Chemical Bonding IPE

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Chemical BondingPrepared by V. Aditya vardhan

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CHEMICAL BONDING

Chemical Bond : The attraction between two atoms or ions is called a chemical bond. Chemical bondis formed either due to sharing or transfer of electrons between atoms.

A chemical bond is formed by an atom inorder to get stability by lowering its potential energy.Atoms (noble gases) with octet configuration in outer shell are stable. Hence every atom tries to getoctet configuration either by losing or gaining or sharing electrons.

Types of chemical bond :1) Ionic bond2) Covalent bond3) Metallic bond

1) Ionic bond :The electrostatic force of attraction between two oppositely charged ions is called ionicbond.

* An ionic bond is formed due to transfer of electrons from one atom to another.* The atom which loses electrons will form a cation and the atom which gains electrons will form an

anion. These oppositely charged ions come closer to each other due to electrostatic force of attractionand thus form an ionic bond.

* An ionic bond is formed between two atoms when their electronegativity difference is greaterthan 1.7.

* Usually an ionic bond is formed between a metal and a nonmetal.E.g., NaCl, LiF, MgCl2 etc.,

2) Covalent Bond : The attraction between two atoms formed due to the sharing of electron pair(s) iscalled covalent bond.

* It is formed when electronegativity difference between two atoms is less than 1.7.* Usually two nonmetals form a covalent bond.E.g., H2, F2, HCl, H2O etc.,

3) Metallic bond: It is the attraction between metal atoms in a metallic crystal. It is formed betweenelectropositive metal atoms.

KOSSELL AND LEWI'S ELECTRONIC THEORY OF CHEMICAL BONDING* The atoms of inert gases are stable due to octet configuration (ns2np6) in the outer shell. Hence atomsmust posses eight electrons in their outer shell to get stability. This is referred to as octet rule.* Helium is also highly stable due to 1s2 configuration.* Hence every atom tries to get nearest inert gas configuration either by losing or gaining or sharingelectrons.* Only the electrons in outer shell participate in bond formation. These electrons are called valenceelectrons.

The electrons in the inner shell are called core electrons and do not participate in bond formation.* An ionic bond is formed due to transfer of electrons between atoms whereas a covalent bond isformed due to sharing of valence electrons.

Electrovalency: The number of valence electrons either lost or gained by an atom during the ionic

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adichemadi @ gmail.combond formation is called electrovalency.

IONIC BOND FORMATION

1) LiF (Lithium fluoride)In the formation of Lithium fluoride, Li loses one electron to get nearest inert gas - Helium's con-

figuration. Fluorine atom gains one electron to get nearest inert gas - Neon's configuration. Thus formedLi+ and F- ions form the ionic compound LiF.1)

+ -

1

Li Li + 1e z=3He 2s He

2)

- -

2 5 2 6

F + 1e Fz=9

He 2s 2p He 2s 2p

3) Li F LiF

* The electrovalencies of Li and F are equal to one.

2) MgCl2 (Magnesium chloride)Magnesium loses two electrons to get Neon's configuration. Chlorine atom gains one electron to

get Argon's configuration. Thus formed Mg2+ and Cl- ions combine together by forming MgCl2.1)

2+ -

2

Mg Mg + 2ez=12Ne 3s Ne

2)

- -

2 5

2 x ( Cl + 1e Cl ) z=17 Ne 3s 3p Ne

2 63s 3p

3) 2+2Mg 2C MgCl l

* The electrovalency of Mg is 2 and that of Cl is 13) AlF3

(Aluminium Fluoride)Aluminium loses 3 electrons and Fluorine atom gains 1 electron to get Neon's configuration. The

formed Al3+ and F- ions are combined to form AlF3.

3+ -

2 1

Al Al + 3e z=13

Ne 3s 3p Ne

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2)

- -

2 5 2 6

3 x ( F + 1e F )z=9

He 2s 2p He 2s 2p

3) 3+ -3Al + 3F AlF

* The electrovalency of Al is 3 and that of F is 1.4) Na2O (Sodium monoxide)

Sodium loses 1 electron and oxygen gains 2 electrons to get nearest inert gas Neon's configuration.Thus formed Na+ and O2- ions combine to give Na2O.1)

+

2 2 6 1 2 2 6

2 x Na Na + 1e z=11 1s 2s 2p 3s 1s 2s 2p

2)

2

2 2 4 2 2 6

O + 2e O8

1s 2s 2p 1s 2s 2pz

3) + 2-22Na + O Na O

* The electrovalency of 'Na' is 1 and that of 'O' is 2.

FACTORS FAVOURING THE FORMATION OF IONIC BONDIonic bond is electrostatic force of attraction between cation and anion. Hence factors

favouring their formation also favour the formation of ionic bond.

Factors favouring the formation of cation :Big atomic size : In bigger atoms, the nuclear attraction over the outer electrons is less. Hence theylose the electrons easily to form cations.

E.g. The ease of formation of cation increases from Li+ to Cs+ in IA group with increase in size.2) Low ionization energy : The removal of electrons is easy from atoms with low ionization energyvalues. Hence these atoms form cation easily.

E.g. IA and IIA group elements readily form cations due to low ionization energies.3) Low charge on cation : As the successive ionization energies are increased, the formation ofcations becomes difficult with increase in charge on them.

E.g. Among Na+, Mg+2 and Al+3 , the order of ease of formation is as followsNa+ > Mg+2 > Al+3

i.e., Na+ is formed more readily than Mg+2 and Al+3

4) Octet electronic configuration : The cations with 8 electrons in the outer shell (octet configura-tion) are highly stable and hence formed readily. Whereas cations with 18 electrons in outer shell(Pseudo inert gas configuration) are comparatively less stable and hence are not formed easily.

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E.g. Ca2+ 2 2 6 2 61s 2s 2p 3s 3p

8

with octet configuration is formed easily.

Zn2+ 2 2 6 2 6 101s 2s 2p 3s 3p 3d

18

with pseudo inert gas configuration cannot be formed

easily.

Factors favouring the formation of anion1) Small atomic size : Atoms with small atomic size have stronger attraction towards electron andhence can form anions readily.2) High electron affinity and electronegativity : The atoms, with high electron affinity and elec-tronegativity, gain electrons easily and hence form anions readily.

E.g., Halogens can form anions readily due to small size, high electron affinity and high electronega-tivity.3) Low charge on anion : The formation of highly charged anions is difficult as the addition of succes-sive electrons to the atoms becomes difficult due to repulsion from electrons in the atom. Hence anionwith low charge is formed readily.

E.g., Among C4-, N3-, O2- and F- ions, the anion with low charge (F-) is formed readily whereas theformation of anion with higher charge (C4-) is difficult.i.e., The order of ease of formation is C4- < N3- < O2- < F-

FAJAN'S RULESThese rules are used to predict the nature of the bond formed by atoms based on their polarizing

power and polarizability.1) Greater the size of cation, greater is the ionic nature.

