CHEMICAL BONDING 1
Chapter Contents
Orbital overlap and covalent bond; Hybridization involving s, p and d orbitals only; Orbital energy diagrams for homonuclear diatomic species; Hydrogen bond; Polarity in molecules; Dipole moment (qualitative aspects only); VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral) and various levels of multiple-choice questions.
CHEMICAL BOND
Chemical bond is the force of attraction that binds two atoms together. A chemical bond balances the force of attraction and force of repulsion at a particular distance.
A chemical bond is formed to: attain the octet state minimize energy gain stability decrease reactivity
When two atoms come close to each other, forces of attraction and repulsion operate between them. The distance at which the attractive forces overcome repulsive forces is called bond distance. The potential energy for the system is lowest, hence the bond is formed.
Types of Bonds
1.1
Following are the six types of chemical bonds. They are listed in the decreasing order of their respective bond strengths.
1. Ionic bond
2. Covalent bond
3. Coordinate bond
4. Metallic bond
5. Hydrogen bond
6. van der Waals bond
Fig.
Metallic bond, hydrogen bond and van der Waals bond are interactions.
Octet Rule
It was introduced by Lewis and Kossel. According to this rule, each atom tries to obtain the octet state, that is, a state with eight valence electrons.
Exceptions to the octet rule
Transition metal ions like Cr3+, Mn2+ and Fe2+ .
Pseudo inert gas configuration cations like Zn2 + and Cd2 +
Contraction of octet state
The central atom is electron deficient or does not have an octet state. For example,
Expansion of octet state
The central atom has more than 8 electrons due to empty d-orbitals. For example, PCl5, SF6, OsF8, ICl3, etc.
P Cl 5 10 S F6 12 Os F 8 16 I CI 3 etc. 10 e
Odd electronic species like NO, NO2 and ClO 2
Interhalogens compounds like IF7 and BrF3.
Compounds of xenon such as XeF2, XeF4 and XeF6
IONIC OR KERNEL BOND
An ionic bond is formed by the complete transfer of valence electrons from a metal to a nonmetal. This was first studied by Kossel. For example,
Na + Cl Na+ C1
(2, 8, 1) (2, 8, 7) (2, 8) (2, 8)
Mg + O Mg+2 O 2
(2, 8, 2) (2, 6) (2, 8) (2, 8)
Al + N Al+3 N 3
(2, 8, 3) (2, 5) (2, 8) (2, 8)
Number of electrons transferred is equal to electro-valency.
Maximum number of electrons transferred by a metal to non-metal is three, as in the case with AlF 3 (Al metal transfers three electrons to F).
During electron transfer, the outermost orbit of metal is destroyed. The remaining portion is called core or kernel, hence this bond is also called kernel bond.
Nature of ionic bond is electrostatic or coloumbic force of attraction. It is a non-directional bond.
Conditions for the Formation of an Ionic Bond
The process of bond formation is exothermic where ΔH= Ve. The essential conditions include the following:
Metal must have low ionization energy. Non-metals must have high electron affinity. Ions must have high lattice energy. Cation should be large with low electronegativity.
Anion must be small with high electronegativity.
Born–Haber Cycle
The formation of an ionic compound in terms of energy can be shown by Born–Haber cycle. It is also used to find lattice energy, ionization energy and electron affinity.
For example,
M(s) +S
Sublimation M(g) +I Ionization M+(g) + e ½ X2 +1/2 D
Decomposition X(g) E Addition of e X (g)
M (g) + + X (g) U Crystal formation MX(g)
ΔHf = S + 1/ 2 D + I E U
Here,
S = Heat of sublimation
D = Heat of dissociation
I = Ionization enthalpy
E = Electron gain enthalpy or electron affinity
U = Lattice energy
For the formation of an ionic solid, energy must be released during its formation, that is, ΔH must be negative. E U > S + ½ D + I
Properties of Ionic Compounds
1. Ionic compounds have solid crystalline structures (flat surfaces), with definite geometry, due to strong electrostatic force of attraction as constituents are arranged in a definite pattern.
