XI 04 Chemical Bonding

CHEMISTRY LECTURE NOTES

CHEMISTRY Chemical Bonding LECTURE # 1 TO 12

CHEMICAL BONDING LECTURE # 1 Starts from lattice energy (Ionic Bond)

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

One of the atom must be electro +ve can easily loose e–s or should have few more e–s than a noble gas. It should have low .E value

(ii)

Other element should be a electro –ve element having high .E. value, and more negative value of electron gain enthalpy, having few e–s less than noble gas.

(iii)

Energy released because of the combination of cation and anion should be high.This energy is also defined in terms of lattice enthalpy. Lattice Enthalpy : The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCI is 788 kJ mol-1. This means that 788 kJ of energy is required to separate one mole of solid NaCI into one mole of Na+1 (g) and one mole of Cl–(g) to an infinite distance. This process involves both the attractive forces between Ions of opposite charges and the repulsive forces between ions of like charge. The solid crystal being three- dimensional; it is not possible to calculate lattice enthalpy directly from the interaction of forces of attraction and repulsion only. Factors associated with the crystal geometry have to be included. Na+(g) + Cl–(g)  NaCl(s) H = Hlattice  –ve (always) 2+ – Mg (g) + 2Cl (g)  MgCl2 (s)

FACTORS AFFECTING L.E. (i) (ii)

Lattice energy (L.E.) 

1 r

r = r+ + r–

= interionic distance L.E.  Z+, Z– Z+  charge on cation in terms electronic charge Z–  charge on anion in terms electronic charge (a) NaCl > KCl (size) (b) NaCl < MgO (size, charge) (c) NaCl < MgCl2 (size) In size and charge, charge will dominate Na2O > NaF NaCl < Na2S (d)

Al2O3 Na2O MgO Al2O3 > MgO > Na2O

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

RESONANCE

CHEMICAL BONDING - 1

Born Haber Cycle for NaCl (s)

Na (s)

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

+

Hsub Na(g) HI.E.

NaCl(s)

HL.E.

HEg

+

-

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

Hf

Hsub

+ve/–ve Generally Q.

+

+

H.E.





+ve

+ve

1 H 2 BE  +ve

+

Heg

+

HL.E.

 +ve/-ve

Born haber cycle for Al2O3(s)

2Al (s) +

Hsub 2Al(g) HI.E

1

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

Al2O3(s)

1

HL.E

+

2Al (g)

–

3O (g)

HI.E

2

 Heg

2+

2Al (g) HI.E

2

3

3+

2–

2Al (g) + 3O (g)

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

3 HBE + 3Heg1 + 3Heg2 + HL.E. 2

Born haber for NaBr

Na (s) Hsub Na(g)

HI.E.

+

1 Br () 2 2 1 2

Hf

NaBr(s)

Hvap

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

H

HEg +

-

Na (g) + Br (g)

Ex.

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

RESONANCE

CHEMICAL BONDING - 2

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

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

Solvation or Hydration :

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



Z+ Z–



 1 1   r 



1   1    r  r 

  (nature of solvent) where  is dielectric constant. r 

Greater the polarity, greater will be r r r

  1/ r  , 1–1/r   S.E.  H2O 81

CH3OH 34

C2H5OH 27

C6H6 34

Applications of Hydration energy : (a) (b) (c)

Size of the hydrated ions: Greater the hydration of the ion greater will be its hydrated radii Li+(aq) > Na+(aq) Mobility of the ion: more is the hydration smaller will be the mobility of the ions Li+(aq) < Na+(aq) < K+(aq) < Rb+(aq) < Cs+(aq). Electrical conductance : is related to mobility so follows the same order

RESONANCE

CHEMICAL BONDING - 3

LECTURE # 2 (d)

Solubility of ionic solids : 1 Now we have

 L.E.  r  r

and

S.E. 

 1 1     r r 

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

1 110

1 120

1 130

1 140

1 1 1 1 1 1 1 1     100 10 100 20 100 30 100 40 decreament in solvation energy > decreament in lattice energy

S.E.

Solubility Orders. Increase difference in radii i.

LiF

ii.

NaF

iii.

LiF

iv.

CsF

v

Li

vi

Be(OH)2

vii

LiOH

viii

BeSO4

 solubility  and viceversa increses  LiI increase  Na increase  CsF decrease  Cs decrease  Cs increase  Ba(OH)2 increase  CsOH decrease  BaSO4 white ppt

ix

LiClO4

 CsClO4

decrease

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

BeF2

xi

AgF

 Be2  Ag

 solubility 

decrease decrease

Q.

NaF Na[BF4]– which is more soluble

Ans.

Na[BF4]– difference will

Q. Ans.

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

Ex.

Predict the product of decomposition of given polyhalides

 solubility 

Rb+ [Cl2]–  RbCl + Cl



halide with stable lattice ( having ions of comparable radii) are produced Li Be F Na Mg Cl Na+ < F–  K+ K Ca Br K+ < Cl–  Rb+ Rb Sr  Cs Br At  KCl + Br K+ [Br Cl]– 

RESONANCE

CHEMICAL BONDING - 4

(g)

Isomerism  onic compound does not show lsomerism due to non-directional nature of ionic bond But onic compound can show

(h)

Polymorphism  If an ionic compound is having two or more than two crystalline structures then it is said to be polymorphus CaCO3 ––– Calcite form Aragonite Zns ––– Zinc blende Wurtzite

(i)

somorphism  If two different ionic compounds are having similar crystalline structures then, these ar known as Isomorphism to each other. For two compounds to be isomorphus both should have similar formula type eg  FeSO4 .7H2O cannot be isomorphus of CuSO4.5H2O Na2CO3 NaNO3 [Not isomorphus] MgSO4.7H2O FeSO4.7H2O ZnSO4 .7H2O (Isomorphus)



a

b

The cations of both compound should be of similar shape or structure ( isostructural) similarly anions of both compound should be isostructural

Ex.



MgCO3 CO32–

NaNO3 NO3–

trigonal planar

trigonal planar

O

O

N

C –

– O

O–

Can be Isomorphus Na2CO3 CO32– SP2

Na2SO3 SO32– SP3

(Not isomorphus)

O

: S C O–

– BaSO4 SO42– Sp3

O –

O–

KMnO4 MnO4– Sp3

O

O

Mn

S O –

O–

NaNO3 Sp2

O

O–

O NaClO4 Sp3

O

O

Cl

N O

RESONANCE

[X]

O–

O O

O–

CHEMICAL BONDING - 5

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

 

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

More distortion of anion, more will be polarisation then covalent character increases. Factors affecting the polarisation : (i)

Small size of cation  polarisation. e.g.

(ii)

BeCl2

MgCl2

Size of cation  Polarisation  Large size of anion  polarisation

CaCl2

SrCl2

BaCl2

Covalent character 

e.g.

(iii)

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

(iv)

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

RESONANCE

CHEMICAL BONDING - 6

LECTURE # 3 Application & Exceptions of Fajan’s Rules : Applications : (i)

Ag2S is less soluble than Ag2O in H2O because Ag2 S is more covalent due to bigger S2– ion.

(ii)

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

(iii)

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



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

(iv)

(v)

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

MP 

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

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

g M

g g Draw Lewis dot structures

• Each bond Is formed as a result of sharing of an electron pair between the atoms. • Each combining atom contributes at least one electron to the shared pair. • The combining atoms attain the outer- shell noble gas configurations as a result of the sharing of electrons.

RESONANCE

CHEMICAL BONDING - 7

TO DECIDE THE CENTRAL ATOM (1) In general the least electronegative atom occupies the central position in the molecule/ion. For example in the NF3 and CO32–, nitrogen and carbon are the centra atoms whereas fluorine and oxygen occupy the terminal positions. (2) Generally the atom which is/are less in number acts as central atom (3) Generally central atom is the atom which can form maximum number of bonds( which is generally equal to the number of electrons present in the valence shell of the atom). (4) Atom of highest atomic number or largest atom atom generally acts as central atom. Hence we can say that Flourine and Hydrogen can never act as central atoms. • After accounting for the shared pairs of electrons for single bonds, the remaining electron pairs are either utilized for multiple bonding or remain as the lone pairs. The basic requirement being that each bonded atom gets an octet of electrons. Limitations of the Octet Rule : The octet rule, though useful, is not universal. It is quite useful for understanding .the structures of most of the organic compounds and it applies mainly to the second period elements of the periodic table. There are three types of exceptions to the octet rule. 1. The incomplete octet of the central atom In some compounds, the number of electrons surrounding the central atom Is less than eight. This is especially the case with elements having less than four valence electrons. Examples are LiCl. BeH2 and BCl3.

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

.

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

Interestingly, sulphur also forms many compounds in which the octet rule is obeyed. In sulphur dichloride, the S atom has an octet of electrons around it. 4. Other drawbacks of the octet theory • It is clear that octet rule is based upon the chemical Inertness of noble gases. However, some noble gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of compounds like XeF2 , KrF2 , XeOF2 etc., • This theory does not account for the shape of molecules. • It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.

RESONANCE

CHEMICAL BONDING - 8

MODERN CONCEPT OF COVALENT BOND (VBT) : Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms. The energy so released is called as bond enthalpy, which is corresponding to minimum in the curve depicted in Fig. 4.8. Conversely. 435.8 kJ of energy is required to dissociate one mole of Hg molecule. H2(g) + 435.8 kJ mol –  H(g) + H(g)

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

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

<



<

[Principal Quantum no. same, n = 2]

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

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

RESONANCE

CHEMICAL BONDING - 9



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

H=

H=

(ii) HCl gas molecule (s-p)

(iii) F2, Cl2, Br2, 2 (p-p) F2 2p-2p

Cl2 3p-3p

Br2 4p-4p

2 5p-5p

Strength of Sigma and pi Bonds : Basically the strength of a bond depends upon the extent of overlapping- In case of sigma bond, the overlaping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond . between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double or triple bonds) Overlapping of Atomic Orbitals : When two atoms come close to each other. there is overlapping of atomic orbitals. This overlap may be positive, negative or zero depending upon the properties of overlapping of atomic orbitals. The various arrangements of s and p orbitals resulting in positive, negative and zero overlap are depicted in Fig. 4.9.



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

(a)

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

(b)

Total overlapping will zero

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

(B)

Negative overlapping :

not allowed not allowed not allowed Q. Ans.

Count the  &  bonds in 6&5

RESONANCE

N  C – C  C – C N

CHEMICAL BONDING - 10

LECTURE # 4 VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY As already explained, Lewis concept is unable to explain the shapes of molecules. This theory provides a simple procedure to predict the shapes of covalent molecules. Sidgwick and Powell in 1940, proposed a simple theory based on the repulsive interactions of the electron pairs in the valence shell of the atoms. It was further developed and redefined by Nyholm and Gillespie (1957). The main postulates of VSEPR theory are as follows : • The shape of a molecule depends upon the number of valence shell electron pairs [bonded or nonbonded) around the central atom. • Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged. • These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them. • The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another. • A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair. • Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure. The repulsive interaction of electron pairs decrease in the order: Lone pair (lp) - Lone pair (Ip) > Lone pair (Ip) - Bond pair (bp) > Bond pair (bp) - Bond pair (bp) Nyholm and Gillespie (1957) refined theVSEPR model by explaining the important difference between the lone pairs and bonding pairs of electrons. While the lone pairs are localised on the central atom, each bonded pair is shared between two atoms. As a result, the lone pair electrons in a molecule occupy more space as compared to the bonding pairs of electrons. This results in greater repulsion between lone pairs of electrons as compared to the lone pair - bond pair and bond pair - bond pair repulsions. These repulsion effects result in deviations from idealised shapes and alterations in bond angles in molecules. For the prediction of geometrical shapes of molecules with the help of VSEPR theory. it is convenient to divide molecules into two categories as (i) molecules in which the central atom has no lone pair and (ii) molecules in which the central atom has one or more lone pairs. Table shows the arrangement of electron pairs about a central atom A (without any lone pairs) and geometries of some molecules/ions of the type AB. Next Table shows shapes of some simple molecules and ions in which the central atom has one or more lone pairs. Next Table explains the reasons for the distortions in the geometry of the molecule. As depicted in first Table, in the compounds of AB2 , AB3 , AB4 , AB5 and AB6 , the arrangement of electron pairs and the B atoms around the central atom A are : linear, trigonal planar, tetrahedral, trigonal-bipyramidal and octahedral, respectively. Such arrangement can be seen in the molecules like BF3 (AB3) , CH4(AB4) PCl5 as depicted below by their ball and stick models.

RESONANCE

CHEMICAL BONDING - 11

Fig. 4.6 The shapes of molecuies in which central atom has no lone pair

Table 4.7 Shape (geometry) of Some Simple Molecules/Ions with Central Ions having One or More Lone Pairs of Electrons (E).

RESONANCE

CHEMICAL BONDING - 12

Table 4.8 Shapes of Molecules containing Bond Pair and Lone Pair

RESONANCE

CHEMICAL BONDING - 13

The VSEPR Theory is able to predict geometry of a large number of molecules, especially the compounds ofp-block elements accurately. It is also quite successful in determining the geometry quite-accuralely even when the energy difference between possible structures is very small. The theoretical basis of the VSEPR theory regarding the effects of electron pair repulsions on molecular shapes is not clear and continues to be a subject of doubt and discussion. Directional Properties of Bonds : As we have already seen the formation of covalent bond depends on the overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of 1s-orbitals of two H atoms, when they combine with each other. In case of polyatomic molecules like CH4 , NH3 and H2O the geometry of the molecules is also important in addition to the bond formation. For example why is it so that CH4 molecule has tetrahedral shape and HCH bond angles are 109.5° ? Why is the shape of NH3 molecule pyramidal ? The valence bond theory explains the formation and directional properties of bonds in polyatomic molecules like CH4 , NH3 and H2O, etc. in terms of overlap and hybridisation of atomic orbitals.

RESONANCE

CHEMICAL BONDING - 14

LECTURE # 5 NEED OF NEW CONCEPT (HYBRIDISATION) : The valance bond theory (overlapping concept) explains satisfactorily the formation of various molecules but it fails to account the geometry and shapes of various molecules. It does not give the explanation why BeCl2 is linear , BF3 is planar, CH4 is tetrahedral , NH3 is pyramidal and water is V– shaped molecule. In order to explain these cases , the valance bond theory has been supplemented by the concept of hybridization. This is a hypothetical concept and was introduced by Pauling & Slater. Let us first consider the CH4 (methane) molecule. The electronic configuration of carbon in its ground state is [He]2s2 2p2 which in the excited stale becomes [He] 2s1 2px1 2px1 2px1. The energy required for this excitation is compensated by the release of energy due to overlap between the orbitals of carbon and the hydrogen.The four atomic orbitals of carbon, each with an unpaired electron can overlap with the 1 s orbitals of the four H atoms which are also singly occupied. This will result in the formation of tour C – H bonds. It will , however , be observed that while the three p orbitals of carbon are at 90° to one another, the HCH angle for these will also be 900 That Is three C – H bonds will be oriented at 900 to one another. The 2s orbital of carbon and the 1s orbital of H are spherically symmetrical and they can overlap in any direction. Therefore the direction of the fourth C – H bond cannot be ascertained. This description does not fit in with the tetrahedral HCH angles of 109.5°. Clearly, it follows that simple atomic orbital overlap does not account for the directional characterisics of bonds in CH4 . Using similar procedure and arguments, it can be seen that in the case of NH3 and H2O molecules, the HNH and HOH angles should be 90°. This is in disagreement with the actual bond angles of 1070 and 104.5° in the NH3 and H2O molecules respectively.

Hybridisation

(i) (ii)

(iii) (iv)

(v)

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

nucleus

bigger lobe will be used for bonding

(vi)

Since hybrid orbitals have been generated for the bonding purpose. So, bond formed by a hybrid orbitals are stronger than bond formed by pure atomic orbitals

(vii)

The orientations of hybrid orbitals generated will be dependent on type of atomic orbitals and on number of atomic orbitals taking part in hybridisation s+p

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

s + 3p

 4 new sp3 hybridised

s + 3p + d  sp3d s + 3p + 2d  sp3d2 s + 3p + 3d  sp3 d3 d + s + 2p  dsp2

RESONANCE

CHEMICAL BONDING - 15

The orientation of hybridised orbitals will be such that there will be minimum repulsion between any two hybridised orbitals

s + p  s + px

Yaxis SP

s + pY

x axis 180º

120º s + 2p  s + px + py

in x - y plane

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

(viii)

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

2

sp -sp bond (unstable)

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

(a) CO2

(S+Px+Pz)

:

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

C

O: S.N. = 3 (SP2)

SP (S+Px)

RESONANCE

(S+Px+Py)

CHEMICAL BONDING - 16

xy plane py

(xy)

(xz)

:

: x axis





:

:

x-z plane

(s + px + pz)

(s + px+ py)

(s + px)

F (b)

SF6

F S F F F

3

S.N. = 4(SP )

S.N. = 6 (SP3d2) F 2

S.N = 3 (SP)

B

(c)

F

S.N.= 4 (SP)3 2

S.N = 3 (SP )

:

F

SO2

:O

:

(d)

N 3 sp

3

E.N order

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

LECTURE # 6 sp2 hybridisation : In this hybridisation there is involvement of one s and two p-orbitals in order to form three equivalent sp2 hybridised orbitals. For example, in BCl3 molecule , the ground state electronic configuration of central boron atom is 1s2 2s2 2p1. In the excited state, one of the 2s electrons is promoled to vacant 2p orbital as

Fig.4.11 Formation of sp2 hybrids and the BCl3 molecule a result boron has three unpaired electrons. These three orbltals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals. The three hybrid orbitals so formed are oriented In a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B – Cl bonds. Therefore, in BCl3 (Fig.4.11), the geometry is trigonal planar with CIBCl bond angle of 120°.

RESONANCE

CHEMICAL BONDING - 19

OTHER EXAMPLES : B

AB3

A

B

trigonal planer

120º

BF3, C6H6, CO32– , B(OH)3, SO3, NO3– ,+CH3, graphite,

B

: A

ALB2 (f)

Bent molcule

B

B

Allotropes of carbon : 1. Graphite - sp2

2.

SO2, SnCl2, NOCl, O3, NO2–

Diamond - sp3

3.

Fullerene - sp2 and sp3

Graphite each C  sp2 – has 1 e in pure P orbital because of this graphite is a sheet like structure (planar structure) . the unpaired e– will be used for  bond formation. These  bonded e– will delocalised in whole of the layer 1.40 Aº

3.40 Aº

The structure of a graphite sheet C - C B.L in sheet = 1.40 Aº (benzene)

Q. Ans.

Graphite is anisotropic with respect to electrical conductivity. Graphite is good conductor of electricity along the layer because of delocalised electrons and poor conductor of electricity perpendicular to the layer because of large distance between the layers. On increasing temperature conductivity increases perpendicular to the layer and decreases along the layer. Therefore it behaves as semiconductor as well as conductor.

Q. Ans.

Graphite is consider as diamagnetic substance. Since e– are paired due to delocalised -bond. So it is diamagnetic in nature (repelled in to the magnetic field)

Q. Ans.

Graphite is used as lubricants. Attraction force between 2 layers is vanderwaal attraction force. So, layess can slide over each other. So, it is solt lubricating substance.

Fullerenes (C60 , C48 , C30 ...........) foot ball like spherical balls of C  Most of the C atom are bonded 3 other C atoms so, these will be sp2 hybridised.  This structure is made up of alternate pentagons and hexagons.  And Because of this all the double bond cannot be in conjugation. Hence, 2 different type of B.L. are present  in fullerenes one is C –C single B.L. and the other is C–C partial double bond length.

Bucky ball (C60)

RESONANCE

Structure of buckminster fullerence CHEMICAL BONDING - 20

BORON TRIHALIDES BF3

BCl 3

BBr3







sp

2

sp

2

sp

B 3  unstable

2

(stearic h indrance ) Discuss the structure and hybridisation inorganic benzene B3N3H6 borazine H N

N

H

B

B

H

B

B

N

N

 N

N H

B

B

H

H

Q.

How many dichloro borazine are possible ?

B

Cl N

N B

B N Cl

Inorganic benzene is chemically more reactive in comparison to benzene because N – B bond is polar in nature while in all the bonds are non-polar HCl + C6H6  no. chemical reactions. HCl + B3N3H6  B3N3H9 Cl3 3

H H

N

Cl

B

sp

Cl

H

H

B N

B

N

H H

H H

Cl H

sp3 hybridisation : This type of hybridisation can be explained by taking the example of CH4 molecule in which there is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp3 hybrid orbital of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp3 hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp3 hybrid orbital is 109.5° as shown in Fig. 4.12.

RESONANCE

CHEMICAL BONDING - 21

The structure of NH3 and H2O molecules can also be explaned with the help of sp3 hybridisation. In NH 3 , the valence shell (outer) electronic configuration of nitrogen in the ground state is 2s2 2px1 2py1 2pz1 having three unpaired electrons in the sp3 hybrid orbitals and a lone pair of electrons is present in the fourth one. These three hybrid orbitals overlap with Is orbitals of hydrogen atoms to form three N-H sigma bonds. We know that the force of repulsion between a lone pair and a bond pair is more than the force of repulsion between two bond pairs of electrons. The molecule thus gets distorted and the bond angle is reduced to 107° from 109.5°. The geometry of such a molecule will be pyramida as shown in Fig. 4.13.

Fig.4.13 Formation of NH3 molecule In case of HgO molecule, the four oxygen orbitals (one 2s and three 2p) undergo sp3 hybridisation forming four sp1 hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These four sp3 hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by hydrogen atoms while the other two by the lone. pairs. The bond angle in this case is reduced to 104.5° from 109.5° (Fig. 4.14) and the molecule thus acquires a V-shape or angular geometry.

Fig .4.14 Formation of H2O molecule

RESONANCE

CHEMICAL BONDING - 22

Examples of sp3 hybridisation AB4 tetrahedral

shape

AB3L pyramidal

AB2L2 V shape (bent)

ABL3 Linear

B

Example

B

B

B

CH4 (109º 28’)

ºB b.p – b.p.

.. A .. – B

..

B

.. A

Ä

109º 28'

..

A

Compounds of berillium 1. [BeF4]– (sp3 hybridised) (already taught) Tetraflouroberrylate (ii) ion 2.

BeCl2(s) (BeCl2)n polymeric solid (already taught) Cl

Cl

Cl

Be Cl

Cl

Cl Be

Be Cl

S.No. = 4 sp3 Cl

Compounds of Boron, Aluminium : [B(OH)4]–, [BF4]–, [H3 N :  BF3] 3

3

sp sp AlCl3 anhydrous exist in dimer form Al2Cl6 Crystaline AlCl3 or Hydrated AlCl3  [Al(H2O)6]Cl3 Al(OH)4– , AlH4–, AlCl4– are sp3 Borax Na2B4O7.10H2O Borax anion [B4O5(OH)4]2–

Compounds of Carbon :  All saturated organic compounds has C in sp3 hybridised state : – ..  C H3 C S.N = 4 3 sp H H H  Diamond Graphite is more reactive than diamond Graphite

   

1500 º C  Diamond 50,000  60,000 atm

Each C is bonded to 4 other C atoms.  sp3 hybridisation. Each C – C B. L. = 1.54 Å. Since all the e– s have taken part in bonding. so, are paired up. Hence it is diamegnetic and a bad conductor of electricity. The tetrahedran are linked together into a three-dimensional giant molecule. The unit cell is cubic. Strong covalent bonds extend in all directions. Thus the melting point is abnormally high (about 3930ºC) and the structure is very hard (In a rare modification of diamond.

RESONANCE

CHEMICAL BONDING - 23

LECTURE # 7 Compounds of Silicon Q.

CO2 is a gas but SiO2 is a high melting solid. O=C=O (discrete molecule) O = Si = O (unstable) Si –– O –– Si (stable)



(will form 3-D structure)

Elements forming Multiple bonds C N O P S Cl

SiO2  Covalently bonded 3D giant network solid like diamond. O

Si O

O

O

O

Si O

O O

Si

Si O

O

O

O

O

Silicates (Si  Sp3 hybridised) Basic repeating unit of silicate [SiO4]4–

O– Si

O

O

–

–

O

–

O

It is not O – O bond (there is no such bond)

O O

O Silicon atom O  O atom

CLASSIFICATION OF SILICATES : (A)

Orthosilicates : These contain discrete [SiO4]4– units i.e., there is no sharing of corners with one another as shown is figure.

(B)

Pyrosilicate : In these silicates two tetrahedral units are joined by sharing oxygen at one corner thereby giving [Si2O7]6– units.



(–) charge will be present on the oxygen atoms which is bonded with one Si atom.

RESONANCE

CHEMICAL BONDING - 24

(C)

Cyclic silicates : If two oxygen atoms per tetrahedron are shared to form closed rings such that the structure with general formula (SiO32–)n or (SiO3)n2n– is obtained, the silicates containing these anions are called cyclic silicates. Si3O96– and Si6O1812– anions are the typical examples of cyclic silicates. O

–

O–

O O

O

O

O

–

O

O O

–

O

O

O–

O–

O

O

O–

O

–

O

–

O

–

–

–

–

O

O

–

O

O

–

O

O

–

O– – O Si6O1812–

Si3O96–

(D)

Chain silicates : Chain silicates may be further classified into simple chain & double chain compounds. In case of simple chains two corners of each tetrahedron are shared & they form a long chain of tetrahedron. Their general formula is also same as the cyclic silicates i.e. (SiO3)n2n–

Similarly, double chain silicates can be drawn in which two simple chains are joined together by shared oxygen. The anions of double chain silicates have general formula (Si4O11)n6n– . –

O

–

O

O

O–

O–

O O

O

Compounds of P :

O

O–

O–

O

O

O

O

O–

O–

–

O

O O

O

O–

O

O

O

O

O–

O––

O–

O

O

–

P4 (White phosphorus)

P

P

P

60º

(P will be on all 4 vertexes of tetrahedron) (nothing at centre)

P

6–P–P bonds P–P–P = 60º S.N. = 4 (109º28’) Since B.A. is 60º so, P4 molecules is a strained molecule. Hence is chemically very reactive. Can catch fire at normal room temperature in air.

RESONANCE

CHEMICAL BONDING - 25

P4O6

no P–P bond, 12 POP bond. O O

P

O

O P4O10

OP O

O P O

P

O

O

P4S3

Oxygen Family : H2O

V-shaped molecule

H3O+ (hydronium ion) Compounds of S : S8 (Crown sulphur)

S32– SCl2 V-shaped S2Cl2 open book like structure SOCl2 (thionyl chloride) (derivative of SO2)

SO2Cl2 (sulphuryl chloride)

RESONANCE

CHEMICAL BONDING - 26

Halogen family : HClO4, HClO3, HClO2, HClO Noble gases Family : XeO4 Xenon tetraoxide

tetrahedral

XeO3 Xenon trioxide

Q.1 Ans.

pyramidal

Discuss the shape, geometry and bond angle of the following molecules. NH3 H2O O3 SO2 NH3 (Ammonia)

.. .. N

shape : pyramidal

H H H Geometry : tetrahedral , bond angle = 107° : S

B.A. < 120º

O

118º O O3 (Ozone)

: O O

O

:

:

:

:O

O

:

O: : O

O 117º O

Bond strength of both bond is equal diamagnetic substance. Ozonide ion (O3–) from KO3 paramagnetic substance. unpaired e

–

: O

:

O

:

:

: :

:

O

O2  Paramagnetic (can be explained on the basis of MOT) Q.2 Ans.

Compair the O–O bond length in H2O2 and O2F2. H2O2 (hydrogen peroxide)

open book like structure O – O BL is more than the expected.

O2F2

open book like structure

O–O B.L. is smaller then H2O2 because of smaller electron density on O atom in comparison to H2O2.

RESONANCE

CHEMICAL BONDING - 27

NOTE: If bond length in any molecule is found to be more than the expected then the main reason will be lone pair lone pair( and lone pair lone pair repulsion is considered when both the bonded atoms are of second period and both are having lone pairs on them) repulsion or decrement of electron density of the bond may be because of sharing or donation For example: H2O2, F2, NH2OH, NF3, The carbon carbon bond length increases in many -bonded complexe which we will study in coordination compounds. such as zeisse’s salt. Q.3

Discuss the shape of NH2– (Amide ion) an

Ans.

– .. .. N

V–shaped like H2O molecule

H H NH2OH (Hydroxylamine) .. N

H H

.. O .. – H

N – O B.L. is larger than the expected (.p. – .p. repulsion) Order of base strength NH3 > NH2 – NH2 > NH2OH

STRUCTURE OF OXY ACIDS OF P, S, Cl, N OXY-ACIDS OF PHOSPHORUS (P, atom is sp3 hybridised) S. NO.

NAME

1.

H3PO2

Hypophosphorus acid (monobasic)

2.

H3PO3

Phosphorus acid (Dibasic)

3.

H4P2O6

Hypophosphoric acid (tetrabasic)

4.

H4P2O7

Pyrophospho ric acid ( tetrabasic )

STRUCTURE

or Dipolyphos phoric acid

5.

H3PO4

Phosphoric acid (tribasic)

6.

H4P2O5

Pyrophosphorus acid

7.

HPO3

Metaphosphoric acid

RESONANCE

CHEMICAL BONDING - 28

8.

9.

Cyclic-metaphosphoric (HPO3)n (n = 2)

Dimeta posphoric acid

n=3

Trimeta phosphoric acid

OXY-ACIDS OF SULPHUR (S atom in sp3 hybridised) A. Sulphurous acid series.

1.

2.

3.

H2SO3

Sulphurous acid

H2S2O4

Dithionous acid

H2S2O5

O || HOS ..  O  H

O O || || H O S . . S . . O  H

Pyrosulphurous acid

O O || || H  O  S S . . O  H || O

Sulphuric acid

O || H  O  S O  H || O

Thiosulphuric acid

O || H  O  S O  H || S

B. Sulphuric acid series.

1.

2.

H2SO4

H2S2O3

Oleum

O O || || HOS OSOH || || O O

3.

H2S2O7

C.

Thionic acid series (General formula H2Sn+2O6, n = 0 to 12 are practicaly known) O O || || H  O  S  (S )n  S  O  H || || O O

H2S2O6 (n = 0)

RESONANCE

Dithionic acid

D.

CHEMICAL BONDING - 29

Peroxy acids of S

1.

2.

H2SO5

H2S2O8

Peroxymonosulphuric acid (caro’s acid)

O || HO O S O H || O

Peroxydisulphuric acid (Marshall’s acid)

O O || || HO S O O S O H || || O O

OXY-ACIDS OF ‘Cl’ (Cl atom is sp3 hybridised)

1.

HClO

Hypochlorous acid

2.

HClO2

Chlorous acid

3.

HClO3

Chloric acid

4.

HClO4

Perchloric acid

OXY-ACIDS OF NITROGEN (N atom sp2 hybridised)

1.

HNO3

Nitric acid (Colourless)

2.

HNO2

Nitrous acid (Pale blue)

3.

H2N2O2

Hyponitrous acid

4.

HNO 4 or HNO 2 (O 2 )

Pernitric acid or Peroxy nitric acid

5.

HNO(O2)

Peroxy nitrous acid

.. ..

RESONANCE

HO . . N  O

CHEMICAL BONDING - 30

LECTURE # 8 Example of sp3d hybridisation :  (s + px + py + pz + dz2) sp3d Steric number = 5 molecular gemetry  TBP Bond angle  120° & 90°

E axial

A E

E axial E = equitorial position

(i) In this type of geometry, . p. or .ps will always occupy E positions. (ii) the more negative atom occupies axial positions (iii) since in this geometry, axial bonds are found to be slightly longer and weaker as composed to equitorial bonds. Therefore db will always occupy equitorial positions All the examples of this category can be classified into following 4 types. Type AB5 AB4L AB3L2 AB2L3 Shape TBP see-saw T-shape Linear Ex.

PF5 :

PF5 SF4 ClF3 XeF2 PCl5 XeO2F2 XeF3+ 3– PBr5 Cl2– P5 PF5 is a gaseous molecule in which all the P-F BL are found to be equal which is due to continuous exchange between the axial and equitorial F atoms at very highrate (1200 times per sec.) Cl Cl 90° 0° 12 P

PCl5 :

Cl

net = 0

Cl Cl

Two types of P-Cl BL axial bonds are longer & weaker PCl5 (g) PCl3 (g) + Cl2(g) PCl5 (s) consist of PCl4+ and PCl6– sp3 sp3d2 + PBr5(s) consist of PBr4 and Br– P5(s) is expected to be P4+ and – Cl F

PF2Cl3

P Cl

net = 0

Cl F

F

F

P PCl2F3

Cl

Cl

=0

F AB4L

RESONANCE

CHEMICAL BONDING - 31

•• F

S

SF4

Seasaw

F

F F

•• F Xe

XeO2 F2

Seasaw

O

O F

AB3L2 •• F

 Cl ••

ClF3

T-shaped

F F

 > 180° Due to .p. – b.p. repulsions therefore FClF is not linear +

•• F

[XeF3]+

••

Xe

F F

It appear that all interhalogens of AB3 type where A is central halogen (not F) will have sp3d hybridisation on A. But it is found that Cl3 exists as a dimer 2Cl6 (bright yellow solid with the following structure. Cl

••

Cl



Cl

••

Cl

••  ••

Cl Cl On the basis of above structure  must be in sp3d2 hybridisation state. •• F

••

Xe

••

AB2 L3 XeF2

F – •• 

••

••



3–



+  3+

S. No. = 4

sp3







bent

–

• • Cl

••



••

Cl2–

••

••

Cl

RESONANCE

CHEMICAL BONDING - 32

Q.

Select the correct structure of ClF3 amongs the following with proper reasoning.

F ••

F

Cl

(A)

F

(B)

F

••

F Cl

• •

F

(C*)

••

Example of sp3d2 hybridisation S.N. = 6 geometry  Octahedral B.A. = 90° Types AB6 Shape Octhedral Ex. SF6 PF6¯ [SiF6]2– [AlF6]3– [XeO6]4– Perxenate ion

AB5L sq. pyramidal ICl4¯ XeOF4 BrF5

AB4L2 Sq. planar XeF4

__

PF6¯

F

F

F

F

F

F

F

[SiF6]2-

P

S SF6

F

F

F

F

F

[AlF6]3-

Si F

F

F

F

2-

F

F

[XeO6]4– XeO3 + NaOH  Na+[HXeO4]¯ Sodium Xenate H2XeO4 S.N. = 5

Xenic Acid

•• O

H

Xe

sp3d

O

O H

O

H4XeO6

Perxenic Acid (No peroxide group) Structure of [XeO6]4– 4-

O O

O

O

O¯

¯O

Xe

Xe

OR O

O O

RESONANCE

O¯

¯O O

CHEMICAL BONDING - 33

AB5L : BrF5

F

F

F Br

F

..

F

Presence of 1 l.p. doesnot create any problem. It can be placed any where because all the 6 comes of the octahedron are equivalents. But in case of XeOF4 d.p. and .p. are kept at 180°.

O

F

F Xe

F

..

F

AB4L2 : Cl4¯

_

..

Cl

Cl 

Cl4¯

µ=0

Cl 2.

..

Cl

l.p. of are found to be at trans position so as to have min. repulsion. ..

F

F Xe

XeF4

µ=0 F

..

F

Examples of sp3d3 hybridisation Type Example Shape

AB7 F7 Pentagonal bipyramidal

AB6L XeF6 Distorted octahedral

F

F F

F

F

90° F

72°

F



Xe

F

F

F

F F

F

But XeF6(s) is found to exist in tetrameric form when [XeF5]+ units are linked by F¯

RESONANCE

CHEMICAL BONDING - 34

 hyb of Xe in [XeF5]+ is found to be sp3d2 +

[XeF5]

Electrostatic Attraction

F¯

F¯ +

+

[XeF5]

[XeF5]

F¯

F¯ +

[XeF5]

BOND LENGTH (IT HAVE FOLLOWING SEQUENCE) (i)

Size of atom (see along the group)  bond length HF < HCl < HBr < HI F – F < Cl – Cl < Br – Br <  –  CH4 < SiH4 < GeH4 < SnH4

(ii)

Multiplicity of bond (nearly same period element) single bond > double bond > triple bond C—C > C=C > CC F—F > O=O > NN

(iii)

Electronegativity difference (See along the period) H—C > H—N > H—O > H—F

HOW TO COMPARE BOND ANGLES Ans.

Bond angle depends on the following factor I. Hybridisaition II. No. of lone pair III. Size or electronegativity of central atom IV. Size or electronegativity of terminal atom 1.

Hybridisaition : sp > sp2 > sp3 180° 120° 109°28’

2.

No. of lone pair : If hybridisaition of the central atom is same but no. of lone pair is diffrent then more in the lone pair len in the bond angle. e.g. CH4 NH3 H2O hybridisaition sp3 sp3 sp3 lone pair .P. = 0 .P. = 1 .P. = 2 B.A. 109°28’ 107° 104°

3.

Size or electronegativity of central atom : When hybridisation same no. of lone pair is same but central atom is different then see the electronegativity of central atom. More is the electronegativity more is the bond angle. e.g. NH3 PH3 AsH3 SbH3 hybridisaition sp3 no no no lone pair .P. = 1 .P. = 1 .P. = 1 .P. = 1 B.A. 107° 93° 92° 91° 3 Note : But if we apply steric no. rule than all have sp hybridisation.

RESONANCE

CHEMICAL BONDING - 35

Drago rule : Element of 3rd period (p-Block) and lower than 3rd period does not allowed hybridisation in molecule when they form compound with less electronegative elements such as hydrogen eg : PH3, SiH4, AsH3 , H2S not have hybridisation 4.

Size or electronegativity of terminal atom : Hybridisation same, lone pair same, central atom same but terminal atom is different then greater is the size of the terminal atom greater will be the bond angle. Only in case of flourine the electronegativity factor is considered, due to greater electronegativity of the flourine atom the bond angle for it comes out to be smallest( due to smaller bond bond pair repulsions) e.g. PF3 PCl3 PBr3 PI3 hybridisaition sp3 sp3 sp3 sp3 lone pair .P. = 1 .P. = 1 .P. = 1 .P. = 1 B.A. 98° 100° 101° 102° Reasion : As the E.N. of x , b.p.–b.p. repulsion will less but l.p. compression will work as usual

Q.4

Compair bond angle of OF2 and Cl2O.

Ans.

OF2

Cl2O because of large size of Cl atom Q.5

Discuss the bond angle in carbonyl halides COF2 , COCl2 , COBr2 , CO2

Ans. Carbonyl Hallide B.A.   COF2 < COCl2 < COBr2 < CO2 Explanation  (A) double bonds require more room than single bonds. Hence C = O group compresses the molecule and bond angle max. in COF2. Later on halogen atom becomes bigger and less (–ve) also so  inter atomic repulsion between the halogens causes  in B.L. As X (halogen) becomes less and less (–ve) b.p.-b.p. repulsion also becomes imp and therefore   Q.6 Ans.

Discuss the bond angle in Thionyl halide. Thionyl hallides

E.N. of halogen   b.p. – b.p. repulsion  l.p. on the S atom and more space required S = O grp will compress the molecule SOF2  b.p – b.p. repulsion min SO Br2  b.p. – b.p repulsion max.

RESONANCE

CHEMICAL BONDING - 36

Q.7 Ans.

Discuss the basic strength and bond angle hydride of nitrogen family. NH3 PH3 AsH3 SbH3 B.A. 107º 92º-93º 92º 91º l.p. availability on the centre atom  Lewis base strength  (Due to fact that l.p. on N in NH3 is in sp3 hybrid orbital which is more diffused and directional on the other hand concept of hybridisation is not applicable in the rest of the molecules. The l.p is present in a S orbital which is contracted and non-directional. This observation that the members of nitrogen family( other than N) and members of Oxygen family( other than O) do not use hybridised orbitals to make bond with hydrogen atom( or any atom with electronegativity less than 2.1) is called Drago’s rule.

Q.8

Comment on the bond angle of the following species.

Ans.

NO2

NO2

NO 2

[O = N = O]+ B.A.

Q.9

ONO

NO 2 < NO2 < NO 2 115º 132º 180º

Compare bond angles in NOCl and NO3–. NOCl

NO 3

x – B.L. of 3c – 2e bond > B.L. of 2c – 2e– bond 2 > 1 B.A.

RESONANCE

CHEMICAL BONDING - 37

Ht  terminal H atom Hb  bridging H atom B. A < Ht BHt > HbBHb 

If all the 4 Ht are considered in one plane then, the 2 3c – 2e– bonds will not be in that plane. These will be in a perpendicular plane.



If substitution (like chlorination) then Ht will be substituted only. Hb cannot be substituted with out breaking the molecule. (BeH2)n – Polymeric solid.

Al2(CH3)6

Ga2(CH3)6

H H H H3C

CH3

C Al

Al

H3C

CH3

C H

H

H

But Al2Cl6  hame covalent bond only no electron deficient bonding. Cl

Cl

Al

Cl Al

Cl

Cl

Cl

(b) Back Bonding: Back bonding generally takes place when out of two bonded atoms one of the atom has vacant orbitals( generally this atom is from second or third period) and the other bonded atom is having some non-bonded electron pair( generally this atom is from the second period) Back bonding increases the bond strength and decreases the bond length * The extent of back bonding the much larger if the orbitals involved in the back bonding are of same size, for example Back bonding in boron trihalides

BORON TRIHALIDES BF3

BCl 3

BBr3







sp

2

sp

2

sp

2

B 3  unstable (stearic h indrance )

Q.

Observed bond length in BF3 molecule in found to less than the expected.

F

B

Ans.

:

:F

There will be p - p back bonding between 2p of B atom & 2p of F atom. Bond length of B

F

- F will decreases

RESONANCE

CHEMICAL BONDING - 38

Q.

Compare B-F bond length in BF3 and [BF4]–.

Ans.

BF3 + KF  K + [BF4]– F [No back bonding]

SP3 hybridised

B F F F

B–F Bond length (order) [BF4]– > BF3 Q. Ans.

Explain all boron trihalides are lewis acids also explain their order. Boron trihalides are electron deficient molecules therefore act as a Lewis acids

Cl But in BCl3 

P -P back bonding 2p of B & 3p of Cl less effective

B Cl

Cl

tendency to accept L.P in BCl3 > BF3. Lewis acid strength BF3 < BCl3 < BBr3. * The extent of back bonding decreases if the atom having vacant orbitals is also having some non-bonded electron pairs on it. So among the atoms of third period the extent of back bonding follows the order Si > P > S > Cl for example, Q.

Trisilylamine is a planar molecular and does not act as a lewis base while trimethyl is a pyramidal and act as lewis base

SiH3

:

N

N | CH3 CH CH3 3

S.N = 3

SiH3

:

SiH3 Planar

N (CH3)3 pyramidal

delocalised .P. S.N. = 3 sp2

Localised .P.

sp3

S.No. = 4

But in a similar compound N(PH2)3 the shape is found to be pyramidal, so N atom must be sp3 hybridised due to much less extent of back bonding into the vacant orbitals of P. Silyl isocyanate (SiH3NCO) is linear but methyl isocyanate (CH3NCO) is bent explain !

Sol.

H H H

C

O

vacant d-orbitals .P. of N can be delocalised (Back bonding) S.N of N = 2 (sp) p – d back bonding.

RESONANCE

:

N

H H C N H No vacant orbitals

:

Si

:

:

Q.

C

O

S.N = 3 (sp2)

CHEMICAL BONDING - 39

(c) Hydrolysis Hydrolysis means reaction with water molecules ultimately leading to breaking of O-H bond into H+ and OH– ions. While the term Hydration means the surrouding of and polar molecule or ions by polar molecules of water. Hydrolysis  complex formation with water molecule or reaction with water molecule. Hydration  cluster formation by water molecule around the ion or molecule of solute. Ionic compounds  will under go hydrolysis st then hydration Covalent compound  mainly hydrolysis will takes place and negligible hydration. Hydrolysis in covalent compounds takes place generally by two mechanisms (a) By H-bond formation: As of Nitrogen Trihalides (b) By Coordinate bond formation : Generally in halides of atoms having vacant d-orbitals or of halides of atoms having vacant orbitals. 2nd period

LiCl

BCl3

Li+

C Cl4

NCl3

2sº 2pº – sp3 hybridisation H

H

OH2

O

O

H

H

Primary layer (hydrolysis)

Li

H2O

H

H2 O

OH 2

O

H H

Be2+

O

H (hydration) many such layer will form.

2sº 2pº = sp3 hybridisation

LECTURE # 10 Hydrolysis via co-ordinate bond formation. Hydrolysis of BCl3( due to presence of vacant orbitals on boron) H

vacant pure Cl p- orbital B Cl Cl

OH

OH B

+ HCl

OH

RESONANCE

Cl

H O H

O

H

B

Cl

H2O

Cl

Cl

OH B

+ HCl

Cl

CHEMICAL BONDING - 40

Hydrolysis of boron trihalides :

Cl

Cl undergo hydrolysis by coordinate

B

bond formation

Cl

Q.

B (OH)3 (Boric acid) + 3HCl

BF3 does not undergo complete hydrolysis OR Hydrolysis of BF3 is only partial ? Explain. BF3 + 3H2O  B(OH)3 + 3HF But the produced HF will react with the reactant BF3 to form BF4– ion BF3 + HF  H+ [BF4]– Hydrogen tetrafluoroborate (III) (stable species) BCl3 + H2O  B (OH)3 + 3HCl BCl3 + HCl  H [BCl4] unstable BCl4 ion is unstable because of steric hindered Ex. of sp2 hybridisation –

CCl4  no vacant orbitals So, will not under go hydrolysis.

:

NCl3  no vacant orbitals. Hydrolysis of NCl3 takes place by H bond formation H

N

O H

Cl Cl Cl

 NH3 + HOCl

By similar mechanism there will be hydrolysis of NBr3 and NI3 but NF3 does not undergo hydrolysis mainly because it is a nonpolar molecule and also the hydrolysis produft FOH is an unstable compound. 

NF3 is almost a non-polar molecular : It has  = 0.23 D much less than  NH3 . In NF3 molecule, the resultant of bond moments is going to oppose the l.p. moment.



NF3 reacts with water very slowly or it undergoes hydrolysis very slowly. Because (a) NF3 is almost non-polar (b) If NF3 is hydrolysised HOF is formed which is unstable. NF3 + 3H2O  NH3 + HOF unstable So above equation is mainly backword shifted.

RESONANCE

CHEMICAL BONDING - 41

3rd period

3

2

sp d

MgCl2

<

NaCl S,P,d

<

SiCl4

AlCl3

PCl5 SCl4 (not found)

Ionic +

–

2+

3+

([Na(H2O)6] Cl )[Mg(H2O)6 ] [Al(H2O)6 ] SiCl4  vacant 3d orbitals

 Si(OH)4 + 4HCl unstable  H2 SiO3 + H2O silicic acid (cannot be isolated,formed only in aq. phase) H2CO3 & H2SiO3 –– both are unstable are found only in aq. solution phase. H2CO3  H2O + CO2 H2SiO3  H2O + SiO2

:

PCl3 ( p cannot form H bond)



Hydrolysis will takes place by coordinate bond formation

P (OH)3 + HCl.

 H3PO3 (Phosphorus acid) PCl5  P(OH)5 + HCl

 H3PO4 (Phosphoric acid) DIPOLE MOMENT

Let B is more –ve than A. Dipole moment of the bond = x unit for DM = esu × cm 1D = 10–10 esu × 10–8 = 10–18 esu × cm 1D = 3.33 ×10–30 coloumb × metre. Resultant DM

P 2  Q 2  2PQ cos 

RESONANCE

CHEMICAL BONDING - 42

SOME IMPORTANT POINTS ABOUT DIPOLE MOMENT :



A polyatomic molecule having polar covalent bonds but zero dipole moment indicates the symmetrical structure of the molecule. e.g. B - F bonds are polar in BF3 but BF3 has  = 0 due to its symmetrical geometry.



If molecule have  = 0. then it should be linear or having symmetrical geometry. e.g. linear – CO2, CS2, BeCl2 (g). symmetrical geometry – BF3 , CH4 , PCl5 , SF6 , IF7 , XeF4



If molecule has  0 then it should be angular or having unsymmetrical geometry. e.g. angular – SnCl2 , PbCl2 , SO2 , Unsymmetrical geometry – NH3, H2O, NF3, SF4, H2S,



% Ionic character =

µExperiment al µTheoretica l

( Observed)  100  100 = (100% Ionic compound)

Following compounds have zero dipole moment BF3 ,CO2 , SO3 , CF4 , PF5 (g) , SF6 , XeF2 , CS2 , CCl4 , PCl5 (g) , XeF4

(i)

Some orders of D.M. HF > HCl > HBr > HI 1.92 D 1.08 0.78 0.38

& HF (aq) weak acid

(ii)

CH3Cl > CH2Cl2 > CHCl3 > CCl4 1.8D 1.6D 1.0 0.0 D

(iii)

Usually net D.M. of cis isomer is more than trans. but there can be exceptions also CH3– CH=CH–Cl

(iv)

For substituted benzene usually the order is o>m>p

CH3

exceptional case 

p > (m/o)

Cl

RESONANCE

CHEMICAL BONDING - 43

(v)

 H  N  F N 3 3

HCN 2.95D Q.

H2O 1.84D

H2S 0.92D

CH3–F

CD3–F

DM = 1.847 D D is more (+ve) than H

D.M. = 1.858D

SO2 1.63D

O3 0.52

CO 0.10D

APPLICATION OF D.M. Q.

Given that the net D.M. of H2O = 1.85 D and B.A. = 104.5ºC (cos 104.5º = –0.25) ignoring the effect of l.p. calculate O–H bone D.M. (1.85)2 = x2 + x2 + 2x2 (–0.25) 3.1025 = 2x2 – 1.5 x2 = 1.5 x2 x2 =

Q.

3.1025 ; x2 = 2.69 ; 1 .5

x=

2.69 = 1.51

You are given a triatomic molecule AB2 B more –ve than A. For which of the following B.A. net = max. (A) 0º (B) 30º (C) 45ºC (D) 90ºC cos  max.  = 30º  # 0 : (diatomic molecule) (ii) Calculation of % ionic character in covalent polar bond  observed  100 % ionic character = u calculated theoritica lly

for calculation of theoritical it is assumed that 1e– is completely transferred from one atom to another. 1

1

 = 4.8 × 10–10 e.s.u. A  B x = given to us as B.L. (cm) Q.

Sol.

For H–Cl gas molecule exp = 1.03D Bond distance = 1.275 Å Calculate the % ionic character. theo = 4.8 ×10–10 × 1.275 × 10–8 = 6.12 × 10–18 % ionic char. =

RESONANCE

1.03  10 18 6.12  10 18

 100 = 17%

CHEMICAL BONDING - 44

Q.

If OH bond distance in water = 0.572 Å Then calculate what fraction of electronic charge is present on the O atom. x = 0.572 10–8 x 1.51 =  × x =

1.51 10 18 0.527  10  8

Fraction of electronic charge = Q.

 4.8  10 10

Net  of SO2 = 1.61 D. B.L. S–O = 0.145 nm and each bond has approx. 25% ionic character. Ignoring the effect of I.P. on S atom. Calculate the B.A. for S–O–bond.

25 =

x  100 4.8  10

10

 0.145  10 7 cm

x = 1.74 D 1.61 =

x 2  x 2  2 x 2 cos  cos  = – 0.572   125 º

The % of ionic character in a covalent bond can also be calculated by Haney & smith eq. % ionic character = 16 + 3.52  = absolute diff. of E.N. of bonded atoms on pauling scale for 50% ionic character  = 2.1 But in many cases even at a diff. of 1.7 50% ionic character is observed.

LECTURE # 11 Introduction to MOT : A molecule is formed from the combination of two or more than two atoms of the same or different elememnts and has characteristic properties of its own. A molecule is formed because it has a lower energy( and hence is stable) as compared to the constituent atoms existing separately. It is convenient to think that the atoms in a molecule are linked by chemical bonds. A chemical bond may be defined as the force that holds the atoms together within a molecule. Two thoeries of chemical bonding, namely valence bond theory and molecular orbital theory have been proposed to describe covalent bond formation and the electronic structure of molecules. Both these theories are quantum mechanical theories of chemical bonding. MOLECULAR ORBITAL THEORY : The molecular orbital theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features are: (i) Just as elctrons of any atom are present in various atomic orbitals, electrons of the molecule are present in various molecular orbitals. We know the names of atomic orbitals like s, p, d and f orbitals but the molecular orbitals will be having different names as well as different electronic distributions. (ii)

Molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper symmetry.

(iii)

While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclie depending upon the number of the atoms in the molecule. Thus an atomic orbital is monocentric while a molecular orbital is polycentric.

(iv)

The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals called bonding molecular orbital and anti-bonding molecular orbital are formed.

(v)

The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.

RESONANCE

CHEMICAL BONDING - 45

(vi)

Just as the electron probability distribution around a nuclues in an atom is given by an atomic orbital , the electron probablity distribution around a group of nuclei in a molecule is given by molecular orbital.

(vii)

The molecular orbitals like the atomic orbitals are filled in accordance with the Aufbau principle obeying the Pauli Exclusion principle and the Hund’s Rule of Maximum Multiplicity. But the filling order of these molecular orbitals is always experimentally decided, there is no rule like (n + l) rule in case of atomic orbitals. ( In Valence Bond Theory we modify the pure orbitals of a single atom first - in hybridisation and then use these modified or hybridised orbitals to form the bonds while in Molecular orbital Theory orbitals of different atom are used to form new kind of orbitals called molecular orbitals.) Formation of Molecular Orbitals: Linear Combination of Atomic Orbitals(LCAO) In principle, Schrodinger equation can be written for any molecule. However, since it can not be solved exactly for any system containing more than one electron, molecular orbitals which are one electron wave functions for the molecules are difficult to obtain directly from the solution of the Schrodinger equation. This difficulty is overcome by resorting to an approximation methid called the Linear Combination of Atomic Orbitals(LCAO) method to form molecular orbitals. Let us consider the application of the LCAO method to the homonuclear diatomic hydrogen molecule. The two hydrogen atoms in the H2 molecule, fot the sake of converience may be labeled as A and B. Each hydrogen atom in ground state has one electron in the 1s orbital. These atomic orbitals may be represented by the wave functions A and B. Mathermatically the formation of molecular orbitals is described by the linear combination (addition or subtraction of the wave functions of the individual atomic orbitals) of A and B as shown below. MO = A ± B Therefore, two molecular orbitals  and * are formed  = A + B * = A –B The molecular orbital  formed by the addition of atomic orbitals is called the bonding molecular orbital and the molecular orbital * formed by the subtraction of atomic orbitals is called antibonding molecular orbital (Fig.)

Atomic orbital

Molecular orbitals

Atomic orbital

A

B







*A – B



Increasing energy

Antibonding orbital higher enrgy than that of atomic orbitals

 =A + B

Bonding orbital lower energy than that of atomic orbital Fig. Formation of bodnding () and antibonding (*) molecular orbitals by the linear combination of atomic orbitals  A and  B centered on two atoms A and B respectively.

Qualitatively, the formation of molecular orbitals can be understood in terms of the constrcutive or destructive interference of the electron waves of the combining atoms. In the formation of bonding melecular orbital, the two electron waves of he bonding atoms reinforce each other (constructive interference) while in th formation of antibonding molecular orbital, these electron waves cancel each other (destructive interference). The result is that in a bonding molecular orbital most of the electron density is located between the nuclei of the bonded atoms and hence the repulsion between the nuclei is very low while in an antibonding molecular orbital, most of the electron denstity is located away from the space between the nuclei, as a matter of fact there is a nodal plane (i.e., plane in which the electron density is zero) between the nuclei. Electrons placed in a bonding molecular orbital tend to hod the nuclei together and stabilize a molecule. A bonding molecular orbital is, therfore, always of lower energy than either of the atomic orbitals that have combined to form it. In contrast, electrons placed in the antibonding molecular orbital destabilize the molecule. The attraction between the electrons and the nuclei is less than the mutual repulsion of electrons in this orbital and this produces a net increase in enrgy. Consequently, the electrons placed in this molecular orbital tend to destabilise the molecule and that is why this orbital is said to be antibonding.

RESONANCE

CHEMICAL BONDING - 46

It needs to be pointed out that the energy of the antibonding orbital is raised above the energy of the atomic orbitals that have combined by an amount more than that by which the energy of the bonding orbital has been lowered. The total energy for two molecular orbitals however remains the same as that of the two orginal atomic orbitals. CONDITIONS FOR THE COMBINATION OF ATOMIC ORBITALS : The linear combination of atomic orbitals to from molecular orbitals takes place only if the following conditions are satisfied : 1.

The combining atomic orbitals must have the same or neraly the same energy. This means that 1s orbital can combine with another 1s orbital but not with 2s orbital because the energy of 2s orbital is appreciably higher than that of 1s orbital. It, therefore, means that only a limited number of combinations of atomic orbitals are possible.

2.

The combining atomic orbitals must have the same symmetry about the molecular axis. By convention z-axis is taken as the molecular axis. It is important to note that atomic orbitals having same or nearly the same energy will not combine if they do not have the same symmetry. For example, 2pz orbital of one atom can combine with 2pz orbital of the other atom but not with the 2px or 2py orbitals because of their different symmetries.

3.

The combining atomic orbitals must overlap to the maximum extent. Greater the extent of overlap, the greater will be the electron-density between the nuclei of a molecular orbital. TYPES OF MOLEUCLAR ORBITALS : Molcular orbitals of diatomic moelcules are designated as (sigma), (delta), etc. In this nomenclature, the sigma () moleuclar orbitals are symmetrical around the bond-axis while pi () molecular orbitals are not symmetrical. For example, the linear combination of 1s orbitals centered on two nuclei produces two molecular orbitals which are symmetrical around the bond-axis. Such molecular orbital’s are of the  type and are designated as 1s and *1s [Fig.(a)]. If internuclear axis is taken to be in the direction, it can be seen that a linear comination of 2pz - orbitals of two atoms also produces two sigma molecular orbitals designated as 2pz and * 2pz. [Fig. (b)] Molecular orbitals obtained from 2px and 2py orbitlas are not symmetrical around the bond axis because of the presence of positive lobes plane. Such molecular orbitals, are labelled as  and * [Fig. (c)]. A bonding MO has large electron density above and below the inter nuclear axis. The * antibonding MO has a node between the nuclei. -type of molecular orbitals are obtained by in involment of d-orbitals into bonding which is not required for us to study. ENERGY LEVEL DIAGRAM FOR MOLECULAR ORBITALS : We have seen that 1s atomic orbitlas on two atoms from two molecular orbitals designated as 1s and * 1s. In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals on two atoms) give rise to the following eight molecular orbitals : Antibonding MO’s *2s *2pz *2px *2py Bonding MO’s 2s 2pz 2px 2py The energy levels of these molecular orbitals have been dtermined expermientally from spectrospic data for homonuclear diatomic molecules of second row elements of the periodic table. The increasing order of energies of various molecular orbitals for O2 and F2 is given below 1s < *1s < 2s < *2s < 2Pz < (2px =2py) < *2px = *2pz < *2pz However, this sequence of energy levels of moleuclar orbitals is not correct for the remaining molecules Li2, Be2, B2, C2, N2. For instance, it has been observed experimentally that for molecules such as B2, C2, N2 etc. the increasing order of energies of various molecular orbitals is 1s < * 1s < 2s < *2s < (2Px = 2py) < 2pz < (*2Px = *2py) < *2pz The important characteristic feature of this order is that the energy of 2pz moleuclar orbital is higher than that of 2px and 2py molecular orbitals. Do tell the students to make up the above energy level diagram second order is for species having number of electrons N < 14 and for species derived from the species having number of e– < 14. While the Ist order is valid for species having number of e– > 15.

RESONANCE

CHEMICAL BONDING - 47

Molecular orbitals

Antibonding sigma molecular orbital

Energy

Atomic orbital

Atomic orbital

+ 1s

+ 1s

+ *1s





*1s

–

1s

Bonding sigma molecular orbital





1s + 1s

 1s

+

+ + 1s

1s

(a)

Molecular orbitals

Atomic orbital

2pz

2Pz

– +

– + –

2pz

2pz

– + – +

*1s Bonding sigma molecular orbital





*2pz



Energy

Atomic orbital

Antibonding sigma molecular orbital



– +

+ + –

2pz

2pz

2pz

– + –

2pz

(b)

Antibonding sigma molecular orbital

Molecular orbitals

Energy

Atomic orbital

Atomic orbital *2px

+ –

–

2pz





2pz

+ –

2px





+ –

(c)

– +

*2px Bonding sigma molecular orbital

2Px

2px

+ –

2px

–

+ –

+ –

2px

2px

Fig. Constours and energies of bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals; (b) 2pz atomic orbitals and (c) 2px atomic orbitals. **

(i) (ii)

Write dawn electronic configurations of H2, He2, Li2, Be2, B2, C2, N2, O2, F2, Na2, H2+, H2–, He2+, He2–, N2+, N2–, O2+, O2–, O22–, CO, NO, CN, CN–, NO+, CO+ Do tell the students to write down all these electronic configurations Ist either in the class or from their homes. ELECTRONIC CONFIGURATION AND MOLECULAR BEHAVIOUR The distribution of electrons among varous molecular orbitals is called the electronic configuration of the moelcule. From the electronic configuraiton of the molecule, it is possible to get important information about the molecule as discussed below. The molecule is stable if Nb is greater than Na, and The molecule is unstable if Nb is less than Na In (i) more bonding orbitals are occupied and so the bonding influence is stroger and a stable molecule results. In (ii) the antibonding influence is strogner and therfore the moleucle is unstable. Na – number of e– in bonding molecular orbitals and Nb – number of e– in antibonding MO’s.

RESONANCE

CHEMICAL BONDING - 48

BOND ORDER Bond order (b.o.) is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals i.e., Bond order (b.o.) = ½ (Nb – Na) The rules discussed above regarding the stability of the moleuclar can be restated in terms of bond order as follows : A positive bond order (i.e., Nb > Na) means a stable molecule while a negative (i.e., Nb < Na) or zero (i.e., Nb = Na) bond order means an unstable moelcule. NATURE OF THE BOND Integral bond order values of 1, 2 or 3 correspond to single, double or tripll bonds respectively as studied in the classical concept. BOND-LENGTH The bond order between two atoms in a molecule may be taken as an approximate measure of the bond length. The bond length decreases as bond order increases. MAGNETIC NATURE If all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic (repelled by magnetic field). However if one or more molecular orbitals are singly occupied it is paramagnetic (attracted by magnetic field), e.g., O2 moleucle. BONDING IN SOME HOWONUCLEAR DIATOMIC MOLECULES In this section we shall discuss bonding in some homonuclear diatomic molecules. 1.

Hydrogen molecule (H2) : It is formed by the combination of two hydrogen atoms. Each hydrogen atoms has one electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule which are present in 1s molecular orbital. So electronic configuration of hydrogen molecule is H2 : (1s)2 The bond order of H2 molecule can be calculated as given below : Nb  Na 2  0  1 2 2 This means that the two hydrogen atoms are bonded together by a single covalent bond. The bond dissociation energy of hydrogen molecule has been found to be 438 kJ mol–1 and bond length equal to 74 pm. Since no unpaired electron is present in hydrogen molecule, therefore, it is diamagnetic.

Bond order =

2.

Helium molecule (He2) : The electronic configuration of helium atoms is 1s2. Each helium atom contains 2 electrons, therefore, in He2 molecule there would be 4 elctrons. These electrons will be accommodated in  1s and *1s molecular orbitals leading to electronic configuration : He2 : (1s)2 – (*1s)2 Bond order of He2 is ½(2 – 2) = 0 He2 molecule is therefore unstable and does not exist. Similarly, it can be shown that Be2 moelcule (1s)2 (*1s)2 (2s)2 (*2s)2 does not exist.

3.

Lithium molecule (Li2) : The electronic configuration of lithium is 1s2, 2s1 . There are six electrons is Li2. The electrons in Li2. The electronic confiugraiton of Li2 molecule, therefore, is Li2 : (1s)2 (*1s)2 (2s)2 The above configuration is also written as KK(2s)2 where KK represents the closed K shell structure (1s)2 (*1s)2. From the electronic configuration of Li2 molecule it is clear that there are four electrons present in bonding molecular orbitals and two electrons present in antibonding moleuclar orbitals. Its bond order, therefore, is ½(4 – 2) = 1. It means that Li2 molecule is stable and since it has no unpaired electrons it should be diamagnetic. Indeed diamagnetic Li2 molecules are known to exist in the vapour phase.

4.

Carbond molecule (C2) : The electronic configuration of carbons is 1s2 2s2 2p2. There are twelve electrons in C2. The electronic configuraiton of C2 molecule, therefore, is C2 : (1s)2 (*1s)2 (2s)2 (*s)2 (2p2x = 2p2y ) or KK (2s)2 (*2s)2 (2p2x = 2p2y ) The bond order of C2 is ½ (8 – 4) = 2 = 2 and C2 should be diamagnetic. Diamagnetic C2 molecules have indeed been detected in vapour phase. It is important to note that double bond in C2 consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals. In most of the other molecules a double bond is made up of a sigma bond and a pi bond.

RESONANCE

CHEMICAL BONDING - 49

5.

Nitrogen molecule (N2) : Electronic configuration : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 or KK (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 1 (10 – 4) = 3. one sigma and two  bonds. 2 N2¯ : Though 15e– but derived from N2, hence EC will be occording to N2 Electronic configuration : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 , *2px)1 or KK (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 , *2px)1 B.O. of N2 =

6.

1 (10 – 5) = 2.5 2 N2+ : B.O. = 2.5, out of N2+ and N2¯, N2¯ is less stable though both have equal bond order but N2– has greater number of antibonding electrons.

B.O. of N2¯ =

With 2s-2p mixing

Without 2s-2p mixing

C2

N2

O2

F2

Ne2





B2

 

 



 



 











































2s







2s







*2s





*2s

2p





2p











2p



Energy

*2p

*2p











2p



*2p





*2p



900

150

159

945

143

131

600

121

110

620

100

498

300

50 290

290

Bond order Magnetic properties

1

2

Paramagnetic 2

Valence electron configuration

Bond length/pm

–1













Bond energy/kJmol

7.

2

(2s) (*2s) (2p)2

Diamagnetic 2

2

(2s) (*2s) (2p)4

3

2

Diamagnetic

Paramagnetic

2

2

(2s) (*2s) (2p)4 (2p)2

2

1

2

(2s) (*2s) (2p)2 (2p)4 2 (*2s)

0

Diamagnetic 2

2

(2s) (*2s) (2p)2 (2p)4 4 (*2s)

– 2

2

(2s) (*2s) (2p)2 ( 2p)4 4 2 (*2s) (*2s)

Fig. MO occupancy and molecular properties for B2 through Ne2.

RESONANCE

CHEMICAL BONDING - 50

8.

Oxygen molecule (O2) : The electronic configuration of oxygen atoms is 1s2 2s2 2p4 . Each oxygen atom has 8 electrons, hence, in O2 moleucle there are 16 electrons. The electronic configuration of O2 molecule, therefore, is O2 : (1s)2 (*1s)2 (2s)2 (*2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y) or 2 2 2   O2 :  KK (22s) (2* 2s) (21p z ) 1   ( 2p x  2p y ),(  * 2p x   * 2p y ) 

From the electronic configuration of O2 moleucle it is clear that ten electrons are present in bonding moleclar orbitals and six electrons are present in antibonding molcular orbitals. Its bond order, therefore, is 1 Nb  Na   1 [10  6]  2 2 2 So in oxgyen molecule, atoms are held by a double bond. Moreover, it may be noted that it contains two unpaired electrons in *2px and *2py molecular orbitals, therefore, O2 molecule should be paramagnetic, a prediction that corresponds to experimental observation. In this way, the theory successfully explains the paramagnetic nature of oxygen. Similarly, the electronic configurations of other homonuclear diatomic molecules of the second row of the periodic table can be written. In Fig. are given the molecular orbital occupancy and molecular properties for B2 through Ne2. The sequence of MOs and their electron population are shown. The bond energy bond length, bond order, magnetic properties and valence electron configuraiton appear below the orbital diagrams. Bond order =

9.

O2+ : B.O. = 2.5 paramagnetic.

10.

O2– : B.O. = 1.5 paramagnetic.

11.

O22– : B.O. = 1 diamagnetic.

HOMO : Highest Occupied Molecular Orbital. LUMO : Lowest Unoccupied Molecular Orbital O22– – No. of unpaired e s 0 B.O. 1.0 B.L.

O2¯ 1 1.5

O2 2 2.0

O2+ 1 2.5

O22+ 0 3.0

B.L.

Ne2

1s2 ,  *1s2 ,  * 2s2 ,  2s2 ,  * 2s2 ,  2 2 s2 (  22py ,  22pz ) (  *2py ,  *2pz )  * 2px 2 (HOMO)

10  10 =0 2 Species H2 H2+ He2+ B.O. =

BE (kJ/mol) 432 255 230

B.L. (pm) 74 106 108

MOT FOR HETERONUCLEAR DIATOMIC MOLECULES : CO is  donor in metal complexes by This is isoelectronic with N2 molecule 

This is isoelectronic with N2 molecule  Its configuration must be simular to N2.

1s2 , 1s2 ,  2s2 , (  2py 2 ,  2pz 2 ) ,

 22px HOMO

B.O. = 3 1 and 2 bond

RESONANCE

CHEMICAL BONDING - 51

The MO configuration of CO has given above cannot be considered to be 100% correct because if CO is converted into CO+ then, its B.O. =

94 = 2.5. 2

 B.L. must 

B.L.CO = 1.128 Å B.L.CO+ = 1.115 Å NO  O2+

(Acc. of spectroscopy)

Total no. of e– = 15

1s2 ,  *1s2 ,  2s2 ,  * 2s2 ,  2px 2 , (  2py 2 ,  2pz 2 ) ,

(  * 2py ' ,  * 2pz  )  * 2px  HOMO

LUMO

10  5 = 2.5 2 1 1.5  bond

Odd e– B.O. =

NO+............................. (  2py 2 ,  2 z 2 )(  * 2py ,  2pz  )

(D)

10  4 =3 2 1 and 2 bond

B.O. =

NO+ > NO 3 2.5 O 2+ > O 2 2.5 2



more stable

Formation of NO+ and O2+ is endothermic (difficult) Diagrams for M.O. 1s + 1s Add

Electron Density or + +

+ 1s 1s Due to presence of electrong density between two nuclei, repulasion between two nucleis rather attraction takes place.   bonding M.O.  type

Nodal plane Add

+ 1s

1s

AMO = a - b *1s

more repulsion (ungerade) GERADE OR UNGERADE ORBITALS If we move equal dist in the opp. direction from the centre of a MO and sign of the wave functions remains the same, the MO will be known as gerade otherwise it will be ungerade.  is gerade and *1s is ungerade. H2  1s2 (g)  *1s0 (u)

RESONANCE

CHEMICAL BONDING - 52

LINEAR COMBINATION OF 2p-ORBITALS (a)

Along the Molecular Axis ()

nodal plane

+ + +

– 2px

–

Addition

–

–

Subn –

2px 2px gerade

(b)

+

–

+

2px ungerade

Of 2py And 2py orbitals

+

+ +

–

+

Addition

–

2py 2py –

ungerade 2py

–

Subn

gerade 2py

Conclusion : s p p

s *p *p

all gerade. MOT

FOR

ungerade

HF

HF

H

*

F

p 2py

2pz (NBD)

2px

2py

2pz

MO

NBD

2s

NBD

1s

2–0 =1 2 Atomic orbitals is always monocentric But molecular orbitals is always polycentric.

B.O. =

RESONANCE

CHEMICAL BONDING - 53

LECTURE # 12 INTERMOLECULAR BONDS : HYDROGEN BOND : Nitrogen, oxygen and fluorine are the higly tronegative elements. When they are died to a hydrogen atom to form covalent bond , the electrons of the covalent bond are ted towards the more electronegative atom. This partially positively charged hydrogen atom forms a bond with the other electronegative atom. This bond is i as hydrogen bond and is weaker than avalent bond. For example, in HF molcule, the hydrogen bond exists between hydrogen atom of one molecule and fluorine atom of another molecule as depicted below : – – – H+ – F – – – – H+ – F– – – – H+ – F– Here , hydrogen bond acts as a bridge between toms which holds one atom by covalent bond and the other by hydrogen bond. Hydrogen bond is represented by a dotted line (– – –) while a solid line represents the covalent bond. Thus, hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. (I) Cause of Formation of Hydrogen Bond When hydrogen is bonded to strongly electronegative element 'X'. the electron pair shared between the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes highly electropositive with respect to the other atom 'X'. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (+) while X' attain fractional negative charge (–). This results in the formation of a polar molecule having electrostatic force of attraction which can be representes as : H+ – X – – – – H+ – X– – – – H+ – F– The magnitude of H – bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong Influence on the structure and properties of the compounds. (ii)

Infact energy of H-bond is usually low but there are examples in which H-bond energy is found to be nearly 220 kJ/mol (comparable to covalent bond) Therefore H-bond can be classified into 3 categories. (a) Weak BE < 25 kJ/mol. (b) Medium 25 – 35 (c) Strong > 35 kJ/mol ex : K+HF2– or [HF2¯] ion energy = 220 kJ/mol.

(iii)

Discovered by Huggius, Latimer and Rodbush.

(iv)

Infact H-bond is due to electrostatics attraction between the pole of H and .p. of e– s on another –ve atom

(v)

Ex.

+ A H ----- :B (-ve) (-ve) H-bond H-bonds are found to be much stronger with F-atoms but in case of other atoms the relative strength of Hbond will depend upon + charge of H( possiblity when atom A is more -ve) and availability of l.p. on B-atom. (B must be relatively less –ve)

Max H-bond strength (A)

+ O

H - - - - - - :N

(B)

O

H - - - - - - :O

(C)

N

H - - - - - - :N

(D)

N

H - - - - - - :O

Order of H-bond strength A > B > C > D (vi)

Ex.

Ans. (A)

A

H- - - - - :B (generally distance x < y) x y But in case of very strong H-bonding x = y,

K+[HF2]–

RESONANCE

CHEMICAL BONDING - 54

TYPES OF H-BONDING. (A) Intermolecular Ex. : H2O, HF, R – OH, CH3COOH / HCOOH, NH3(R), R – NH2, R NH, HCN, H BO (s) etc. 3 3

R

1.74Å 1.01Å

O- - -H | H

H

O | H

F 120° to 140°

H F

F

H

H

H

O

H

F

H3C

O

C

C O

H

H

O

HCO3– in KHCO3 In dimeric form but HCO3– in NaHCO3 is in linear polymeric form.

O

H

O C=O

O=C O¯

H

O H | O

H O | B

H3BO3(s)

H O

(B)

O

B | O

H

H

H

O

Intramolecular : O

O (i)

O N

O-nitrophenol

O

(ii)

Chloral-hydrate  CCl3CH(OH)2 CCl3CHO (Chloral)

H

Cl

O C

Cl

C

H

O

Cl H

2 – OH groups are stabilised on the same C-atom.

RESONANCE

CHEMICAL BONDING - 55

(iii)

Maleic Acid :

H H

(iv)

C || C

COOH

H

COOH

H

C || C

O || C

O + H+

H C || O

 K2 0ºC

In this temperature region some of the ice melts and hence some H2O molecule go into the cages of remaining ice structure. (ix)

(x)

 V  d becoming max at 4ºC but beyond this temp thermal effects become dominating V   d  Water can extinguish the burning alcohol but not the burning petrol or a Hydrocarbon (in general). H2O(l) + R–OH(l)  solution (miscible) H2O(l) + Petrol or hydrocarbon (l)  Not miscible (No H bonding) density of open chain HC < dH2O lighter than water.  continue burning even in presence of H2O Which is a stronger base why? (a)

Trimethyl ammonium hydroxide

(b)

Tetramethyl ammonium hydroxide NCH3 4  OH 

b is more basic (CH3)3N: - - - - H – O Due to nitramolecular H bonding, release of OH– ion become difficult  it is a weaker base. (xi)

C2H2 is most soluble in (A) H2O (B) C2H5OH

RESONANCE

(C*) CH3COCH3

(D) none or equally

CHEMICAL BONDING - 58

C2H2 is highly soluble in acetone due to H bonding

None of the above 2 has H-bonding individually. But C2H2 is not soluble in water because water molecules already so much associated through H-bond that it is almost impossible for C2H2 molecules to break that association.  it is not soluble in H2O (l) (xii)

B.pt of o-nitrophenol is less than m-nitrophenol or p-nitrophenol and this compound is much less soluble in water than the remaining 2. Due to intromalecular H-bonding in O – Due to intermolecular H-bonding in m-p-

INTERMOLECULAR FORCES (VAN DER WAAL’S FORCES) : Intermolecular attractions hold two or more molecules together. These are weakest chemical forces and can be of following types. (a) Ion-dipole attraction : (b) Dipole-dipole attraction : (c) Ion-induced dipole attraction : (d) Dipole-induced dipole attraction : (e) Induced dipole-Induced dipole : (Dispersion force or London forces) strength of van der Waal’s forces a > b > c > d > e Note : According to NCERT ion-dipole attraction is not considered as a van der Waal’s forces (a) Ion-dipole attraction : Exists between an ion and a polar molecule. Its strength depends on (i) size of ion (ii) charge on the ion (iii) dipole moment of the polar molecule.

Na

H

–

+

O H

+

This force is responsible for hydration.

+

(b) Dipole-dipole attraction : Electrostatic attractions between the oppositively charged ends of permanent dipoles. Exists between polar molecules and due to this force gas can be liquified. 

