Chemical+Bonding+-A

CHEMISTRY LECTURE NOTES COURSE : VIKAAS (A) (LECTURE No. 1 TO 20)

TOPIC : CHEMICAL BONDING

LECTURE # 1 NOTE : All faculty members are required to read unit 4 : Chemical Bonding and molecular stucture from NCERT text book CHEMICAL BOND : A force that acts between two or more than two atoms to hold them together as a stable moleculeis known as a chemical bond. Why chemical bond forms between atoms (A) Potential energy concept : Every system tries to minimise their energy. Here system of two or more atom undergoes chemical bond formation to get minimise their P.E.

r = distance between two nucleus of different atom. (B)

KOSSEL - LEWIS CONCEPT

*

Lewis pictured the atom in terms of a positively charged 'Kernel' (the nucleus plus the inner electrons) and the outer shell that could accommodate a maximum of eight electrons.

*

Lewis found that octet of electrons, represents a particularly stable electronic arrangement. To achieve octet of electron atom can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons. This is known as octet rule

Lewis Symbols: In the formation of a molecule, only the outer shell electrons take part in chemical combination and they are known as valence electrons. For example, the Lewis symbols for the elements of second period are as under:

Ionic bond (Electrovalent bond) (i)

It involves complete transfer of one or more electrons from the valence shell to the other atom. Ionic bond is always form between electro positive and electro negative elements.

or Mg+2 (F–)2

Page # 2

(ii)

The capacity of an element is terms of number of loss of e– or gain of electron to form ionic bond is termed electrovalency

Cavalent bond (i) A covalent bond is formed by mutual sharing of one of more electron pairs between two atoms to aquire nearest noble gas configuration. (ii) Sharing of one e–  single bond ’’ two e–  Double bond ’’ three e–  Triple bond (iii)

Covalent bond energy arises due to forces of attraction between nuclei and shared pair of electron.

or Cl – Cl 8e–

8e–

• Thus in water and carbon tetrachloride molecules, formation of covatent bonds can be represented as: –

Cl Cl H –

2e

H

O 8e–

8e

–

2e–

8e

C

Cl

Cl

8e

–

–

8e

H atom attain a duplet of electrons and O , the octet

Each of the four Cl atoms along with the C atom attains octet of electrons

For example, in the carbon dioxide molecule, we have two doublebonds between the carbon and oxygen atoms. Similarly in ethene molecule the two carbon atoms are joined by a double bond.

O

C

O

or O

8e–

8e–

8e–

C

O

Double bonds in CO2 molecule

H C H

8e

–

C

or

C 8e

–

H

H

H

C H

H

H

C2H4 molecule When combining atoms share three electron pairs as In the case of two nitrogen atoms in the N2 molecule and the two carbon atoms in the ethyne molecule, a triple bond is formed.

N

N

–

C 8e–

N

–

8e

H

or

8e N2 molecule

C

H

or

8e–

C2H2 molecule

Page # 3

Co-ordinate bond or dative bond is special type of covalent bond is which shared bond pair of electron is donated by one atom, called doner and other atom is acceptor. + H | H–N: | H

or [ A  B]



F | B–F | F



H F     | | H – N  B – F   | |  h F 

NH3 + H+ = NH3  H+

or

NH4+

Ex. Let’s try Lewis dot structure of following (a) O2

O

(b) O3 

O

(c) NF3

O

O

F

F

O

C

(d) CO32–

N F

(f) CO C

(e) HNO3 H–O – N –

O

–

O

O

O

O

(g) NO2–

(h) CaF2

N O

(i) NH4NO3

–

(j) CuSO4 Pro. 7 Write the Lewis dot structure of CO molecule. Sol. Step 1. Count the total number of valence electrons of carbon and oxygen atoms. The outer (valence) shell configuration of carbon and oxygen atoms are : 2s2 2p2 and 2s2 2p4 , respectively. The valence electrons available are 4 + 6 = 10. Step 2. The skeletal structure of CO is written as : CO Step 3. Draw a single bond (one shared electron pair) between C and O complete the octet on O , the remaining two electrons are the lone pair on C. This does not complete the octed on carbon and hence we have to resort to multiple bonding (in this case a triple bond) between C and O atoms. This satisfies the octet rule condition for both atoms.