2. These compounds are hard in nature. Hardness ∝ Electrostatic force of attraction
∝ Charge on ion
∝ 1 Ionic radius
3. Ionic compounds have high value of boiling point, melting point and density due to strong electrostatic force of attraction. Boiling point, melting point ∝ Electrostatic force of attraction
Volatile nature
∝ 1 Electrostatic force of attraction
4. Ionic compounds show isomorphism, that is, they have same crystalline structure. For example, all alums, NaF and MgO, ZnSO4 7H2O and FeSO4⋅7H2O.
5. These are conductors in fused, molten or aqueous state due to the presence of free ions. In solid state these are non-conductors as no free ions are present.
6. They show fast ionic reactions as activation energy is zero for ions.
7. They do not show space isomerism due to non-directional nature of ionic bond.
8. They have high latice energy (U). It is released during the formation of an ionic solid molecule from its constituent ions.
Lattice Energy
Lattice energy is the energy needed to break an ionic solid molecule into its constitutent ions. It is denoted by U.
U ∝ Charge on ion
∝ 1 Size of ion
Hence, lattice energy for the following compounds increases in the order shown under:
NaCl < MgCl2 < AlCl 3 < SiCl4
As charge on a metal atom increases, its size decreases.
In case of univalent and bivalent ionic compounds, lattice energy decreases as follows:
Bi-bi > Uni-bi or Bi-uni > Uni-uni
For example, MgO > MgCl2 > NaCl.
Some orders of lattice energy
(i) LiX > NaX > KX > RbX > CsX
(ii) LiF > LiCl > LiI
(iii) AgF > AgCl > AgI
(iv) BeO > MgO > CaO > SrO
9. Ionic compounds are soluble in polar solvents like water due to high dielectric constant of these solvents. The force of attraction between ions is destroyed and hence they dissolve in the solvent.
Facts Related to Solubility
If ΔH (hydration) > Lattice energy then the ionic compound is soluble.
If ΔH (hydration) < Lattice energy then the ionic compound is insoluble
If ΔH (hydration) = Lattice energy then the compound is at equilibrium state
Some Solubility Orders
a. LiX < NaX < KX < RbX < CsX
b. LiOH < NaOH < KOH < RbOH < CsOH
c. BeX 2 < MgX2 < CaX 2 < BaX 2
d. Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Ba(OH)2
e. BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4
f. AIF 3 > AlCl 3 > AlBr 3 > AlI 3
Crystals of high ionic charges are less soluble. For example, compounds of CO 3–2, SO4–2, PO4–3 are less soluble.
Compounds Ba+2, Pb+2 are insoluble as lattice energy > ΔHhy.
Compounds of Ag (salts) are insoluble as lattice energy > ΔH hy
Presence of common ions decrease solubility. For example, solubility of AgCl decreases in presence of AgNO3 or KCl, due to the presence of common ions, that is, Ag+ and Cl respectively.
Note: The concept of ionic bond is not a part of IIT-JEE syllabus but has been discussed for the better understanding of the chapter.
COVALENT BOND
A covalent bond is formed by the equal sharing of electrons between two similar or different atoms.
If atoms are same or their electronegativity is same, the covalent bond between them is non-polar. For example, X – X, O = O, N ≡ N
If atoms are different or have different value of electronegativity, the covalent bond formed between them is polar. For example, +δ δ +δ, +δ δ H O H, H X
The number of electrons shared or covalent bonds represent covalency.
One atom can share maximum three electrons with another atom. For example, in ammonia the covalency of nitrogen atom is three.
Properties of Covalent Compounds
1. Covalent compounds mostly occur in liquid and gaseous state but if molecular weight of the compound is high they may occur in solid state too.
For example, F2, Cl 2 Br 2 I 2 ‘g’ ‘g’ ‘l’ ‘s’
Glucose Sugar
Molecular wt. 180 342 (less solid) (more solid)
2. Solubility of these compounds follows the concepts ‘like dissolves like’, that is, nonpolar solute dissolves in non-polar solvent. For example, CCl4 dissolves in organic solvents. Similarly, polar solutes dissolves in polar solvent. For example, alcohol and ammonia dissolve in water.