E.g. In IA group elements ionic nature increases with increase in the size of cation from Li+ - Cs+.i.e, Increasing order of ionic nature : Li+ < Na+ < K+ < Rb+ < Cs+

less more ionic ionic

2) Greater the size of anion, greater is the covalent nature.E.g. Among the halides of the calcium the covalent nature increases from CaF2 to CaI2 with in-

crease in the size of anion.Increasing order of covalent nature : CaF2 < CaCl2 < CaI2

3) Greater the charge on cation, greater is the covalent nature.E.g. In case of Na+, Mg+2 and Al+3 , the covalent nature of cations increases with increase in the

charge as follows.Na+ < Mg+2 < Al+3

i.e., The compounds of Al3+ are more covalent.4) The cations with octet configuration in the valence shell exhibit more ionic nature whereas cation withpseudo inert gas configuration exhibit more covalent nature in their compounds.

E.g. CaCl2 is more ionic [ Ca2+ has octet configuration]Whereas CuCl, AgCl, ZnCl2 are more covalent. [Cu+, Ag+, Zn2+ have Psuedo inert gas configu-

ration].

CRYSTAL LATTICE ENERGY (U)The amount of energy liberated when one mole of the crystalline substance is formed from

the gaseous ions is called lattice energy (U) of the crystal.

ENERGY CHANGES IN IONIC BOND FORMATION

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The crystal lattice energy of an ionic crystal can be calculated by using Born-Haber cycle. Forexample the lattice energy of NaCl can be calculated as follows:

BORN-HABER CYCLE : The principle involved in Born-Haber cylcle is Hess's law of constantheat summation which can be stated as follows.

Hess's Law : The total energy change in a reaction remains same whether the reaction takesplace in one step or in several steps.

Calculation of lattice energy of NaClNaCl can be obtained from sodium metal and chlorine gas either in one step or in several steps as

shown below.

Direct step : NaCl crystals are formed by combining Na metal with chlorine gas in one step. The

energy evolved during this reaction is called heat of formation fΔH .

fs 2 g s1Na Cl NaCl ; ΔH = -410.5 kJ/mol2

Indirect method : The formation of NaCl crystals may occur in several steps as follows.i) Sublimation of sodium : Solid sodium is first converted gaseous sodium by absorbing 108.7 kJ/

mole of energy. This is called sublimation energy sΔH or + S .

ss gNa Na ; ΔH = + S = +108.7 kJ/mol

ii) Ionization of sodium : Gaseous sodium atoms are ionized by absorbing 492.82 kJ / mole of

energy. It is called ionization energy iΔH or + I .

+ -

ig gNa Na + 1e ; ΔH = + I = + 492.82 kJ/mol

iii) Dissociation of chlorine molecule : One mole of gaseous chlorine molecules are dissociated intotwo moles of chlorine atoms by absorbing energy equal to 239.1 kJ/mole. This is called dissociation

energy dΔH or D .

g2 dgCl 2Cl ; ΔH = +D = 239.1 kJ/mol

But the energy required to get one mole of chlorine atoms is equal to+D 139.2= = +119.55 kJ/mol2 2

2 dg g1 1 +DCl Cl ; ΔH = = +119.55 kJ/mol2 2 2

Chloride ion formation : The gaseous chlorine atoms are added with electrons to get gaseous chlo-ride ions. The energy liberated in this process is called electron affinity e(ΔH or E) .

- -

2 eg gCl + e Cl ; ΔH = - E = -361.6 kJ/mol

Formation of NaCl Crystals : The gaseous Na+ and Cl- ions unite to form one mole of NaCl crys-

tals. The energy liberated during this process is called lattice energy uΔH or -U . This value can becalculated by using Hess's law as follows.

+ -

ug g sNa + Cl NaCl ; ΔH = -U

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adichemadi @ gmail.comAccording to Hess's law the energy change in the steps involved in indirect method.

i.e.12f s f d e uH H H H H H

(or)

2fDH S I E U

i.e., 2d

f S i eHU H H H H

1410.5 108.7 492.82 119.55 ( 361.6) 769.97 .U kJ mol

Na(s) + 1/2Cl2(g) NaCl(s)

Na(g) Cl(g)

Na+(g) Cl

-(g) +

Hf

Hs= +S

Hi= +I

2dH

He

-U= +E

Born-Haber cycle

BORN LANDE EQUATIONThe lattice energy of an ionic crystal is equal to the sum of attractive and repulsive forces in the crystal.It can be calculated by using Born Lande equation as follows:

2.o o n

AZ Z BeU N Nr r

attractive repulsiveforce force

Where A = Madelung constant (which depends on geometry of the crystal)2 1. nAZ Z e rB

n

= Repulsion coefficient

(This depends on the structure and approximately proportional to the number of nearestneighbours)No = Avogadro's numberZ+ & Z- = Charges on the positive and negative ions respectively.e = Charge of an electron.r = Distance between the oppositely charged ionsn = Born exponent (a constant which is usually taken as 9)

Born Lande equation can be written as follows

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adichemadi @ gmail.com2. 11o

o

AZ Z e NUr n

CRYSTAL STRUCTURESUnit cell : The smallest part of the crystal which produces entire crystal upon repeating three dimen-sionally is called unit cell.

Coordination number : Maximum number of nearest oppositely charged ions surrounding any par-ticular ion in ionic crystal is called the coordination number of that ion.

Limiting radius ratio (rr

): The ratio of radius of positive ion to that of negative ion is called limiting

radius ratio.

The coordination number and crystal structure of an ionic crystal can be predicted from the limitingradius ratio value.

Limiting radius ratio (rr

) Coordination number Structure

less than 0.155 2 linear 0.155 - 0.225 3 Planar triangle 0.225 - 0.414 4 Tetrahedral 0.4142 - 0.732 4 Squared planer 0.4142 - 0.732 6 Octahedral 0.732 - 0.999 8 Body centered cubic

1) Crystal structure of Cesium chloride

* Limiting radius ratio ( Cs

Cl

rr

)= 0.93.

* The coordination number is equal to 8.* Crystal structure is body centered cubic (bcc).* Each unit cell contains 1 Cs+ and 1 Cl- ions.

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Cs+

Cl-

Cl- Cl-

Cl- Cl-

Cl-

Cl-

Cl-

There are eight Cl- ions at the corners each of which contributes only 1/8th part and one Cs+ at the centre. Hence

-

+

1Number of Cl ions = 8 x = 18

Number of Cs ions = 1

2) Crystal structure of sodium chloride

* Limiting radius ratio ( Na

Cl

rr

)= 0.52

* The coordination number is equal to 6.* The crystal structure is face centered cubic (fcc).* Each unit cell contains 4 Na+ and 4 Cl- ions.

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Na+

Na+

Na+

Na+

Na+

Na+

Na+ Na+

Na+

Na+

Na+

Na+

Na+

-

-

-

+

+

1Number of Cl ions at corners = 8 x = 18

1Number of Cl ions at the centre of faces = 6 x = 32

Total number of Cl ions = 1+3 = 4

1Number of Na ions on the edges = 12 x = 34

Number of Na ions at the c+

entre = 1Total number of Na ions = 3+1 = 4

Properties of ionic compounds1) Ionic compounds contain oppositely charged ions which are strongly attracted to each other.

Hence these are hard substances with high melting and boiling points.2) Ionic bond is direction less and the electrostatic forces of attraction are present in all directions

around an ion. Hence there is no isolated discrete molecule in the ionic crystal. Entire crystal is consid-ered as the giant molecule.

3) In the solid state, ions cannot move freely and hence they do not conduct electricity . But infused state or in aqueous solutions, they exhibit electrical conductivity as the ions are free to move.

4) Ionic compounds are polar in nature and hence they are soluble in polar solvents like water.These compounds do not dissolve in non polar solvent like benzene, carbon tetrachloride etc.,

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adichemadi @ gmail.com5) The reactions between ionic compounds occur very fast due to presence of separate ions.e.g. A white precipitate of AgCl is formed instantly when aqueous solution of NaCl and AgNO3 are

mixed.

(aq) 3(aq) 3(aq)NaCl + AgNO NaNO + AgCl

6) Ionic compounds do not exhibit isomerism due to non directional nature of ionic bond.

Formula weight: Ionic compounds contain only ions and there are no molecules in it. Hence theirmolar mass is expressed as formula weight instead of molecular weight.

COVALENT BOND FORMATION & LEWI'S DOT STRUCTURESA covalent bond is formed by sharing of pair of electrons between two atoms.It is formed between two atoms when their electronegativity difference is less than 1.7 . Usually a

covalent bond is formed between two nonmetals.The formation of covalent bonds can be explained based on Lewi's dot structures. Atoms contrib-

ute their valence electrons for the bond formation and get octet or nearest inert gas configuration.Covalency : The number of electrons contributed by the atom of an element in the formation of

covalent compound is known as covalency of that element.

Examples1) H2 molecule* Electronic configuration of hydrogen is 1s1.* Each hydrogen atom contribute one electron to form a pair of electrons, which is shared in betweenthe two atoms. Thus a covalent bond is formed.* Thus in H2 molecule, each hydrogen atom gets its nearest inert gas - Helium's configuration.* Covalency of hydrogen is 1.

H . H.+ H H..

H Hor

2) Cl2 molecule

* The electronic configuration of Cl is 2 2 6 2 51s 2s sp 3s 3p

7.

* In order to get the nearest inert gas- Argon's configuration, each chlorine contributes one electron forthe bond formation.* Covalency of chlorine is 1.

. .Cl. .. . .

. .Cl. .

. ..+. .Cl. .. . . . . . .Cl. ..

. .Cl. .. .

. . . .Cl. .

or

3) Hydrogen chloride (HCl)* Electronic configuration of hydrogen is 1s1.

* Electronic configuration of chlorine is 2 2 6 2 51s 2s sp 3s 3p

7.

* In the formation of HCl molecule, the hydrogen and chlorine atoms contribute one electron each for

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adichemadi @ gmail.comthe bond formation. Thus both of them get their nearest inert gas configurations.

H X

. .Cl. .

. .. H. .Cl. .

. .or

4) Methane (CH4)

* The electronic configuration of carbon is 2 2 21s 2s 2p

4.

* The electronic configuration of hydrogen is 1s1.* The carbon atom forms four covalent bonds by contributing four of its valence electrons for the bondformation. It forms 4 bonds with four hydrogen atoms. Thus it gets octet configuration.* Covalency of carbon is 4.

H x .C.. .H

H

Hx

x orx

CH

HH

H

5) Ammonia (NH3) :

* The electronic configuration of nitrogen is 2 2 31s 2s 2p

5

.

* The electronic configuration of hydrogen is 1s1,* In the formation of Ammonia molecule, nitrogen atom contributes 3 of its electrons to form threebond pairs which are shared with hydrogen atoms. Thus nitrogen forms 3 single bonds with threehydrogen atoms and gets the configuration of Neon.* Covalency of nitrogen is 3.

N .....

H x

Hx

Hx N..

H

H

Hor

5) H2O molecule* Electronic configuration of oxygen is 2 2 41s 2s 2p

6

.

* Electronic configuration of Hydrogen is 1s1.* In the formation of water molecule, oxygen atom contributes two electrons to form two bond pairswhich are shared with hydrogen atoms. Thus two bonds are formed by oxygen atom to get the configu-ration of neon. There are also two lone pairs on oxygen atom.* Covalency of oxygen is 2.

....

H x HxO.. or H O..

..H

6) Oxygen molecule (O2)

* The electronic configuration of oxygen is 2 2 41s 2s 2p

6.

* In the formation of oxygen molecule, each oxygen atom contributes 2 electrons to form 2 bond pairs.

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adichemadi @ gmail.comThus a double bond is formed between oxygen atoms. Thus each oxygen atom gets Neon's configura-tion.

or..

O O....

..

.. O O....

..

....

7) Nitrogen molecule (N2)

* The Electronic configuration of nitrogen is 2 2 31s 2s 2p

5

.

* In the formation of Nitrogen molecule, each nitrogen atom contributes 3 electrons to form 3 bondpairs. Thus a triple bond between nitrogen atoms is formed. Each nitrogen atom gets Neon's configura-tion.

orN N.. ........ N.. N ..

8) Carbon dioxide (CO2)

* Electronic configuration of carbon is 2 2 21s 2s 2p

4.

* Electronic configuration of oxygen is 2 2 41s 2s 2p

6.

* In carbon dioxide, carbon atom forms double bond with each oxygen. Thus both oxygen and carbonatoms get the octet configuration in their valence shell.

orO .. O ....C.. .. O.. C O ....

..

.. .. ..

9) Ethylene molecule (C2H4)

* Electronic configuration of carbon is 2 2 21s 2s 2p

4* Electronic configuration of hydrogen is 1s1

C C

H

H

H

H

C C

H

H

H

H. .. .

*

* *

*or.. ..

Molecules violating octet rule:

BeCl2 (Beryllium chloride)

* Electronic configuration of Beryllium is 2 21s 2s2

* Electronic configuration of Chlorine is 2 2 6 2 51s 2s sp 3s 3p

7.

* In the formation of Beryllium chloride, the beryllium atom contributes its two valence electrons andforms two bond pairs. These are shared with chlorine atoms.* BeCl2 is a stable molecule, even though beryllium gets only four electrons in its valence shell. This is

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adichemadi @ gmail.comthe violation of octet rule.* Covalency of beryllium is 2.

or.. .. ..Be Be.. ..

. . ..x x.. ..

.. .. .. ..Cl Cl Cl Cl..

BCl3 (Boron trichloride)

* Electronic configuration of Boron is 2 2 11s 2s 2p

3.


7.

* In BCl3 molecule, boron contributes 3 of its valence electrons and forms three bond pairs withchlorine atoms. There are only six electrons in the valence shell of boron atom in BCl3. But still it isstable. It is an electron deficient compound. It is also the violation of octet rule.* Covalency of boron is 3.

or.. .. ..B B... . .. ..x x..