Pro.4.2 Write the Lewis structure of the nitrite ion , NO2 – . Sol. Step 1. Count the total number of valence electrons of the nitrogen atom , the oxygen atoms and the additional one negative charge (equal to one electron). N (2s2 2p3) , O (2s2 2p4) 5 + (2 × 6) + 1 = 18 electrons Step 2. The skeletal structure of NO2 – is written as : O

N

O

Step 3. Draw a single bond (one shared elecron pair) between the nitrogen and each of the oxygen atoms completing the octets on oxygen atoms. This , however , does not complete the octet on nitrogen if the remaininig two electrons constitute lone pair on it.

IONIC BOND (ELECTROVALENT BOND) : The chemical bond formed between two or more atoms as a result of transfer of one or more electrons between them.

Page # 4

FAVOURABLE CONDITION : (i) One of the atom must be electro +ve can easily loose e–s or should have few more e–s than a noble gas. It should have low .E value (ii) Other element should be a electro –ve element having high .E. value, and more negative value of electron gain enthalpy, having few e–s less than noble gas. (iii) Energy released because of the combination of cation and anion should be high.This energy is also defined in terms of lattice enthalpy. Lattice Enthalpy The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCI is 788 kJ mol-1. This means that 788 kJ of energy is required to separate one mole of solid NaCI into one mole of Na+1 (g) and one mole of Cl–(g) to an infinite distance. This process involves both the attractive forces between Ions of opposite charges and the repulsive forces between ions of like charge. The solid crystal being three- dimensional; it is not possible to calculate lattice enthalpy directly from the interaction of forces of attraction and repulsion only. Factors associated with the crystal geometry have to be included. Na+(g) + Cl–(g)  NaCl(s)

H = Hlattice  –ve (always)

Mg2+(g) + 2Cl– (g)  MgCl2 (s)

LECTURE # 2 FACTORS AFFECTING L.E. (i)

Lattice energy (L.E.) 

1 r

r = r+ + r– = interionic distance

(ii)

L.E.  Z+, Z– Z+  charge on cation in terms electronic charge Z–  charge on anion in terms electronic charge

(iii)

L.E.  cordination number (C.N.) C.N. of cation is no. of anion surrounding the cation. C.N. of anion is no. of cation surrounding the anion C.N. factor is important only in case of sulphates and carbonates of alkaline earth metals and coordination number is calculated by experimently or by radius ratio. BeSO4 < MgSO4 < CaSO4 < SrSO4 < BaSO4 Size of cation  C.N.  L.E.  (a) NaCl > KCl (size) (b) NaCl < MgO (size, charge) (c) NaCl < MgCl2 (size) In size and charge, charge will dominate Na2O > NaF NaCl < Na2S (d)

Al2O3 Na2O MgO Al2O3 > MgO > Na2O

Note : Dont discuss melting point of ionic compound here. CALCULATION OF L.E. : Indirect methods : Born-Haber Cycle ( Hess’s law) Hess’s Law  the net enthalpy change of a chemical reaction or of any process always remain same whether the reaction takes place in 1 step or many step

Page # 5

Born Haber Cycle for NaCl (s)

Na (s)

Hf 1 Cl (g) 2 2 1( H ) BE 2 Cl(g)

+

Hsub Na(g) HI.E.

NaCl(s)

HL.E.

HEg

+

-

Na (g) + Cl (g) Hf  enthalpy of formation of any compound is defined as the enthalphy change when 1 mol of that compound is formed form the elements in their standard states =

Hf +ve/–ve

Generally Q.

+

Hsub

H.E.

1 H 2 BE

+







+ve

+ve

+ve

+

Heg

+

HL.E.