3. Covalent compounds have lower boiling point and melting point values than those of ionic compounds. This is because covalent bond is a weak van der Waals force in nature.
For example, KOH > HX Strong Weak ionic force van der of attarction Waals forces
Boiling point and melting point ∝ Hydrogen bonding ∝ Molecular weight
For example,
H2O, H2Te H 2 Se H 2 S
HF > HI > HBr > HCl
Due to As molecular H-bonding weight decreases
4. Covalent compounds are non-conductors due to absence of free ions, but graphite is a conductor, as the free electrons are available in its hexagonal sheet like structure. In case of diamond, the structure is tetrahedral hence free electrons are not available. Therefore, it is not a conductor.
5. Covalent bond is directional, hence these compounds can show structural and space isomerisms.
6. The reactions involving covalent bond are slow as these need higher activation energy.
COORDINATE OR DATIVE SEMI-POLAR BOND
Coordinate bond is a special type of bond which is formed by donation of electron pair from the donor to the receiver, that is, it involves partial transfer or unequal sharing of electrons. It is denoted as ( ) from donor to receiver.
A: + B (A B)
Donor or Receiver Lewis base Lewis acid
Coordinate bond is intermediate between ionic and covalent bonds, but more closely resembles a covalent bond. The properties of coordinate compounds are more close to covalent compounds. For example,
Sugden Linkage
Sugden or singlet linkage is formed by donation of one electron and is denoted by ( ). For example, PCl5, SF6, IF7.
It was studied for the first time by Lorry and Sidwick.
Fig. 1.2
MODERN CONCEPT OF COVALENT BOND
The nature of covalent bond is explained on the basis of Heitler–London’s Valence bond theory,
Pauling and Slater’s overlapping theory and Hund, Mullikan’s theory.
Valence Bond Theory or Heitler–London Theory
Orbital concept of covalent bond was introduced by Heitler and London. According to this concept, “A covalent bond is formed due to the half-filled atomic orbitals having electrons with opposite spin to each other”.
Features
1. The atoms should have unpaired electrons to form a covalent bond.
2. A covalent bond is formed by the pairing of electrons.
3. The maximum electron density lies between the bonded atoms.
4. There is a tendency to form close shells of the atom, though the octet is not attained in BeCl2, BF3, etc., and is exceeded in PCl5, SF6, IF7, etc.
Due to overlapping the potential energy of system decreases.
The internuclear distance with maximum overlapping and greater decrease of potential energy is known as bond length.
Limitations
1. It cannot explain the formation of odd electron molecule such as ClO3, NO, H2+ , etc. In such molecules, electron pairing does not take place.
2. It cannot explain the formation of coordinate bond where only one atom donates a lone pair of electrons to the other atom, molecule or ion.
3. It cannot explain the formation of π-bond.
4. It cannot explain the stereochemistry of the molecules and ions.
Pauling and Slater’s Theory
It deals with the directional nature of the bond formed and is simply an extension of Heitler–London theory.
1. Greater the overlapping, stronger will be the bond formed. It means bond strength depends upon the overlapping and is directly proportional to the extent of overlapping.
2. A spherically symmetrical orbit, say, s-orbital will not show any preference in direction whereas non-spherical orbitals, say, p- or d-orbitals will tend to form a bond in the direction of maximum electron density with the orbital.
REMEMBER
The overlapping of the orbitals of only those electrons which take part in the bond formation will occur and not with the electrons of other atoms.
The wave function of an electron of s-orbital is spherically symmetrical, therefore such an electron exhibits no directional preference in bond formation. The two orbitals having similar energy level, the one which is more directionally concentrated, will form a stronger bond.
Strongness of overlapping ∝ mode of overlapping.
For example, linear overlapping is stronger than lateral overlapping.
Strongness of overlapping ∝ directional nature of orbital.
For example, p-p > s-p > s-s > p-p
Linear or axial Lateral overlapping overlapping
Types of Overlapping
1. s-s Overlapping: Overlapping between s-s electrons of two similar or dissimilar atoms is called s-s overlapping and forms a single covalent bond.
Fig. 1.3 Formation of hydrogen molecule by s-s overlapping.
