultant dipole moment may be calculated using vectorial method.

${\mu }_R=\sqrt{{\mu }_1^2+{\mu }_2^2+2{\mu }_1{\mu }_2\cos \theta }$

when $\theta$ = 0 the resultant is maximum

${\mu }_R={\mu }_1+{\mu }_2$

when, $\theta$= 180°, the resultant is minimum

${\mu }_R={\mu }_1\sim {\mu }_2$

Example 6. The compound which has zero dipole moment is

(A) CH₂Cl₂ (B) NF₃

(C) PCl₃F₂ (D) ClO₂

Solution: (C)

Example 7. Sketch the bond moments and resultant dipole moment in

(i)$S{O}_2$ (ii)$Cis-C_2{H}_{2}C{l}_2$ and (iii) $trans-C_2{H}_{2}C{l}_2$

Resultant m = 0

PERCENTAGE OF IONIC CHARACTER

Every ionic compound having some percentage of covalent character according to Fajan's rule. The percentage of ionic character in a compound having some covalent character can be calculated by the following equation.

The percent ionic character = \dfrac{Observed\stackrel{\scriptscriptstyle\to}{\leftarrow}dipole\stackrel{\scriptscriptstyle\to}{\leftarrow}moment}{Calculated\stackrel{\scriptscriptstyle\to}{\leftarrow}dipole\stackrel{\scriptscriptstyle\to}{\leftarrow}moment\text{ }assu\min g\stackrel{\scriptscriptstyle\to}{\leftarrow}100%\stackrel{\scriptscriptstyle\to}{\leftarrow}ionic\text{ bond}}\times 100

Example 8. Dipole moment of KCl is 3.336 x 10^–29 coulomb metre which indicates that it is highly polar molecule. The interatomic distance between k⁺ and Cl^– is

2.6 x 10^–10 m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.

Solution: Dipole moment $\mu$ = e x d coulomb metre

For KCl d = 2.6 x 10^–10 m

For complete separation of unit charge

e = 1.602 x 10^–19 C

Hence $\mu$ = 1.602 x 10^–19 2.6 x 10^–10 = 4.1652 x 10^–29 Cm

m_KCl = 3.336 x 10^–29 Cm

$\therefore$ % ionic character of KCl = $\frac{3.336\times {10}^{-29}}{4.165\times {10}^{-29}}\times 100$ = 80.09%

BOND CHARACTERISTICS

1. Bond Length: The distance between the nuclei of two atoms bonded together is termed as bond length or bond distance. It is expressed in angstrom $\left(\stackrel{\circ }{A}\right)$units or picometer (pm).

$\left[1\stackrel{\circ }{A}={10}^{-8}cm;1pm={10}^{-12}m\right]$

Bond length in ionic compound =$r_c^{+}+r_a^{-}$

Similarly, in a covalent compound, bond length is obtained by adding up the covalent (atomic) radii of two bonded atoms.

Bond length in covalent compound (AB) = $r_A+r_B$

The factors such as resonance, electronegativity, hybridization, steric effects, etc., which affect the radii of atoms, also apply to bond lengths.

Important features

(i) The bond length of the homonuclear diatomic molecules are twice the covalent radii.

(ii) The lengths of double bonds are less than the lengths of single bonds between the same two atoms, and triple bonds are even shorter than double bonds.

Single bond > Double bond > Triple bond (decreasing bond length)

(iii) Bond length decreases with increase in s-character since s-orbital is smaller than a

p – orbital.

$s{p}^{3}C-H=1.112A^{\circ };$ $s{p}^{2}C-H=1.103A^{\circ };$ $spC-H=1.08A^{\circ };$

(25% s-character as in alkanes) (33.3% s-character as in alkenes) (50% s-character as in alkynes)

(iv) Bond length of polar bond is smaller than the theoretical non-polar bond length.

2. Bond Energy or Bond Strength: Bond energy or bond strength is defined as the amount of energy required to break a bond in molecule.

Important features

(i) The magnitude of the bond energy depends on the type of bonding. Most of the covalent bonds have energy between 50 to 100 kcal $mo{l}^{-1}$(200-400 kJ$mo{l}^{-1}$). Strength of sigma bond is more than that of a $\pi$-bond.

(ii) A double bond in a diatomic molecules has a higher bond energy than a single bond and a triple bond has a higher bond energy than a double bond between the same atoms.

$C\equiv C>C=C>C-C$ (decreasing bond length)

(ii) A double bond in a diatomic molecules has a higher bond energy than a single bond and a triple bond has a higher bond energy than a double bond between the same atoms.

$C\equiv C>C=C>C-C$ (decreasing bond length)

(iii) The magnitude of the bond energy depends on the size of the atoms forming the bond, i.e. bond length. Shorter the bond length, higher is the bond energy.

(iv) Resonance in the molecule affects the bond energy.

(v) The bond energy decreases with increase in number of lone pairs on the bonded atom. This is due to electrostatic repulsion of lone pairs of electrons of the two bonded atoms.

(vi) Homolytic and heterolytic fission involve different amounts of energies. Generally the values are low for homolytic fission of the bond in comparison to heterolytic fission.

(vii) Bond energy decreases down the group in case of similar molecules.

(viii) Bond energy increase in the following order:

\begin{align}  \,\,\,\,\,\,\,\,\,\,\,s<p<sp<s{{p}^{2}}<s{{p}^{3}} \\ C-C>N-N>O-O \\ \left( No\,\,lone\,\,pair \right)\left( One\,\,lone\,\,pair \right)\left( Two\,\,lone\,\,pair \right) \\\end{align}

3. Bond angles: Angle between two adjacent bonds at an atom in a molecule made up of three or more atoms is known as the bond angle. Bond angles mainly depend on the following three factors:

(i) Hybridization: Bond angle depends on the state of hybridization of the central atom

Generally s- character increase in the hybrid bond, the bond angle increases.

(ii) Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.

(iii) Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.