 +ve/-ve

Born haber cycle for MgCl2(s) & calculate its lattice energy in terms HSub(Mg), HI.E1 (Mg),  HI.E 2 (Mg), Heg(Cl); HB.E.(Cl2(g)) + Hf(Mg Cl2,(s))

Mg (s) + Cl2(g) Hsub Mg(g) HI.E

MgCl2(s)

HB.E 2Cl(g)

1

HL.E

+

Mg (g) HI.E

Hf

2 Heg 2

2+

Mg (g) + 2Cl–(g) Hf = Hsub + HI.E1 +  HI.E 2 + HBE + Heg + HL.E. Q.

Born haber cycle for Al2O3(s)

2Al (s) + Hsub 2Al(g) HI.E

1

2Al (g)

3O (g) 2

2Al (g)

3+

1

–

2+

HI.E

Al2O3(s)

HL.E

+

HI.E

3 O (g) Hf 2 2 3 H B.E 2 3O(g) Heg

Heg

2

3

2–

2Al (g) + 3O (g)

Hf = 2Hsub + 2 HI.E1 + 2  HI.E 2 + 2HI.E3 +

3 HBE + 3Heg1 + 3Heg2 + HL.E. 2

Page # 6

Q.

Born haber for NaBr

Na (s)

+

Hsub

1 Br () 2 2 1 2

Hf Hvap

1 Br (g) 2 2 1 H B.E. 2 Br(g)

Na(g)

HI.E.

NaBr(s)

H

HEg +

-

Na (g) + Br (g)

Born haber for Na

Na (s) Hsub

+

Hf 1 I (s) 2 2 1 H sub(I (s)) 2

NaI(s)

2

Na(g)

Ex.

1 I2 (g) 2 1 H B.E. 2 I(g)

MgO  found as Mg2+O2– but not as Mg +O– Mg2+ O2– L.E. is very large (-ve) Mg+O– L.E. is –ve

General characteristics of ionic compounds : (a) Physical state (c) Conductance (e) Brittlrness (g) Isomorphism

(b) Melting and boiling points (d) Crystal structure (f) Solubility

Solvation or Hydration :

Whenever any compound generally ionic or polar covalent is dissolved in an polar solvent or in water then., different ions of the compound will get separated and will get surrounded by polar solvent molecules. This process is known as solvation or hydration. Energy released in this process is known as solvation energy or hydration energy The ionic compound will be soluble only if solvation energy (H.E.) is more than the lattice energy

Page # 7

Factors affecting solvtion energy or hydration energy. S.E.  Z+ Z– 1   1     r  r  

 1  1    (nature of solvent) where r is dielectric constant. r   Greater the polarity, greater will be r r

  1/ r  , 1–1/r   S.E. 

r

H2O 81

CH3OH 34

C2H5OH 27

C6H6 34

Applications of Hydration energy (a)

Size of the hydrated ions: Greater the hydration of the ion greater will be its hydrated radii Li+(aq) > Na+(aq)

(b)

Mobility of the ion: more is the hydration smaller will be the mobility of the ions Li+(aq) < Na+(aq) < K+(aq) < Rb+(aq) < Cs+(aq)

(c)

Electrical conductance : is related to mobility so follows the same order

LECTURE # 3 Now we have



1 L.E.  r  r  

and

 1 1 S.E.      r r 

From these 2 equations, it can be mathematically proved that if the difference between the radii of Cation & Anion is large then, solvation energy will dominate and if radii are comparable or difference is small, then lattice energy will dominate Greater the difference between radii, greater will be solubility 100 10 100 20 100 30 100 40 Cs F > Cs Cl > Cs Br > Cs  L.E.

1 110

1 120

1 130

1 140

S.E.

1 1  100 10

1 1  100 20

1 1  100 30

1 1  100 40

decreament in solvation energy > decreament in lattice energy Solubility Orders. Increase difference in radii

 solubility  and viceversa

i.

LiF

ii.

NaF

iii.

LiF

iv

CsF

v

Li

vi

Be(OH)2

vii

LiOH

viii

BeSO4

 LiI  Na  CsF  Cs  Cs  Ba(OH)2  CsOH  BaSO4

ix

LiClO4

 CsClO4

increses increase increase decrease decrease increase increase decrease

white ppt decrease

Page # 8

Explain by Fazan’s rule, because of covalent character x

BeF2

 Be2  Ag

xi

AgF

Q.