2. s-p Overlapping: Overlapping between s and p electrons is called s-p overlapping. NH3 is formed by the overlapping between three electrons of nitrogen (px, py and pz) with three electrons of three hydrogen atoms.
7N = 1s2, 2s2, 2px1 py1 pz1
1 H = 1s1
Strong bond can be formed only when hydrogen electrons approach in the direction of X, Y and Z axis at right angles to each other.
s — p
Fig. 1.4
17 Cl =1s2, 2s2 2p6, 3s2, 3px2, 3py2, 3pz1 p p
Fig. 1.5
Some Important Features of Bond Bond Length
Bond length is the average distance between the centers of the nuclei of the two bonded atoms.
It is determined by X-ray diffraction and spectroscopic methods.
In case of ionic compounds, it is the sum of ionic radius of cation and anion, while in case of covalent compounds, it is sum of their covalent radius.
Factors affecting bond length
Bond length ∝ Size of atom. For example, HF < HCI < HBr < HI (Atomic size)
Since F < CI < Br < I
Bond length ∝ 1 Bond order or multiplicity
For example, C – C > C = C > C ≡ C 1.54Å 1.34Å 1.32Å
Bond length ∝ 1 s%
that is, sp3 > sp2 > sp s% 25% 33% 50%
Bond length ∝ Electronic repulsion
3. p-p Overlapping: p-p overlapping is formed by the overlapping of the p-orbitals of the atoms. In case of chlorine molecule, it is formed by the overlapping of the 3pz orbitals of two chlorine atoms.
For example, H2 > H 2 +
Resonance and hyperconjugation also change bond length.
For example, in benzene, C – C bond length is 1.39 Å, that is, in between C – C and C=C.
Bond Energy
It is the energy needed to break one mole of bond of a particular type, so as to separate them into gaseous atoms. It is also called bond dissociation energy.
Bond energy can also be defined as the energy released during the formation of one mole of a particular bond.
Factors affecting bond energy
Bond energy ∝ Bond order or multiplicity
For example, C ≡ C > C = C > C – C
Bond energy ∝ 1 Bond length or size of atom
For example,
HF > HCl > HBr > HI
Bond energy ∝ s% or s-orbital character involved in hybridisation
For example, sp, sp2, sp3
50% 33% 25%
Bond energy ∝ 1 Lone pair of electrons/ electronic repulsion
For example, C C > N N > O O > F F lp e
Some diatomic molecules in order of bond energy are
C = O > N ≡ N > C ≡ N > C ≡ C
Bond Angle
It is the angle between the lines representing the directions of the bonds or the orbitals having bonding pair of electrons.
Factors affecting bond angle
Bond angle ∝ Bond order ∝ s%
1
Bond length
For example, 180
Bond angle is also affected by electronic repulsion (see VSEPR theory).
For example, NH4+ > NH3 > NH2 no lp 1 lp e 2 lp e
Bond angle ∝ 1 Size of terminal atom
For example, I2O > Br 2 O > Cl 2 O > OF 2
Bond angle ∝ 1 Size of central atom/electronegativity
Normally, bond angle decreases when we move down the group, as electronegativity decreases.
For example, NH3 > PH 3 > AsH 3 > BiH 3 H 2 O > H 2 S > H 2 Se > H 2 Te BF 3 > PF 3 > ClF 3
Bond angle ∝ Electronegativity of terminal atom
For example, PF3 > PCl 3 < PBr 3 < PI 3
PF 3 has more bond angle than PCl3 due to pπ-dπ bonding.
REMEMBER
PF 3 has greater bond angle when compared to PH 3 due to resonance in PF3, where a double bond character develops. NH3 has more bond angle value than NF3 as F-atom pulls the bpe– away from N-atom in NF3
POLORIZATION AND FAJAN’S RULE
When cation and anion are close to each other, the shape of anion is distorted by the cation. This is known as polarization. Due to this, covalent nature develops in an ionic molecule.
