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MEDICAL ENTRANCE

CHEMICAL BONDING

I N D E X

Topic Page No.

INORGANIC CHEMISTRY

CHEMICAL BONDING

1. Introduction 02

2. Electron-Dot (Lewis) Structure 03

3. Exceptions of Octet Rule 04

4. Glesspie and Nyhom theory or VSEPR theory 10

5. Shapes of molecules based on VSEPR theory 11

6. Covalent Bond 12

7. Variable valency in covalent bonds 13

8. Valence Bond Theory 14

9. Hybridisation 16

10. Difference between hybridisation & overlapping 17

11. Determination of hybridisation state 18

12 Bond Parameters 19

13. Molecular Orbital Theory 23

14. Energy Level Diagram of molecular orbital 24

15. Bonding inMolecules 27

MEDICAL ENTRANCE

CHEMICAL BONDING

Topic Page No.

16. Dipolemoment 28

17. Application of Dipole moment 30

18. Vander waal’s Forces 31

19. Hydrogen Bond 32

20 Effect of IntramolecularH–Bonding 36

16. Summary 37

17. Exercise - 1 39

Exercise - 2 50

Exercise - 3 56

Exercise - 4 58

18. Answer Key 60

19. Hints/Solution 66

BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-5

ACC_CHEMISTRY_Chemical Bonding 1

Syllabus :

• Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character ofcovalent bond, valence bond theory, resonance, geometry of molecules, VSEPR theory, concept ofhybridization involving s, p and d orbitals and shapes of some simple molecules, molecular orbitaltheory of homonuclear diatomic molecules (qualitative idea only). Hydrogen bond.

Contents :

• Introduction • Chemica bond

• Cause of chemical combimation • Octet rule (lewis-kossel rule)

• Electron-dot(lewis) structure • Exceptions of octet rule

• Ionic bond • Covalent bond

• Bond pair (bp) and lone pair (lp)

• Coordinate bond or dative bond or coordinate covalent bond

• Ionic character in covalent compounds

• Covalent character in ionic compounds (Fajan’s rule)

• Glesspie and nyhom theory or vsepr theory • Resonance

• Shapes of molecules based on vsper theory • Orbital concept of covalent bond

• Variable valency in covalent bonds • Applications of variable valency

• Valence bond theory • Hybridisation

• Determination of hybridisation state • Bond parameters

• Factors affecting bond length • Molecula orbital theory (mot)

• Energy level diagram of molecular orbital • Bondinginmolecules

• Dipolemoment • Application of dipole moment

• Vander waal’s forces • Hydrogen bond

• Effect of hydrogen bond on physical properties • Effect of intramolecular h-bonding

CHEMICAL BONDING



CHEMICAL BONDING

INTRODUCTION

(i) Most of the elements exist as molecules which are cluster of atoms. How do atoms combine to

form molecules and why do atoms form bonds. Such doubts will be discussed in this chapter.

(ii) A molecule will only be formed if it is more stable and has a lower energy, than the individual

atoms.

CHEMICAL BOND

(i) A force that acts between two or more atoms to hold them together as a stable molecule.

(ii) It is union of two or more atoms involving redistribution of electron among them.

(iii) This process accompanied by decrease in energy.

(iv) Decrease in energy Strength of the bond.

(v) Therefore molecules are more stable than atoms.

CAUSE OF CHEMICAL COMBINATION

1. Tendency to acquire noble gas configuration : (classical concept)

(i) Atom combines to acquire noble gas configuration.

(ii) Only outermost electron i.e. ns, np and (n-1)d electrons participate in bond formation.

(iii) Inert gas elements do not participate, as they have stable electronic configuration and hence

minimum energy. (Stable electronic configuration : 1s2 or ns2np6)

CHEMICAL BONDS

(A)Ionicbond

(Energy released >42 KJ/mole)

(B)Covalent

bond

(C)Co-ordinate

bond

(D)Metalicbond

STRONG BOND(Inter atomic)

(E)Hydrogen

bond

(Energy 2 - 42 KJ/mole)

(F)Vander waal's

bond

WEAK BOND(Inter Molecular)

(8 - 42 KJ/mole) (2 - 8 KJ/mole)

2. Tendency to acquire minimum energy :(modern concept)

(i) When two atoms approaches to each other. Nucleus of one atom attracts the electron of another

atom.

(ii) Two nuclei and electron of both the atoms repells each other.

(iii) If net result is attraction, the total energy of the system (molecule) decreases and a chemical

bond forms.

(iv) So, Attraction 1/energy Stability.



1. OCTET RULE (LEWIS-KOSSEL RULE)

Noble gases have 8 electron in their outermost shell (complete octet) and outermost configuration is

ns2p6. Every atom has a tendency to complete its octet by losing or gaining or by sharing electron.

Three different types of bonds—ionic, covalent, and metallic—are formed depending on theelectropositive or electronegative character of the atoms (of an element) involved in a bond

formation. Atoms of various elements may be electropositive or electronegative depending on

the type of element. Elements may be classified as follows:

(i) Electropositive elements. These elements readily lose one or more electrons.

(ii) Electronegative elements. These elements accept electrons and achieve a stable electronic configuration.

(iii) Elements that have a little tendency to lose or gain electrons.

ELECTRON-DOT(LEWIS)STRUCTURE

For single central atom Lewis structure drawing

Rules for Lewis Structure Drawing :

(i) Calculate n1 = no. of valence shell electrons of all atoms

+ no. of negative charge (if any)

– no. of positive charge (if any)

(ii) Calculate n2 = (no. of H-atom × 2) + (no. of atoms other than H-atoms × 8)

(iii) Calculate n3 = n2–n1 no. of shared electrons

i.e. No. of bonds =2

n3......(a)

(iv) Calculate n4 = n1 – n3 no. of unshared electrons

i.e. no. of lone pairs =2

n4......(b)

Using the informations (a) & (b) the structure is to be assigned as follows:

(i) Find out the central atom first (i.e. either least in number or more electro positive)

(ii) Allocate the surrounding atoms around the central atom with the help of bonds available in (a).

(iii) To fulfill the octet of each atom. utilise the lone pairs available in (b).

(iv) Finally calculate the formal charge for each atom and assign on the atoms according the formula

given.

F.C. (of an atom) = Valence shell electron of that atom

– no. of bonds associated with it

– no. of unshared electrons on it.

In order to write the Lewis structure of a molecule, the points given below need to be kept in mind.



(i) Add the valence electrons of all atoms. In case of an ion, add electrons equal to negative charge (for

negative ion) or subtract the number of electrons equal to the positive charge (for a positive ion).

(ii) Write the formula of the compound, and join atoms by single bonds.

(iii) Complete the octet of all the atoms (or two electrons for hydrogen) which are bonded to the central atom.

(iv) Fix up the remaining electrons, if any, on the central atom.

(v) In case the central atom does not have the requisite number of electrons to complete its octet, attach it by

multiple bonds.

____________________________________________________________________________

S.NO. CO2 NH3 PCl4+ NO2

–

____________________________________________________________________________

1. n1 16 8 32 18

2. n2 24 14 40 24

3. n2 – n1= n3 8 6 8 6

4. n3/2 4 3 4 3

5. n3 – n1= n4 8 2 24 12

6. n4/2 4 1 12 6

O = C = O

::

::

NH H H

:

Cl

P

Cl

::

:

::

:

Cl

::

:Cl

:

::+

O

N

:

:

:: O: :

(1) (2)

Formal change C = zero N = zero P = +1 N = zero

O = zero H = zero Cl = zero O(1) = zero

O(2) = –1

____________________________________________________________________________

1 EXCEPTIONS OF OCTET RULE

(a) Incomplete octet molecules : or (electron defficient molecules) or Hypovalent molecules

Compound in which octet is not complete in outer most orbit of central atom.

Examples - Halides of IIIA groups, BF3, AlCl3, BCl3, hydride of III A/13th group etc.

B× ×

×

Cl

Cl

Cl

selectron6onlyhasBoron

BClIn 3



Other examples - BeCl2 (4 electron), Ga(CH3)3 (6 electron)

(b) Expansion of octet or (electron efficient molecules) or Hypervalent molecules

Compound in which central atom has more than 8 electron in outermost orbits.

Ex. PCl5, SF6, IF7, the central atom P, S and I contain 10, 12,

and 14 electrons respectively.

P Cl××

×

×Cl ×Cl

ClCl

Electron dot formula of PCl5

Cl

Cl

(c) Pseudo inert gas configuration : -

(I) Cations of transition metals, which contains 18 electrons in outermost orbit

Examples : Ga+3, Cu+, Ag+, Zn+2, Cd+2, Sn+4, Pb+4 etc.

Electronic configuration of Ga - 1s2, 2s22p6, 3s23p63d10, 4s24p1

Electronic configuration of Ga+3 - 1s2, 2s2 2p6, 3s23p63d10

18 electron

(d) Odd electron molecules : -

Central atom have an unpaired electron or odd number (7 electron, 11 electron, etc) of electrons

in their outer most shell.

Examples : NO, NO2 ClO2, ClO3 etc.

e.g., NO molecule ×ON

×××

×

7 electron 8 electron

2. IONIC BOND

(i) Atoms get established by ion formation in formation of an ionic bond. One atom forms a cation by losing

electrons and other forms anion by accepting electrons.

(ii) The octets of the atoms are completed by transfer of electrons.

(iii) Cations and anions are bonded by electrostatic force of attraction in ionic bond.

(iv) Ionic bond is formed between metals (cations) and nonmetals (anions).

(v) During the formation of an ionic bond, cation can attain one of the following configurations.

(a) Inert Configuration : IAand IIAgroup metals (octet configuration) Na+1, Mg+2, K+1, Rb+1 attain

ns2np6 inert configuration.

(b) Pseudo Inert Configuration : Metals of I B and II B groups – Cu+, Zn+2,Ag+1,Au+1, Cd+2, Hg+2.