NaF Na[BF4]– which is more soluble

Ans.

Na[BF4]– difference will

Q. Ans.

Which is more soluble Cs3 or [N(CH3)4]+ 3– Cs3

 solubility 

decrease decrease

 solubility 

Note : It has generally observed that larger cation will be thermally stable with the larger anion and smaller will be stable with the smaller anion. e.g. [N(CH3)4]+ I3– more stable than CsI3. Facts about solubility in H2O : a All the salts of NH4+ , alkalimetals are soluble except LiF, Li3 PO4, Li2C2O4 , Li2CO3 b All the NO3–, ClO4–, CH3COO– are soluble c S2– ( sulphides) of NH4+, alkalimetals or alkaline earth metal are soluble rest are insoluble d

Sulphates of all elements are soluble except Ba 2 , Sr 2 , Ca 2 ,Pb 2

 whiteppt .

+

e f

Hydroxides of NH4 , alkalimetals and Ba(OH)2 are soluble rest are insoluble Halides are generally soluble except AgC(white), AgBr(pale yellow), Ag(yellow) and Pb2(yellow)

Ex.

Which salt is insoluble in the following reaction (i) BaCl2 + Na2SO4  BaSO4 + 2NaCl (ii) BaCl2 + 2AgNO3

 2AgCl + Ba(NO3)2 (iii) (NH4)3 PO4 + 3LiCl  3NH4Cl + Li3PO4 (iv) Na2CO3 + (CH3COO)2 Pb  PbCO3 + 2CH3COONa (v) Ba(OH)2 + MgCl2  Mg(OH)2 + BaCl2 Ans.

(i) BaSO4 (ii) AgCl (iii) Li3PO4 (iv) PbCO3 (v) Mg(OH)2

Ex.

Predict the product of decomposition of given polyhalides Rb+ [Cl2]–  RbCl + Cl halide with stable lattice ( having ions of comparable radii) are produced Li Be F Na Mg Cl Na+ < F–  K+ K Ca Br K+ < Cl–  Rb+ Rb Sr  Cs Br At



 K+ [Br Cl]–  KCl + Br

COVALENT CHARACTER IN IONIC COMPOUNDS (FAJAN’S RULE) :

 

There is no compound which is 100% ionic. Covalent character in ionic compound can be explained with the help of Fajan’s rule. According to Fajan’s rules, covalent character will be more if Cation Anion (i) Small size (i) Large size (ii) More charge (ii) More charge (iii) Pseudo inert gas configuration of cation

More distortion of anion, more will be polarisation then covalent character increases. Page # 9

Factors affecting the polarisation : (i)

Small size of cation  polarisation. e.g. BeCl2 MgCl2 Size of cation 

(ii)

Polarisation 

CaCl2

SrCl2

BaCl2

Covalent character 

Large size of anion  polarisation

e.g.

(iii)

Charge on cation or anion  polarisation. (a) Charge on cation : NaCl MgCl2 AlCl3 Na+ Mg+2 Al+3 – Charge of cation  – Polarisation  – Covalent character  (b) Charge on anion : AlF3 Al2O3 AlN F– , O–2 , N–3 – Charge on anion  – Polarisation  – Covalent character 

(iv)

Cation which has pseudo inert gas configuration, shows more polarising power in comparison of cation that has inert gas configuration. CuCl > NaCl (Covalent character) Cu   [Ne] 3s 2p 6 d10 Na   1s 2 ,2s 2p 6 – 18e 8e – Pseudo inert inert gas configurat ion gas configurat ion

Application & Exceptions of Fajan’s Rules : Applications : (i) Ag2S is less soluble than Ag2O in H2O because Ag2 S is more covalent due to bigger S2– ion. (ii)

Fe(OH)3 is less soluble than Fe(OH)2 in water because Fe+3 is smaller than Fe+2 and thus charge is more.  Fe(OH)3 is more covalent than Fe(OH)2 .