+



–



+

H Cl H Cl– (c) Ion-induced dipole attraction : Exists between ion and non-polar molecules Na +



–



Cl

+

Cl

(d) Dipole-induced dipole attraction : Exists between polar and non-polar molecules. +



H

–





Cl

+



Cl

–

Cl

(e) Instantaneous dipole- Instantaneous induced dipole attraction : Exists among the non-polar molecules like H2 , O2 , Cl2 etc. in solid or liquid states

e.g. ,



Cl

–



Cl

RESONANCE

+



Cl

–



+

Cl

CHEMICAL BONDING - 59

 

Strength of vander waal force  molecular mass.

Q.1

Give the order of boiling point of following Cl2 , HCl

Ans.

Cl2 – Cl2 < dispersion force

Q.2

Arrange the inert gases, according to their increasing order of boiling points

Ans.

He < Ne < Ar < Kr < Xe Boiling point Because strength of van der Waal’s force increases down the group with increase in molecular mass.

van der Waal’s force  Boiling point.

HCl – HCl dipole-dipole attraction

(B.P.)

METALLIC BOND : Nature– electrical attraction between delocalised electrons (i.e. mobile electrons) and the positive part of the atom (i.e. kernel). Strength of metallic bond depends upon following factors : (i)

No of metallic bonds per atom : Larger the number of metallic bonds (number of valence electrons) per atom, stronger is the metallic bonding.



Alkali metals-only one valence electron and therefore, one metallic bond per atom and thus gives weakest metallic bonding. Hence their melting points are low.



Alkaline earth metals-two valence electrons and form two metallic bonds per atom and, therefore, gives a comparatively bit stronger metallicbonding. Transition metals may use inner -d-electrons along with the outers electrons for bonding as (n–1) d and ns have nearly same energy. So in them number of metallic bonds per atoms is quite large (more than two always). Hence their melting points are higher.

(ii)

Type of hybrid orbitals participating in metallic bonding. More directional and diffused hybrid orbitals form more stronger metallic bonds. ds hybrid orbitals (in transition metals) are more directional and diffused than sp hybrid orbitals (in alkali & alkaline earth metals) Therefore, much stronger metallic bonding takes place in transition metals.

(iii)

Radius of metallic atom : Smaller the radius of atom, shorter is the bond length, hence greater is the bond strength.

RESONANCE

CHEMICAL BONDING - 60

REVISION LECTURE # 1 Naming of Inorganic compounds

 

Name of inorganic compounds mainly consist of two parts st part - name of cationic part or name of less electronegative atom or part Name of cationic part is generally now modified and generally this name ends in - ium

Ex.

NaCl sodium chloride (NH4)2SO4 ammonium sulphate 2nd part - name of anionic part generally follow name of cationic part. The name of anionic part is modified and depending of nature of anion if generally ends with – ide – ate – ite Elemental or non – oxoanion Name of these anions ends with – ide

Ex.

Halogen family

F– – flouride Cl– – Chloride Br– – Bromide – – iodide

Oxygen family O2– – oxide

O 22 O2

– peroxide

– S – Se–2 – Te–2 – Nitrogen family N3– – 2–

N3

superoxide sulphide selenide telluride nitride

– azide

3–

P – phosphide Carbon family C4– – carbide Combined elemental anions OH – – hydroxide SH – – bisulphide CN– – cyanide NC – – isocyanide If both element are non-metallic then more electronegative element is anionic part As2O3 – arsenic (III) oxide OF2 – oxygen di flouride (1) (4)

Ca3P2 ClF3

Write down the formula of (1) Sulphur hexafluoride (4) dioxygen di fluoride

1.

2.

(2) (5)

Ba(CN)2 SF4

(3)

Na2S

(2) (5)

Lithium nitride barium azide

(3)

stroncium chloride

Naming of oxoanions The anions of oxyacids of different element are called oxoanions. The names of oxoanions are derived from the names of their parent oxyacids. If name of acid ends with ic then name of anion will end with – ate If name of acid ends with –us then name of anion will end with – ite Name of oxyacid should end with – ic or – us is decided by oxidation no of element in the oxyacid. The following naming procedure is applied for naming oxyacids If an element forms only one oxyacid then name of that generally ends with ic H3BO3 – boric acid (only oxyacid of boron + 3 ) H2CO3 – carbonic acid ( + 4 = O.No.) H4SiO4 – Silica acid ( + 4 = O.No) If any element forms oxyacids in two oxidation. States. Then acid of higher O.No. ends with – ic & acid with lower O.No. ends with – us like sulphur H2SO4 – Sulphuric (+6  O.No.) H2SO3 – sulphurus (+4  O.N0.)

RESONANCE

CHEMICAL BONDING - 62

3.

If any element forms oxyacids in 3 O.No. Then for highest – ic middle one – us Nitrogen + 5 HNO3 nitric acid +3 HNO2 nitric acid +1 H2N2O2 hypo nitrius acid Phosphorous + 5 H3PO4 phosphoric acid + 3 H3PO3 phosphorus acid + 1 H2PO2 hypo phosphorus acid

4.

If element forms oxyacid Then highest – per next lower next lower lowest – hypo Chlorine +7 HClO4 +5 HClO3 +3 HClO2 +1 HClO

in four. O.No. – ic – ic – us – us. perchloric acid chloric acid chlorus acid hypo chlorus acid

per – ic – ic – only one oxyacid } two oxyacids ] three oxyacids } four oxiacids – us hypo – us There can also be some derived oxyacids such as – pyro acids – meta acids pyro name is attached with acid if it is derived by removing one water molecule from two acid molecules.  H2 O Two acid molecules   pyro acid  Cl2O7 not oxiacid it is an oxide 2HClO4  H O 2

so there will not be any pyro or metra acid of oxiacids of chlorine. Sulphur  H S O pyro sulphuric acid (Oleum) 2H2SO4  H 2 O 2 2 7   2H2SO3  H S O pyro sulphurous acid H 2 O 2 2 5 Nitrogen No pyro acid (as an oxide will be obtained) phosphorous  H P O pyro phospheric acid 2H3PO4  H 2 O 4 2 7   2H3PO3  H P O pyro phosphoros acid H 2 O 4 2 5 Carbon family – pyro – not found  H6SiO7 Silicon 2H4SiO4  pyro silicic acid H 2 O   Boron 2H3BO3  H B O pyroboric acid 4 2 5 H O 2

meta acid: If one water molecule is derived from one acid molecule then meta acid is obtained.  meta acid One acid molecule  H O 2

Chlorine – no meta acid Sulphur – no meta acid Nitrogen – no meta acid  HPO3 phosporus H3PO4  H 2 O    H2SiO3 Silicon – H4SiO4 H O

meta boric acid

Carbon – Boron –

meta boric acid

2

no meta acid  HBO2 H3BO3  H O

meta phosphoric acid

2

Naming of oxoanions – ic acid  – ate ;

– us acid  – ite

Chlorine HClO4  ClO4– perchlorate HClO2  ClO2– chlorite

; ;

RESONANCE

HClO3 HClO

 ClO3– chlorate  ClO– hypochlorite

CHEMICAL BONDING - 63

sulphur H2SO4   H2SO3   H2S2O7 

SO42– sulphate HSO4– bisulphate (hydrogen sulphate) SO32– – sulphite HSO3– – bisulphite (hydrogen sulphite) S2O72– pyrosulphate

Nitrogen HNO3  NO3– – nitrate HNO2  NO2– – nitrite H2N2O2  N2O22– – hyponitrite phosphorous H3PO4  N2O23– – phosphate H3PO3  PO43– – phosphite H3PO3  PO23– – hypo phosphite H4P2O7  P2O74– – pyrophosphate HPO3  PO3 – meta phosprous Silicon H4SiO4 H6Si2O7 H2SiO3

Q.

:  SiO44– – silicate  Si2O76– – pyro silicate  SiO32– meta silicate

Carbon H2CO3  CO32– – Carbonate  HCO3– – bicabonate { hydrogen carbonate) Boron H3BO3  BO33– borate H4B2O5  B2O54– pyroborate HBO2  BO2– meta borate Write the formula of (1) barium perchlorate (2) sodium hypochlorite (4) magnesium pyro phosphate

(3) calcium phosphate (5) Copper () metaborate

Q.

Write the name of (1) Co(BO2)2 (2) Sc2Si2O7 (3) Na2S2O7 (4) Ba(NO3)2 (5) Na2SiO3 There are some more anions which are very common like + 6 CrO42– – Chromate ( name is derived from SO42– sulphate as all features are same) + 6 FeO42– – ferrate + 6 MoO42– – molybolate + 6 WO42– – tungstate + 6 MnO42– – manganate corresponding acids can be H2CrO4 – chromic acid H2MnO4 – manganic acid



higher oxidation state of manganese  MnO 4 So called permanganate HMnO4– permanganic acid



Polysulphides

7

S2x ( x = 2, 3, 4, 5..........) S22 –S – S– disulphide

structures S

S23

trisulphide

–

S

S

S

S

–

S24

–

S

tetra sulphide S

–

Sodium disulphide  NO2S2 Sulphate & thiosulphate (hypo) when ever oxygen of normal compound is replaced with sulphur then thio word is used before name of normal compound alcohol – OH Thioalcohol – SH ether – O – Thioether– S – sulphate SO42– thriosulphate (S2O32–)

RESONANCE

CHEMICAL BONDING - 64

O || S || O

S || S || O

– O O There oxygens can not be replaced

O

–

*

–

O

–

cyanate ion & Thiocyanate ion cyanic acid (HOCN)

N  C  O

cyanate ion 



NCO

N  C  S

Thio cyanate ion 



Resonating structure Resonating structure

NCS

•

Chromate & dichromate dichronate ion is stable in acidie medium while chromate is stable in basic medium Cr2O72– + 2OH– (not a redox char) H2O + 2CrO42–

*

Cations – higher oxidation state of Cations ends with ic & leaver by – us Fe3+ – ferric Cu2+ – cupric 2+ Fe ferrous Cu2 2+ – cuprous Hg2+ – mercuric Hg22+ – mercurous

*

xenon perxenic acid H4xeO6 XeO64– – perxenate ion O || Xe || O

O

–

–

O

–

O

O–

Xenic acid H2XeO4 – Xenic acid XeO42– – Xenate ion (c) Hydrolysis : Hydrolysis means reaction with water molecules ultimately leading to breaking of O-H bond into H+ and OH– ions. While the term Hydration means the surrouding of and polar molecule or ions by polar molecules of water. Hydrolysis  complex formation with water molecule or reaction with water molecule. Hydration  cluster formation by water molecule around the ion or molecule of solute. Ionic compounds  will under go hydrolysis st then hydration Covalent compound  mainly hydrolysis will takes place and negligible hydration. Hydrolysis in covalent compounds takes place generally by two mechanisms (a) By H-bond formation: As of Nitrogen Trihalides (b) By Coordinate bond formation : Generally in halides of atoms having vacant d-orbitals or of halides of atoms having vacant orbitals. 2nd period LiCl

C Cl4

NCl3

2sº 2pº – sp3 hybridisation H

H

Li+

BCl3

OH2

O

O

H

H

Primary layer (hydrolysis)

Li

H2O

H

O

H2 O

OH 2

H

H

Be2+

O

H (hydration) many such layer will form.

2sº 2pº = sp3 hybridisation

RESONANCE

CHEMICAL BONDING - 65