Polarization ∝ Covalent nature ∝ 1 Ionic nature
Polarization
Distorted anoin
Fig. 1.6
Fajan’s Rule
Polarization or covalent nature is explained by the following rules:
Charge on Cation Polarization, covalent nature or polarizing power of a cation ∝ charge on cation. That is greater the charge on cation, greater will be its polarizing power and more will be covalent nature.
For example,
SiCl4 > AlCl 3 > MgCl2 > NaCl
Size of Cation When the charge is same and the anion is common, consider that the covalent nature ∝ 1
Size of cation
That is, smaller cation has more polarizing power.
For example,
LiCl > NaCl > KCl > RbCl > CsCl
Max. covalent
Least ionic
Li+ < Na+ < K+ < Rb+ < Cs+
Max. ionic
Least covalent
Smallest Largest in size in size
Size of Anion This property is taken into account when the charges are same and the cation is common.
Polarization or covalent nature ∝ size of anion. Hence, larger anions are more polarized.
For example, LiF < LiCl < LiBr > LiI
Since, F < Cl < Br < I Larger the size of anion, easier will be its polarization.
A cation with 18 valence electrons has more polarizing power than a cation with 8 valence electrons.
For example,
Group IB > Group IA
Cu+ Na+
Αg+ K+
Group IIB > Group IIA
Zn+2 Mg+2
For example, ZnO > MgO
Zn+2 Mg+2
2, 8, 18 2, 8
REMEMBER
As the covalent nature increases, the intensity of the colour increases. For example, FeCl 3 is reddish-brown while FeCl2 is greenish-yellow.
Dipole Moment
Fig. 1.7
Dipole moment is used to measure the polarity in a molecule. It is denoted by μ. Mathematically, it is given as
μ = q × r coulomb metre
μ = e × d esu cm
1 debye = 1 × 10 18 esu cm.
It is represented by ( ) from electropositive to electronegative species or less electronegative to more electronegative species. For example, AX3
In case of para forms M μnet is positive if both the species are different. For example,
0
In case a molecule has more than one polar bonds μnet is given as follows:
μnet= √ μ12+μ22 +2μ1,μ 2 Cos θ
Dipole moment ∝ Electronegativity difference. For example, HF > HCI > HBr > HI
Dipole moment ∝ Number of lone pair of electrons.
For example, HF > H2O > NH3
Fluorine has 3 lone pair, oxygen has 2 lone pair, and ammonia has 1 lone pair of electron.
Dipole moment ∝ 1 θ
ortho > meta > para
For example,
Homoatomic molecules like X2, N2, O2 and molecules having normal shapes according to hybridization like linear, trigonal, tetrahedral will be non-polar, as for them, the dipole moment is zero. For example, BX3, CH4, CCl4, SiCl4, PCl5
Here, μ (net) = 0 as C = O bonds are in opposing directions.
Molecules in which the central atom has lone pair of electrons or have distorted shapes, like angular, pyramidal, sea-saw shapes will have some value of dipole moment and will be polar in nature. For example, H2O, H2S, OF2, NH3, PH3, PCl3, SCl4, SO2, SnCl2 etc.
(net) = 1.82D
Ammonia has more dipole moment than NF3 as in ammonia μ (net) is in the direction of lone pair electrons i.e., it is additive while in NF3 μ (net) is opposite to lone pair i.e., substractive. Dipole moment of a cis-alkene is more than trans-alkene. In trans-alkenes, it is zero due to symmetry in most of the cases.
Dipole Moment of Some Common Molecules
Dipole moment
Molecule
Dipole moment 1.021.460.170.921.840.551.150.582.933.91.860000000 cis-but-2-enetrans-but-2-ene
Exception: Unsymmetric alkenes with odd number of carbon atoms have some value of dipole moment.
For example, trans-2-pentene
Specific cases of dipole moment
→ CH 3 Cl > CH 2 Cl 2 > CHCl 3 > CCl4 Highly polar Non-polar → CH 3 Cl > CH 3 F > CH 3 Br > CH 3 I
Uses
To find geometry of a complex/molecule etc. To find ionic character or nature in a covalent species.