(iii)

The colour of some compound can be explained on the basis of polarisation of their bigger negative ions. For ex : AgCl is white AgBr, Ag, Ag2CO3 are yellow The bigger anions are more polarised. and hence their electrons get excited by partial absorption of visible light



similarly, SnCl2 is white but Sn2 is red. PbCl2 is white but Pb2 is yellow.

Page # 10

(iv)

Variation of M.P. [M.P. of covalent < M.P. of ionic] : BeCl2 , MgCl2 , CaCl2, SrCl2, BaCl2 ––––––––––––––> –––––––––––––––––––––> – ionic charater  ,  r+ ion & r – ion = constant  CaF2 , CaCl2 , CaBr2 , Ca2 ––––––––––––––> ––––––––––––––––––> – – Covalent character  M.P.   r ion  & r+ ion = constant

(v)

MP 

Thermal stability of carbonates  ionic character Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3  Li2CO3  Li2O + CO2

LECTURE # 4 COVALENT BOND : Theories explaning the nature of covalent bond are as follows

g M g g Draw Lewis dot structures

• Each bond Is formed as a result of sharing of an electron pair between the atoms. • Each combining atom contributes at least one electron to the shared pair. • The combining atoms attain the outer- shell noble gas configurations as a result of the sharing of electrons. TO DECIDE THE CENTRAL ATOM (1) In general the least electronegative atom occupies the central position in the molecule/ion. For example in the NF3 and CO32–, nitrogen and carbon are the centra atoms whereas fluorine and oxygen occupy the terminal positions. (2) Generally the atom which is/are less in number acts as central atom (3) Generally central atom is the atom which can form maximum number of bonds( which is generally equal to the number of electrons present in the valence shell of the atom). (4) Atom of highest atomic number or largest atom atom generally acts as central atom. Hence we can say that Flourine and Hydrogen can never act as central atoms. • After accounting for the shared pairs of electrons for single bonds, the remaining electron pairs are either utilized for multiple bonding or remain as the lone pairs. The basic requirement being that each bonded atom gets an octet of electrons. Lewis representations of a few molecules/ions are given in the following Table

Page # 11

Formal Charge Lewis dot structures, in general, do not represent the actual shapes of the molecules. In case of polyatomic ions, the net charge is possessed by the ion as a whole and not by a particular atom. It is, however, feasible to assign a formal charge on each atom. The formal charge of an atom in a polyatomic molecule or ion may be defined as the difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure. It is expressed as :

The counting is based on the assumption that the atom in the molecule owns one electron of each shared pair and both the electrons of a lone pair. Let us consider the ozone molecule (O3 ). The Lewis structure of O3, may be drawn as :

The atoms have been numbered as 1. 2 and 3. The formal charge on: 1 • The central O atom marked 1 = 6 –2 – (6) = + 1 2 1 • The end O atom marked 2 =6–4– (4) = 0 2 1 • The end O atom marked 3 =6– 6– (2) = – 1 2 Hence, we represent O3 along with the formal charges as follows:

We must understand that formal charges do not indicate real charge separation within the molecule. Indicatng the charges on the atoms in the Lewis structure only helps in keeping track of the valence electrons in Page # 12

themolecule. Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms. The formal charge Is a factor based on a pure covalent view of bonding in which electron pairs are shared equally by neighbouring atoms. Ex.

Calculate the formal charge on the all atoms present in the molecules. (a) NO3– (b) NH4+ (c) N3–

Ans.

(a)

(b)







(c) N = N = N

Limitations of the Octet Rule The octet rule, though useful, is not universal. It is quite useful for understanding .the structures of most of the organic compounds and it applies mainly to the second period elements of the periodic table. There are three types of exceptions to the octet rule. 1. The incomplete octet of the central atom In some compounds, the number of electrons surrounding the central atom Is less than eight. This is especially the case with elements having less than four valence electrons. Examples are LiCl. BeH2 and BCl3.

Li. Be and B have 1,2 and 3 valence electrons only. Some other such compounds are AlCl3 and BF3. 2. Odd-electron molecules In molecules with an odd number of electrons like nitric oxide. NO and nitrogen dioxide. NO2, the octet rule is not satisfied for all the atoms

.

Cl O 2 3. The expanded octet/ super octet / hypervalent compound Elements in and beyond the third period of the periodic table have, apart from 3s and 3p orbitals, 3d orbitals also available for bonding. In a number of compounds of these elements there are more than eight valence electrons around the central atom. This is termed as the expanded octet. Obviously the octet rule does not apply in such cases. Some of the examples of such compounds are: PF5 SF6 , H2SO4 and a number of coordination compounds.

Interestingly, sulphur also forms many compounds in which the octet rule is obeyed. In sulphur dichloride, the S atom has an octet of electrons around it.

4. Other drawbacks of the octet theory • It is clear that octet rule is based upon the chemical Inertness of noble gases. However, some noble gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of compounds like XeF2 , KrF2 , XeOF2 etc., • This theory does not account for the shape of molecules. • It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.

Page # 13

MODERN CONCEPT OF COVALENT BOND (VBT) : (Do not take much time here to discuss with the students) As we know that Lewis approach helps in writing the structure of molecules but it fails to explain the formation of chemical bond. It also does not give any reason for the difference in bond dissociation enthalpies and bond lengths in molecules like H2 (435.8 kJ mol – , 74 pm) and F2 (150.6 kJ mol– , 42 pm). although in both the cases a single covalent bond is formed by the sharing of an electron pair between the respective atoms. It also gives no idea about the shapes of polyatomic molecules.Similarly the VSEPR theory gives the geometry of simple molecules but theoretically, it does not explain them and also it has limited applications. To overcome these limitations the two Important theories based on quantum mechanical principles are Introduced. These are valence bond (VB) theory and molecular orbital (MO) theory. Valence bond theory was introduced by Heitler and London (1927) and developed further by Pauling and others. A discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configurations of elements (Units 2), the overlap criteria of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and superposition. A rigorous treatment of the VB theory in terms of these aspects is beyond the scope of this book. Therefore, for the sake of convenience, valence bond theory has been discussed in terms of qualitative and non-mathematical treatment only. To start with, let us consider the formation of hydrogen molecule which is the simplest of all molecules. Consider two hydrogen atoms A and B approaching each other having nuclei N^ and Ny and electrons present in them are represented by e and e^. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms approach each'other, new attractive and repulsive forces begin to operate. Attractive forces arise between: (i) nucleus of one atom and its own electron that is NA – eA and NB – eB. (ii) nucleus of one atom and electron of other atom i.e., NA – eB. NB– eA. Similarly repulsive forces arise between (i) electrons of two atoms like eA – eB , (ii) nuclei of two atoms NA – NB. Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart (Fig. 4.7).

Fig- 4.7 Forces of attraction and repulsion during the formation of H2 molecule. Experimentally it has been found that the magnitude of new attractive force is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Ultimately a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm. Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms. The energy so released is called as bond enthalpy, which is corresponding to minimum in the curve depicted in Fig. 4.8. Conversely. 435.8 kJ of energy is required to dissociate one mole of Hg molecule. H2(g) + 435.8 kJ mol –  H(g) + H(g)

Page # 14

Fig. 4.8 The potential energy curve for the formation of H2 molecule as a function of internuclear distance of the H atoms. The minimum in the curve corresponds to the most stable state of H2 . 4.5.1 Orbital Overlap Concept In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are so near that their atomic orbltals undergo partial interpenetration. This partial merging of atomic orbitals is called overlapping of atomic orbitals which results in the pairing of electrons. The extent of overlap decides the strength of a covalent bond. In general, greater the overlap the stronger is the bond formed between two atoms. Therefore, according to orbital overlap concept, the formation of a covalent bond between two atoms results by pairing of electrons present In the valence shell having opposite spins. Main points of valency bond theory : (i) A covalent bond is formed by partial overlapping of two atomic orbitals

(ii) More is the extent of overlapping between the two atomic orbital, stronger will be bond.

<



<

[Principal Quantum no. same, n = 2]

() () () s orbital are spherical in nature so they are least diffused hence it will provide less area for overlapping.

(iii) Orbitals which are undergoing overlapping must be such that (a) Each orbital should have one electron with opposite spin (for formation of covalent bond) (b) One orbital have pair of electron and the other orbital have no electron (for formation of co-ordinate bond) (iv) If the overlapping is along the molecular axis then bond will be sigma () & in the perpendicular direction, it will be pi() bond.

Page # 15



Examples of overlapping of pure atomic orbitals. (i) H2 (s–s)

H=

H=

(ii) HCl gas molecule (s-p)

(iii) F2, Cl2, Br2, 2 (p-p)

F2 2p-2p

Cl2 3p-3p

Br2 4p-4p

2 5p-5p

4.5.4 Types of Overlapping and Nature of Covalent Bonds (NCERT Theory) The covalent bond may be classified into two types depending upon the types of overlapping : (i) Sigma() bond, and (ii) pi () bond (i) Slgma() bond : This type of covalent bond is formed by the end to end (hand-on) overlap of bonding orbitals along the intemuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals.  s-s overlapping : In this case, there is overlap of two half filled s-orbitals along the intemuclear axis as shown below :

 s-p overlapping: This type of overlap occurs between half filled s-orbitals of one atom and half filled porbitals of another atom.

 p-p overlapping : This type of overlap takes place between half filled p-orbitals of the two approaching atoms.

(iii) pi() bond : In the formation of  bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the intemuclear axis. The orbitals formed due to sidewise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms. Page # 16

4.5.5 Strength of Sigma and pi Bonds Basically the strength of a bond depends upon the extent of overlapping- In case of sigma bond, the overlaping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond . between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double or triple bonds) 4.5.3 Overlapping of Atomic Orbitals When two atoms come close to each other. there is overlapping of atomic orbitals. This overlap may be positive, negative or zero depending upon the properties of overlapping of atomic orbitals. The various arrangements of s and p orbitals resulting in positive, negative and zero overlap are depicted in Fig. 4.9. The criterion of overlap, as the main factor for the formation of covalent bonds applies uniformly to the homonuclear/heteronuclear diatomic molecules and polyatomic molecules. In the case of polyatomic molecules like CH4, NH3 and H2O, the VB theory has to account for their characteristic shapes as well. We know that the shapes of CH4 , NH3 , and H2O molecules are tetrahedral, pyramidal and bent respectively. It would be therefore interesting to find out if these geometrical shapes can be explained in terms of the orbital overlaps.



Following overlappings are not allowed. (A) Zero overlapping :

(a)

(i) (ii) Also in an s-orbital,  is positive throughout but in p-orbital it is positive and negative 

(b)

Total overlapping will zero

this type of overlapping is not allowed. Because it is neither along the molecular axis nor  to it.

(B)

Negative overlapping :

not allowed not allowed not allowed Q. Ans.

Count the  &  bonds in 6&5

N  C – C  C – C N

Page # 17

LECTURE # 5 4.5.2 Directional Properties of Bonds As we have already seen the formation of covalent bond depends on the overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of 1s-orbitals of two H atoms, when they combine with each other. In case of polyatomic molecules like CH4 , NH3 and H2O the geometry of the molecules is also important in addition to the bond formation. For example why is it so that CH4 molecule has tetrahedral shape and HCH bond angles are 109.5° ? Why is the shape of NH3 molecule pyramidal ? The valence bond theory explains the formation and directional properties of bonds in polyatomic molecules like CH4 , NH3 and H2O , etc. in terms of overlap and hybridisation of atomic orbitals.

NEED OF NEW CONCEPT (HYBRIDISATION) : The valance bond theory (overlapping concept) explains satisfactorily the formation of various molecules but it fails to account the geometry and shapes of various molecules. It does not give the explanation why BeCl2 is linear , BF3 is planar, CH4 is tetrahedral , NH3 is pyramidal and water is V– shaped molecule. In order to explain these cases , the valance bond theory has been supplemented by the concept of hybridization. This is a hypothetical concept and was introduced by Pauling & Slater. Let us first consider the CH4 (methane) molecule. The electronic configuration of carbon in its ground state is [He]2s2 2p2 which in the excited stale becomes [He] 2s1 2px1 2px1 2px1. The energy required for this excitation is compensated by the release of energy due to overlap between the orbitals of carbon and the hydrogen.The four atomic orbitals of carbon, each with an unpaired electron can overlap with the 1 s orbitals of the four H atoms which are also singly occupied. This will result in the formation of tour C – H bonds. It will , however , be observed that while the three p orbitals of carbon are at 90° to one another, the HCH angle for these will also be 900 That Is three C – H bonds will be oriented at 900 to one another. The 2s orbital of carbon and the 1s orbital of H are spherically symmetrical and they can overlap in any direction. Therefore the direction of the fourth C – H bond cannot be ascertained. This description does not fit in with the tetrahedral HCH angles of 109.5°. Clearly, it follows that simple atomic orbital overlap does not account for the directional characterisics of bonds in CH4 . Using similar procedure and arguments, it can be seen that in the case of NH3 and H2O molecules, the HNH and HOH angles should be 90°. This is in disagreement with the actual bond angles of 1070 and 104.5° in the NH3 and H2O molecules respectively.

Hybridisation

(i) (ii)

(iii) (iv)

(v)

Inter Mixing of pure atomic orbitals before bonding to produce new hybrid orbitals, specially for bonding purpose Postulates It is a hypothetical concept Only those orbitals can take part in hybridisation which have comparable ( almost equal ) energies. So, orbitals must be having same principal quantum number or these can be a maximum different of unity (if d orbitals are involved) The number of hybrid orbitals generated will be equal to the number of pure atomic orbitals taking part in hybridisation. All three type of orbitals (having a pair of e– s or having a unpaired e– or completely empty can take part in hybridisation. empty orbitals are used in coordination compounds The hybrid orbital generated will be represented by

nucleus (vi) (vii)

bigger lobe will be used for bonding

Since hybrid orbitals have been generated for the bonding purpose. So, bond formed by a hybrid orbitals are stronger than bond formed by pure atomic orbitals The orientations of hybrid orbitals generated will be dependent on type of atomic orbitals and on number of atomic orbitals taking part in hybridisation

Page # 18

s+p

 2 new sp hybridised orbitals (sp hybridisation)

s +2p  3 new sp2 hybridised (sp2 hybridisation) s + 3p

 4 new sp3 hybridised

s + 3p + d  sp3d s + 3p + 2d  sp3d2 s + 3p + 3d  sp3 d3 d + s + 2p  dsp2 The orientation of hybridised orbitals will be such that there will be minimum repulsion between any two hybridised orbitals

s + p  s + px

Yaxis SP s + pY

x axis 180º

120º

s + 2p  s + px + py in x - y plane

s + py + pz  in y– z plane s + px + pz  in x– z plane sp3  s + px + py + pz  directed along the 4 corners of tetrahedron (viii)

hybridised orbitals will be generally used for making  bond and for  bond pure p-orbitals will be used 2

2

sp -sp bond (unstable)

Calculation of state of Hybridisation : Steric number rule Steric number of any atom in a molecule = number of atoms bonded to an atom + number of lone pair left on the atom after bonding S.N = 2 SP. S.N. = 3 SP2 S.N. = 4 SP3 S.N. of central atom will decide the shape or structure of the molecule Page # 19

Ex.

(a) CO2

(S+Px+Pz) :

:

:O (SP2) S.N. = 3 S.N of C=2

O:

C

S.N. = 3 (SP2) SP (S+Px)

(S+Px+Py)

xy plane py (xy)

(xz)

:

: x axis 

:



(s + px + pz)

(b)

:

x-z plane

(s + px+ py)

(s + px)

SF6

F F S F F F

S.N. = 4(SP3)

S.N. = 6 (SP3d2) F 2

S.N = 3 (SP)

B

(c)

F

3

S.N.= 4 (SP) 2

S.N = 3 (SP )

:

F

SO2

:O :

N 3 sp

3

E.N order

more E.N., lesser will be tendency to donate the l.p.

Page # 22