(c) Transition metals : e.g. Ti+3 (9), Ti+2 (10).

(vi) Energies involved in ionic bond formation :

(b) Ionisation Energy – The energy required to remove an electron from the outermost shell of metal in

gaseous state.



(d) ElectronAffinity / Electron Gain Enthalpy (Heg

) – The energy required during formation of an

anion by addition of an electron to a nonmetal in gaseous state.

(e) Lattice Energy –

(i) The energy released during the formation of ionic bond.

(ii) Cations and anions form crystal lattice of ionic crystal in space by electrostatic force of attraction.

(iii) Lattice Energy –rr1

charge on Cation /Anion

X(g)X+(g) + 1e–(Ionisation Energy)

Y(g) + 1e–Y–(g) (ElectronAffinity)

X+(g) + Y–(g)XY(s) (Lattice Energy)

(f) Ionic compound is formed when the energy required ( Ionisation energy) is less than energy released

(Electron affinity + Lattice energy)

2.2 Characteristic of Ionic Compounds

(i) Solid and Crystalline Structure :

Ionic compounds do not show molecular structure.

Ionic compounds have definite crystal structures.

CsCl has body centered cubic (BCC) structure.

(ii) Melting and Boiling Point :

Ionic compounds have high meltingand boiling points.

(iii) Hardness :

Ionic compounds are solid with brittle nature.

(iv) Conductivity :

Ions are not free in solid state so ionic compounds are bad conductor of electric current in solid state but

in solution and fused state electric current passes through and the ionic compound becomes good

conductor of electricity

(v) Solubility of Ionic Compounds :

Ionic compounds are soluble in solvents which have high dielectric constant like water.

(vi) Solvation of Hydration Energy :

The energy released by the attraction of ions and solvent molecules is called solvation energy and when

solvent is water this is called hydration energy.

Solubility in waterHydration energy > Lattice energy



3. COVALENT BOND

A covalent bond is formed by mutual sharing of electrons between two nonmetal atoms. e.g. in H–H,

O=O and NN.

A B

Sharedelectrons

Covalentbond

or A —— B

Formation of a covalnet bond

A B

(a) Nonpolar Covalent Bonds :

Covalent bond between two identical atoms (e.g. H–H, O=O) electronegativity of both atom is same.

(b) Polar Covalent Bond :

When covalent bond is formed between two different atoms with different electronegativity

e.g. HCl, HF

H – Cl –

5. BOND PAIR (bp) AND LONE PAIR (lp)

(i) In covalent compounds, the number of bond pairs is equal to the number of shared pairs of electron.

Number of bond pairs = Number of bonds

(ii) The electron pair, which does not participate in bond formation, is called lone pair.

(iii) N N (N2) 2 lone pairs and 3 bond pairs. Ratio of lp : bp in N2 is 2 : 3

(iv) Cl – Cl: :. .

. .

. .

. . (Cl2) 6 lone pairs and 1 bond pairs.

B

F: : : F: : :

F:

:

:

Central atom = B , Surrounding atom = F

lp at central atom is zero but present on surrounding atom.



6. COORDINATE BOND OR DATIVE BOND OR COORDINATE COVALENT BOND

(i) Coordinate bond is formed by unequal sharing of electron pair between two atoms, one of which, called

the donor atom, provides the electron pair and the other atom, called the acceptor atom, receives that

electron pair.

(ii) The donor should have complete octet with a lone pair of electrons and acceptor atom should have

incomplete octet and a vacant orbital to accept the electron pair.

(a)

A B+Donor

atom witha lone pair

Acceptor atomshor of twoelectrons

shortA B

CoordinateCovalent Bond

Coordinatecompound

or A B

It is also known as Dative Bond, according to Scientist, Menzie.

(iii) Lone pair donors are called Lewis base.

Example : NH3, H

2O etc.

Lone pair acceptor are called Lewis acid.

Example :All cation , Hypovalent molecule etc.

(iv) This bond is different from a covalent bond, in the latter one electron provided by each of the bonding

atoms.

(v) Coordinate bond is also different from ionic bond because electrons are not completely transferred and

it is known as complex covalent bond

N

O

O O—

N+

O O—

O—

(vi) In this bond, electrons are donated only in the form of pairs.

(vii) Electron pair is provided by only one atom and shared equally by both the atoms. So according to

Sugden it is a polar bond.

(viii) Once a coordinate bond is formed, there remains no difference between the coordinate and covalent

bonds. So the property of coordinate compounds are similar to those of covalent compounds.

6.2 Compounds Having Ionic, Covalent and Coordinate Bonds

Some compounds have all the three types of bonds, i.e. ionic, covalent and coordinate bond, e.g.

NH4Cl, CuSO4, K4[Fe(CN)6], KNC, Na3PO4, KNO3, etc.

(i) NH4Cl – Ionic bond between NH

4+ and Cl– and covalent bonds and a coordinate bond are present in



7. IONIC CHARACTER IN COVALENT COMPOUNDS

(i) Paulinguseddipolemoment fordepictingpercentageofpolarityand ioniccharacterof thebond.According

to Pauling, a bond can never be 100% ionic.

(ii) When electronegativity difference between two atoms is 2.1, there is 50% ionic character in the bond.

(iii) When electronegativity difference is zero (identical atoms), the bond will be 100% covalent.

According to Haney and Smith, the percentage of ionic character of a polar covalent bond can be

calculated with help of the following expression.

% Ionic character = (0.16 + 0.0352) × 100

Where= Electronegativity difference between bonded atoms.

8. COVALENT CHARACTER IN IONIC COMPOUNDS (FAJAN’S RULE)

Deformation of anion due to cation is known as polarisation, due to polarisation covalent character

increases and ionic character decreases.

+ +

(+) (–)

+ +

(+) (–)

Polarisation

(i) Size of Cation : Size of cation must be small

Covalent character1

Size of cation

ionic characterSize of cation

Covalent character : LiCl > NaCl > KCl > RbCl > CsCl

(ii) Size ofAnion : Size of anion must be large

Covalent characterSize of anion

Ionic character anionofSize1

Covalent character : LiF < LiCl < LiBr < LiI

(iii) Charge on Cation andAnion :

Covalent characterCharge on cation and anion

Covalent character : LiF < BeF2

, Na2O > NaF, NaF < MgO

(iv) Inert and Pseudo Inert Structures :

Configuration of cations must be equal to pseudo inert type [ns0 (n – 1) d10 ] e.g. Cu+,Ag+,Au+, Zn2+,

Cd+2, Hg+2, Pb+4 , Covalent character of pseudo inert > inert configuration. eg. NaCl < NaF


ACC_CHEMISTRY_Chemical Bonding 1 0

GLESSPIE AND NYHOM THEORY OR VSEPR THEORY :

(Valence shell electron pair repulsion theory)

(i) If the central atom possess only bonded pairs of electrons along with identical atoms then shape

of the compound is symmetrical and according to Sidgwick & Powe l.

eg. BF3 – 120° trigonal planar

CH4 – 109°28 tetrahedral

CO2 – 180° linear

(ii) If the central atom possess bonded pair of electrons as well as lone pair of electron, then shape

of the molecule will be unsymmetrical ie. the original bond angle will disturbed due to repulsion

between lone pair of electrons.

similarly on having different type of side atoms, molecule becomes unsymmetrical due to unequal

force of repulsion between electron.

Order of repulsion is .p, – .p > .p – b.p. > b.p. – b.p.

Bond angle electronofpairloneofNo.1

RESONANCE

(i) The concept of resonance was introduced by heisen berg (1920) , and later developed by pauling

and in gold, to explain the properties of certain molecules,

(ii) It has been found that the observed properties of certain compounds cannot be satisfactorily

explained by writing a single lewis structure. The molecule is then supposed to have many structures,

each of which can explain most of the properties of the molecule but none can explain all the

properties of the molecules. The actual structure is in between of all these contributing structures

and is called resonance hybrid and the different individual structures are called resonating structures

or canonical forms. This phenomenon is called resonance.

(iii) Let us discuss resonance in ozone, according to its resonating structure it should have one single

bond (OO = 1.48Å) but experiments show that both the bonds are same which can be proved

by its resonance hybrid as shown below.O

O OO

O OO

O O

Resonance hybrid

To calculate bond order in the polyatomic molecule or ion use following formula :

Bond order = StructuresResonating

moleculeainbondsofnumberTotal

eg. O CO¯

O¯CO Bond order

3

41.33

O O¯P

O¯

O¯

PO Bond order 4

51.25

O O¯Cl

O

O

ClO Bond order 4

71.75



SHAPES OF MOLECULES BASED ON VSEPR THEORY

Total no. No. of b.p. No. of General Type of Stereo Shape Exam.of hybrid (bondpairs) unshared formula hybridisation chemicalorbitals pair i.e. l

pformula/str.

2 2 0 AB2 sp B—A—B180° linear BeCl2

3 3 0 AB3 sp2

B B

B

A

120°

Trigonalplanar

3 2 1 AB2 sp2

B B

A



COVALENT BOND

(i) A covalent bond is formed by the mutual sharing of electrons between two atoms of electronegative

elements to complete their octet. (Except H which completes its duplet).

H H

H – H

H molecule2

O

O molecule2

O

O O

N

N molecule2

N

N N

(ii) The shared pair of electrons should have opposite spins, and are localised between two atoms

concerned.

(iii) Shairing of electrons may occurs in three ways –

No. of electrons shared Electron pair Bond

between two atoms

2 1 Single bond ( )

4 2 Double bond ( )

6 3 Triple bond ( )

Ex. – H N H

H

Three single bonds (not triple bond)

N N , Triple bond. (not three single bond)

O = O, Double bond (Not two single bond)

H – O – H , (Two single bonds.)

ORBITAL CONCEPT OF COVALENT BOND :

(i) One orbital can accommodate at the most 2 electrons with opposite spins

(ii) Half filled orbital or unpaired electron orbital accepts one electron from another atom, to complete

its orbitals.

(iii) Tendency to complete orbital or to pair the electron is an essential condition of covalent bond.

Completion of octet is not the essential condition of covalent bond.

(iv) Covalency : It is defined as the number of electrons contributed by an atom of the element for

shairing with other atoms to achieve noble gas configuration.

(v) If the outermost orbit has empty orbitals then covalent bonds are formed in excited state.



VARIABLE VALENCY IN COVALENT BONDS :

(i) Variable valencies are shown by those elements which have empty orbitals in outermost shell.

(ii) Lone pair electrons gets excited in the subshell of the same to form the maximum number of

unpaired electrons. Maximum covalency is shown in excited state.

(iii) The energy required for excitation of electrons is called promotion energy.

(iv) Promotion rule – Excitation of electrons in the same orbit.

Example :

(a) Phosphorus Ground state

3s 3p

Covalency 3 (PCl3)

Phosphorus Excited state

3s 3p 3d

Covalency – 5 (PCl5)

(b) Sulphur Ground state.

3s 3p 3dCovalency – 2 (SF2)

Sulphur Excited state

1st excited state

3s 3p 3d

Covalency – 4 (SF4)

2nd excited state

3s 3p 3d

Covalency – 6 (SF6)

So variable covalency of S is 2, 4 & 6.

(c) Iodine has three lone pair of electrons

(Ground state)

5s 5p 5d

So it shows three excited states – Maximum number of unpaired electrons = 7

Variable Valencies are 1, 3, 5, 7

Note : Maximum covalency is correspond to the maximum number of unpaired electron in

ground state or excited state of the element.



APPLICATIONS OF VARIABLE VALENCY :

To explain existence of molecules :

NCl3 – exists

NCl5 – doesn’t exists (due to absence of d-orbitals in Nitrogen.) While PCl3 and PCl5 both exist

because 3d orbitals are present in phosphorus.

OF2 –– exists, but OF4 and OF6 doesn’t exists due to absence of d-orbitals, while SF4 and

SF6 exists due to presence of d-orbital, present in its valence shell.

Wave mechanical concept of chemical bonding – (Overlapping)

To explain the nature of covalent bond two theories based on quantum mechanics have been

proposed.

(1) Valence bond theory (VBT) (2) Molecular orbital theory (MOT)

VALENCE BOND THEORY

(i) It was presented by Heitler & London to explain how a covalent bond is formed.

It was extended by Pauling & Slater.

(ii) The main points of theory are –

(a) To form a covalent bond overlapping occurs between half filled valence shell orbitals of the two

atoms.

(b) Resulting bond acquires a pair of electrons with opposite spins to get stability.

(c) Orbitals come closer to each other from the direction in which there is maximum overlapping

(d) So covalent bond has directional character.

(e) Extent of overlapping strength of chemical bond.

(f) Extent of overlapping depends on two factors.

(i) Nature of orbitals – p, d and f are directional orbitals more overlapping

s-orbitals non directional – less overlapping

(ii) Nature of overlapping – Co-axial overlapping - extent of overlapping more.

Collateral overlapping - extent of overlapping less

Order of strength of Co - axial overlapping –

p - p > s - p > s - s (When intramolecular distance remains same)

p - p p - s s - s

(g) As the value of n increases, bond strength decreases.

1 - 1 > 1 - 2 > 2 - 2 > 2 - 3 > 3 - 3

(h) If n is same bond strength order will be following

2p - 2p > 2s - 2p > 2s - 2s

1s - 2p > 2s - 2p > 3s - 3p



(i) Electron which is already paired in valency shell can enter into bond formation, it they can beunpaired first and shifted to vacant orbitals of slightly higher energy of the same energy shell.

(j) This point can explain the trivalency of boron, tetravalency of carbon, penta-valency ofphosphorus etc.

(k) Two types of bonds are formed on account of overlapping.

(A) Sigma () bond (B) Pi () bond

(A) Sigma () bond :

(i) Bond formed between two atoms by the overlapping of half filled orbitals along their axis (end to

end overlap) is called sigma bond.

(ii) bond is directional.

(iii) bond do not take part in resonance.

(iv) Free rotation is possible about a single bond.

(v) Maximum overlapping is possible between electron clouds and hence it is strong bond.

(vi) There can be only bond between two atoms.

Sigma bond are formed by three types of overlapping

(a) s - s overlapping (H2) - Two half filled s-orbitals overlap along the internuclear axis.

+

1s 1s

bond

1s 1s

(Formation of H molecule)2(1s orbital of hydrogen) (1s orbital of hydrogen)

(b) s - p overlapping (Formation of HF) – When half fill s-orbital of one atom overlap with half filledp-orbital of other atom.

1s orbital of Hydrogen 2p of Fluorineorbital H–F

+

(c) p–p overlapping – (Coaxial) – It involves the coaxial overlapping between half filled p-orbitals of

two different atoms.

eg. Formation of Cl2, F2, Br2

– + + + – –

p orbital p orbital



(B) Pi () bond

(i) The bond formed by sidewise (lateral) overlapping are known as bonds.

(ii) Lateral overlapping is only partial, so bonds formed are weaker and hence more reactive than

bonds (Repulsion between nucleus is more as orbitals have to come much close to each other for

bond formation)

Example: Formation of O2 molecule –

+

O molecule2

Pz

Py Py

Pz

Atomic orbitalof oxygen

Atomic orbitalof oxygen

Only py and pz of oxygen atom have unpaired electron in each orbital for bonding.

Electronic configuration of oxygen is – 1s22s22px2 2py

12pz1

(iii) Free rotation about a bond is not possible.

(iv) bond is weaker than bond (Bond energy difference is 63.5 KJ or 15 K cal/mole)

(v) bonds are non-directional, so do not determine the shape of a molecule.

(vi) bond takes part in resonance.

(vii) bond formed by pure or unhybrid orbitals.

HYBRIDISATION

Consider an example of Be compound

If it is formed without hybridisation then - Cl Be Clp – s p – p

both the Be–Cl bonds should have different parameters and p-p bond strength is greater than

s–p bond strength.

Practically bond strength and distance of both the Be-Cl bonds are same.

This problem may overcome if hybridisation of s and p-orbital occurs.

(i) It is introduced by pauling, to explain equivalent nature of covalent bonds in a molecule.

(ii) Definition : Mixing of different shapes and approximate equal energy atomic orbitals, and

redistribution of energy to form new orbitals, of same shape & same energy. These new orbitals

are called hybrid orbitals. and the phenomenon is called hybridisation.

Now after considering s-p hybridisation in BeCl2

Cl Be Clp – sp sp – p

bond strength of both the bonds will be equal.

Characteristic of Hybridisation :

(i) Hybridisation is a mixing of orbitals and not electrons. Therefore in hybridisation full filled, half filled

and empty orbitals may take part.



(ii) Number of the hybrid orbitals formed is always be equivalent to number of atomic orbital which

have taken part in the process of hybridisation.

Structure ofhybrid orbital

(iii) Each hybrid orbital having two lobes, one is larger and other is smaller. Bond will be formed from

large lobe.

(iv) The number of hybrid orbitals on central atom of a molecule or ion = number of bonds + lone pair

of electron.

(a) The 1st bond between two atoms will be sigma.

(b) The other bond between same two atoms will be bond.

(c) The electron pair of an atom which do not take part in bond formation called as lone pair of

electron.

(v) One element can represent many hybridisation state depending on experimental conditions for

example, Carbon showing sp, sp2 and sp3 hybridisation in its compounds.

(vi) Hybrid orbitals are differentiated as sp, sp2, sp3 etc.

(vii) The order of repulsion between p & bp is : p - p > p - bp > bp - bp

(viii) The directional properties in hybrid orbital is more than atomic orbitals. Therefore hybrid orbitals

form stronger sigma bond. The directional property of different hybrid orbitals will be in following

order. sp < sp2 < sp3 < sp3d < sp2d2 < sp3d3

DIFFERENCE BETWEEN HYBRIDISATION & OVERLAPPING

Overlapping Hybridisation

1. It occurs between orbitals of two atoms 1. It occurs among orbitals of the same atom

2. Only half filled orbitals takes part in 2. Any type of orbital can participates

overlapping

3. It occurs during bond formation bond 3. Process, just before overlapping.

formed after hybridisation

4. Orbital of different energies may participates 4. It may takes place in ground or in excited

in excited states. state

In ground state– NH3, NCl3, PH3, PCl3,



DETERMINATION OF HYBRIDISATION STATE :

Method (I) :

Count the following pair of electron around the central atom :

(a) Count all pure bonded electron pairs (or bonds)

(b) Count all lone pair of electron

(c) Count Co-ordinate bond

Method (II) :

To predict hybridisation following formula may be used :

Number of hybrid orbital =2

1[Total number of valence electron in the central atom + total number

of monovalent atoms – charge on cation + charge on anion]

eg. 4NH : H = 21

[5 + 4 – 1] = 4 : sp3 hybridisation.

SF4 : H = 21

[ 6 + 4 ] = 5 : sp3d hybridisation.

–2O4S : H = 21

[ 6 + 2 ] = 4 : sp3 hybridisation.

(‘O’ is divalent so add only charge on anion)

–O3N : H = 21

[5 + 1] = 3 : sp2 hybridisation.

Where, H is the number of hybrid orbitals.

If such type of electron pairs are –

Type & Shape of d-orbital and participation of d-orbital in Hybridise molecule

Hybridised orbital Participation orbital Geometrysp s, pz linearsp2 s, px, py trigonal planarsp3 s, px , py, pz tetrahedrald3s dxy, dyz, dzx, s square planarsp3d s, px , py , pz , dz

2 trigonalbipyramidalsp3d2 s, px , py , pz , dx

2–y2, dz2 octahedral

sp3d3 s, px , py , pz , dx2–y2, dz

2, dxy pentagonal bipyramidal



BOND PARAMETERS :

(i) Bond Length (Bond distance) (ii) Bond Angle (ii) Bond Energy

(i) Bond Length :- The average distance between the nucleus of two atoms is known as bond length,

normally it is represented in Å. eg. A B

It depends mainly on electronegativities of constituent atoms

Case –I If electronegativity difference is zero, then

Bond length = rA + rBor dA –B = rA + rB

where, rA is covalent radius of A

rB is covalent radius of B

If rA = rB , then, Bond length = 2rA or 2rBCase II If electronegative difference is not equal to zero then-

Bond length is given by shomaker & Stevenson formula, Bond length = rA + rB – 0.09 (XA – XB)

XA is electronegativity of A

XB is electronegativity of B

where (XA – XB) Difference in electronegativities of A and B

FACTORS AFFECTING BOND LENGTH :

(a) Electronegativity difference :- Bond length DifferenceativityElectroneg

1

( While Bond Energy Electronegativity difference)

H–F < H–Cl < H–Br < H–I

(b) Bond order or number of bonds :- Bond length orderbondorbondsofNumber1

Bond energy Number of bonds

e.g. C–C, C = C, C C

Bond length 1.54 Å 1.34 Å 1.20 Å gsinincrea

Bond energy 80 140 180–200K.Cal. gsinincrea

C—O C O C O

1.43 Å 1.20 Å 1.13Å

C N C N C N

1.47 Å 1.28 Å 1.15 Å

(c) Resonance :- (due to resonance bond length affected)

eg.1. Benzene

C—C bond length 1.54Å but bond length is between

C=C bond length 1.34 Å single & double bond is = 1.39 Å



eg.2. Bond length of C—O in CO2 is 1.15 Å, Resonance occurs in CO2 as follows-

O C O O — C O– + O C — O+ –

Bond length = 1.15 Å (Between double & triple bond)

(d) Hybridisation : - Bond length characters%

1

3 3sp spEthane 1.54 Å

3 2sp sp 1.51 Å

3sp sp 1.47 Å2sp sp2 1.46 Å2sp sp 1.42 Å

sp sp 1.37 Å

%s-c

ha

racte

rin

cre

ases

(ii) Bond Angle :- The angle between any two adjacent bond is known as bond angle. It is

represented in degree (°), min () and second ()

Factors affecting the bond angle-

(a) Number of bond : Bond angle Number of bonds (Bond order)

109° 120° 180°

C C C C C C

(b) Hybridisation :-

Case I

When hybridisation is same, bonded atoms are same but central atom and lone pair are different.

Then bond angle pairloneofNumber1

Example :- CH4 3HN•• ••

••2OH

Hybridisation sp3 sp3 sp3

Bond angle 109° > 107° > 105°

no lone pair one lone pair two lone pair

Case-II

When hybridisation is same, bonded atoms are same, lone pair is same but central atom isdifferent. Then bond angle electronegativity of central atom

Example:-3HN

••

3HP••

3HAs••

Bond angle 107° 93° 91°

– Electronegativity decreasing

– Bond angle will decrease



Case-III

When hybridisation is same, lone pair are same, Central atom is same but bonded atoms are

different.

sp3 OF2 103 – 105° Electronegativity

sp3 Cl2O 109 – 111° of bonded atom is

sp3 Br2O 116 – 118° decreasing

Here, bond angle atombondedofativityelectroneg

1size of side atom

(iii) Bond Energy (BE) : Bond energy may be defined as-

(a) Bond formation energy : Energy released when any bond is formed is known as bond

formation energy or bond energy.

(b) Bond dissociation energy : Energy required to dissociate any bond is known as bond

dissociation energy.

Calculation of released energy is more difficult than the dissociation energy, therefore dissociation

energy of bond is calculated and is assumed as bond energy or bond formation energy.

Case–I In diatomic molecule :

Bond energy = bond dissociation energy

eg : N2 > O2 > H2 > F2

Case–II For polyatomic molecule :-

eg:|

|

H

H

HCH Bond energy = 99.5 K. Cal/mole (per CH bond)

Theoritical values of bond dissociation energy (D) of individual CH bonds CH4 are given below-

D (CH3H) = 102 K Cal/mole

D (CH2H) = 105 K Cal/mole

D (CHH) = 108 K Cal/mole

D (CH) = 83 K Cal/mole

Hence, bond energy per CH bond in methane4

398 = 99.5 K Cal/mole.

Bond dissociation energy (D) is related to the state of hybirdisation.

Factors affecting the bond energy : -

(a) Electronegativity difference (b) Bond order (c) Atomic size (d) Bond polarity

(e) Resonance (f) Hybridisation (g) Lone pair electron

(a) Electronegativity difference :- Bond energy Electronegativity difference

eg. HF > HCl > HBr > HI



(b) Bond order :- Bond energy Bond order

eg. CC > C = C < – C C –

79 K. Cal, 143.3 K. Cal., 199.0 K. Cal.

(c) Atomic size :- Bond energy sizeAtomic1

eg. CC < CN < NN

Exception :- In case of halogen group, order of bond energy is-

Cl Cl > Br Br > F F > I I

Because of higher electron density and small size of F atoms, repulsion between of two F atom,

weakens the bond energy.

Other eg. S – S > O – O

C – C > Si – Si > Ge – Ge

(d) Bond Polarity :- Bond energy polarity

eg. HF > HCl > HBr > HI

(e) Resonance :- Bond energy increases due to resonance

eg. In benzene, bond energy of CC increases due to electrons of C = C.

(f) Hybridisation :- Bond energy s-character in hybrid orbitals.

eg. spsp > sp2sp2 > sp3sp3

% s-character 50% 33.3% 25%

(g) Lone pair of electrons :- Bond energy electronsofpairlone1

C C > N N > O O > F F××

××

××

××××

××

××

××

××

××

× ×× ×

Size of F and O atoms small so their bond energy should be high (small atomic radius) but it isactually less due to lone pair of electrons present on F and O atoms, which repells each other inFF and OO type of bonds.

Characteristic of Covalent Compound :

(i) Physical state :- Covalent compounds are found in all the three states - Gas, Solid & Liquid.

Separate molecules – In gaseous state

Associate molecules – In liquid & solid state

(Due to strong vander waal’s force and hydrogen bonding among the molecules.)

As the molecular weight increases, physical state changes :

eg. F2 and Cl2 Br2 I2, At2

gas liquid solid

Top to bottom in a group, Vander waal’s force increases between the molecules.



(ii) Covalent solid : Those solids in which atoms are linked together by covalent bonds, forms infinite

three dimensional giant structure.

e.g. Diamond, Graphite, AN, SiC, SiO2 etc.

Molecular solid : Discrete (separate) molecules are formed by covalent bonds and then the

molecules associated due to intermolecular force of attraction. (Vander-waal’s force)

eg. Solid I2, dry ice (Solid CO2) etc.

(iii) Conductivity : Mostly covalent compounds are bad conductor of electricity. But few polar covalent

compounds due to self ionisation can conduct electricity.e.g. H2O, liq. NH3 etc.

H2O + H2O H3O+ + OH–

2NH3 NH4+ + NH2

–

Free ions are formed which can conduct electricity.

Exceptions : Graphite, HCl in water.

(iv) Solubility : Non polar compound are soluble in non polar solvents. Non polar compounds forms

Vander-waal’s bond with non polar solvent molecules.

(v) Isomerism : Covalent bond is rigid and directional, so it shows isomerism.

eg. Organic compounds.

(vi) Reaction : Reaction between covalent compounds are slow. Because it involves breaking of old

bonds and formation of new bonds.

MOLECULAR ORBITAL THEORY (MOT):

MOT put forward by Hund & Mulliken, which can be applied to explain the properties, which the

old VBT (Valence bond theory) was unable to explain eg. Paramagnetic nature of O2 molecule,

as per VBT (:O: :O:), it should be diamagnetic.

Definition:

The atomic orbital lose their identity during molecule formation (overlapping) and form new orbitals

termed as molecular orbitals.

Characteristic of molecular orbitals:

(i) Molecular orbital formed by overlapping of atomic orbital of same energy

(ii) Number of molecular orbital formed = number of atomic orbital involved in overlapping

(iii) Half of the molecular orbital have lower energy are called Bonding molecular orbital.

(iv) Half are of higher energy is termed as Antibonding molecular orbital

(v) Electronic configuration in various molecular orbital are governed by same three rules.

(a) Aufbau’s rule (b) Hund’s rule (c) Pauli’s exclusion principle



Table : Comparision of Bonding molecular orbital & Antibonding molecular orbital :

Bonding molecular orbital (BMO) Antibonding Molecular orbital (ABMO)

Bonding MO is the result of the linear ABMO is result of linear combination ofAO

combination of AO when their wave when their wave function are substracted

function are addedb=

A+

B

a=

A–

B

BA2B

2A

2BA 2)( BA

2B

2A

2BA 2)(

It does not have node. It always have a node between two nuclei of

bonded atom.

Charge density increase between two Charge density decrease in between two nuclei,

nuclei resulting attraction between two leads to repulsion between two atoms.

atoms

Energy of BMO is less, hense stable Energy ofABMO is high, hence unstable

Notation of molecular orbitals:

As atomic orbitals are known by letters s, p, d and f depending on their shapes. Similarly for

molecular orbital.

For bonding molecular orbital- etc.

For antibonding molecular orbital- etc.

are used for different shapes of electron cloud.

ENERGY LEVEL DIAGRAM OF MOLECULAR ORBITAL :

On the basis of Aufbau’s rule - increasing order of energies of various molecular orbitals is-

(1s) < (1s) < (2s) < (2s) < (2pz) < (2px) =(2py) < 2px) = (2py) <

(2pz)

Energy level diagram for homonuclear diatomic molecules like, O2, F2, Ne2

For O2 molecule-



2px)

( )2pz

2p 2p2px)

2py)

2pz)

2p

Bond order : O = 22O = 2.52

+

O = 1.52–

O = 1.022–

Stability order – O2 > O O O2 2 2> >+ – 2–

Bond length – O >O2 2 > O O2 2>2– – +

2s)

2s2s2s)

1s1ss)

Incr

easi

ng

ene

rgy

Having two unpairedelectrons so paramagnetic

Bond order = 2

* 2py)*

*

*

s)*

Atomic orbitalof oxygen atom

Atomic orbitalof oxygen atom

Molecular orbitalof oxygen Molecule



ENERGY LEVEL DIAGRAM FOR B2, C2 AND N2 MOLECULES –

(1s) < (1s) < (2s) < (2s) < 2px) = (2py) < (2pz) < (2px) = (2py) <

(2pz)

For N2 molecule

*(2p )x

*(2p ) z

*(2p )y

2p 2p

(2p )x

2p

*(2s)

2s2s2s)

*(1s)

1s1ss)

2p )y

Incr

easi

ng

energ

yin

Nm

ole

cule

2

Atomic orbitalof nitrogen atom

Atomic orbitalof nitrogen atom

Molecular orbitalof nitrogen molecule

Cause of exceptional behavior of molecular orbital in B2, C2 and N2:

Energy of 2s and 2p atomic orbitals lie fairly close

Due to small energy difference between 2s and 2pz orbitals, the interaction between them is

quite large.

This results in loss of energy by (2s) and (2s) and thus (2s) and (2s) becomes more

stable at the cost of (2px) and (2px) which gets unstablised (Higher energy).

Electronic configuration of molecules and their related properties :-

for writing electronic configuration of diatomic molecules following two rules to be followed-

– Count the number of electrons present in two atoms and then fill in the appropriate energy level

diagram according to Aufbau rule.

– The pairing in (2px) and (2py) or (2px) and (2py) will take place only when each

molecular orbital of identical energy has one electron.

– After writing the molecular orbital diagram following parameter about molecules/ion may be

predicted.



(i) Bond order :- Bond order = ½ [Nb–Na]

Nb – Number of electron in bonding molecular orbital

Na – Number of electron in antibonding molecular orbital

(ii) Bond length :- (distance between two nuclei)orderBond

1lengthBond

If existmoleculethen:NN ab

existnotdoesMoleculeNNNN

ab

ab

(iii) Stability of molecules - stability of molecule Bond order of molecule

(iv) Dissociation energy - Bond dissociation energy Bond order

(v) Magnetic property -

(a) When electron in Molecular orbital are paired – then the molecule is diamagnetic.

(b) When electron in Molecular orbital are unpaired – the molecule is paramagnetic.

BONDING IN MOLECULES :

(i) Hydrogen molecule-

Having two H atoms with one electron each (1s)1

1s)

1s(Atomic orbital ofhydrogen atom)

1s)

Molecular orbitalof H molecule2

1s(Atomic orbital ofhydrogen atom)

E

Molecular orbital (M.O.) configuration of H2 = (1s)2 (1s)°

Bond order = ½ [Nb – Na] = ½ [2 – 0] = 1 i.e. single bond

Having paired electron, so diamagnetic.

Stability quite stable (having single bond)

(ii) H2+ ion –

1s)

1s(Atomic orbitalof H-atom)

1s)


+

1s(Atomic orbitalof H -atom)

E

Molecular orbital (M.O.) configuration of Configuration of H2+ = (1s)1 (1s)°

One electron in bonding molecular orbital.



So paramagnetic

Bond order = ½ [1– 0] = ½

Less stable

(iii) H2– anion -

Molecular orbital (M.O.) configuration - (1s)2 (1s)1

Paramagnetic

Bond order = ½ [2 – 1] = ½

1s)

1s(Atomic orbital

of H-atom)

1s)


–

1s(Atomic orbital

of H -atom)–

E

Stability is less than [H2+] because H2

– Contain antibonding molecular orbital (ABMO) electron

(iv) Helium molecule (He2) :

– Molecular orbital (M.O.) configuration (1s)2 (1s)2

– Diamagnetic

– Bond order = ½ [2 – 2] = 0 (zero)

– Bond order zero indicates no linkage

between He atoms. Hence He2 molecule does not exist

– Stability (He2) Highly unstable molecule

1s)

1s(Atomic orbital

of He-atom)

1s)Molecular orbitalof He molecule2

1s(Atomic orbital

of He-atom)

E

DIPOLE MOMENT :

(Ionic Nature in Covalent Bond)

(i) Polarity of any polar covalent bond or molecule is measured in terms of dipole moment.

(ii) For measurement of extent of polarity, Pauling introduced the concept of dipole moment ().

The product of positive or negative charge (q) and the distance (d) between two poles is called

dipole moment.

Here : = q × d (magnitude of charge × distance)



(iii) Dipole moment is a vector quantity i.e. it has both magnitude as well as direction.

(iv) Direction of dipole moment is represented by an arrow pointing from electropositive to

electronegative element and from central atom to lone pair of electrons.

central atom lone pair of electronor

(v) Unit of dipole moment is Debye

1 Debye = 1 × 10–18 esu cm.

= 1.6 × 10–29 coulomb metre

(vi) In the diatomic molecule dipole moment () depends upon difference of Electronegativity i.e.

dipole moment () Electronegativity difference

order of dipole moment () : H–F > H–Cl > H–Br > H–I

dipole moment () = 0 for H–H, F–F, Cl–Cl, Br–Br, O=O

(vii) For polyatomic molecules dipole moments () depends on the vector sum of dipole moments of

all the covalent bonds.

(viii) For PCl5 and SF6 , etc. dipole moment () = 0 due to their regular geometry.

(ix) Benzene, naphthalene, biphenyl have dipole moments () = 0 due to planar structure.

(x) If the vector sum is zero, than compound is non-polar compound or symmetrical compound (and

it is not essential that individual dipole moments () of every bond should be zero).

Example : (A) BX3, CCl4, SiCl4, CH4, CO2, CS2, PCl5, SiH4 etc.

In these examples the bond B–F, C–Cl, C–H, C–O, P–Cl etc. are polar even though compounds

are non-polar.

(B) ••N H

H3

1

2

=1.47D

4

NH3••

H

••P

H

H

H

Electronegativity of P H

4

PH3••

••N F

F3

1

2

=0.24 D

4

NF3••

F

(xi) Dipole moment of H2O is 1.85 D which is resultant dipole moment ()of two O–H bonds.

H

O

H

dipole moment ()of H2O is more than dipole moment ()of H2S because electronegativity

of oxygen is higher than sulphur.

(xii) Angular structure of molecule have greater dipole moment.



APPLICATION OF DIPOLE MOMENT :

(i) To determine polarity and geometry of molecule :

If dipole moment ()= 0 compound is non polar and symmetrical

eg. CO2, BF3, CCl4, CH4. BeF2 etc.

If dipole moment ()0 compound will be polar and unsymmetrical.

eg. H2O, SO2, NH3, Cl2O, CH3Cl, CHCl3 etc.

(ii) To calculate % ionic character :

100)(momentdipoleofvaluelTheoritica

)(momentdipoleofvaluealExperimentcharacterIonic%

(iii) To distinguish cis form or trans form:-

(a) Dipole moment of cis isomers is normally higher than trans isomers.

eg. :

Cl—C—H

Cl—C—H

cis-form

Polar ( )

Cl—C—H

H—C—Cl

Trans-form

Non Polar ( )

(b) If two groups have opposite inductive effect than trans-isomer will have greater dipole moment-

eg. :

H

C

C

CH3 H

Cl H

C

C

CH3H

Cl

(iv) To locate position of substituents in aromatic compounds.

angleBond

1)(momentDipole

(a) If same substituents are present in the symmetrical position dipole moment (m) of benzene ring

compounds will be zero.Cl

ClAngle 180°p-dichloro benzene

ClCl Cl

ClCl

Cl

Angle 120°m-dichloro benzene

Angle 60°o-dichloro benzene

(b) As angle between substituents decrease value of dipole moment increase



Some important orders of dipole moments ()

HF > H2O > NH3 > NF3 H2O > H2S

CH3Cl > CH3F > CH3Br > CH3I BF3 < NF3 < NH3HF > H2O > SO2 > NH3 H2O < H2O2

VANDER WAAL’S FORCES

(a) This type of attractive forces occurs in case of non-polar molecules such as H2, O2, Cl2, CH4,

CO2 etc.

(b) The existence of weak attractive forces among the non-polar molecule was first proposed by dutch

scientist J.D. Vander Waal

(c) Vander waal’s force molecular weight

Atomic weight

Boiling point

Types of Vander Waal’s force : -

(i) Dipole-Dipole attraction - It is again in between two polar molecules such as HF and HCl

+

HF HCl

(ii) Dipole - Induced dipole attraction : In this case a neutral molecule is induced as a dipole by

another dipole as shown in fig.

e.g. HCl Cl2

Before induction +

After induction + +

(iii) Induced dipole - induced dipole attraction or London dispersion force

between two non polar molecules as in Cl2, He etc.

+

+

Cl2 Cl2



HYDROGEN BOND

Definition :

(i) It is an electrostatic attractive force between covalently bonded hydrogen atom of one molecule

and an electronegative atom (F, O, N)

(ii) It is not formed in ionic compounds

(iii) Hydrogen bond forms in polar covalent compounds, (not in non-polar)

(iv) It is also known as dipole-dipole attraction

H— F– .... H+ — F–...... H+ — F–

Main condition for Hydrogen bonding :

(i) Hydrogen should be covalently bonded with high electronegative element like F, O, N

(ii) Atomic size of electronegative element should be small.

Decreasing order of atomic size is–

N > O > F

Decreasing order of electronegativity is –

F > O > N

(4.0) (3.5) (3.0)

(iii) Strength of Hydrogen bond Electronegativity of element elementofsizeatomic

1

(iv) Hydrogen bonding occurs in HCN, due to (–C N) triple bond (sp hybridisation), electronegativities

of carbon and nitrogen increases.

NC—H..........NC—H.........NC—H ––

Types of Hydrogen Bonding

Inter Molecular Intra Molecular

Homo Inter Molecular Hetero Inter Molecular

(A) Intermolecular Hydrogen bond

Hydrogen bond formation between two or more molecules of either the same or different compounds

known as Inter molecular Hydrogen bonding

These are two types.

(i) Homointermolecular :- Hydrogen bond between molecules of same compounds.

eg.

+H

+H

+H

+H

+H

+H +H

+H

O– O– O– +H +H +H

F–

F–

F–



(ii) Hetro intermolecular :- Hydrogen bond between molecules of different compounds.

eg. alcohol, water

O — H + – + – + – + O — H O — H O — H

R H R H

alcohol Water alcohol Water

(B) Intra molecular Hydrogen bond :- It takes place within the molecule.

(i) Hydrogen bonded with electronegative elements of a functional group, form Hydrogen bond

with another electronegative element present on nearest position on the same molecule.

(ii) This type of Hydrogen bond is mostly occurred in organic compounds.

(iii) It result in ring formation (Chelation).

eg.

O

N

H+

O–

O

o–nitrophenol

C

O

H

H–

Salicylaldehyde

O+

O

H+

o–fluorophenol

O

H

O

2, 6-dihydroxyl benzoate

C

H

OO

EFFECT OF HYDROGEN BOND ON PHYSICAL PROPERTIES :

(i) Solubility :

(A) Inter molecular Hydrogen bonding

(a) Few organic compounds (Non-polar) are soluble in water (Polar solvent) due to Hydrogen

bonding. eg. alcohol, acetic acid etc. are soluble in water.

–O — H O — H + – +

R H

Other examples–Glucose, Fructose etc, dissolve in water.

(b) Ketone, ether, alkane etc. are insoluble (no Hydrogen bond)

(c) Solubility order– CH3OCH3 < CH3OH

Primary amine > secondary amine > tertiary amine

(B) Intra molecular Hydrogen bonding:

(a) It decreases solubility as it form chelate by Hydrogen bonding,

so Hydrogen is not free for other molecule.



(b) It can not form H–bond with water molecule so can not dissolves.

C

O–

O

H+H

(Salicylaldehyde)

(C) Inter molecular Hydrogen bond

O — H O C — H O — H+– +

p–hydroxy benzaldehyde

O HH

C C

O — H O+ –

It can form Hydrogen bond with water molecule so it can dissolved

(ii) Viscosity:

Hydrogen bond associates molecules together, so viscosity increases

CH — OH2CH — OH2

CH OH3

CH — OH2

CH CH3 3— O —CH3 — OHH O2

water

<

>

<

>

alcohol ether

CH — OH

CH3 — OH

(iii) Melting point and boiling point

(a) Due to intermolecular Hydrogen bond Melting Point & Boiling Point of compounds increases.

H2O > CH3OH > CH3 — O—CH3

(b) Trihydric alcohol > dihydric alcohol > monohydic alcohol

Monocarboxylic acid form stronger Hydrogen bond than alcohol of comparable molecular

weight. Therefore Boiling Point of carboxylic acid is higher than alcohol.

(c) Decreasing order of Melting Point & Boiling Point isomer amines-

1°–amine > 2°–amine > 3°–amine

R — NH2 > R — NH — R > R — N

R|

— R (no hydrogen with nitrogen atom)



(d) Boiling points of VA, VIA, VIIA hydrides decreases on decreasing molecular weights.

VA VIA VIIA

NH3 H2O HF Boiling point HF > HI > HBr > HCl

PH3 H2S HCl H2O > TeH2 > SeH2 > H2S

AsH3 SeH2 HBr SbH3 > NH3 > AsH3 > PH3

SbH3 TeH2 HI

(e) But sudden increase in boiling point of NH3, H2O and HF is due to Hydrogen bonding

H2O > HF > NH3

Intramolecular Hydrogen bonding gives rise to ring formation, so the force of attraction among

these molecules are vander waal’s force. So, Melting Point and Boiling Point are low.

(iv) Molecular weight :

Molecular weight of CH3COOH is double of its molecular formula, due to dimer formation occur

by Hydrogen bonding

R — C C — R

O H —O+–

O — H O+ –

(v) Physical state:

H2O is liquid while H2S is gas.

Water and Ice:- Both have Hydrogen bonding even then density of ice is less than water.

Volume of ice is more because of open cage like crystal structure, from by association of water

molecules with the help of Hydrogen bond.

H2O becomes solid (Ice) due to four hydrogen bond among water molecule are formed in tetrahedral

manner.

H H H H

13

42

H H H H

O O O O

(vi) Base strength:

CH3NH2, (CH3)2 NH, (CH3)3 N, form Hydrogen bond with water. So, less hydrolysis i.e. it gives

OH– ions. While (CH3)4 N+ OH

–(ammonium compound) will give OH– ion in large amount due

to no Hydrogen bonding.

H

— NCH H — O

H H

3

H

— NCH — H + OH

H

3

–+

CH3

CH N CH

CH

3 3

3

(Ammonium compound.)NO Hydrogen atom bonded directly with Nitrogen atomso no hydrogen bonding occurs.

+



EFFECT OF INTRAMOLECULAR H–BONDING

(i) Strength of acid

(a) The formation of intramolecular H–bonding in the conjugate base of an acid gives extra stabilityto conjugate base and hence acid strength increases eg. Salicylic acid is stronger than benzoicacid and 2, 6 – dihydroxy benzoic acid > salicylic acid.

C C

O O–

O

O O + H+

H

Conjugate base

2, 6-dihydroxyl benzoate ion

O

H H

H

+ H

C

O O

–1/2 –1/2O

O+

H

(b) C2H5SH is more acidic than C2H5OH. In C2H5OH, Hydrogen bond forms, so H+ is not free

(c) HF is weaker acid than HI, due to Hydrogen bond in H – F, H+ is not free

(ii) Stability of chloral hydrate:-

If two or more OH group on the same atom are present it will be unstable, but chloral hydrateis stable (due to Hydrogen bonding).

Cl

Cl

Cl C C H Chloral hydrateO

OH

H

(iii) Maleic acid (cis) is stronger acid than fumaric acid (trans).

H H

H

O O

C C

C

H H

HOOC

C C

C C

C

C C

OH O–

OH O

+ H+

Stable conjugate base of maleic acid

O

H

COOH

Fumaric acid (No-intramolecular Hydrogen bonding)

O

(Maleic acid)

H

Note: The relative strength of various bonds is as follows

Ionic bond > Covalent bond > Metallic bond > Hydrogen bond > Vander waal’s bond



SUMMARY

Kösselís first insight into the mechanism of formation of electropositive and electronegative ions related

the process to the attainment of noble gas configurations by the respective ions. Electrostatic attraction

between ions is the cause for their stability. This gives the concept of electrovalency.

The first description of covalent bonding was provided by Lewis in terms of the sharing of

electron pairs between atoms and he related the process to the attainment of noble gas configurations by

reacting atoms as a result of sharing of electrons. The Lewis dot symbols show the number of valence

electrons of the atoms of a given element and Lewis dot structures show pictorial representations of

bonding inmolecules.

An ionic compound is pictured as a three-dimensional aggregation of positive and negative ions

in an ordered arrangement called the crystal lattice. In a crystalline solid there is a charge balance between

the positive and negative ions. The crystal lattice is stabilized by the enthalpy of lattice formation.

While a single covalent bond is formed by sharing of an electron pair between two atoms,

multiple bonds result from the sharing of two or three electron pairs. Some bonded atoms have additional

pairs of electrons not involved in bonding. These are called lonepairs of electrons.ALewis dot structure

shows the arrangement of bonded pairs and lone pairs around each atom in a molecule. Important

parameters, associated with chemical bonds, like: bond length, bond angle, bond enthalpy,

bond order and bond polarity have significant effect on the properties of compounds.

A number of molecules and polyatomic ions cannot be described accurately by a single Lewis

structure and a number of descriptions (representations) based on the same skeletal structure are written

and these taken together represent the molecule or ion. This is a very important and extremely useful

concept called resonance. The contributing structures or canonical forms taken together constitute the

resonance hybrid which represents the molecule or ion.

The VSEPR model used for predicting the geometrical shapes of molecules is based on the

assumption that electron pairs repel each other and, therefore, tend to remain as far apart as possible.

According to this model, molecular geometry is determined by repulsions between lone pairs and lone

pairs ; lone pairs and bonding pairs and bonding pairs and bonding pairs. The order of these repulsions

being : lp-lp > lp-bp > bp-bp

The valence bond(VB)approach to covalent bonding isbasicallyconcerned with the energetics

of covalent bond formation about which the Lewis and VSEPR models are silent. Basically the VB

theory discusses bond formation in terms of overlap of orbitals. For example the formation of the H2

molecule from two hydrogen atoms involves the overlap of the 1s orbitals of the two H atoms which are

singly occupied. It is seen that the potential energy of the system gets lowered as the two H atoms come

near toeachother.At theequilibriuminter-nucleardistance (bonddistance) theenergytouchesaminimum.

Any attempt to bring the nuclei still closer results in a sudden increase in energy and consequent

destabilizationof themolecule.Becauseoforbitaloverlap theelectrondensitybetween thenuclei increases

which helps in bringing them closer. It is however seen that the actual bond enthalpy and bond length

values are not obtained by overlap alone and other variables have to be taken into account.

Forexplaining the characteristic shapes of polyatomic molecules Pauling introduced the concept

of hybridisation of atomic orbitals. sp,sp2, sp3 hybridizations of atomic orbitals of Be, B,C, N and O

are used to explain the formation and geometrical shapes of molecules like BeCl2, BCl

3, CH

4, NH

3and

H2O. They also explain the formation of multiple bonds in molecules like C

2H

2and C

2H

4.



The molecular orbital (MO) theory describes bonding in terms of the combination and

arrangment of atomic orbitals to form molecular orbitals that are associated with the molecule as a

whole. The number of molecular orbitals are always equal to the number of atomic orbitals from which

they are formed. Bonding molecular orbitals increase electron density between the nuclei and are lower

inenergy than the individual atomicorbitals.Antibondingmolecularorbitals have a region of zero electron

density between the nuclei and have more energy than the individual atomic orbitals.

Theelectronicconfigurationof themolecules iswrittenbyfillingelectrons in themolecularorbitals

in the order of increasing energy levels.As in the case of atoms, the Pauli exclusion principle and Hundís

rule are applicable for the filling of molecular orbitals. Molecules are said to be stable if the number of

elctrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals.

Hydrogen bond is formed when a hydrogen atom finds itself between two highlyelectronegative

atoms such as F, O and N. It may be intermolecular (existing between two or more molecules of the

same or different substances) or intramolecular (present within the same molecule). Hydrogen bonds

have a powerful effect on the structure and properties of many compounds.



EXERCISE-1

[SINGLE CORRECT CHOICE TYPE]

Q.1 The octet rule is not obeyed in –(1) CO2 (2) BCl3 (3) PCl5 (4) Both (2) and (3)

Q.2 Weakest bond is –(1) ionic bond (2) covalent bond (3) coordinate bond (4) hydrogen bond

Q.3 Which follows octet rule -(1) FeCl2 (2) AgCl (3) CaCl2 (4) CuCl

Q.4 The compound completing its octet by transfer of electrons is(1) MgO (2) H

2S (3) PH

3(4) CCl

4

Q.5N N N

(I) (II) (III)

Calculate the formal charge of each I, II and III nitrogen atom respectively(1) +1, –1, 0 (2) –1, –1, +1 (3) –1, +1, –1 (4) 0, –1, –1

Q.6 According to Lewis structure the number of lone pair & bond pair of electrons in SO42– ion.

(1) 12, 6 (2) 12, 4 (3) 10, 8 (4) 8, 8

Q.7 Which condition favours the bond formation:(1)Maximum attraction and maximum potential energy(2)Minimum attraction and minimum potential energy(3) Minimum potential energy and maximum attraction(4) None of the above

Q.8 ElementAhas 3 electrons in the outermost orbit and element B has 6 electrons in the outermost orbit.The formula of the compound formed byAand B has following property :(1) High melting points and non-directional bonds(2) High melting points and low boiling points(3) Directional bonds and low boiling points(4) High solubilities in polar and non-polar solvent

Q.9 An ionic bond BA is most likely to be formed when :(1) the ionization energy of A is high and the electron gain enthalpy of B is low(2) the ionization energy of A is low and the electron gain enthalpy of B is high(3) the ionization energy of A and the electron gain enthalpy of B both are high(4) the ionization energy of A and the electron gain enthalpy of B both are low

Q.10 The compound which has the highest Lattice energy is(1) LiF (2) LiCl (3) NaCl (4) MgO

Q.11 Compound of a metal ‘M’ is M2O3. The formula of its nitride will be:(1) M3N (2) MN (3) M3N2 (4) M2N3



Q.12 Ionic reactions occur in :(1)Aqueous solution and organic solvents of high polarity(2) Non-polar or solvents of low polarity(3) Gaseous state(4) Solid state

Q.13 Electrovalent compounds or ionic compounds do not show stereoisomerism. The reason is:(1) Presence of ions (2) Strong electro static force of attraction(3) Brittleness (4) Non-directional nature of ionic bond

Q.14 The total number of valence electrons in the PO43– ion is -

(1) 26 (2) 28 (3) 24 (4) 32

Q.15 PCl5 exists but NCl5 does not because :(1) Nitrogen has no vacant 2d-orbitals (2) NCl5 is unstable(3) Nitrogen atom is much smaller than P (4) Nitrogen is highly inert

Q.16 An element X forms compounds of formula XCl3, X2O5 and Mg3X2 but does not formXCl5 then X is :(1)Aluminium (2) Phosphorus (3) Nitrogen (4) Boron

Q.17 Which of the following is not a characteristic of a covalent compound :(1) It has low melting point and boiling point(2) It is formed between two atoms having very small electronegativity difference(3) They have no definite geometry(4) They are generally insoluble in water

Q.18 Correct statement regarding this reactionBF3 + NH3 [F3B NH3]

(1) Hybridisation of N is changed (2) Hybridisation of B is changed(3) NH3 act as a lewis base (4) (2) & (3) both

Q.19 The pair of compounds which can form a co-ordinate bond is:(1) (C2H5)3 B and (CH3)3N (2) HCl and HBr(3) BF3 and NH3 (4) (1) and (3) both

Q.20 The possible structure(s) of monothiocarbonate ion is :

(1)

C

O

S

O

(2)

C

O

S

O

2–

(3)

S

C

OO

(4)

S

O

C

O

Q.21 In Co-ordinate bond, the acceptor atoms must essentially contain in its valency shell an orbital:(1) With paired electron (2)With single electron(3) With no electron (4)With three electron

Q.22 The octet rule is not obeyed in :(1) CO2 (2) CH4 (3) PCl5 (4) SiF4



Q.23 Correct order of covalent character of alkaline earth metal chloride in:(1) BeCl2 < MgCl2 < CaCl2 < SrCl2 (2) BeCl2 < CaCl2 < SrCl2 < MgCl2(3) BeCl2 > MgCl2 > CaCl2 > SrCl2 (4) SrCl2 > BeCl2 > CaCl2 > MgCl2

Q.24 Which of the compound is least soluble in water:(1)AgF (2)AgCl (3)AgBr (4)Agl

Q.25 Among LiCl, BeCl2, BCl3 and CCl4, the covalent bond character follows the order:(1) LiCl < BeCl2 > BCl3 > CCl4 (2) LiCl > BeCl2 < BCl3 < CCl4(3) LiCl < BeCl2 < BCl3 < CCl4 (4) LiCl > BeCl2 > BCl3 > CCl4

Q.26 Polarization of ions is governed by -(1)Hybridization (2) VSEPR theory (3) Fajan's rules (4) Paulingrule

Q.27 A sigma bond is formed by the overlap of atomic orbitals of atoms A and B. If the bond is formed alongthe x-axis, which of the following overlaps is acceptable ?

(1) s orbital of A and zp orbital of B (2) xp orbital of A and yp orbital of B

(3) zp orbital of A and xp orbital of B (4) xp orbital of A and s orbital of B

Q.28 The strength of bonds by 2s – 2s, 2p – 2p and 2p – 2s overlapping has the order:(1) s – s > p – p > s – p (2) s – s > p – s > p – p(3) p – p > s – p > s – s (4) p – p > s – s > p – s

Q.29 A-bond is formed by two xp orbitals each containing one unpaired electron when they approach each

other along :(1) x - axis (2) y - axis (3) z - axis (4) any direction

Q.30 Which is not characteristic of -bond:(1) -bond is formed when a sigma bond already formed(2) -bond are formed from hybrid orbitals(3) - bond may be formed by the overlapping of p-orbitals(4) - bond results from lateral overlap of atomic orbital

Q.31 Number of and bonds present in : CHC—CHCH—CH3 are:

(1) 10, 3 (2) 102 (3) 9, 2 (4) 8, 3

Q.32 The correct order of bond length is(1) C – C < C = C < C C (2) C C < C = C < C – C(3) C = C < C C < C – C (4) C = C < C – C < C C

Q.33 Cyanogen, 2)(CN , has a ____ shape/structure :

(1) Linear (2) Zig-zag (3) V-shape (4) Cyclic

Q.34 The d-orbitals involved in sp3d hybridisation in trigonal bipyramidal geometry :

(1) 22 yxd (2) 2zd (3) xyd (4) xzd



Q.35 A sp3 hybrid orbital contains:

(1)4

3 s–character (2)4

1p–character (3)

4

3p–character (4)

2

1s – character

Q.36 Which of the set of species have same hybridisation state but different shapes:

(1) 222 NO,NO,NO (2) 444 XeF,SF,CIO (3) 234 OF,OH,NH

(4) 43

42

4 ClO,PO,SO

Q.37 Among the following compounds the one that is planar and has the central atom with sp2 hybridisationis:(1) NH4

+ (2) SiF4 (3) BF3 (4) ClO4–

Q.38 In an octahedral structure, the pair of d orbitals involved in d2sp3 hybridization is :

(1) dxy, dyz (2) 2z2y2xd,d

(3) 2y2xxzd,d

(4) xz2zd,d

Q.39 Which of the following is true ?

(1) energybondlengthbond

1orderBond (2)

energybond

1lengthbondorderBond

(3)energybond

1

lengthbond

1orderBond (4) energybondlengthbondorderBond

Q.40 Which of the following is the correct order of bond-orders of the given species are such that -(1) O2

– > O2+ > O2

2– > O2 (2) O2+ > O2 > O2

– > O22–

(3) O2+ > O2

2– > O2– > O2 (4) O2

2– > O2 > O2+ > O2

–

Q.41 The number of antibonding electron pairs in 22O molecular ion on the basis of molecular orbital theory is

(at no. O = 8) :(1) 2 (2) 3 (3) 4 (4) 5

Q.42 In which of the following set, the value of bond order will be 2.5:(1) O2

+, NO, NO+2, N2+ (2) CN, NO+2, CN–, F2

(3) O2+, NO2+, O2

2+, CN– (4) O22–, O2

–, O2+, O2

Q.43 How many unpaired electrons are present in 2N :

(1) 1 (2) 2 (3) 3 (4) 4

Q.44 Which set of molecules is polar:(1) XeF4, IF7, SO3 (2) PCl5, C6H6, SF6 (3) SnCl2, SO2, NO2

– (4) CO2, CS2, C2H6

Q.45 BeF2 has zero dipole moment whereas H2O has dipole moment because:(1) Water is linear (2) H2O is bent(3) F is more electronegative than O (4) Hydrogen bonding is present in H2O

Q.46 The correct order of decreasing polarity is:(1) HF > SO2 > H2O > NH3 (2) HF > H2O> SO2 > NH3(3) HF > NH3 > SO2 > H2O (4) H2O > NH3 > SO2 > HF



Q.47 The correct order of dipole moment is:(1) CH4 < NF3 < NH3 < H2O (2) NF3 < CH4 < NH3 < H2O(3) NH3 < NF3 < CH4 < H2O (4) H2O < NH3 < NF3 < CH4

Q.48 The H bond in solid HF can be best represented as:

(1) FHFHFH ........ (2)H

FH

FH

FH

(3) H

F

H

F

H

F

H (4) F

H

F

H

F

H

F

Q.49 Intramolecular H-bond:(1) Decreases Volatility (2) Increases melting point(3) Increases viscosity (4) Increases volatility

Q.50 The hydrogen bond is strongest in:(1) O – H - - - S (2) S – H - - - O (3) F – H - - - F (4) O – H - - - O

Q.51 Which of the following has strongest intra molecular hydrogen bonding:

(1)

OH

COOH

(2)

OH

COOH(3)

OH

COOH

(4)

OCH3

COOH

Q.52 The boiling point of p-nitrophenol is higher than that of o-nitrophenol because:(1) NO2 group at p-position behaves in a different way from that at o-position(2) intramolecular hydrogen bonding exists in p-nitrophenol(3) there is intermolecular hydrogen bonding in p-nitrophenol(4) p-nitrophenol has a higher molecular weight than o-nitrophenol

Q.53 Out of the two compounds shown below, the vapour pressure of (2) at a particular temperature isexpected to be:

NO2(1)

OH

and

OH

NO2

(2)

(1) Higher than that of (1)(2) Lower than that of (1)(3) Same as that of (1)(4) Can be higher or lower depending upon the size of the vessel

Q.54 KF combines with HF to form KHF2. The compound contains the species:(1) K+, F– and H+ (2) K+, F– and HF (3) K+ and [HF2]

– (4) [KHF]+ and F2



Q.55 Density of ice is less than that of water because of :(1) presence hydrogen bonding(2) crystal modification of ice(3) open porous structure of ice due to hydrogen bonding(4) different physical states of these

Q.56 A hybrid orbital formed from s-and p-orbital can contribute to:

(1) bond only (2) bond only (3) Either or bond (4) None of these

Q.57 In a change from PCl3PCl5, The hybrid state of P change from:

(1) sp2 to sp3 (2) sp3 to sp2 (3) sp3 to sp3d (4) sp3 to dsp2

Q.58 Which of the following has been arranged in order of decreasing dipole moment :

(1) CH3Cl > CH3F > CH3Br > CH3I (2) CH3F > CH3Cl > CH3Br > CH3I

(3) CH3Cl > CH3Br > CH3l > CH3F (4) CH3F > CH3Cl > CH3I > CH3Br

Q.59 Which of the following statement is not correct -

(1) CH3+ shows sp2-hybridisation whereas CH3

– shows sp3-hybridisation

(2) NH4+ has a regular tetrahedral geometry

(3) sp2-hybridised orbitals have equal s and p character

(4) Hybridisation orbitals always form -bonds

Q.60 The magnitude of the lattice energy of a solid increases if:

(1) The ions are of large size (2) The ions are of small size

(3) The ions are of equal size (4) Charges on the ions are small

Q.61 The bond angle in H2O molecule is less than that of NH3 molecule because:

(1) The hybridisation of O in H2O and N in NH3 is different

(2) The atomic radii of N and O are different

(3) There is one lone pair of electrons on O and two lone pairs of electrons on N

(4) There are two lone pairs of electrons on O and one lone pairs of electrons on N

Q.62 The bond angles of NH3,4NH and

2NH are in the order:

(1) NH2– > NH3 > NH4

+ (2) NH4+ > NH3 > NH2

–

(3) NH3 > NH2– > NH4

+ (4) NH3 > NH4+ > NH2

–

Q.63 Which one of the following molecules has highest dipole moment:

(1) H2S (2) CO2 (3) CCl4 (4) BF3

Q.64 The nature of intermolecular forces among benzene (C6H6) molecules is:

(1) Dipole-dipole attraction (2) London dispersion force

(3) Ion-dipole attraction (4) Hydrogen bonding



Q.65 Amongst NH3, BeCl2, CO2 and H2O, the non-linear molecules are :

(1) BeCl2 and H2O (2) BeCl2 and CO2 (3) NH3 and H2O (4) NH3 and CO2

Q.66 The number of unpaired electrons in paramagnetic tetrachloromagnate (II) anion is:

(1) 5 (2) 2 (3) 3 (4) 6

Q.67 In the series, ethane, ethylene & acetylene, the carbon-hydrogen bond length is:

(1) Equal in all the three (2) Largest in ethane (3) Smallest in ethylene (4) Largest in acetylene

Q.68 The maximum covalency is equal to(1) the number of unpaired p-electrons(2) the number of paired d-electrons(3) the number of unpaired s and p-electrons(4) total number of unpaired electron in ground state or in excited state

Q.69 The cationic part of solid Cl2O6 is having the ___________shape.(1) linear (2) angular (3) Tetrahedron (4) undefined

Q.70 Molecular shapes of SF4, CF4 and XeF4 are:(1) the same, with 2, 0 and 1 lone pair of electrons respectively(2) the same, with 1, 1 and 1 lone pair of electrons respectively(3) different with 0, 1 and 2 lone pair of electrons respectively(4) different with 1, 0 and 2 lone pair of electrons respectively

Q.71 Pick out the incorrect statement?(1) N2 has greater dissociation energy than N2

+ (2) O2 has lower dissociation energy than O2+

(3) Bond length in N2+ is less than N2 (4) Bond length in NO

+ is less than in NO.

Q.72 Which of the following molecules has a square pyramidal structure :

(1) XeO2F2 (2) XeOF2 (3) XeO3F2 (4) XeOF4

Q.73 Which of the following has been arranged in order of decreasing bond length :(1) P–O > Cl–O > S–O (2) P–O > S–O > Cl–O

(3) S–O > Cl–O> P–O (4) Cl–O > S–O > P–O

Q.74 Which of the followinghybridizations involves dxy orbitals :

(1) sp3d (2) sp3d2 (3) dsp2 (4) sp3d3

Q.75 Among LiCl, BeCl2, BCl3 and CCl4, the covalent bond character varies as -

(1) LiCl < BeCl2 > BCl3 > CCl4 (2) LiCl > BeCl2 < BCl3 < CCl4(3) LiCl < BeCl2 < BCl3 < CCl4 (4) LiCl > BeCl2 > BCl3 > CCl4

Q.76 The formation of which of the following ions is not possible -

(1) [SiF6]2– (2) [AlF6]

3– (3) [BF4]– (4) [BF6]

3–

Q.77 The ratio between bond and bond in tetracyano ethylene :

(1) 2 : 1 (2) 1 : 1 (3) 1 : 2 (4) None of these



Q.78 The boiling point of ICl is nearly 40°C higher than that of Br2 although the two subtances have the same

relative molecular mass. This is because:

(1) ICl is ionic compound

(2) I-Cl bond is stronger than Br-Br bond

(3) ICl is polar covalent molecular while Br2 is non polar

(4) Ionization energy IP of Iodine is less than that of Br

Q.79 Choose the molecules in which same hybridisation occurs in the ground state:

(i) BCl3 (ii) NH3 (iii) PCl3 (iv) BeF2The correct answer is:

(1) i, ii, iv (2) i, ii, iii (3) ii, iii (4) iii, iv

Q.80 Four elements A (with one valence electron), B (with three valence electrons), C (with five valence

electrons) and D (with seven valence electrons) are lying in the second period in periodic table

which of the following is/are diatomic at room temperature :

(1) Only C2 (2) Only A2 (3) C2 and D2 (4) Only B2

Q.81 Which of the following statement is wrong:

(1) Hybridisation is the mixing of atomic orbitals prior to their combining into

molecular orbitals

(2) sp2 hybrid orbital are formed from two p atomic orbitals and one s atomic orbital

(3) dsp2 hybrid orbitals are all at 90° to one another

(4) sp3d2 hybrid orbitals are directed towards the corners of a regular tetrahedron

Q.82 The order of increasing adjacent bond angle in the molecules BeCl2, BCl3, CCl4 and SF6 is:

(1) SF6 < CCl4 < BCl3 < BeCl2 (2) BeCl2 < BCl3 < CCl4 < SF6(3) SF6 < CCl4 < BeCl2 < BCl3 (4) BCl3 < BeCl2 < SF6 < CCl4

Q.83 Which of the following has more dipole moment :

(1)

Cl

(2)

Cl

Cl

(3)

Cl

Cl

Cl

Cl (4)All has same dipole moment

Q.84 The order of increasing bond length in F2, N2, Cl2 and O2 is:

(1) N2 < O2 < Cl2 < F2 (2) N2 < O2 < F2 < Cl2(3) O2 < N2 < Cl2 < F2 (4) N2 < Cl2 < O2 <

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· 2019. 12. 24. · BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-5 ACC_CHEMISTRY_Chemical Bonding 1 Syllabus : • Valenceelectrons,ionicbond