Chemical Bonding Ipe

Chemical Bonding

1

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VISIT THE FOLLOWING SITE FOR OTHER FILES AND UPDATES IF ANY http://www.adichemadi.com CHEMICAL BONDING Chemical Bond : The attraction between two atoms or ions is called a chemical bond. Chemical bond is 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 get octet configuration either by losing or gaining or sharing electrons.

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Types of chemical bond : 1) Ionic bond 2) Covalent bond 3) Metallic bond

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1) Ionic bond :The electrostatic force of attraction between two oppositely charged ions is called ionic bond. * 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 attraction and thus form an ionic bond. * An ionic bond is formed between two atoms when their electronegativity difference is greater than 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) is called 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 between electropositive 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 atoms must 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 sharing electrons. * Only the electrons in outer shell participate in bond formation. These electrons are called valence electrons. 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 is formed 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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bond 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 configuration. Fluorine atom gains one electron to get nearest inert gas - Neon's configuration. Thus formed Li+ and F- ions form the ionic compound LiF. Li +

 

 He

1e -  

F +  z=9  He 2s 2 2p5

F



F-

 He 2s2 2p6

 

LiF

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3) Li

1e-

+

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2)

Li z=3    He 2s1

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1)

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

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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) Mg   Mg 2+ + 2e z=12   Ne 3s 2  Ne

2) 2 x (

3) Mg 2+

Cl  z=17   Ne 3s 2 3p5



2Cl 

+

1e -

Cl-

 

)

 Ne 3s2 3p6  

MgCl2

* The electrovalency of Mg is 2 and that of Cl is 1 3) 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. Al  z=13 Ne   3s2 3p1

 

Al3+

 Ne

+

3e-

Chemical Bonding

2)

3 x(

F +  z=9  He   2s 2 2p5

3) Al3+

+

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3

3F -

1e -  

F-

)

 He 2s2 2p6

 

AlF3

O 2

 

1s2 2s 2 2p 6

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O 2-

2e 

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3) 2Na + +

+

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O  z  8 1s 2 2s 2 2p 4

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2)

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* 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) 2x  Na   Na + + 1e    z=11 1s 2 2s2 2p 6 3s1 1s2 2s 2 2p 6

 

Na 2O

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

FACTORS FAVOURING THE FORMATION OF IONIC BOND Ionic 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 they lose 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 energy values. 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 of cations 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 follows Na+ > 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 configuration) 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.

Chemical Bonding

E.g.

Ca2+

4

1s 2 2s 2 2p 6 3s 2 3p 6

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with octet configuration is formed easily.

8 Zn2+

1s 2 2s 2 2p 6 3s 2 3p6 3d10 with pseudo inert gas configuration cannot be formed 18

easily.

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Factors favouring the formation of anion 1) Small atomic size : Atoms with small atomic size have stronger attraction towards electron and hence can form anions readily. 2) High electron affinity and electronegativity : The atoms, with high electron affinity and electronegativity, gain electrons easily and hence form anions readily. E.g., Halogens can form anions readily due to small size, high electron affinity and high electronegativity. 3) Low charge on anion : The formation of highly charged anions is difficult as the addition of successive electrons to the atoms becomes difficult due to repulsion from electrons in the atom. Hence anion with low charge is formed readily. E.g., Among C4-, N3-, O2- and F- ions, the anion with low charge (F-) is formed readily whereas the formation of anion with higher charge (C4-) is difficult. i.e., The order of ease of formation is C4- < N3- < O2- < F-

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FAJAN'S RULES These 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 increase 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 with pseudo 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 configuration].

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. For example 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 constant heat summation which can be stated as follows. Hess's Law : The total energy change in a reaction remains same whether the reaction takes place in one step or in several steps. Calculation of lattice energy of NaCl NaCl can be obtained from sodium metal and chlorine gas either in one step or in several steps as shown below.

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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  ΔH f  . 1 Na  s   Cl2 g    NaCls  2

;

ΔH f = -410.5 kJ/mol

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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/

; ΔHs = + S = +108.7 kJ/mol

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Na  s    Na  g 

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mole of energy. This is called sublimation energy  ΔH s or + S  .

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ii) Ionization of sodium : Gaseous sodium atoms are ionized by absorbing 492.82 kJ / mole of energy. It is called ionization energy  ΔH i or + I  . Na  g    Na +g  + 1e-

; ΔH i = + I = + 492.82 kJ/mol

iii) Dissociation of chlorine molecule : One mole of gaseous chlorine molecules are dissociated into two moles of chlorine atoms by absorbing energy equal to 239.1 kJ/mole. This is called dissociation energy  ΔH d or  D  .