Ionic nature % = μobserved μcalculated (q × r ) × 100
To distinguish between cis and trans alkenes
Cis-but 2-ene > trans but-2-ene μ=+ve μ=0
Illustrations
1. The experimentally determined dipole moment of KF is 2.87 × 10 29 cm. The distance between the centers of charge in a KF dipole is 2.66 × 10 10 m. Calculate the percentage ionic character of KF.
Solution
μ = e × d coulombs meter
For KF = 2.66 × 10 10 m
For complete separation of a unit charge (electronic charge), e = 1.602 × 10 19 C
μ = 1.602 × 10 19 × 2.66 × 10 10 = 4.26 × 10 29 cm
% of ionic character of a KCl × × = × 29 29 2.8710 100 4.2610 = 67.4%.
2. The dipole moment of KCl is 3.336 × 10 29 coulomb meter which indicates that it is a highly polar molecule. The interatomic distance between K+ and Cl in this molecule is 2.6 × 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
μ = e × d coulombs meter
For KCl d = 2.6 × 10 10 m
For complete separation of unit charge (electronic charge), e = 1.602 × 10 19 C
μ = 1.602 × 10 19 × 2.6 × 10 10 = 4.1652 × 10 29 cm
μ (KCl) = 3.336 x 10 29 cm
Per cent ionic character of KCl = 3.336 × 10 29 4.1652 × 10 29 × 100 = 80.09%
SIGMA AND PI BONDS
Sigma (σ)
Bond
Sigma bond is formed by axial or headtohead or linear overlapping between two s – s or s – p or p – p orbitals.
Sigma(σ) bond S – SS PPP σ bond σ bond
Fig. 1.8
1. Sigma bond is stronger and therefore less reactive, due to more effective and stronger overlapping than σ bond.
2. The minimum and maximum number of σ bonds between two bonded atoms is 1.
3. Stability ∝ Number of sigma bonds.
4. Reactivity ∝ 1 σ
5. In sigma bond, free rotation of the atoms is possible.
6. Sigma bond determines the shape of molecules.
Pi (π) Bond
Pi bond is formed by lateral or sidewise overlapping between the two p orbitals. π bond ++
Fig. 1.9
1. It is a weak or less stable bond, and therefore more reactive, due to less effective overlapping.
2. Minimum and maximum number of π bonds between two bonded atoms are 0 and 2, respectively.
3. Stability ∝ 1 Number of π bonds .
4. Reactivity ∝ Number of π bonds.
5. In case of a π bond, free rotation is not possible.
6. It does not determine the shape of a molecule but shortens the bond length.
(C − C > C = C > C ≡ C) π-bonds 1 2
Strength of s- and π-Bonds
The strength of a bond depends upon the extent of overlapping of half-filled atomic orbitals. The extent of overlapping between two atoms is always greater when there is end-to-end overlapping of orbitals. Therefore, a σ-bond is always stronger than π-bond.
REMEMBER
It is the number of unpaired s- or p-electrons present in its atom in the ground state. Thus, covalency of hydrogen atom is 1.
To find sigma and pi bonds in a molecule
Single bond = 1σ
Double bond = 1σ, 1π
Triple bond = 1σ, 2π
For example,
1. Enolic form of acetone
It has 9 σ and 1 π bond, and 2 lone pairs of electrons.
2. C2(CN)4 NC NC CC NC NC
It has 9 σ and 9 π bond, and 4 lone pairs of electrons.
It has 12σ and 3π bonds.
4. Buta-1, 3-diene
CCC
It has 9σ and 2π bonds.
5. CaC2 Ca2+ (C ≡ C)2
It has 1σ, 2π bonds.
HYDROGEN BOND
Hydrogen bond was introduced by Latimer and Rode-bush.
It is a weak interaction denoted by dotted (…) lines between hydrogen and a highly electronegative and small sized atom like F, O and N. Here, the hydrogen atom is covalently bonded to any of these.
The nature of a hydrogen bond is either dipole–dipole type, ion–dipole type or dipole–induced dipole type
HCl has no H-bonding as chlorine is large in size.
H-bond strength for the following order is 10 kcal per mole, 7 kcal per mole and 2 kcal mole, respectively. HF > H2O > NH3
Hydrogen bonding is of the following two types:
