Chemical Bonding Class11th by PS Sir IIT JEE

Chemical Bonding Class11th By PS Sir IIT JEE /PMT Course

 

BY: LEARN FOR FREE IN: CHEMISTRY

Chemical

bond

is

an

attractive force which keeps tow atoms or ions together in a molecule.A molecule is formed if it is more stable and has lower energy than the individual atoms. Normally only electrons in the outermost shell of an atom are involved in bond formation and in this process each atom attains a stable electronic configuration of inert gas. Atoms may attain stable electronic configuration in three different ways by losing or gaining

electrons by sharing electrons. The attractive forces which hold various constituents (atoms, ions etc) together in different chemical species are called chemical bonds. Elements may be divided into three classes.



Electropositive elements, whose atoms give up one or more electrons easily, they have low ionization potentials.



Electronegative elements, which can gain electrons. They have higher value of electronegativity.



Elements which have little tendency to loose or gain electrons.

Three different types of bond may be formed depending on electropositive or electronegative character of atoms involved.

Electropositive element (electrovalent bond)

+

Electronegative

element

=

Ionic

bond

Electronegative element + Electronegative element = Covalent bond

or less electro positive + Electronegative element = Covalent bond

Electropositive + Electropositive element = Metallic bond.

The Lewis Theory The octet rule:- The Lewis theory gave the first explanation of a covalent bond in terms of electrons that was generally accepted. If two electrons

are shared between two atoms, this constitutes a bond and binds the atoms together. For many light atoms, a stable arrangement is attained when the atom is surrounded by eight electrons.

This octet can be made up from some electrons which are totally owned and some electrons which are ‘shared’. Thus atoms continue to form bonds until they have made up an octet of electrons. This is called the ‘octet rule’. The octet rule explains the observed valencies in a large number of cases. There are exceptions to the octet rule; for example, hydrogen is stable with only two electrons. Other exceptions are discussed later.Today, the conventional Lewis structure representation of a pair of bonded electrons is by means of a ‘dash’ (-) usually called a ‘bond’. Lone pairs or ‘non-bonded’ electrons are represented by ‘dots’. Some structures are represented below:

It

is

therefore

exceptions:

useful

to

remember

some

important

categories

of



Odd-electron species like NO, NO2, O2 etc. Since it is impossible to distribute an

odd number

of electrons

into pairs, these species

necessarily violate the octet rule. 

The Valence Bond theory handles such species rather clumsily (oddelectron bonds etc.) the molecular orbital theory copes much better.



Species in which the central atom `expands' its octet (so to speak) PCl5, SF6, many transition metal compounds etc.



Electron -deficient species like BCl 3, BeCl2, AlCl3, B2H6 etc. in which the central atom has fewer than eight electrons.



Electrovalency

This

type

of

valency

involves transfer of electrons from one atom to another, whereby each atom may attain octet in their outermost shell. The resulting ions that are formed by gain or loss of electrons are held together by electrostatic force of attraction due to opposite nature of their charges. The reaction between potassium and chlorine to form potassium chloride is an example of this type of valency.

Here potassium has one electron excess of it’s octet and chlorine has one deficit of octet. So potassium donates it’s electron to chlorine forming an ionic bond.

Here the oxygen accepts two electrons from calcium atom. It may be noted that ionic bond is not a true bond as there is no proper overlap of orbitals.



Criteria for Ionic Bond Formation

One of the species must have electrons in excess of octet while the other should be deficit of octet. Does this mean that all substance having surplus electron and species having deficient electron would form ionic bond? The answer is obviously no. Now you should ask why? The reasoning is that in an ionic bond one of the species is cation and the other is anion. To form a cation from a neutral atom energy must be supplied to remove the electron and that energy is called ionization energy. Now it is obvious that lower the ionization energy of the element the easier it is to remove the electron. To form the anion, an electron adds up to a neutral atom and in this process energy is released. This process is called electron affinity. So for an ionic bond one of the species must have low ionization energy and the other should have high electron affinity. Low ionization energy is mainly exhibited by the alkali and alkaline earth metals and high electron affinity by the halogen and chalcogens. Therefore this group of elements are predominant in the field of ionic bonding.



Energy Change During the Formation of Ionic Bond

The formation of ionic bond can be consider to proceed in three steps (a) Formation of gaseous cations A(g) + I.E. → A+ (g) + e– The energy required for this step is called ionization energy (I.E) (b) Formation of gaseous anions X(g) + e– → X–(g) + E.A The energy released from this step is called electron affinity (E.A.) (c) Packing of ions of opposite charges to form ionic solid A+(g) + X–(g) → AX(s) + energy The energy released in this step is called lattice energy.

Now for stable ionic bonding the total energy released should be more than the energy required. From the above discussion we can develop the factors which favour formation of ionic bond and also determine its strength. These factors have been discussed below: (a) Ionization energy: In the formation of ionic bond a metal atom loses electron to form cation. This process required energy equal to the ionization energy. Lesser the value of ionization energy, greater is the tendency of the atom to form cation. For example, alkali metals form cations quite easily because of the low values of ionization energies. (b) Electron affinity: Electron affinity is the energy released when gaseous atom accepts electron to form a negative ion. Thus, the value of electron affinity gives the tendency of an atom to form anion. Now greater the value of electron affinity more is the tendency of an atom to form anion. For example, halogens having highest electron affinities within their respective periods to form ionic compounds with metals very easily. (c) Lattice energy: Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three dimensionally in a definite geometric pattern to form ionic crystal. Since the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied with the release of energy referred to as lattice energy. Lattice energy may be defined as the amount of energy released when one mole of ionic solid is formed by the close packing of gaseous ion. In short, the conditions for the stable ionic bonding are: (a) I.E. of cation forming atom should be low: (b) E.A. of anion – forming atom should be high; (c)

Lattice

energy

should

be

high.





Determination of Lattice Energy

The direct calculation of lattice enthalpy is quite difficult because the required

data

is

often

not

available.

Therefore

lattice

enthalpy

is

determined indirectly by the use of the Born – Haber cycle. The cycle uses ionization enthalpies, electron gain enthalpies and other data for the calculation of lattice enthalpies. The procedure is based on the Hess’s law,

which states that the enthalpy of a reaction is the same, whether it takes place in a single step or in more than one step. In order to understand it let us consider the energy changes during the formation of sodium chloride from metallic sodium and chlorine gas. The net energy change during the process is represented by ΔHf.

For example: Born- Haber Cycle for Soidum Chloride

Solved Problem on Born- Haber Cycle Calculate the lattice enthalpy of MgBr2. Given that Enthalpy of formation of MgBr2 = -524 kJ mol–1 Some of first & second ionization enthalpy (IE1 + IE2 ) = 148 kJ mol-1 Sublimation energy of Mg = +2187 kJ mol–1

Vaporization energy of Br2(I) = +31kJ mol–1 Dissociation energy of Br2(g) = +193kJ mol–1 Electron gain enthalpy of Br(g) = -331 kJ mol–1 Solution: ΔHof = S + I.E. + ΔHvap + D + 2 × E.A. + U Or U = ΔHof – (S + I.E. + ΔHvap + D + 2 × E.A.) Or U = -524 - [2187 - 148 + 31 + 193 + 2 × (-331)] =



–

524

–

1897

=

–2421

kJ

mol –1

Characteristics of Ionic Compounds

The following are some of the general properties shown by these compounds (i) Crystalline nature : These compounds are usually crystalline in nature with constituent units as ions. Force of attraction between the ions is nondirectional and extends in all directions. Each ion is surrounded by a number of oppositely charged ions and this number is called co-ordination number. Hence they form three dimensional solid aggregates. Since electrostatic forces of attraction act in all directions, therefore, the ionic compounds do not posses directional characteristic and hence do not show stereoisomerism. (ii) Due to strong electrostatic attraction between these ions, the ionic compounds have high melting and boiling points. (iii) In solid state the ions are strongly attracted and hence are not free to move. Therefore, in solid state, ionic compounds do not conduct electricity. However, in fused state or in aqueous solution, the ions are free to move and hence conduct electricity. (iv) Solubility : Ionic compounds are fairly soluble in polar solvents and insoluble in non-polar solvents. This is because the polar solvents have high values of dielectric constant which defined as the capacity of the solvent to weaken the force of attraction between the electrical charges immersed in that solvent. This is why water, having high value of dielectric constant, is one of the best solvents. The solubility in polar solvents like water can also be explained by the dipole nature of water where the oxygen of water is the negative and hydrogen being positive, water molecules pull the ions of the ionic

compound from the crystal lattice. These ions are then surrounded by water dipoles with the oppositely charged ends directed towards them. These solvated ions lead an independent existence and are thus dissolved in water. The electrovalent compound dissolves in the solvent if the value of the salvation energy is higher than the

lattice energy of that

compounds. AB + Lattice energy → A+ + B– These

ions

are

surrounded

by

solvent

molecules.

This

process

is

exothermic and is called salvation. A+ + x(solv.) → [A(solv.)x]+ + energy B– + y(solv.) → [B(solv.)y]– + energy The value of solvation energy depend on the relative size of the ions. Smaller the ions more is the solvation. The non-polar solvents do not solvate ions and thus do not release energy due to which they do not dissolve

ionic

compounds. (v) Ionic reactions: Ionic compound furnish ions in solutions. Chemical reactions are due to the presence of these ions. For example Na2SO4 → 2Na+ + SO42– BaCl2 →

Ba2+ +

2Cl–





Covalancy

This type of valency involves sharing of electrons between the concerned atoms to attain the octet configuration with the sharing pair being contributed by both species equally. The atoms are then held by this common pair of electrons acting as a bond, known as covalent bond. If two

atoms share more than one pair then multiple bonds are formed. Some examples of covalent bonds are



Valence Bond Theory (VBT)

A covalent bond is formed by overlapping of valence shell atomic orbitals of the two atoms having unpaired electron. As a result of overlapping, there is maximum electron density between the bonding atoms and large part of bonding force arises due to electrostatic force of attraction between

accumulated

electron

cloud

and

two

nuclei.

Greater

the

overlapping of atomic orbitals higher is the strength of chemical bond. The paired electron of valence shell

of an atom can take part in covalent

bonding subject to availability of vacant orbitals of slightly higher energy of the same main energy shell and availability of energy for unpairing of paired electron and their shifting to vacant orbitals. This point explains the trivalency of boron, tetravalency of carbon, pentavalency of phosphorous hexavalency of S and hepta valency of Cl, Br, I. Depending on type of overlapping atomic orbitals covalent bond can be classified into two types 1.



Sigma

(s)

2.

Pi

(p)

bond

Sigma and Pi Bonding

When two hydrogen atoms form a bond, their atomic orbitals overlap to produce a greater density of electron cloud along the line connecting the two nuclei. In the simplified representations of the formation of H 2O and NH3 molecules, the O—H and N—H bonds are also formed in a similar manner, the bonding electron cloud having its maximum density on the lines connecting the two nuclei. Such bonds are called sigma bonds (σbond).

A covalent bond established between two atoms having the maximum density of the electron cloud along the line connecting the centre of the bonded atoms is called a σ-bond. A σ-bond is thus said to possess a cylindrical symmetry along the internuclear axis. Let us now consider the combination of two nitrogen atoms. Of the three singly occupied p-orbitals in each, only one p-orbital from each nitrogen (say, the px may undergo “head –on” overlap to form a s-bond. The other two p-orbitals on each can no longer enter into a direct overlap. But each p-orbital may undergo lateral overlap with the corresponding p-orbital on the neighbour atom. Thus we have two additional overlaps, one by the two py orbitals, and the other by the two p z orbitals. These overlaps are different from the type of overlap in as-bond. For each set of p-orbitals, the overlap results in accumulation of charge cloud on two sides of the internuclear axis. The bonding electron cloud does no more posses an axial symmetry as with the s-bond; instead, it possess a plane of symmetry. For the overlap of the p z atomic orbital, the xy plane provides this plane of symmetry; for the overlap of the p yatomic orbitals, the zx plane serves the purpose. Bonds arising out of such orientation of the bonding electron cloud are designated as π-bonds. The bond formed by lateral overlap of two atomic orbitals having maximum overlapping on both sides of the line connecting the centres of the atoms is called a π-bond. A π-bond possess a plane of symmetry, often referred to as the nodal plane.

σ-Bond : When covalent bond is formed by overlapping of atomic orbitals along the same axis it is called s - bond. Such type of bond is symmetrical about the line joining the two nuclei e.g.

(a) s-s overlapping

(b) s-p overlapping

(c) p-p overlapping

π - Bond: This type of bond is formed by the sidewise or lateral overlapping of two half filled atomic orbitals.

|The strength of a bond depends upon the extent of overlapping of halffilled atomic orbitals. The extent of overlapping is between two atoms is always greater when there is end to end overlapping of orbitals than, when there is sidewise overlapping of oritals. Hence s-bond is always stronger than p-bond. The average distance between the nuclei of the two bonded atoms in a molecule is called bond length and the energy required to break one mole of bonds of particular type in gaseous state is called Bond energy or Bond strength. The same amount of energy is released in formation of one mol of particular bond. Limitation: VBT cannot explain the paramagnetic properties of B2,O2 etc.



Co-ordinate Covalency

A covalent bond results from the sharing of pair of electrons between two atoms where each atom contributes one electron to the bond. It is also possible to have an electron pair bond where both electrons originate from one atom and none from the other. Such bonds are called coordinate bond or dative bonds. Since in coordinate bonds two electrons are shared

by two atoms, they differ from normal covalent-bond only in the way they are formed and once formed they are identical to normal covalent –bond. It is represented as [——→] Atom/ion/molecule donating electron pair is called Donor or Lewis base. Atom / ion / molecule accepting electron pair is called Acceptor or Lewis acid, [——→] points donor to acceptor NH4+, NH3 has three (N – H) bond & one lone pair on N – atom. In NH4+ formation this lone pair is donated to H + (having no electron) NH3 + H+ → NH4+

Properties of the coordinate compounds are intermediates of ionic and covalent



compounds.

Conditions for Formation of Covalent Bond 1.

The electronegativity deference between the two atoms should be less.

2.

The covalent bond is formed between the two non metals.

3.

Each combining atom must contribute at least one electron to the shared pair.

4.

The combining atoms should attain the noble gas configuration after bond formation.



Polar and Non-Polar Covalent Bonds

Covalent bonds can be classified into following two groups depending on the electronegativity difference between the bonded atoms..

1.

Polar covalent bond.

2.

Nonpolar covalent bond.

Polar covalent bond is formed between two atoms which have large difference in electronegativity. The electronegativity difference disturbs the distribution of shared pair of electrons between the two atoms as the electron density would be more toward the element which is more electronegative.

This

will

develop

partial

positive

charge

on

more

electronegative element and partial positive charge on less electronegative one. For example, bond between H and F would be polar covalent bond.

Non-polar covalent bonds are formed between two like atoms i.e. the atoms which have almost same electronegativity. Due to almost same electronegativity, both atoms attract electron pair equally and no charge appears on any atom, and the whole molecule becomes neutral. For example bond between two H atoms would be non-polar.



Maximum Covalency

Elements

which

have

vacant

d-orbital

can

expand

their

octet

by

transferring electrons, which arise after unpairing, to these vacant dorbital e.g. in sulphur.

In excited state sulphur has six unpaired electrons and shows a valency of six e.g. in SF6. Thus an element can show a maximum covalency equal to its group number e.g. chlorine shows maximum covalency of seven.



Dipole Moment

Difference in polarities of bonds is expressed on a numerical scale. The polarity of a molecule is indicated in terms of dipole moment (μ). To measure dipole moment, a sample of the substance is placed between two electrically charged plates. Polar molecules orient themselves in the electric field causing the measured voltage between the plates to change. The dipole moment is defined as the product of the distance separating charges of equal magnitude and opposite sign, with the magnitude of the charge. The distance between the positive and negative centres called the bond length. Thus, = μ = electric charge × bond length = q × d As q is in the order of 10–10 esu and d is in the order of 10–8 cm, μ is the order of10–18 esu cm. Dipole moment is measured in ‘Debye’ unit (D) 1D = 10–18 es cm = 3.33 × 10–30 coulomb metre Note: 1.

Generally

as

electronegativity

difference

increase

in

diatomic

molecules, polarity of bond between the atoms increases therefore value of dipole moment increases.

2.

Dipole moment is a vector quantity.

3.

A symmetrical molecule is non-polar even though it contains polar bonds. For example, CO2, BF3, CCl4 etc. because summation of all

bond

moments

present

in

the

molecules

cancel

each

other. 4.

Unsymmetrical non-linear polyatomic molecules have net value of dipole moment. For example, H2O, CH3OH, NH3 etc.?

Calculation of Resultant Bond Moments

Let AB and AC are two polar bonds inclined at an angle θ their dipole moments are μ1 and μ2. Resultant dipole moment may be calculated using vectorial method. μ = √μ12 + μ22 + 2μ1μ2 cos θ

when θ = 0 the resultant is maximum μR = μ1 + μ2 when, θ = 180°, the resultant is minimum μR = μ1 ∼ μ2 For example, CO2 has got dipole moment of zero. The structure of CO2 is.This is a highly symmetrical structure with a plane of symmetry passing through the carbon. The bond dipole of C–O is directed towards oxygen as it is the negative end. Here two equal dipoles acting in opposite direction

cancel

each

other

and

therefore

the

dipole

moment

is

zero. Similarly dipole moment of CCl4 is zero while that of CHCl 3 is non zero. . Explanation is

again

in

geometry

of

the

molecules both

CCl4 &

CHCl3 have tetrahedral structure but CCl 4 is symmetrical while CHCl3 is non-symmetrical.

Due to the symmetrical structure of CCl4 the resultant of bond dipoles comes out to be zero. But in case of CHCl3 it is not possible as the presence of hydrogen introduces some dissymmetry.

Bond Length

The distance between the nuclei of two atoms bonded together is termed as bond length or bond distance. It is expressed in angstrom (Å) units or picometer (pm). [1Å = 10–8 cm; 1 pm = 10–12 m] Bond length in ionic compound = rc+ + ra– Similarly, in a covalent compound, bond length is obtained by adding up the covalent (atomic) radii of two bonded atoms. Bond length in covalent compound (AB) = rA + rB The factors such as resonance, electronegativity, hybridization, steric effects, etc., which affect the radii of atoms, also apply to bond lengths.

Important Features of Bond Length 

The bond length of the homonuclear diatomic molecules are twice the covalent radii.



The lengths of double bonds are less than the lengths of single bonds between the same two atoms, and triple bonds are even shorter than double bonds.Single bond > Double bond > Triple bond (decreasing bond length)



Bond length decreases with increase in s-character since s-orbital is smaller than a p – orbital.



sp3 C – H = 1.112Å: = 1.08Å;

sp2 C – H = 1.103Å;

sp C – H



(25% s-character as in alkanes) (33.3% s-character as in alkenes) (50% s-character as in alkynes)



Bond length of polar bond is smaller than the theoretical non-polar bond length.

Bond Energy or Bond Strength Bond energy or bond strength is defined as the amount of energy required to break a bond in molecule. Important features of bond energy: 

The magnitude of the bond energy depends on the type of bonding. Most of the covalent bonds have energy between 50 to 100 kcal mol– 1

(200-400 kJ mol–1). Strength of sigma bond is more than that of a π-

bond. 

A double bond in a diatomic molecules has a higher bond energy than a single bond and a triple bond has a higher bond energy than a double bond between the same atoms. C ≡ C > C = C > C – C (decreasing bond length)



The magnitude of the bond energy depends on the size of the atoms forming the bond, i.e. bond length. Shorter the bond length, higher is the bond energy.



Resonance in the molecule affects the bond energy.



The bond energy decreases with increase in number of lone pairs on the bonded atom. This is due to electrostatic repulsion of lone pairs of electrons of the two bonded atoms.



Homolytic and heterolytic fission

involve different amounts

of

energies. Generally the values are low for homolytic fission of the bond in comparison to heterolytic fission. 

Bond energy decreases down the group in case of similar molecules.



Bond energy increase in the following order:

s < p < sp < sp2 < sp3 C – C pair)

(One lone pair)

>

N – N

>

O – O

(No lone

(Two lone pair)

Bond Angles Angle between two adjacent bonds at an atom in a molecule made up of three or more atoms is known as the bond angle.

Bond angles mainly depend on the following three factors: (i) Hybridization: Bond angle depends on the state of hybridization of the central atom Hybridization

Bond angle

Example

sp3

109o28'

CH4

sp2

120o

BCl3

sp

180o

BeCl2

Generally s- character increase in the hybrid bond, the bond angle increases. (ii) Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle. (iii) Electronegativity: If

the

electronegativity

of

the

central

atom

decreases, bond angle decreases.

Resonance There may be many molecules and ions for which it is not possible to draw a single Lewis structure. For structures of O3.

example we can write two electronic

In (A) the oxygen - oxygen bond on the left is a double bond and the oxygen-oxygen bond on the right is a single bond. In B the situation is just opposite. Experiment shows however, that the two bonds are identical. Therefore neither structure A nor B can be correct. One of the bonding pairs in ozone is spread over the region of all the three atom rather than associated with particular oxygen-oxygen bond. This delocalised bonding is a type of bonding in which bonding pair of electrons is spread over a number of atoms rather than localised between two.

Structures (A) and (B) are called resonating or canonical structures and C is the resonance hybrid. This phenomenon is called resonance, a situation in which more than one plausible structure can be written for a species. Atoms gain or lose electrons to attain a more stable noble gas - like electron configuration (octet rule). There are two ways in which atoms can share electrons to satisfy the octet rule: Ionic Bonding - occurs when two or more ions combine to form an electrically-neutral compound The positive cation "loses" an electron (or 2 or 3) The negative anion "gains" the electron (or 2 or 3) The anion steals the electrons from the cation. Covalent Bonding - occurs when two or more atoms combine to form an electrically-neutral compound The electrons are shared between the two atoms. Both atoms don't have charge in the beginning and the compound remains with zero charge. The chemical activity of an atom is determined by the number of electrons in its valence shell. With the help of concept of chemical bonding one can define the structure of a compound and is used in many industries for manufacturing products in which the true structure cannot be written at all. Some other examples (i) CO32– ion

Example (ii) Carbon-oxygen bond lengths in carboxylate ion are equal due to resonance.

(iii) Benzene

(iv) Vinyl Chloride

Difference in the energies of the canonical forms and resonance hybrid is called resonance stabilization energy and provides stability to species.

Rules for writing Resonating Structures Only electrons (not atoms) may be shifted and they may be shifted only to adjacent atoms or bond positions. The number of unpaired electrons should be same in all the canonical form. The positive charge should reside as far as possible on less electronegative atom and negative charge on more electronegative atom. Like charge should not reside on adjacent atom The larger the number of the resonating structures greater the stability of species. Greater number of covalency adds to the stability of the molecule. Example: Out of the following resonating structures for CO 2 molecule, which are important for describing the bonding in the molecule and why?

Solution: Out of the structures listed above, the structure (III) is wrong since the number of electron pairs on oxygen atoms are not permissible. Similarly, the structures (II) has very little contribution towards the hybrid because one of the oxygen atoms (electronegative) is show to have

positive charge. Carbon dioxide is best represented by structures (I) and (IV). The mixing or merging of dissimilar orbitals of similar energies to form new orbitals is known as hybridization and the new orbitals formed are known as hybrid orbitals.

Important Characteristics of Hybridization? 1.

Orbitals belonging to the same atom or ion having similar energies get hybridized.

2.

Number of hybrid orbitals is equal to the no. of orbitals taking part in hybridization.

3.

The hybrid orbitals are always equivalent in energy and shape.

4.

The hybrid orbitals form more stable bond than the pure atom orbital.

5.

The reason hybridization takes place is to produce equivalent orbitals which give maximum symmetry.

6.

It is not known whether actually hybridization takes place or not. It is a concept which explains the known behaviour of molecules.

7.

The hybrid orbitals are directed in space in same preferred direction to have some stable arrangement and giving suitable geometry to the molecule.

Depending upon the different combination of s and p orbitals, these types of hybridization are known. sp3 hybridization: In this case, one s and three p orbitals hybridise to form four sp3 hybrid orbitals. These four sp3 hybrid orbitals are oriented in a tetrahedral arrangement.

sp2 hybridization: In this case one s and two p orbitals mix together to form three sp2 hybrid orbitals and are oriented in a trigonal planar geometry.

The remaining p orbital if required form side ways overlapping with the other unhybridized p orbital of other C atom and leads to formation of p2C = CH2 bond as in H sp hybridization: In this case, one s and one p orbital mix together to form two sp hybrid orbitals and are oriented in a linear shape.

The

remaining

two

unhybridised

p

orbitals

overlap

with

another

unhybridised p orbital leading to the formation of triple bond as in HC

hape

Hybridisation

inear

sp

rigonal planar

sp2

etrahedral

sp3

rigonal bipyramidal

sp3d

ctahedral

sp3d2

entagonal bipyramidal

sp3d3

CH.

Rule for Determination of total Number of Hybrid Orbitals Detect the central atom along with the peripheral atoms. Count the valence electrons of the central atom and the peripheral atoms. Divide the above value by 8. Then the quotient gives the number of s bonds and the

remainder gives the non-bonded electrons. So number

of lone pair = non bonded electrons/2 . The number of s bonds and the lone pair gives the total number of hybrid orbitals. An Example Will Make This Method Clear:- SF 4 Central atom S, Peripheral atom F ∴ total number of valence electrons = 6 + (4 × 7) = 34 Now 34/8 = 4 2/8 ∴ Number of hybrid orbitals = 4σ bonds + 1 lone pair)

So, five hybrid orbitals are necessary and hybridization mode is sp 3d and it is trigonal bipyramidal (TBP). Note:Whenever there are lone pairs in TBP geometry they should be placed in equatorial position so that repulsion is minimum.

Compound and its Hybridization

Molecular geometry

1. NCl3 Total valence electrons = 26 Requirement = 3σ bonds + 1 lone pair Hybridization = sp3 Shape = pyramidal 2. BBr3 Total valence electron = 24 Requirement = 3σ bonds Hybridization = sp2 Shape = planar trigonal 3. SiCl4 Total valence electrons = 32 Requirement = 4σ bonds Hybridization = sp3 Shape = Tetrahedral 4. CI4 Total valence electron = 32 Requirements = 4σ bonds Hybridization = sp3 Shape = Tetrahedral 5. SF6 Total valence electrons = 48 Requirement = 6σ bonds Hybridization = sp3d2 Shape = octahedral/square bipyramidal 6. BeF2 Total valence electrons : 16

F – Be – F

Requirement : 2σ bonds Hybridization : sp Shape : Linear 7. ClF3 Total valence electrons : 28 Requirement: 3σ bonds + 2 lone pairs Hybridization : sp3d Shape : T – shaped 8. PF5 Total valence electrons : 40 Requirement : 5σ bonds Hybridization : sp3d Shape : Trigonal bipyramidal (TBP) 9. XeF4 Total valence electrons : 36 Requirement:4σ bonds+ 2 lone pairs Hybridisation : sp3d Shape : Square planar

Here three arrangements are possible out of which A and B ar

be inter converted by simple rotation of molecule. The basic d

(C) is that in (B) the lone pair is present in the anti position wh

repulsion which is not possible in structure (C) where the lone

So in a octahedral structure the lone pairs must be placed at th minimize repulsion. So both structure (A) and (B) are correct. 10. XeF2 Total valence electrons : 22 Requirements : 2σ bonds + 3 lone pairs Hybridisation: sp3d Shape : Linear 11. Total valence electrons : 32 Requirement : 4σbonds Hybridisation: sp3 Shape: tetrahedral

Here all the structures drawn are resonating structures with O– double bonded oxygen.

12. NO2– Total valence electron: 18 Requirement : 2σ bonds + 1 lone pair Hybridisation: sp2 Shape: angular 13. CO32– Total valence electrons: 24

But C has 4 valence electrons of these 3 form s bonds \ the res

Requirement = 3σ bonds Hybrdisation = sp2 Shape: planar trigonal

In the structure one bond is a double bond and the other 2 are s

of the double bonds keeps changing in the figure. Since periph

isovalent, so contribution of the resonanting structures are equ

that none of the bonds are actually single or double. The actua

14. CO2 Total valence electrons:16 Requirement: 2σ bonds Hybridisation: sp Shape: linear 15. Total valence electrons = 32 Requirement= 4σ bonds Hybridisation: sp3 Shape: Tetrahedral 16. Total valence electron = 26 Requirement = 3σ bond + 1 lone pair Hybridization: sp3 Shape: pyramidal

O=C=O

17. XeO2F2 Total valence electrons : 34 Requirement: 4σ bonds +1 lone pairs Hybridization : sp3d Shape: Distorted TBP (sea-saw geometry) 18. XeO3 Total valence electrons : 26 Requirement: 3σ bonds + 1 lone pair Hybridization: sp3 Shape: Pyramidal 19. XeOF4 Total valence electrons : 42 Requirement: 5σ bonds + 1 lone pair Hybridization: sp3d2 Shape: square pyramidal. 20. PF2Br3 Total valence electrons : 40 Requirements : 5σ bonds Hybridisation: sp3d Shape : trigonal bipyramidal

Regular Geometry of Molecules Geometry of the molecules in which the central atom has no lone pairs are regular and can be predicted simply. Bond ange of any molecule with regular geometry = 360 o /Number of bond pairs

Number of electron pairs

Arrangement of electrons

Molecular geometry

Examples

B – A– B

2

Linear

3

BeCl2, HgCl2

BF3, AlCl3

θ = 120°

θ = 120°

4

CH4, NH4+, SiF4

5

PCl5, PF5

6

SF6

Irregular Geometry of Molecules and VSPER Theory Geometry of the molecules having predicted

simpley

using

above

lone pair of electrons can not be mentioned

method. The

geometric

arrangement of atoms in molecules and ions may be predicted by means of the

valence-shell

electron-pair

repulsion

(VSEPR)

theory.

This

type

includes molecules which may or may not obey the octet rule but have only single bonds. Postulates of VSEPR theory: 

The shape of the molecule is determined by repulsions between all of the electron pairs present in the valence shell.



A lone pair of electrons takes up more space around the central atom than a bond-pair, since the lone pair is attracted to one nucleus whilst the bond pair is shared by two nuclei. It follows that repulsion between two lone pairs is greater than repulsion between a lone pair and a bond pair, which in turn is greater than the repulsion between two bond pairs. Thus the presence of lone pairs on the central atom causes slight distortion of the bond angles from the ideal shape. If the angle between a lone pair, the central atom and a bond pair is increased, it follows that the actual bond angle between the atoms must be decreased.The descending order of repulsion is



(lp – lp) > (lp – bp) > (bp – bp)



where lp-Lone pair; bp-bond pair



The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms.



Double bonds cause more repulsion than single bonds and triple bonds cause more repulsion than double bonds.



A brief summary of molecular shapes resulting from different configurations of electrons pairs is presented below:



With very few exceptions, the predictions based on the VSEPR theory have been shown to be correct.

No. Arrangemen No. of of Shape Molecul t of Example Bondin lon (Geometr e Type electrons s g pairs e y) pairs pair AB2E

2

1

Bent

SO2, O3

AB3E

3

1

Trigonal pyramidal

NH3

AB2E2

2

2

Bent

H2O

See saw

SF4

T – shaped

CIF3

Square pyramidal

BrF5

AB4E

4

1

AB3E2

3

2

AB5E

5

1

AB4E2

4

2

Square planar

XeF4

To find the shape of a molecule follow the steps given below: 

Identify the central atom and count the number of valence electrons.



Add to this, number of other atoms.



If it is an ion, add negative charges and subtract positive charges. Call it total N



Divide N by 2 and compare the result with chart I and obtain the shape.

Shape of molecule or ion

Total N/2

Example

2

Linear

HgCl2/BeCl2

3

Triangular planar

BF3

3

Angular

SnCl2, NO2

4

Tetrahedral

CH4, BF4-

4

Trigonal Pyramidal

NH3, PCl3

4

Angular

H2O

5

Trigonal bipyramidal

PCl5, PF5

5

Irregular tetrahedral

SF4, IF4+

5

T-shaped

CIF3, BrF3

5

Linear

XeF2, I3-

6

Octahedral

SF6, PF6-

6

Square Pyramidal

IF5

6

Square planar

XeF4, ICI4

Introduction to Molecular Orbital Theory One of the most important theories developed is the wave-particle, duality of particles. Electrons can be considered as particles and waves also. Based on this, it can be concluded that electrons behaving as waves can interact with each other and the process is called interference. As in waves, two types of interferences are possible: (1) constructive and (2) Destructive.

In molecules the atomic orbitals of all the atoms are assumed to interfere with each other in the form of waves and depending on the nature of interferences, two molecular orbitals result. The one which results from constructive interference is called bonding molecular orbital and the one which

results

from

destructive

interference

is

called

anti-bonding

molecular orbital. Obviously anti-bonding MO is of higher energy than Bonding MO. In Molecular Orbital Theory (MOT) the atoms in a molecule are supposed to loose their individual control over the electrons. The nuclei of the bonded atoms are considered to be present at equilibrium inter-nuclear positions. The orbitals where the probability of finding the electrons is maximum are multicentred orbitals called molecular orbitals extending over two or more nuclei. In MOT the atomic orbitals loose their identity and the total number of electrons present are placed in Mo’s according to increasing energy sequence (Auf Bau Principle) with due reference to Pauli’s Exclusion Principle and Hund’s Rule of Maximum Multiplicity. As mentioned above, when a pair of atomic orbitals combine they give rise to a pair of molecular orbitals, the bonding and the anti-bonding. The number of molecular orbitals produced must always be equal to the number of atomic orbitals involved. Electron density is increased for the bonding MO’s in the inter-nuclear region but decreased for the antibonding MO’s, Shielding of the nuclei by increased electron density in bonding MO’s reduces inter nuclei repulsion and thus stabilizes the molecule whereas lower electron density even as compared to the individual

atom

in

anti-bonding

MO’s

increases

the

repulsion

and

destabilizes the system. In simple homonuclear diatomic molecules the order of MO's based on increasing energy is :

For molecules including O2 and above, the order is

This order is true except B2, C2 & N2. If the molecule contains unpaired electrons in MO’s it will be paramagnetic but if all the electrons are paired up then the molecule will be diamagnetic.

Bond Order Bond order is a number which indicates the no. of bonds a molecule possesses and the stability of the molecule in comparison to another. An integral value implies that so many bonds exist in the molecule. Anything fractional indicates that the bond is intermediate. Bond-order = 1/2 (no. of bonding electrons - No. of antibonding electrons).

Application of MOT to Homonuclear Diatomic Molecules H2+ molecule ion. Total no. of electrons = 1 

Arrangement : s 1s1



An

unpaired

electron

always

indicates

that

the

molecule

is

paramagnetic. H2 molecule, Total no. of electrons = 2 

Arrangement : s 1s2



Paired electrons , so diamagnetic.



BO = 1/2 (2-0) = 1



Therefore No. of bonds = 1

He2+ molecule ion, Total no. of electrons = 3, 

Arrangement : s1s2, s* 1s1



Unpaired electron, so paramagnetic



BO = 1/2 (2-1) = 1/2



Bond existing by virtue of a single electron.

He2 molecule. Total No. of electrons = 4, 

Arrangement : s1s2, s* 1s2



Diamagnetic BO = 0



Molecule does not exist.

Example: Which diatomic molecule of second period besides O2 should be paramagnetic? Solution:

As, paramagnetism arises due to unpaired electron. Therefore B 2is paramagnetic molecule.

M.O. of Some Diatomic Heteronuclei Molecules The molecular orbitals of heteronuclei diatomic molecules should differ from those of homonuclei species because of unequal contribution from the participating atomic orbitals. Let’s take the example of CO. The M.O. energy level diagram for CO should be similar to that of the isoelectronic molecule N2. But C & O differ much in electronegativity and so will their corresponding atomic orbitals. But the actual MO for this species is very much complicated since it involves a hybridisation approach between the orbital of oxygen and carbon. HCl Molecule: Combination between the hydrogen 1s A.O’s. and the chlorine 1s, 2s, 2p & 3s orbitals can be ruled out because their energies are too low. The combination of H 1s1 and 3p1x gives both bonding and antibonding orbitals, and the 2 electrons occupy the bonding M.O. leaving the anti-bonding MO empty. NO Molecule: The M.O. of NO is also quite complicated due to energy difference of the atomic orbitals of N and O. As the M.O.’s of the heteronuclei species are quite complicated, so we should concentrate in knowing the bond order and the magnetic behaviour.

Molecules/Ions

Total No. of Electrons

Magnetic Behaviour

CO

14

Diamagnetic

NO

15

Paramagnetic

NO+

14

Diamagnetic

NO–

16

Diamagnetic

CN

13

Paramagnetic

CN–

14

Diamagnetic

Inert Pair Effect Heavier p-block and d-block elements show two oxidation states. One is equal to group number and second is group number minus two. For example Pb(5s25p2) shows two OS, +II and +IV. Here +II is more stable than +IV which arises after loss of all four valence electrons. Reason given for more stability of +II O.S. that 5s 2 electrons are reluctant to participate in chemical bonding because bond energy released after the bond

formation is less than that required to unpair these electrons (lead forms a weak covalent bond because of greater bond length). Example: Why does PbI4 not exist? Solution: Pb(+IV) is less table than Pb(+II) due to inert pair effect and therefore Pb(+IV) is reduced to Pb(+II) by I – which changes to I2(I– is a good reducing agent)

Covalent Character in Ionic Compounds Fajan's Rule Although atomic bond in a compound like M+X- is considered to be 100% ionic, actually it also has some covalent character. An explanation for the partial covalent character of an ionic bond has been given by Fajan. According to Fajan, if two oppositely charged ions are brought together, the nature of the bond between them depends upon the effect of one ion on the other. When two oppositely charged ions (say A+ and B- ) approach each other the positive ion attracts electrons on the outermost shell of the anion and repels

its

positively

charged

nucleus.

This

results

in

the

distortion,deformation or polarization of the anion. If the polarization is quite small, an ionic bond is formed, while if the degree of polarization is large, a covalent bond results. Thus the power of an ion (cation) to distort the other ion is known as its polarization power and the tendency of the ion(anion) to get polarized by the other ion is known as its polarisability. Greater the polarization power or polarisability of an ion, greater will be its tendency to form a covalent bond. The polarising power, or polarisability and hence formation of covalent bond is favoured by the following factors: 

Small Positive Ion (Cation): Due to greater concentration of positive charge on a small area, the smaller cation has high polarising power. This explains why LiCl is more covalent than KCl.



Large Negative Ion (Anion): The larger the anion, the greater is its polarisability, i.e. susceptibility to get polarised. It is due to the fact that the outer electrons of a large anion are loosely held and hence can be more easily pulled out by the cation. This explains why iodides, among halides, are most covalent in nature.



Large Charge on Either of the Two Ions: As the charge on the ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases, with the result its ability for

forming the covalent bond increases. Thus covalency increases in the order : Na+ Cl-, Mg2+ (Cl2)2-, Al3+ (Cl3)3 

Electronic Configuration of the Cation : For the two ions of the same size and charge, one with a pseudo noble gas configuration (i.e., 18 electrons in outer-most shell) than a cation with noble gas configuration (i.e. 8 electrons in outermost shell) will be more polarising. Thus copper (I) chloride is more covalent than sodium chloride although

Cu+ ion

(0.96A°) and Na+ ion (0.95A°) have same size and charge. The orbital overlapping involved in covalency reduces, the charge on each ion and so weakens the electrovalent forces throughout the solid, as is evident from the melting point of lithium halides. LiF = 870°C

LiCl = 613°C

LiBr = 547°C

LiI

From

the

above

= 446°C discussion,

we

find

that greater

the

possibility

of

polarisation, lower is the melting point and heat of sublimation and greater is the solubility in non-polar solvents. Example: The melting point of KCl is higher than that of AgCl though the crystal radii of Ag+ and K+ ions are almost the same. Solution : Now whenever any comparison is asked about the melting point of the compounds which are fully ionic from the electron transfer concept it means that the compound having lower melting point has got lesser amount of ionic character than the other one. To analyse such a question first find out the difference between the 2 given compounds. Here in both the compounds the anion is the same. So the deciding factor would be the cation. Now if the anion is different, then the answer should be from the variation of the anion. Now in the above example, the difference of the cation is their electronic configuration. K + = [Ar]; Ag+ = [Kr] 4d10. This is now a comparison between a noble gas core and pseudo noble gas core, the analysis of which we have already done.

So try to finish off this

answer.

Percentage of Ionic Character Every ionic compound having some percentage of covalent character according to Fajan’s rule. The percentage of ionic character in a compound having some covalent character can be calculated by the following equation. The percent ionic character = Observed dipole moment/Calculated dipole moment assuming 100% ionic bond × 100

Example: Dipole moment of KCl is 3.336 × 10–29 coulomb metre which indicates that it is highly polar molecule. The interatomic distance between k+ and Cl– is 2.6 ×10–10 m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl. Solution: Dipole moment μ = e × d coulomb metre For KCl d = 2.6 × 10–10 m For complete separation of unit charge e = 1.602 × 10–19 C Hence μ = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 Cm μKCl = 3.336 × 10–29 Cm ∴ % ionic character of KCl = 3.336×10–29/4.165×10–29 = 80.09% Example. Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as it’s observed dipole moment. Solution: Tocalculate μ considering 100% ionic bond = 4.8 × 10–10 × 0.83 × 10–8esu cm = 4.8 × 0.83 × 10–18 esu cm = 3.984 D ∴ % ionic character = 1.82/3.984 × 100 = 45.68 The example given above is of a very familiar compound called HF. The % ionic character is nearly 43.25%, so the % covalent character is (100 – 43.25) = 56.75%. But from the octet rule HF should have been a purely covalent compound but actually it has some amount of ionic character in it, which is due to the electronegativity difference of H and F. Similarly knowing the bond length and observed dipole moment of HCl, the % ionic character can be known. It was found that HCl has 17% ionic character. Thus it can be clearly seen that although we call HCl and HF as covalent compounds but it has got appreciable amount of ionic character. So from now onwards we should call a compound having more of ionic less of covalent and vice versa rather than fully ionic or covalent.

Metallic Bonding Metals are characterised by bright, lustre, high electrical and thermal conductivity, malleability, ductility and high tensile strength. A metallic crystal consists of very large number of atoms arranged in a regular pattern. Different model have been proposed to explain the nature of metallic bonding two most important modules are as follows

The electron sea Model In this model a metal is assumed to consist of a lattice of positive ion (or kernels) immersed in a sea of mobile valence electrons, which move freely within the boundaries of a crystal. A positive kernel consists of the nucleus of the atom together with its core on a kernel is, therefore, equal in magnitude to the total valence electronic charge per atom. The free electrons shield the positively charged ion cores from mutual electrostatic repulsive forces which they would otherwise exert upon one another. In a way these free electrons act as ‘glue’ to hold the ion cores together. The forces that hold the atoms together in a metal as a result of the attraction between positive ions and surrounding freely mobile electrons are known as metallic bonds. Through the electron sea predated quantum mechanics it still satisfactorily explains certain properties of the metals. The electrical and thermal conductivity of metals for example, can be explained by the presence of mobile electrons in metals. On applying an electron field, these mobile electrons conduct electricity throughout the metals from one end to other. Similarly, if one part of metal is heated, the mobile electrons in the part of the metals acquire a large amount of kinetic energy. Being free and mobile, these electrons move rapidly throughout the metal and conduct heat to the other part of the metal. On the whole this model is not satisfactory.

Hydrogen Bonding An atom of hydrogen linked covalently to a strongly electronegative atom can establish an extra weak attachment to another electronegative atom in the same or different molecules. This attachment is called a hydrogen bond. To distinguish from a normal covalent bond, a hydrogen bond is represented by a broken line eg X – H …Y where X & Y are two electronegative atoms. The strength of hydrogen bond is quite low about 2-10 kcal mol–1 or 8.4–42 kJ mol–1 as compared to a covalent bond strength 50–100 kcal mol–1 or 209 –419 kJ mol–1

Conditions for Hydrogen Bonding

Hydrogen should be linked to a highly electronegative element. The size of the electronegative element must be small. These two criteria are fulfilled by F, O, and N in the periodic table. Greater the electronegativity and smaller the size, the stronger is the hydrogen bond which is evident from the relative order of energies of hydrogen bonds.

Types of Hydrogen Bonding Intermolecular hydrogen bonding:This type of bonding takes place between two molecules of the same or different types. For example, H H H | | | O—H—O—H—O—H— Intermolecular hydrogen bonding leads to molecular association in liquids like water etc. Thus in water only a few percent of the water molecules appear not to be hydrogen bonded even at 90°C. Breaking of those hydrogen bonds throughout the entire liquid requires appreciable heat energy. This is indicated in the relatively higher boiling points of hydrogen bonded liquids. Crystalline hydrogen fluoride consists of the polymer (HF) n. This has a zig-zag chain structure involving H-bond.

Intramolecular hydrogen bonding: This type of bonding occurs between atoms of the same molecule present on different sites. Intramolecular hydrogen bonding gives rise to a closed ring structure for which the term chelation is sometimes used. Examples are o-nitrophenol, salicylaldehyde.

Importance of Hydrogen Bonding in Biological Systems Hydrogen bonding plays a vital role in physiological systems. Proteins contain chains of amino acids. The amino acid units are arranged in a spiral form somewhat like a stretched coil spring (forming a helix). The N-H group of each amino acid unit and the fourth C=O group following it along the chain, establishes the N–H---O hydrogen bonds. These bonds are partly

responsible for the stability of the spiral structure. Double helix structure of DNA also consists of two strands forming a double helix and are joined to each other through hydrogen bond.

Effect of Hydrogen Bonding Hydrogen bonding has got a very pronounced effects on certain properties of the molecules. They have got effects on 

State of the substance



Solubility of the substance



Boiling point



Acidity of different isomers

These can be evident from the following examples. Example. H2O is a liquid at ordinary temperature while H 2S is a gas although both O and S belong to the same group of the periodic table. Solution: H2O is capable of forming intermolecular hydrogen bonds. This is possible due to high electronegativity and small size of oxygen. Due to intermolecular H-bonding, molecular association takes place. As a result the effective molecular weight increases and hence the boiling point increases. So H2O is a liquid. But in H 2S no hydrogen bonding is possible due to large size and less electronegativity of S. So it’s boiling point is equal to that of an isolated H2S molecule and therefore it is a gas. Example.Ethyl alcohol (C2H5OH) has got a higher boiling point than dimethyl ether (CH3-O-CH3) although the molecular weight of both are same. Solution: Though ethyl alcohol and dimethyl ether have the same molecular weight but in ethyl alcohol the hydrogen of the O-H groups forms intermolecular hydrogen bonding with the OH group in another molecule. But in case of ether the hydrogen is linked to C is not so electronegative to encourage the hydrogen to from hydrogen bonding. —H—O—H—O—H—O— | | | C2 H 5 C2H5 C2H5 Due to intermolecular H-bonding, ethyl alcohol remains in the associated form and therefore boils at a higher temperature compared to dimethyl ether.

Solved Problems of Chemical Bonding

Prob 1. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3 Solution: N2 < ClF3 < SO2 < K2O < LiF Prob 2. Arrange the following in order of increasing ionic character: C-H, F-H, Br-H, Na-I, K-F and Li-Cl Solution: C-H < Br-H < F-H < Na-I < Li-Cl < K-F Prob 3. Predict the shapes of the following ions (a) BeF3-

(b) BF4-

(c) IF4-

(d) IBr2-

Solution: (a) Triangular (b) Tetrahedral (c) Square planar (d) Linear Prob 4. Arrange the following in increasing order of stability O 2, O2+, O2-, O22Solution: O22- < O2- < O2 < O2+ Calculate first the bond order which is as follows O2 → 2,

O2+ → 2.5,

O2- → 1.5,

O22- → 1

&

then

arrange

according

to

increasing bond order. Prob 5. Arrange the following: (i) N2, O2, F2, Cl2 in increasing order of bond dissociation energy. (ii) Increasing strength of hydrogen bonding (X – H – X): O, S, F, Cl, N Solution: (i) F2 < Cl2 < O2 < N2 (ii) Cl < S < N < O < F Prob 6. Explain the following o - hydroxy benzaldehyde is liquid at room temperature while p - hydroxy benzaldehyde is high melting solid. Solution: There is intramolecular H bonding in o - hydroxy benzaldehyde while intermolecular hydrogen bonding in p-hydroxy benzaldehyde

Prob 7. Explain why ClF2- is linear but ClF2+ is a bent molecular ion? Solution: Chlorine atom lies in sp3d hybrid state. Three lone pairs are oriented along the corners of triangular plane

Chlorine atom lies in sp3 hybrid state. Two lone pairs are oriented along two corners of tetrahedral

Prob 8. AlF3 is ionic while AlCl3 is covalent. Solution: Since F– is smaller in size, its polarisability is less and therefore it is having more ionic character. Whereas Cl being large in size is having more polarisability and hence more covalent character. Prob 9. Write down the resonance structure of nitrous oxide

Solution: Prob 10. Explain why BeH2 molecule has zero dipole moment although the Be-H bonds are

polar.

Solution: BeH2 is a linear molecule (H-Be-H) with bond angle equal to 180 o. Although the Be-H bonds are polar on account of electronegativity difference between Be and H atoms, the bond polarities cancel with each other. The molecule has resultant dipole moment of zero. Prob 11. Why the bond angle of H – C – H in methane (CH 4) is 109° 28’ while H – N – H bond angle in NH 3 is 107° though both carbon and nitrogen are sp3 hybridized Solution: In CH4 there are 4 bond pair of electrons while in NH 3 are 3 bond pair of electrons and 1 lone pair of electrons. Since bond pair bond pair repulsion is less than lone pair bond pair repulsion, in NH 3 bond angle is reduced from 109°28’ to 107°. Prob 12. Why bond angle in NH3 is 107° while in H2O it is 104.5°? Solution: In NH3, central nitrogen atom bears only one lone pair of electrons whereas in H2O central oxygen atom bears two lone pair of electrons. Since the repulsion between lone pair-lone pair and lone pair – bond pair is more than that between bond pair-bond pair, the repulsion in H 2O is much greater than that in NH3 which results in contraction of bond angle from 109°28” to 104.5° in water while in NH 3 contraction is less i.e. from 109°28” to 107°.

“If the electronegativity of the peripheral atoms is more, then the bond angle will be less”. For example if we consider NH 3 and NF3, F – N – F bond angle will be lower than H – N – H bond angle. This is because in NF 3 the bond pair is displaced more towards F and in NH 3 it is displaced more towards N. So accordingly the b.p. – b.p. interaction is less in NF 3 and more in NH3. Prob 13. The bond angle of H2O is 104° while that that of F2O is 102°. Solution:Both H2O and F2O have a lone pair of electrons. But fluorine being highly electronegative, the bond pair of electrons are drawn more towards F in F2O, whereas in H2O it is drawn towards O. So in F 2O the bond pairs being displaced away from the central atom, has very little tendency to open up the angle. But in H2O this opening up is more as the bond pair electrons are closer to each other. So bond Ð of F2O is less than H2O.

Prob 14. Predict the hybridization for the central atom in POCl 3, OSF4, OIF5. Solution: Total No. of V.E. = 5+6+21/8 = 32/8 = 4 So, hybridization = sp3 OSF4 = 6+6+28/8 = 40/8 = 5 So, hybridization of s = dsp3 OIF5 = 6+7+35/8 = 48/8 = 6 So, hybridization of I = d2sp3 Prob 15. Out of the three molecules XeF 4, SF4 and SiF4 one which has tetrahedral structures is (A) All of three (B) Only SiF4 (C) Both SF4 and XeF4 (D) Only SF4 and XeF4 Solution: Hybridization of XeF4 = sp3d2, SF4 = sp3d, SiF4 = sp3 Hence (B) is correct. Prob 16. Among the following compounds in which case central element uses d-orbital to make bonds with attached atom (A) BeF2

(B) XeF2

(C) SiF4

(D) BF3

Solution: In XeF2. Xe atom has sp3d hybridisation. Hence (B) is correct. Prob 17. When NH3 is treated with HCl, state of hybridisation on central nitrogen

(A) Changes from sp3 to sp2 (B) Remains unchanged (C) Changes from sp3 to sp3d (D) Changes from sp3 to sp Solution: On NH4+ state of hybridisation on central nitrogen atom is sp 3 as in NH3.

Hence (B) is the correct answer.