Cl 2g    2Cl g 

; ΔH d = +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/mol 2 2 1 1 +D Cl 2 g    Cl g  ; ΔH d = = +119.55 kJ/mol 2 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 (ΔH e or  E) . Cl 2 g  + e-   Cl- g 

; ΔH e = - E = -361.6 kJ/mol

Formation of NaCl Crystals : The gaseous Na+ and Cl- ions unite to form one mole of NaCl crystals. The energy liberated during this process is called lattice energy  ΔH u or -U  . This value can be calculated by using Hess's law as follows. Na +g  + Cl- g    NaCl s 

;

ΔH u = -U

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According to Hess's law the energy change in the steps involved in indirect method. 1 H f  H s  H f  H d  H e  H u 2 (or)

i.e.

H f   S  I 

D  E U 2

U  H f  H S  H i 

i.e.,

H d  H e 2

U  410.5  108.7  492.82  119.55  (361.6)  769.97 kJ .mol 1

1

+

Hf

Na(g)

Cl(g)

-U

He= +E

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Hi= +I +

-

Cl (g)

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Na (g)

NaCl(s)

H d 2

Hs= +S

+

/2Cl2(g)

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Na(s)

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Born-Haber cycle

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BORN LANDE EQUATION The 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:  AZ  .Z  Be 2  U   N o  No n  r r   attractive repulsive force force

Where

A = Madelung constant (which depends on geometry of the crystal)

 AZ  .Z  e 2 r n 1  B  = Repulsion coefficient n   (This depends on the structure and approximately proportional to the number of nearest neighbours) No = Avogadro's number Z+ & Z- = Charges on the positive and negative ions respectively. e = Charge of an electron. r = Distance between the oppositely charged ions n = Born exponent (a constant which is usually taken as 9) Born Lande equation can be written as follows

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U

AZ  .Z  e 2 N o ro

 1 1  n 

CRYSTAL STRUCTURES Unit cell : The smallest part of the crystal which produces entire crystal upon repeating three dimensionally is called unit cell. Coordination number : Maximum number of nearest oppositely charged ions surrounding any particular ion in ionic crystal is called the coordination number of that ion.

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

radius ratio.

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Limiting radius ratio (

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

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1) Crystal structure of Cesium chloride * Limiting radius ratio (

rCs  rCl 

Coordination number

Structure

2 3 4 4 6 8

linear Planar triangle Tetrahedral Squared planer Octahedral Body centered cubic

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less than 0.155 0.155 - 0.225 0.225 - 0.414 0.4142 - 0.732 0.4142 - 0.732 0.732 - 0.999

r ) r

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Limiting radius ratio (

)= 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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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

-

Cs+

Number of Cl- ions = 8 x Cl-

Cl-

Cl-

1 =1 8

Number of Cs+ ions = 1

2) Crystal structure of sodium chloride * Limiting radius ratio (

rNa rCl 

)= 0.52

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Cl-

Cl-

Na+

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Cl-

Na+ Na+

Cl-

Cl-

Na+

Na+

Cl-

Na+

Na+

1 =3 2

Total number of Cl- ions = 1+3 = 4

Number of Na + ions on the edges = 12 x Cl-

Na+

Cl-

1 =1 8

Number of Cl- ions at the centre of faces = 6 x

Na+

Cl-

Number of Cl- ions at corners = 8 x

Cl-

Na+

Cl-

Cl-

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Na+

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Na+

Cl-

Cl-

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* 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.

Na+

1 =3 4

Number of Na + ions at the centre = 1 Total number of Na + ions = 3+1 = 4

Cl-

Properties of ionic compounds 1) 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 considered as the giant molecule. 3) In the solid state, ions cannot move freely and hence they do not conduct electricity . But in fused 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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5) 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.

NaCl(aq) + AgNO3(aq)  NaNO3(aq) + 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 their molar mass is expressed as formula weight instead of molecular weight.

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COVALENT BOND FORMATION & LEWI'S DOT STRUCTURES A 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 contribute 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.

2) Cl2 molecule

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+

.H

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H.

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Examples 1) H2 molecule * Electronic configuration of hydrogen is 1s1. * Each hydrogen atom contribute one electron to form a pair of electrons, which is shared in between the 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 or H H

2 2 6 2 5 * The electronic configuration of Cl is 1s 2s sp 3s 3p .

7

..

.. . Cl ..

.. .. . Cl .. . . . Cl or .. .. Cl .. . . Cl

..

..

+

..

.. . Cl ..

..

..

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

3) Hydrogen chloride (HCl) * Electronic configuration of hydrogen is 1s1. 2 2 6 2 5 * Electronic configuration of chlorine is 1s 2s sp 3s 3p .

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

Chemical Bonding

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the bond formation. Thus both of them get their nearest inert gas configurations. or

H

.. Cl ..

..

..

.. H X . Cl ..

4) Methane (CH4) 2 2 2 * The electronic configuration of carbon is 1s 2s 2p .

4 1

* The electronic configuration of hydrogen is 1s . * The carbon atom forms four covalent bonds by contributing four of its valence electrons for the bond formation. It forms 4 bonds with four hydrogen atoms. Thus it gets octet configuration. * Covalency of carbon is 4. H .x . H x C. . x H x H

H or

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H C H H

5) Ammonia (NH3) :

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* The electronic configuration of nitrogen is 1s2 2s 2 2p3 .

5

1

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

..

H

x

.N. .

x

or

H

H

.. N

H

x

H

5) H2O molecule

H

* Electronic configuration of oxygen is 1s2 2s 2 2p 4 .

6 1

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

H .O. x

..

x

H

or

H

.. O ..

H

6) Oxygen molecule (O2) 2 2 4 * The electronic configuration of oxygen is 1s 2s 2p .

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

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Thus a double bond is formed between oxygen atoms. Thus each oxygen atom gets Neon's configuration.

.. . O .. .

. .. .O ..

.. O ..

.. O ..

or

7) Nitrogen molecule (N2) * The Electronic configuration of nitrogen is 1s2 2s 2 2p3 .

5

N

N

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8) Carbon dioxide (CO2)

N

..

or

.. ... ... ..

N

..

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

2 2 2 * Electronic configuration of carbon is 1s 2s 2p .

4

2 2 4 * Electronic configuration of oxygen is 1s 2s 2p .

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9) Ethylene molecule (C2H4)

.. O

C

.. O

..

or

..

.. O

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C

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

.. O

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

2 2 2 * Electronic configuration of carbon is 1s 2s 2p

4

* Electronic configuration of hydrogen is 1s1 H H *. . . .* .C . . C. * * H H

H or

H C

C

H

H

Molecules violating octet rule: BeCl2 (Beryllium chloride) * Electronic configuration of Beryllium is 1s 2 2s 2 2 2 2 6 2 5 * Electronic configuration of Chlorine is 1s 2s sp 3s 3p .

7 * In the formation of Beryllium chloride, the beryllium atom contributes its two valence electrons and forms 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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the violation of octet rule. * Covalency of beryllium is 2. or

.. Cl ..

Be

.. Cl ..

..

x

..

Be

x

.. . Cl ..

..

..

.. Cl . ..

BCl3 (Boron trichloride) 2 2 1 * Electronic configuration of Boron is 1s 2s 2p .

3 2 2 6 2 5 * Electronic configuration of chlorine is 1s 2s sp 3s 3p .

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..

B

.. Cl ..

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PCl5 (Phosphorous pentachloride)

.. Cl ..

or

..

B

x

x

..

..

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..

.. . Cl ..

x

.. Cl

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.. . Cl ..

.. Cl .

Pr ep a..r ed

..

..

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

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2 2 6 2 3 * Electronic configuration of phosphorus is 1s 2s 2p 3s 3p .

5

2 2 6 2 5 * Electronic configuration of chlorine is 1s 2s sp 3s 3p .

7

x

. .. Cl ..

..

P

Cl ..

..

..

or

Cl

..

..

..

Cl .. .. Cl ..

SF6 (Sulphur hexafluoride) : * Electronic configuration of Sulphur is 1s 2 2s2 2p 6 3s 2 3p 4 .

6

..

Cl ..

..

P

.. Cl . x ..

..

x

..

..

x .. . Cl ..

..

..

..

.. Cl . .. x

.. Cl .

..

..

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

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* Electronic configuration of Fluorine is 1s2 2s 2 2p5

7 * Sulphur contributes six of its valence electrons to form 6 bonds with six fluorine atoms. Thus there are 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.

F F

F

F

* * * S * * *

F

F

or

F S

F

F

F

F

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F

Conclusion: Lewi's electronic theory could not explain the shapes and bond angles of molecules. It also could not explain why some molecules are stable eventhough they violate octet rule. Some more examples: 1) H2S

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2) CO32-

H

YA

..

or

Hydrogen sulphide

x _

_

Carbonate ion

H C

..

..

O

O

..

..

..

..

..

..

..

Cl ..

OR

O

H

6) HNO3

x

N x

x

.. . F ..

F

N

O

F

or

F

F

+

+

xx

..

xx

.

..

C

Formic acid

5) NF3

N O H O

`

Nitrogen trifluoride

7) CO

H

O H

Silicon tetrachloride

.. . F ..

O

O

..

..

or

C

or

4) HCOOH

Cl .. .. Cl Si Cl Cl ..

..

Ox

..

..

..

Cl .. .. .x x. . x Cl Si Cl .x .. ..

C xx O

x

..

3) SiCl4

O

x

H

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H

x

.. S ..

x

O

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..

H .S. x

_

_

x

O

Nitric acid

8) O3

O N O -

H

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14

:

8) NO2

9) NH4

N O

+

+

+

or

O

Ozone molecule

-1

O

:O

:

:

Carbon monoxide molecule

O:

: :

O

:

O

or

:

O

: :

::: :

O

:

C

:

or

:

O

:

C

:

:

:

Chemical Bonding

H

H

..

O N O

H

x

.N. .

x

H

or

H

N

H

x

Nitrite ion

H

H

VA by RD AD I.C HA OM N

Formal charge (Qf) It is the charge on an atom in a molecule assigned by assuming all the atoms have same electronegativity and the electron pairs are shared equally. It is calculated as follows: 1 N BP 2 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.

V. AD W IT W

YA

Examples: 1) PH3 molecule Lewi's dot structure:

W .A DI CH EM

Where

Pr ep ar ed

Qf = N A - N LP -

..

H

.P. .

x

x

x

H

.. P

H

H

or

H

H 1 6 N BP  5  2   0 2 2

Formal charge on 'P' = N A - N LP -

1 2 N BP  1  0   0 2 2 2) N2O molecule: It exists in following two resonance forms. For the following first resonance form: Formal charge on 'H' = N A - N LP -

(2 )

O

or

(1 )

:

+ : N : :N : :O :

:

: (1 )

-

:

:N

+ N

:

-

(2 )

Formal charge on first nitrogen 'N(1)' = N A - N LP -

1 4 N BP  5  4   1 2 2

Formal charge on second nitrogen 'N(2)' = N A - N LP -

1 8 N BP  5  0   1 2 2

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Formal charge on 'O' = N A - N LP -

1 4 N BP  6  4   0 2 2

For the following second resonance form:

(1 )

O

:

+ N

(2 )

:

: :

:

:N

: ::

(2)

or

:

(1 )

O

:

:N

+ N

Formal charge on first nitrogen 'N (1)' = N A - N LP -

1 6 N BP  5  2   0 2 2

Formal charge on second nitrogen 'N (2) ' = N A - N LP -

1 2 N BP  6  6   1 2 2

VA by RD AD I.C HA OM N

Formal charge on 'O' = N A - N LP -

1 8 N BP  5  0   1 2 2

Hence the formal charges vary with structural environment.

V. AD W IT W

YA

W .A DI CH EM

Pr ep ar ed

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 the electron 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 pair 3) 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 symmetrical structure, 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 than single bonds. 8) The repulsion between electron pairs increases with increase electronegativity of central atom and hence the bond angle increases. 9) Shapes of molecules can be predicted from the number of electron pairs in the valence shell of central atom as follows:

Chemical Bonding

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THE GEOMETRY OF MOLECULES CONTAINING ONLY BOND PAIRS IN THE CENTRAL ATOM Number of Formula Molecular geometry Examples bond pairs 2 AB2 Linear BeCl2, BeF2 B A B B

3

AB3

Trigonal planar

BF3, BCl3

A B

B

B

4

AB4

Tetrahedral

A B

CH4, CCl4

B

VA by RD AD I.C HA OM N

B

B

B

PCl5, PF5

B

B

Octahedral

YA

A

B

W .A DI CH EM

AB6

B

Trigonal bipyramidal

Pr ep ar ed

6

AB5

V. AD W IT W

5

B

B

SF6

A

B

B B

Chemical Bonding

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17

YA

W .A DI CH EM

Pr ep ar ed

VA by RD AD I.C HA OM N

GEOMETRY OF MOLECULES CONTAINING CONTAINING ONE OR MORE LONE PAIRS IN CENTRAL ATOM Total number Number Number Shape of of of bond of lone Formula Bond angle Examples molecule electron pairs pairs pairs o 3 2 1 AB2E Angular 120 SO2 o Trigonal 107 48’ NH3 3 1 AB3E Pyramidal 102o30’ NF3 4 Angular 104o28’ H2O 2 2 AB2E2 o (V- shaped) 103 F2O 4 1 AB4E See-Saw SCl4 , SF4 3 2 AB3E2 T-Shape 90o ClF3 5 o 2 3 AB2E3 Linear 180 XeF2, I3 Square o 5 1 AB5E 90 BrF5 pyramidal 6 Square o 4 2 AB4E2 90 XeF4 planar Where A = central atom B = atom linked to the central atom E = Lone electron pair

V. AD W IT W

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

Cl

180o Be

Cl

Linear molecule

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

B Cl

120o

Cl

Trigonal planar shape

3) CH4: The valence shell of the central atom - carbon contains only four bond pairs. Hence it is tetrahadral in shape with 109o28' bond angle. The bond pairs are arranged tetrahedral symmetry so as to minimize repulsions. If the bond pairs

Chemical Bonding

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18

are arranged in square plane, the angles between them will be only 90o and the repulsions will be more than in case of tetrahedral arrangement. Hence tetrahedral structure is more favorable than square planar structure.

H 109o28' C H

H H

Tetrahedral structure of methane

VA by RD AD I.C HA OM N

4) NH3: There are three bond pairs and one lone pair in the central atom, nitrogen. The bond angle is decreased from 109o28' to 107o48' due to repulsion caused by lone pair.

N

Pr ep ar ed

H

H

H

107o48'

YA

W .A DI CH EM

Trigonal pyramidal structure of ammonia molecule

V. AD W IT W

5) H2O: There are two bond pairs and two lone pairs in the central atom, oxygen. The bond angle is decreased from 109o28' to 104o28' due to repulsion caused by two lone pairs.

O H o

H

104 28'

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 electrons with 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 overlapping. This confers stability to the molecule. * Greater the extent of overlapping, stronger is the bond formed. The direction of the covalent bond is decided 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 nucleus axis 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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bond. In this bond, the electron clouds are present above and below the internuclear axis. It is a weaker bond. Note: The 's' orbitals can only form  bonds, whereas the p, d & f orbitals can form both  and π bonds. Examples 1) H2 molecule * Electronic configuration of hydrogen is 1s1. * In the formation of hydrogen molecule, two half filled 1s orbitals of hydrogen atoms overlap along the internuclear axis and thus form a  s-s bond.

+

1s orbital

H

H

H

VA by RD AD I.C HA OM N

H

1s orbital

 s  s bond

+

Cl

Cl

3pz orbital

Cl

YA

W .A DI CH EM

3pz orbital

Pr ep ar ed

2) Cl 2 molecule * Electronic configuration of Cl is 1s2 2s2 2p6 3s2 3px2 3py2 3pz1 * The two half filled 3pz atomic orbitals of two chlorine atoms overlap along the internuclear axis and thus by forming a  p-p bond. Cl

 p p bond

V. AD W IT W

3) HCl molecule * Electronic configuration of hydrogen is 1s1. * Electronic configuration of Cl is 1s2 2s2 2p6 3s2 3px2 3py2 3pz1 * The half filled 1s orbital of hydrogen overlap with the half filled 3pz atomic orbital of chlorine atom along the internuclear axis to form a  s  p bond. H

+

Cl

3pz orbital

1s orbital

Cl

H

 s  p bond

4) O2 molecule * Electronic configuration of O is 1s2 2s2 2px2 2py1 2pz1 * 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 double bond (one  and one  ) is formed between two oxygen atoms.

 p  p bond  p  p bond O

+

O

O

O

5) N2 molecule * Electronic configuration of N is1s2 2s2 2px1 2py1 2pz1 *  p-pbond is formed between two nitrogen atoms due to overlapping of half filled 2px atomic

Chemical Bonding

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orbitals axially. The remaining half filled 2py and 2pz orbitals form two π p-p bonds due to lateral overlapping. Thus a triple bond (one  and two  )is formed between two nitrogen atoms. 

p p





N

p p

p p

N

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

VA by RD AD I.C HA OM N

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

V. AD W IT W

YA

W .A DI CH EM

Pr ep ar ed

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 the hybridization. 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 repulsion between them and thus get maximum stability. 6) The filling of electrons into hybrid orbitals follows Pauli's exclusion principle and Hund's rule of maximum 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 degenerate 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

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

sp hybrid orbitals

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* Electronic configuration of 'Be' in excited state is 1s 2 2s1 2p1x

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 two half filled 'sp' hybrid orbitals are formed. These are arranged linearly.

 sp  p Cl

 sp  p Be

Cl

Be

Pr ep ar ed

180o Cl

VA by RD AD I.C HA OM N

* 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

Hybridization Shape -

sp linear

Bond angle

180o

-

YA

W .A DI CH EM

Linear molecule

BeCl2

V. AD W IT W

2) Acetylene (C2H2) * Electronic configuration of 'C' in ground state is * Electronic configuration at 'C' in excited state is

1s2 2s2 2p2 1s2 2s1 2p3

2s

2px 2py 2pz

sp hybridization

(2σ bonds) pure orbitals form

2 bonds

* Each carbon atom undergoes 'sp' hybridization by using one 2s and one 2p orbitals in the excited state. 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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 p p  p p  sp s H

 sp  sp C

C

H

C2 H 2 - Acetylene

180o H

C

C

Hybridization Shape -

sp linear

Bond angle

180o

-

H

VA by RD AD I.C HA OM N

Linear structure of acetylene molecule

V. AD W IT W

s orbital

YA

+

+

W .A DI CH EM

Pr ep ar ed

sp2 hybridization : * Intermixing of one 's' and two 'p' orbitals of almost equal energy to give three identical and degenerate 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.