.. .. .. ..x.

.. .. .. ..Cl

Cl

Cl Cl

Cl

Cl

..

..

..

PCl5 (Phosphorous pentachloride)

* Electronic configuration of phosphorus is 2 2 6 2 31s 2s 2p 3s 3p

5.


7.

In the formation of PCl5 molecule, phosphorus contributes five electrons in it's valence shell andforms five bonds with chlorine atoms. There are 10 electrons in the valence shell of phosphorus in thismolecule. It is a stable molecule and violates octet rule.* Covalency of phosphorus in this molecule is 5.

or

.... ... .x x..

.. ..x.

.. ..

.... ... x

...x

..

ClCl Cl

P

..

Cl Cl.. ..

PCl

Cl

Cl

ClCl

..

.. ..

..

.. ..

....

..

..

.. ....

..

..

SF6 (Sulphur hexafluoride) :* Electronic configuration of Sulphur is 2 2 6 2 41s 2s 2p 3s 3p

6

.

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* Electronic configuration of Fluorine is 2 2 51s 2s 2p

7* Sulphur contributes six of its valence electrons to form 6 bonds with six fluorine atoms. Thus thereare 12 electrons in the valence shell in sulphur atom.

In violates octet rule and is still stable.* Covalency of sulphur in this molecule is 6.

SF

F

F

F

FF

** * *

**S

FF

F

F

FF

or

Conclusion: Lewi's electronic theory could not explain the shapes and bond angles of molecules. Italso could not explain why some molecules are stable eventhough they violate octet rule.

Some more examples:1) H2S 2) CO3

2-

....

H x HxS.. or H S..

..H

Hydrogen sulphide

Carbonate ion

C O

O

O

xx

x

x

xx

_

_

C O

O

O

_

_

x

or

x

3) SiCl4 4) HCOOH

Cl x .Si.. .x

x or

x

SiCl

ClCl

Cl

Cl

Cl

Cl

Silicon tetrachloride

.. ..

.. .. .. ..

..

..

....

....

.. .... .... ..

..

....

..

.... HO

OH C H C

O

O

H

OR

Formic acid

5) NF3 6) HNO3

N .. ...F x

Fx

Fx NF

F

For

Nitrogen trifluoride

..

x x x x.. .... ..

`

OO

ON H+

-

O

O

ON H+

-Nitric acid

7) CO 8) O3

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C O

:: C O

::

or

: : :

Carbon monoxide molecule

OOO: : :

:

: OOO: : :

:

:

: : :: :or

Ozone molecule

8) NO2-1 9) NH4

+

O ON-

O ON

Nitrite ion

-or

N .....

H x

Hx

Hx NH

H

Hor

HH

+ +

Formal charge (Qf)It is the charge on an atom in a molecule assigned by assuming all the atoms have same electrone-

gativity and the electron pairs are shared equally. It is calculated as follows:

f A LP BP1Q = N - N - N2

Where NA = Number of electrons in the valence shell of free atom.NLP = Number of electrons in the lone pairs (unshared pairs).NBP = Number of electrons in the bond pairs.

Examples:1) PH3 molecule

Lewi's dot structure:

P .....

H x

Hx

Hx P..

H

H

Hor

A LP BP1 6Formal charge on 'P' = N - N - N 5 2 02 2

A LP BP1 2Formal charge on 'H' = N - N - N 1 0 02 2

2) N2O molecule: It exists in following two resonance forms.For the following first resonance form:

N1

N2

O or:

: : : N1

N2

O:

: : :: : : :+- - +

( ) ( ) ( ) ( )

(1) A LP BP1 4Formal charge on first nitrogen 'N ' = N - N - N 5 4 12 2

(2) A LP BP1 8Formal charge on second nitrogen 'N ' = N - N - N 5 0 12 2

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A LP BP1 4Formal charge on 'O' = N - N - N 6 4 02 2

For the following second resonance form:

N1

N2

O or

::

::-+ :

:

::-+

N1

N2

O: : : :

( ) ( ) ( ) ( )

(1) A LP BP1 6Formal charge on first nitrogen 'N ' = N - N - N 5 2 02 2

(2) A LP BP1 8Formal charge on second nitrogen 'N ' = N - N - N 5 0 12 2

A LP BP1 2Formal charge on 'O' = N - N - N 6 6 12 2

Hence the formal charges vary with structural environment.

VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY* Following are the important postulates of VSEPR's theory proposed by Nyholm, Gillespie.1) The shape of a molecule can be determined from the arrangement and repulsions between theelectron pairs present in the valence shell of central atom of that molecule.2) There are two types of valence shell electron pairs viz.,

i) Bond pair and ii) Lone pair3) The electron pairs in the valence shell the repel each other and determines the shape of the molecule.The magnitude of the repulsion depends upon the type of electron pair.4) The bond pair is attracted by nuclei the occupies less space and hence it causes less repulsion.Whereas, the lone pairs are only attracted by one nucleus and hence occupy more space. As a result,the repulsion caused by them is greater.

The order of repulsion between different types of electron pairs is as follows :Lone pair - Lone pair > Lone Pair -Bond pair > Bond pair- Bond pair

5) When the valence shell of central atom contains only bond pairs, the molecule gets symmetricalstructure, whereas the symmetry is distorted when there are lone pairs along with bond pairs.6) The bond angle decreases due to the presence of lone pairs.7) The repulsion increases with increase in the number of bonds between two atoms.

E.g. Triple bond causes more repulsion then double bond which in turn causes more repulsion thansingle bonds.8) The repulsion between electron pairs increases with increase electronegativity of central atom andhence the bond angle increases.9) Shapes of molecules can be predicted from the number of electron pairs in the valence shell of centralatom as follows:

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THE GEOMETRY OF MOLECULES CONTAINING ONLY BOND PAIRS IN THE CENTRAL ATOM Number of bond pairs Formula Molecular geometry Examples

2 AB2 Linear A BB BeCl2, BeF2

3 AB3 Trigonal planar

B

B B

A

BF3, BCl3

4 AB4 Tetrahedral A

B

B

BB

CH4, CCl4

5 AB5 Trigonal

bipyramidal AB

BB

B

B

PCl5, PF5

6 AB6 Octahedral AB

B

B

B

B

B

SF6

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GEOMETRY OF MOLECULES CONTAINING CONTAINING ONE OR MORE LONE PAIRS IN CENTRAL ATOM

Total number

of electron

pairs

Number of bond

pairs

Number of lone pairs

Formula Shape of molecule Bond angle Examples

3 2 1 AB2E Angular 120o SO2

4 3 1 AB3E Trigonal

Pyramidal 107o48’ 102o30’

NH3

NF3

2 2 AB2E2 Angular

(V- shaped) 104o28’

103o H2O F2O

5 4 1 AB4E See-Saw SCl4 , SF4 3 2 AB3E2 T-Shape 90o ClF3 2 3 AB2E3 Linear 180o

XeF2, I3-

6 5 1 AB5E Square

pyramidal 90o BrF5

4 2 AB4E2 Square planar

90o XeF4

Where A = central atom B = atom linked to the central atom E = Lone electron pair

Explanatory examples:1) BeCl2: The valence shell of central atom, beryllium contains only two bond pairs. Hence it is linearin shape with 180o bond angle.

Cl ClBe

180o

Linear molecule

2) BF3: The valence shell of the central atom - boron contains only three bond pairs. Hence it's shapeis trigonal planar with 120o bond angle.

120o

B

Cl Cl

Cl

Trigonal planar shape

3) CH4: The valence shell of the central atom - carbon contains only four bond pairs. Hence it istetrahadral in shape with 109o28' bond angle.

The bond pairs are arranged tetrahedral symmetry so as to minimize repulsions. If the bond pairs

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adichemadi @ gmail.comare arranged in square plane, the angles between them will be only 90o and the repulsions will be morethan in case of tetrahedral arrangement. Hence tetrahedral structure is more favorable than squareplanar structure.

109o28'

C

H

HH

HTetrahedral structure of methane

4) NH3: There are three bond pairs and one lone pair in the central atom, nitrogen. The bond angle isdecreased from 109o28'

to 107o48' due to repulsion caused by lone pair.

107o48'

NH H

H

Trigonal pyramidal structure of ammonia molecule

5) H2O: There are two bond pairs and two lone pairs in the central atom, oxygen. The bond angle isdecreased from 109o28'

to 104o28' due to repulsion caused by two lone pairs.

104o28'

OH

H

Angular shape of water molecule

VALENCE BOND THEORY (VBT)This theory was proposed by Heitler and London to explain the formation of covalent bond. The

main postulates are as follows:* A covalent bond is formed by the overlapping of two half filled atomic orbitals containing electronswith opposite spins.* Thus formed electron pair is shared between two atoms.* The electron density along the internuclear axis between two bonded atoms increases due to over-lapping. This confers stability to the molecule.* Greater the extent of overlapping, stronger is the bond formed. The direction of the covalent bond isdecided by the direction of overlapping.* There are two types of covalent bonds based on the pattern of overlapping as follows:

(i) bond:- The covalent bond formed due to overlapping of atomic orbital along the inter nucleusaxis is called bond.

It is a stronger bond with cylindrical symmetry.(ii) π bond :- The covalent bond formed by sidewise overlapping of atomic orbitals is called π -

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adichemadi @ gmail.combond.

In this bond, the electron clouds are present above and below the internuclear axis. It is a weakerbond.

Note: The 's' orbitals can only form bonds, whereas the p, d & f orbitals can form both and πbonds.

Examples1) H2 molecule* Electronic configuration of hydrogen is 1s1.* In the formation of hydrogen molecule, two half filled 1s orbitals of hydrogen atoms overlap along theinternuclear axis and thus form a s-s bond.

H H+ H H

bonds s 1s orbital 1s orbital

2) Cl 2 molecule* Electronic configuration of Cl is 1s2 2s2 2p6 3s2 3px

2 3py2 3pz

1

* The two half filled 3pz atomic orbitals of two chlorine atoms overlap along the internuclear axis andthus by forming a p-p bond.

Cl

3pz orbital

+Cl

3pz orbital

ClCl

bondp p

3) HCl molecule* Electronic configuration of hydrogen is 1s1.* Electronic configuration of Cl is 1s2 2s2 2p6 3s2 3px

2 3py2 3pz

1

* The half filled 1s orbital of hydrogen overlap with the half filled 3pz atomic orbital of chlorine atomalong the internuclear axis to form a s p bond.

H

1s orbital

Cl

3pz orbital

+ H Cl

bonds p

4) O2 molecule* Electronic configuration of O is 1s2 2s2 2px

2 2py1 2pz

1

* The half filled 2px orbitals of two oxygen atoms overlap along the internuclear axis and form p-p bond. The remaining half filled 2pz orbitals overlap laterally to form a π p-p bond. Thus a doublebond (one and one ) is formed between two oxygen atoms.

+O O O Obondp p

bondp p

5) N2 molecule* Electronic configuration of N is1s2 2s2 2px

1 2py1 2pz

1

* p-pbond is formed between two nitrogen atoms due to overlapping of half filled 2px atomic

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adichemadi @ gmail.comorbitals axially. The remaining half filled 2py and 2pz orbitals form two π p-p bonds due to lateraloverlapping. Thus a triple bond (one and two )is formed between two nitrogen atoms.

N Np p

p p

p p

Valence bond theory could not explain the structures and bond angles of molecules with more thanthree atoms. E.g. It could not explain the structures and bond angles of H2O, NH3 etc.,

Inorder to explain the structures and bond angles of molecules, Linus Pauling modified the valencebond theory by proposing hybridization concept.

HYBRIDIZATIONThe intermixing of atomic orbitals of almost equal energies of an atom to give an equal

number of identical and degenerate hybrid orbitals is called hybridization.

Characteristics of hybridization :1) Pure atomic orbitals of same atom should participate in the hybridization.2) The energies of atomic orbitals participating in hybridization must be nearly same.3) The number of hybrid orbitals formed is equal to the number of atomic orbitals participating in thehybridization.4) The hybrid orbitals formed are identical in shape and degenerate.5) These hybrid orbitals are symmetrically arranged around the nucleus so as to minimize the repulsionbetween them and thus get maximum stability.6) The filling of electrons into hybrid orbitals follows Pauli's exclusion principle and Hund's rule ofmaximum multiplicity.7) A hybrid orbital may be empty or half-filled or full filled.8) Usually hybrid orbitals form sigma bonds only.

TYPES OF HYBRIDIZATION

1) 'sp' hybridization* Intermixing of one 's' and one 'p' orbitals of almost equal energy to give two identical and degen-

erate hybrid orbitals is called 'sp' hybridization.* These sp-hybrid orbitals are arranged linearly at 180o of angle.* They possess 50% 's' and 50% 'p' character.

+

s orbital p orbital sp hybrid orbitals

Examples:1) BeCl2* Electronic configuration of 'Be' in ground state is 1s2 2s2

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* Electronic configuration of 'Be' in excited state is 2 1 1x1s 2s 2p

2s 2px 2py 2pz

sp hybridization(2σ bonds)

* In the excited state, beryllium undergoes 'sp' hybridization by using a 2s and a 2p orbitals. Thus twohalf filled 'sp' hybrid orbitals are formed. These are arranged linearly.* These half filled sp-orbitals form two sp p bonds with two 'Cl' atoms.* Thus BeCl2 is linear in shape with the bond angle of 180o.

Cl ClBe

Cl ClBe

180o

sp p sp p

Linear molecule

2

o

BeCl Hybridization - spShape - linearBond angle - 180

2) Acetylene (C2H2)* Electronic configuration of 'C' in ground state is 1s2 2s2 2p2

* Electronic configuration at 'C' in excited state is 1s2 2s1 2p3

2s 2px 2py 2pz

sp hybridization

pureorbitals

form2 bonds

(2σ bonds)

* Each carbon atom undergoes 'sp' hybridization by using one 2s and one 2p orbitals in the excitedstate. Thus two half filled 'sp' orbitals are formed. These are arranged linearly.

The two carbon atoms form one sp sp bond with each other by using sp-orbitals. They also form

two p p bonds by overlapping half filled pure p-orbitals ( 2py and 2pz). Thus a triple bond (1 & 2 ) is formed between carbon atoms.

Each carbon also forms a sp s bond with the hydrogen atom.Thus acetylene molecule is linear with 180o bond angle.

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C CH H

p p

p p

sp sp sp s

C C HH

180o

Linear structure of acetylene molecule

2 2

o

C H - AcetyleneHybridization - spShape - linearBond angle - 180

sp2 hybridization :* Intermixing of one 's' and two 'p' orbitals of almost equal energy to give three identical and degener-ate hybrid orbitals is known as sp2 hybridization.* The three sp2

hybrid orbitals are oriented in trigonal planar symmetry at angles of 1200 to each other.* sp2 hybrid orbitals have 33.3% 's' character and 66.6% 'p' character.

+

s orbital px orbital

+

sp2 hybrid orbitalspy orbital

Examples:1) BCl3* Electronic configuration of 'B' in ground state is 1s2 2s2 2p1

* Electronic configuration of 'B' in excited state is 1s2 2s12px12py

1

2s 2px 2py 2pz

sp2 hybridization(3σ bonds)

In the excited state, Boron undergoes sp2 hybridization by using a 2s and two 2p orbitals. Thusthree half filled sp2 hybrid orbitals which are oriented in trigonal planar symmetry are obtained.

Boron forms three 2sp -pσ bonds by using sp2 hybrid orbitals with three chlorine atoms. Each

chlorine atom uses it's half filled p-orbital for the bond formation.Thus the shape of BCl3 is trigonal planar with Cl-B-Cl bond angles equal to 120o.

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B

ClCl

Cl

120o

B

Cl Cl

Cl

Trigonal planar structure of BCl3

32

o

BCl - Boron trichlorine

hybridization - spshape - Trigonal planarBond angle - 120

2) Ethylene (C2H4).* Electronic configuration of 'C' in ground state is 1s2 2s2 2p2


2s 2px 2py 2pz

sp2 hybridization(3σ bonds)

pureorbitalforms

one bond

* In the excited state, each carbon in ethylene undergoes sp2 hybridization by mixing 2s and two 2porbitals. Thus three half filled sp2 hybrid orbitals are formed in trigonal planar symmetry. There is also ahalf filled 2pz orbital on each carbon.* The carbon atoms form a 2 2sp -sp

bond with each other by using hybrid orbitals.

The remaining pure atomic orbitals overlap laterally and form a p-p bond.

* There is a double bond (1 & 1 ) between two carbon atoms.

* Each carbon atom also forms two 2sp -s bonds with two hydrogen atoms.

* Thus ethylene molecule is planar with H-C-H & C-C-H bond angles equal to 120o.* All the atoms are present in one plane.

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H

H

H

H

CC

2 2sp sp

p p

2sp s

C C

H

H H

H

120o

Planar structure of ethylene molecule

2 22

o

C H - Ethylene

Hybridization - spShape - planarBond angle - 120

sp3 hybridization:-* Intermixing of one 's' and three 'p' orbitals of almost equal energy to give four identical and degener-ate hybrid orbitals is called sp3 hybridization.* Thus formed four sp3 hybrid orbitals are oriented in tetrahedral symmetry with 109028' angle witheach other.* In these orbital the ‘s’ character is 25% and ‘p’ character is 75%.

+

s orbital px orbital

+

sp3 hybrid orbitalspy orbital

+

pz orbital

Examples:Methane (CH4)* Electronic configuration of 'C' in ground state is 1s2 2s2 2p2


2s 2px 2py 2pz

sp3

hybridization(4σ bonds)

* In the excited state, the carbon atom undergoes sp3 hybridization by mixing one ‘2s’ and three 2porbitals. Thus four half filled sp3 hybrid orbitals are formed in tetrahedral symmetry.

* Each of these form 3sp - sσ bond with hydrogen atom. Thus carbon forms four bonds with four

hydrogen atoms.* Methane molecule is tetrahedral in shape with 109028' bond angle.

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109o28'

C

H

HH

H

C

H

HH

H

3sp s

Tetrahedral structure of methane

43

o

CH - MethaneHybridization - spShape - TetrahedralBond angle - 109 28'

2) Ethane (C2H6)* Electronic configuration of 'C' in ground state is 1s2 2s2 2p2


2s 2px 2py 2pz

sp3


* In the excited state, each carbon atom undergoes sp3 hybridization by using one 2s and three 2porbitals. Thus four half filled sp3 hybrid orbitals are formed in tetrahedral symmetry around each car-bon.

* The two carbon atoms form a 3 3sp - spσ bond with each other by overlapping sp3 hybrid orbitals

axially.Each carbon atom also forms three 3sp -s

σ bonds with hydrogen atoms.

* Thus there is tetrahedral symmetry around each carbon with HCH and HCC bond angles equalto 1090 28' .

Structure of Ethane molecule

C C

H

H

H H

H

H

C C

H

H

H

H

HH

109o28'109o28'3 3sp sp

3sp s

2 63

o

C H - EthaneHybridization - spBond angle - 109 28 '

3) Ammonia (NH3)* Electronic configuration of nitrogen is 1s2 2s2 2px

1 2py1 2pz

1

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2s 2px 2py 2pz

sp3


* Nitrogen undergoes sp3 hybridization by using one 2s and three 2p orbitals. Thus formed four sp3

orbitals are arranged in tetrahedral symmetry. Among them three are half filled and one is full filled.

* Nitrogen forms 3 3sp -sσ bonds with three hydrogen atoms by using three half filled hybrid orbitals.

The bond angle is decreased from 1090 28'to 1070 48' due to the repulsion caused by lone pair.Thus ammonia acquires trigonal pyramidal shape.

NH

HH

107o48'

NH H

H

Trigonal pyramidal structure of ammonia molecule

3sp s

33

o

NH - AmmoniaHybridization - spShape - TrigonalpyramidalBond angle - 107 48'

4) H2O (Water molecule)* Electronic configuration of oxygen is 1s2 2s2 2px

2 2py1 2pz

1

2s 2px 2py 2pz

sp3


* Oxygen atom undergoes sp3 hybridization by mixing a 2s and three 2p orbitals and forms four sp3

hybrid orbitals arranged in tetrahedra symmetry. Among thenm two are half filled and the remaining twoare completely filled.

Oxygen forms two 3sp -sσ bonds with hydrogen atoms by using half filled hybrid orbitals.

The bond angle is decreased from 1099 28' to 1040 28' due to repulsions caused by two lonepairs. Thus water molecule gets angular shape (V shape).

O

H

H104o28'

OH

H

Angular shape of water molecule

3sp s

23

o

water - H OHybridization - spshape - AngularBond angle - 104 28'

sp3d Hybridization* The intermixing of one 's', three 'p' and one 'd' orbitals to give five identical and degenerate hybrid

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adichemadi @ gmail.comorbitals is called sp3d hybridization.* Thus formed sp3d orbitals are arranged in trigonal bipyramidal symmetry. Among them, three arearranged in trigonal plane and the remaining two orbitals are present perpendicularly above and belowthe trigonal plane.

sp3d hybrid orbitals

Examples:1) PCl5 (phosphorus pentachloride)* E.configuration of 'P' in ground state is 1s2 2s2 2p6 3s2 3p3

* E.configuration of 'P' in excited state is 1s2 2s2 2p6 3s1 3p3 3d1

3s 3p

sp3dhybridization(5σ bonds)

3d

* In the excited state, phosphorus undergoes sp3d hybridization by using a 3s, three 3p and one 3dorbitals which are arranged in trigonal bipyramidal symmetry.* By using these half filled sp3d orbitals, phosphorous forms five 3sp d - p

σ bonds with chlorine atoms.

Each chlorine atom makes use of half filled 3pz orbital for the bond formation.* The shape of PCl5 molecule is trigonal bipyramidal with 1200 and 900 of Cl - P - Cl bond angles.

P

Cl

Cl

Cl

Cl

Cl

120o

90o 53

o o

PCl (Phosphorous pentachloride)Hybridization - sp dShape - Trigonal bipyramidalBond angles - 120 & 90

sp3d2 hybridization* Intermixing of one 's', three 'p' and two 'd' orbitals of almost same energy by giving six identical anddegenerate hybrid orbitals is called sp3d2 hybridization.* These six sp3d2 orbitals are arranged in octahedral symmetry by making 900 angles to each other.

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sp3d2 hybrid orbitals

Examples:1) Sulfur hexa flouride (SF6)* E.configuration of 'S' in ground state is 1s2 2s2 2p6 3s2 3p4

* E.configuration of 'S' in 2nd excited state is 1s2 2s2 2p6 3s1 3p3 3d2

3s 3p

sp3d2


3d

* In the second excited state, sulfur under goes sp3d2 hyrbidization by mixing a 3s, three 3p and two 3dorbitals. Thus formed six half filled sp3d2 hybrid orbitals are arranged in octahedral symmetry.

Sulfur atom forms six 3 2sp d - pσ bonds with 6 fluorine atoms by using these sp3d2 orbitals. Each

flourine uses is half-filled 2pz orbitals for the bond formation.SF6 is octahedral in shape with F-S-F bond angles equal to 90o.

S

F

F

F

FF

F

63 2

o

SF - Sulfur hexafluoride Hybridization - sp d shape - Octahedral Bond angle - 90

Properties of covalent compounds1) In covalent compounds, weak vander Wall's forces of attraction are present between the

molecules. Hence they have low melting and boiling points.2) Covalent compounds do not conduct electricity because of absence of either free electrons or

ions in them.3) Covalent compounds are mostly non polar and hence are more soluble in non polar solvents like

benzene, carbon tetrachloride etc.But some covalent compounds are polar in nature like glucose, fructose etc., and hence are soluble

in polar solvents like water and alcohol.4) The reactions between covalent compounds involve bond breaking and bond making. As the

bond breaking requires energy, the reactions between them occur slowly.5) Covalent bond has specific direction. Hence these compounds exhibit isomerism.

Isomerism is the phenomenon exhibited by different compounds possessing the samemolecular formula.

e.g. Following are the isomers with the molecular formula C2H6O.C2H5OH - Ethyl alcohol

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adichemadi @ gmail.comCH3-O-CH3 - Diethyl ether

COORDINATE COVALENT BONDThe covalent bond formed due to the donation of a pair of electrons by one atom to the

other is called coordinate covalent bond.

* The atom which donates the electron pair is called as donor, whereas the atom which accepts thatpair is called as acceptor.* Both donor and acceptor will share the electron pair.* Coordinate bond is also known as dative bond.* It is represented by an arrow pointing towards the acceptor.

Examples:1) Formation of Ammonium ion (NH4

+)* Ammonium ion is formed when ammonia reacts with hydrogen ion. In the formation of ammoniumion, the sp3 orbital with a lone pair on nitrogen overlaps with the empty 1s orbital of hydrogen ion toform a coordinate covalent bond.

H3N: + H+ [H3N H]+ (or) [NH4]+

Donor Acceptor Ammonium ion

* Ammonium ion contains 4 bond pairs in the valence shell of nitrogen atom and hence it is tetrahedralin shape with 109o28' bond angles. The hybridization of nitrogen will remain sp3 only.

Tetrahedral shape of NH4+

109o28'N

H

HH

H

+

Note: In NH4Cl, there are three covalent bonds, one coordinate covalent bond and an ionic bond.Ionic bond is present between NH4

+ and Cl- ions.

2) Formation of Ammonia-boron trifluoride* The ammonia molecule donates a lone pair of electrons on nitrogen atom to the empty 'p' orbital ofboron in BF3. Thus a coordinate bond is formed. During the bond formation, the sp3 orbital containingthe lone pair on nitrogen overlaps the empty 'p' orbital of boron.

H3N: + BF3 [H3N BF3]Donor Acceptor

* The hybridization in boron is changed from sp2 to sp3 during the formation of coordinate covalentbond. Hence the geometry around boron in the complex formed will be tetrahedral.

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N B

H

H

H

H

H

H

3) Formation of Hydronium ion* The sp3 orbital containing lone pair on oxygen in water molecule overlaps with the empty 1s orbital onhydrogen ion to form a coordinate covalent bond.

H2O: + H+ [H2O H]+ or H3O+

Donor Acceptor Hydronium ion

: :

2) Formation of [AlCl4]-

* The chloride ion (Cl-) donates a lone pair of electrons to the empty 'p' orbital of aluminium in AlCl3.Thus a coordinate bond is formed. During the bond formation, the 'p' orbital containing the lone pair onchloride ion overlaps the empty 'p' orbital of aluminium.

AlCl

Cl

Cl

Cl:

::

: AlCl

Cl

Cl

Cl:

::

_

_+

Acceptor donor AlCl4-

Planar Tetrahedral

* The hybridization in aluminium is changed from sp2 to sp3 during the formation of coordinate covalentbond. Hence the geometry around aluminium in AlCl4

- will be tetrahedral.

More examples:SO2 and SO3 also considered to have coordinate covalent bonds according to octet rule as shownbelow.

S

O O

:

S

O O

O

Properties of compounds containing coordinate covalent bondProperties of compounds having coordinate covalent bond are similar to covalent compounds.

They do not conduct electricity because of absence of either free electrons or ions in them. They aresoluble mostly in non polar solvents like benzene, carbon tetrachloride etc., . But less soluble in polarsolvents like water and alcohol.

But their boiling and melting points are niether very high as in case of ionic compounds nor verylow like in case of covalent compounds.

MOLECULAR ORBITAL THEORY (MOT)* Molecular orbital theory was proposed by Hund and Mulliken.

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adichemadi @ gmail.com* The important postulates of this theory are given below.

1) According to this theory, the orbitals in a molecule are different from those of atoms. Theelectrons of all the bonded atoms in a molecule revolve under the influence of all the nuclei in molecularorbitals.

2) The atomic orbitals (AO's) of the bonded atoms combine linearly to form molecular orbitals(MO's) which are occupied by the electrons of bonded atoms.

Molecular orbital is a region around the nuclei of all the bonded atoms in a molecule wherethe probability of finding electrons is maximum.

3) Only those atomic orbitals with almost same energy and symmetry with respect to internuclearaxis can combine to form molecular orbitals.

4) The number of molecular orbitals formed is numerically equal to the number of atomic orbitalscombining.

5) The shapes of molecular orbitals depend on the shapes of atomic orbitals.6) Each molecular orbital can accommodate a maximum of only two electrons.7) Each molecular orbital is associated with certain amount of energy and are arranged in their

increasing order of energy.8) The degenerate molecular orbitals posses same energy and are filled with electrons according to

Hund's rule. But they may be arranged in different directions.9) The MO's which have lower energy than AO's are called bonding orbitals, whereas those with

higher energy are known as anti bonding orbitals. The number of bonding orbitals formed is equal tothe number of anti bonding orbitals.

The orbitals which are not involved in the combination are called non bonding orbitals.The order of energies of different types of MO's is:Bonding orbitals < Non bonding orbitals < Anti bonding orbitals

Schematic diagram of MO's

Bonding orbitals

Anti bonding orbitals

AO's AO'sNon bonding orbitals

Energy

10) The bond order of the molecule can be calculated by using the following formula.

number of bonding electrons - number of antibonding electronsBond order (B.O) =

2

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Differences between Atomic and Molecular orbitals S.No Atomic orbitals Molecular orbitals 1. They belong to one specific atom only. They belong to all the atoms in a

molecule. 2. Characteristic of atoms. Formed when atomic orbitals of almost

same energy combine. 3. Simple shapes. Complex shapes. 4. Named as s, p, d, f… Named as , , .... 5. Less stable than bonding and more

stable than anti bonding orbitals. Either more or less stable than atomic orbitals.

Bonding and Anti bonding orbitalsBonding molecular orbitals are formed when the orbitals with same signs of wavefunction are

combined, whereas anti bonding molecular orbitals are formed when the orbitals with different signs ofwavefunctions are combined.

Depending on the pattern of overlapping, number of nodes and symmetry, the molecular orbitalsare again divided into two types viz., & .σ -Molecular orbitals

-Molecular orbitals are formed due to linear combination of atomic orbitals along the inter-nuclear axis of bonded atoms. They have cylindrical symmetry about the axis. Bonding orbital isdesignated as , whereas antibonding orbital is designated as * .Illustrations:Combination of two 's' orbitals along the internuclear axis

+s s s

bonding orbital

+ + + + ++

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+s s

anti bonding orbital

+ - + - + -

*s

Combination of 's' and 'p' orbitals along the internuclear axis

+ -+p

bonding orbitals

++ -

+ + -+

s p

+ -+p

bonding orbitals

+ - -- -

-+

*s p

Combination of two 'p' orbitals along the internuclear axis

+- + -+ +- + - - -++

p p

bonding orbitalp

+- +-+ +- +- - +

p panti bonding

orbital

*p

-+

π -Molecular orbitals -Molecular orbitals are formed due to sidewise overlapping of atomic orbitals on either sides of

the internuclear axis of bonded atoms. The electron density is concentrated on either side of the axis.Bonding orbital is designated as , whereas antibonding orbital is designated as * .

Illustrations:

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adichemadi @ gmail.comCombination of two 'p' orbitals above and below the internuclear axis

+

+

-

+

-

+

-

+

- -

+

pbonding orbital

-

+

+

+

- +- +

- +

-

anti bonding orbital

+

+-

-

*p

DIFFERENCES BETWEEN σ AND π MOLECULAR ORBITALS S.No σ -Molecular orbital π -Molecular orbital 1. Formed by end on overlapping along

the internuclear axis. Formed by sidewise overlapping perpendicular to the internuclear axis.

2. Large overlapped region. Small overlapped region. 3. Rotation about the internuclear axis is

symmetrical. Rotation about the internuclear axis is unsymmetrical.

4. Strong bonds are formed. Weak bonds are formed.

ELECTRONIC CONFIGURATIONS OF MOLECULES1) The MO's are arranged in their increasing order of energy and filled with electrons.2) The lowest available MO is filled first (Aufbau principle).3) Each MO can accomodate a pair of electrons with opposite spins (Pauli's exclusion principle).4) Pairing occurs only after all the degenerate MO's are filled with one electron each (Hund's rule).5)The order of energy of homonuclear diatomic MO's upto nitrogen molecule is:

2 2

1 1 2 2 2 22 2

** * *

*y y

x x

z z

p ps s s s p p

p p

But for elements heavier than nitrogen i.e., from oxygen onwards, the order will be as follows:

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2 2

1 1 2 2 2 22 2

** * *

*y y

x x

z z

p ps s s s p p

p p

MOLECULAR ORBITAL ENERGY LEVEL DIAGRAMS (MOED)

1) MOED of H2

H H

1s

1* s

H2

Energy1s1s

* Electronic configuration of H2 molecule is 2 21 1

*s s

number of bonding electrons - number of antibonding electronsBond order (B.O) = 2

2-0= 12

* Hence there is a single bond between two atoms in H2 molecule.

2) MOED of N2

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Energy

1s

1* s

1s 1s

2s 2s

2s

2* s

2px2py2pz 2px 2py 2pz

2 xp

2*xp

2 yp 2 zp

2*yp 2*

zp

N atom N atomN2 molecule

* Electronic configuration of N2 molecule is 2 2 2 2 2 2 21 1 2 2 2 2 2

* *y z xs s s s p p p

* 10 - 4Bond order (B.O) = 3

2

* Hence there is a triple bond in N2 molecule.

2) MOED of O2

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Energy

1s

1* s

1s 1s

2s 2s

2s

2* s

2px2py2pz 2px 2py 2pz

2 xp

2*xp

2 yp 2 zp

2*yp 2*

zp

O atom O atomO2 molecule

* Electronic configuration of O2 molecule is 2 2 2 2 2 2 2 1 11 1 2 2 2 2 2 2 2

* * * *x y z y zs s s s p p p p p

* 10 - 6Bond order (B.O) = 2

2

* Hence there is a double bond in O2 molecule.* It is paramagnetic due to presence of two unpaired electrons.

HYDROGEN BONDINGThe electrostatic force of attraction between partially positively charged hydrogen in a polar

molecule and an electronegative atom is called hydrogen bond.It is represented by dotted line.

Characteristics of Hydrogen bond1) The hydrogen must be bonded to highly electronegative atom and should posses sufficient positivecharge to make hydrogen bond.2) The electronegative atom should be smaller in size and possess high charge density to attract hydro-gen.

Documents

Chemical Bonding IPE