px orbital

sp2 hybrid orbitals

py orbital

Examples: 1) BCl3 * Electronic configuration of 'B' in ground state is * Electronic configuration of 'B' in excited state is

1s2 2s2 2p1 1s2 2s12px12py1

2s

2px 2py 2pz

sp2 hybridization

(3σ bonds) In the excited state, Boron undergoes sp2 hybridization by using a 2s and two 2p orbitals. Thus three half filled sp2 hybrid orbitals which are oriented in trigonal planar symmetry are obtained. Boron forms three σ sp2 -p 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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Cl

Cl

B

B

120o

Cl

Cl

Cl

Cl

VA by RD AD I.C HA OM N

Trigonal planar structure of BCl3

BCl3 - Boron trichlorine

sp 2 Trigonal planar

Bond angle

120o

-

W .A DI CH EM

Pr ep ar ed

hybridization shape -

V. AD W IT W

YA

2) Ethylene (C2H4). * Electronic configuration of 'C' in ground state is * Electronic configuration at 'C' in excited state is

1s2 2s2 2p2 1s2 2s1 2p3

2s

2px 2py 2pz

sp2 hybridization

(3σ bonds) pure orbital forms

one  bond

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

H

 sp 2  sp 2 C

H

C

H

H C

 sp 2  s

C2 H 2 - Ethylene o

C 120

H

H

H

Hybridization Shape -

sp 2 planar

Bond angle

120o

-

Planar structure of ethylene molecule

YA

px orbital

py orbital

+

W .A DI CH EM

s orbital

Pr ep ar ed

+

+

VA by RD AD I.C HA OM N

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

V. AD W IT W

Examples: Methane (CH4) * Electronic configuration of 'C' in ground state is * Electronic configuration at 'C' in excited state is

sp3 hybrid orbitals

pz orbital

1s2 2s2 2p2 1s2 2s1 2p3

2s

2px 2py 2pz

sp3 hybridization

(4σ bonds) * In the excited state, the carbon atom undergoes sp3 hybridization by mixing one ‘2s’ and three 2p orbitals. Thus four half filled sp3 hybrid orbitals are formed in tetrahedral symmetry. * Each of these form σsp3 - s 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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H

H

 sp3  s

109o28' CH 4 - Methane Hybridization sp3 Shape Tetrahedral Bond angle 109o 28 '

C

C H

H

H

H H

H

Tetrahedral structure of methane

1s2 2s2 2p2 1s2 2s1 2p3

VA by RD AD I.C HA OM N

2) Ethane (C2H6) * Electronic configuration of 'C' in ground state is * Electronic configuration at 'C' in excited state is

2px 2py 2pz

2s

sp3 hybridization

Pr ep ar ed

(4σ bonds)

YA

W .A DI CH EM

* In the excited state, each carbon atom undergoes sp3 hybridization by using one 2s and three 2p orbitals. Thus four half filled sp3 hybrid orbitals are formed in tetrahedral symmetry around each carbon. * The two carbon atoms form a σsp3 - sp3 bond with each other by overlapping sp3 hybrid orbitals

V. AD W IT W

axially. Each carbon atom also forms three σsp3 -s bonds with hydrogen atoms. * Thus there is tetrahedral symmetry around each carbon with HCH and HCC bond angles equal to 1090 28' .

H

C

C

H

C

H H

H

H

H 109o28'

 sp3  sp 3

 sp3  s H

H

H Structure of Ethane molecule

C 2 H 6 - Ethane Hybridization sp3 Bond angle 109o 28 '

3) Ammonia (NH3) * Electronic configuration of nitrogen is 1s2 2s2 2px1 2py1 2pz1

109o28'

C

H H

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2px 2py 2pz

2s

sp3 hybridization

(3σ bonds) * 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 σsp3 -s bonds with three hydrogen atoms by using three half filled hybrid orbitals.

N H

 sp3  s H

N H

H

H

107o48'

NH 3 - Ammonia Hybridization sp3 Shape Trigonal pyramidal Bond angle 107o 48'

Pr ep ar ed

H

VA by RD AD I.C HA OM N

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.

Trigonal pyramidal structure of ammonia molecule

V. AD W IT W

YA

W .A DI CH EM

4) H2O (Water molecule) * Electronic configuration of oxygen is 1s2 2s2 2px2 2py1 2pz1

2px 2py 2pz

2s

sp3 hybridization

(2σ bonds)

* 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 two are completely filled. Oxygen forms two σsp3 -s 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 lone pairs. Thus water molecule gets angular shape (V shape).

O

 sp 3  s

O H

H H

o

H

104 28'

water - H 2 O Hybridization sp3 shape Angular Bond angle 104o 28'

Angular shape of water molecule

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

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orbitals is called sp3d hybridization. * Thus formed sp3d orbitals are arranged in trigonal bipyramidal symmetry. Among them, three are arranged in trigonal plane and the remaining two orbitals are present perpendicularly above and below the trigonal plane.

sp3d hybrid orbitals

VA by RD AD I.C HA OM N

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

3p

3d

sp3d hybridization

(5σ bonds)

YA

W .A DI CH EM

Pr ep ar ed

3s

V. AD W IT W

* In the excited state, phosphorus undergoes sp3d hybridization by using a 3s, three 3p and one 3d orbitals which are arranged in trigonal bipyramidal symmetry. * By using these half filled sp3d orbitals, phosphorous forms five σsp3d - p 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. Cl

Cl

90o Cl

P

120o Cl

PCl5 (Phosphorous pentachloride) Hybridization sp3d Shape Trigonal bipyramidal Bond angles 120o & 90o

Cl

sp3d2 hybridization * Intermixing of one 's', three 'p' and two 'd' orbitals of almost same energy by giving six identical and degenerate 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

VA by RD AD I.C HA OM N

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

3d

sp3d2 hybridization

(6σ bonds)

Pr ep ar ed

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

W .A DI CH EM

Sulfur atom forms six σsp3d 2 - p bonds with 6 fluorine atoms by using these sp3d2 orbitals. Each

V. AD W IT W

YA

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. F

F

F

S

F

F

SF6 - Sulfur hexafluoride Hybridization sp3d 2 shape Octahedral Bond angle 90o

F

Properties of covalent compounds 1) 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 same molecular formula. e.g. Following are the isomers with the molecular formula C2H6O. C2H5OH - Ethyl alcohol

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CH3-O-CH3

- Diethyl ether

COORDINATE COVALENT BOND The 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 that pair 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.

H3N:

+

H+

H]+

[H3N

Acceptor

(or)

[NH4]+

Ammonium ion

Pr ep ar ed

Donor

VA by RD AD I.C HA OM N

Examples: 1) Formation of Ammonium ion (NH4+) * Ammonium ion is formed when ammonia reacts with hydrogen ion. In the formation of ammonium ion, the sp3 orbital with a lone pair on nitrogen overlaps with the empty 1s orbital of hydrogen ion to form a coordinate covalent bond.

V. AD W IT W

YA

W .A DI CH EM

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

H

109o28'

N

H

H

H

Tetrahedral shape of NH4+

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 of boron in BF3. Thus a coordinate bond is formed. During the bond formation, the sp3 orbital containing the lone pair on nitrogen overlaps the empty 'p' orbital of boron. H3N: Donor

+

BF3

[H3N

BF3]

Acceptor

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

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30 H

H

H

H N

B

H

H

H2O:

H+

+

Donor

:

:

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

[H2O

Acceptor

H3O+

Hydronium ion

VA by RD AD I.C HA OM N

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 on chloride ion overlaps the empty 'p' orbital of aluminium. _

Cl

donor

YA

Acceptor

Cl : :

Al

Cl

AlCl4-

Tetrahedral

V. AD W IT W

Planar

Cl

W .A DI CH EM

+

Al

Pr : : ep ar ed

Cl

_ : Cl :

:

Cl

Cl

* The hybridization in aluminium is changed from sp2 to sp3 during the formation of coordinate covalent bond. 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 shown below.

:

O

S O

S O

O

O

Properties of compounds containing coordinate covalent bond Properties 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 are soluble mostly in non polar solvents like benzene, carbon tetrachloride etc., . But less soluble in polar solvents like water and alcohol. But their boiling and melting points are niether very high as in case of ionic compounds nor very low like in case of covalent compounds. MOLECULAR ORBITAL THEORY (MOT) * Molecular orbital theory was proposed by Hund and Mulliken.

Chemical Bonding

31

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Energy

YA

V. AD W IT W

Schematic diagram of MO's

W .A DI CH EM

Pr ep ar ed

VA by RD AD I.C HA OM N

* 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. The electrons of all the bonded atoms in a molecule revolve under the influence of all the nuclei in molecular orbitals. 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 where the probability of finding electrons is maximum. 3) Only those atomic orbitals with almost same energy and symmetry with respect to internuclear axis can combine to form molecular orbitals. 4) The number of molecular orbitals formed is numerically equal to the number of atomic orbitals combining. 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 to the 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

Anti bonding orbitals

AO's

Non bonding orbitals

AO's

Bonding orbitals

10) The bond order of the molecule can be calculated by using the following formula. Bond order (B.O) =

number of bonding electrons - number of antibonding electrons 2

Chemical Bonding

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

s

+

YA

+

+

V. AD W IT W

+

W .A DI CH EM

Pr ep ar ed

VA by RD AD I.C HA OM N

Bonding and Anti bonding orbitals Bonding 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 of wavefunctions are combined. Depending on the pattern of overlapping, number of nodes and symmetry, the molecular orbitals are again divided into two types viz.,  &  . σ -Molecular orbitals  -Molecular orbitals are formed due to linear combination of atomic orbitals along the internuclear axis of bonded atoms. They have cylindrical symmetry about the axis. Bonding orbital is designated as  , whereas antibonding orbital is designated as  * . Illustrations: Combination of two 's' orbitals along the internuclear axis

s

+

+ +

s bonding orbital

Chemical Bonding +

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33 +

-

+

s

-

-

+

s *

s

anti bonding orbital

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

+

+ -

+

+ +

-

+

p

s

-

 s p bonding orbital

+

-

-

+

+

p

s

-

VA by RD AD I.C HA OM N

-

-

+

-

 s p *

bonding orbital

+

+

-

+ p

-

-

++

+ p

V. AD W IT W

-

+

-

+

p

-

+-

+ +

-

p bonding orbital

YA

p

-

W .A DI CH EM

-

Pr ep ar ed

Combination of two 'p' orbitals along the internuclear axis

+

-

+

-

+

p* anti bonding orbital

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:

Chemical Bonding

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34

+

+ +

+

+

+

Combination of two 'p' orbitals above and below the internuclear axis

-

- -

-

-

+ -

p bonding orbital

-

-

+

+

+

-

+

-

+

-

+

-

VA by RD AD I.C HA OM N

+ p*

Pr ep ar ed

anti bonding orbital

V. AD W IT W

YA

W .A DI CH EM

DIFFERENCES BETWEEN σ AND π MOLECULAR ORBITALS σ -Molecular orbital π -Molecular orbital S.No Formed by end on overlapping along Formed by sidewise overlapping 1. the internuclear axis. perpendicular to the internuclear axis. Large overlapped region. Small overlapped region. 2. Rotation about the internuclear axis is Rotation about the internuclear axis is 3. symmetrical. unsymmetrical. Strong bonds are formed. Weak bonds are formed. 4.

ELECTRONIC CONFIGURATIONS OF MOLECULES 1) 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 py    *2 p y   1s   *1s   2 s   *2 s      2 px      *2 px 2p    *2 p   z  z  But for elements heavier than nitrogen i.e., from oxygen onwards, the order will be as follows:

Chemical Bonding

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35

  2 py  1s   *1s   2 s   *2 s   2 px   2p  z

   *2 p y      *2 px    *2 p    z 

MOLECULAR ORBITAL ENERGY LEVEL DIAGRAMS (MOED) 1) MOED of H2

 *1s Energy

H

H 1s

VA by RD AD I.C HA OM N

1s

 1s H2

Pr ep ar ed

* Electronic configuration of H2 molecule is  1s 2  *1s 2

2) MOED of N2

V. AD W IT W

YA

W .A DI CH EM

number of bonding electrons - number of antibonding electrons 2 2-0 = 1 2 * Hence there is a single bond between two atoms in H2 molecule. Bond order (B.O) =

Chemical Bonding

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36

 *2 p x  *2 p y  *2 pz 2px 2py 2pz

2pz 2py 2px

 2 px

VA by RD AD I.C HA OM N

 2 p y  2 pz

 *2 s

Energy

YA

V. AD W IT W 1s

N atom

2s

 2s

W .A DI CH EM

Pr ep ar ed

2s

 *1s 1s

 1s N2 molecule

N atom

* Electronic configuration of N2 molecule is  1s 2  *1s 2  2 s 2  *2 s 2  2 p y 2  2 p z 2  2 p x2 10 - 4 3 2 * Hence there is a triple bond in N2 molecule.

* Bond order (B.O) =

2) MOED of O2

Chemical Bonding

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37

 *2 p x  *2 p y  *2 pz 2px 2py 2pz

2pz 2py 2px

 2 p y  2 pz

VA by RD AD I.C HA OM N

 2 px

 *2 s

Energy

YA

V. AD W IT W 1s

O atom

2s

 2s

W .A DI CH EM

Pr ep ar ed

2s

 *1s 1s

 1s O2 molecule

O atom

* Electronic configuration of O2 molecule is  1s 2  *1s 2  2 s 2  *2 s 2  2 px 2  2 p y 2  2 pz 2  *2 p y1  *2 pz1 10 - 6 2 2 * Hence there is a double bond in O2 molecule. * It is paramagnetic due to presence of two unpaired electrons.

* Bond order (B.O) =

HYDROGEN BONDING The 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 bond 1) The hydrogen must be bonded to highly electronegative atom and should posses sufficient positive charge to make hydrogen bond. 2) The electronegative atom should be smaller in size and possess high charge density to attract hydrogen.