NCERT Solutions for Class 11 Chemistry Part 1 chapter 4 Chemical Bonding and Molecular Structure

A chemical bond is defined as an attractive force that holds the constituents (atoms, ions, etc.) together in a chemical species. 

Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory. 

A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble gases was because of their fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transfer of electrons from one atom to another.

 

Mg: There are two valence electrons in an Mg atom. Hence, the Lewis dot symbol for Mg is: Mg¨

Na: There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure is: Na˙

B: There are 3 valence electrons in Boron atom. Hence, the Lewis structure is: .B˙⋅

O: There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure is:  :O¨:

N: There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure is: :N¨⋅

Br: There are seven valence electrons in bromine. Hence, the Lewis dot structure is: :Br⋅¨⋅

 

a) BeCl2

Ans: Cl:BeCl

The central atom has no lone pair and there are two bond pairs. I.e. BeCl2 is of the type AB2 . Hence, it has a linear shape.

b) BCl3 

Ans: C⋅⋅lCl:Be:Cl

The central atom has no lone pair and there are three bond pairs. Hence, it is of the type AB3 . Hence, it is trigonal planar.

3. SiCl4

 

 

Ans: 

The central atom has no lone pair and there are four bond pairs. Hence, the shape of SiCl4 is tetrahedral being the AB4 type molecule.

4. AsF5 

Ans: 

The central atom has no lone pair and there are five bond pairs. Hence, AsF5 is of the type AB5 . Therefore, the shape is trigonal bipyramidal.

5. H2S 

Ans: 

 

The central atom has one lone pair and there are two bond pairs. Hence,H2S is of the type AB2E . The shape is Bent

6. PH3 

Ans: 

The central atom has one lone pair and there are three bond pairs. Hence,

PH3 is of the AB3E type. Therefore, the shape is trigonal bipyramidal.

 

1. H2S 

Ans: 

2. SiCl4 

Ans: 

3. BeF2

Ans: 

 

4. CO32− 

Ans: 

 

The octet rule or the electronic theory of chemical bonding was developed by Kossel and Lewis. According to this rule, atoms can combine either by transfer of valence electrons from one atom to another or by sharing their valence electrons in order to attain the nearest noble gas configuration.

 

The octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.

Limitations of the Octet Theory: 

The Following are the Limitations of the Octet Rule:

1. The rule failed to predict the shape and relative stability of molecules. 

2. It is based upon the inert nature of noble gases. However, some noble gases like xenon and krypton form compounds such as

3. XeF2 ,KrF2  etc.

4. The octet rule cannot be applied to the elements in and beyond the third period of the periodic table. The elements present in these periods have more than eight valence electrons around the central atom. For example: PF5, SF6 etc.

 

An ionic bond is formed by the transfer of one or more electrons from one atom to another. Hence, the formation of ionic bonds depends upon the ease with which neutral atoms can lose or gain electrons. The bond formation also depends upon the lattice energy of the compound formed. 

Hence, favourable factors for ionic bond formation are as follows:

1. Low ionization enthalpy of metal atoms. 

2. High electrons gain enthalpy of a nonmetal atom. 

3. The high lattice energy of the compound formed

 

The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The greater the difference, the greater will be the ionic character of the molecule. 

On this basis, the order of increasing ionic character in the given molecules is

N2<SO2<ClF3<K2O<LiF

 

The molecular geometry of NH3 and H2O can be shown as:

 

The central atom (N) in NH3 has one lone pair and three bond pairs. In H2O, there are two lone pairs and two bond pairs. 

The two lone pairs present on the oxygen atom of H2O molecule repel the two bond pairs. This repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the nitrogen atom. 

Since the repulsions on the bond pairs in H2O molecule are greater than that in

NH3, the bond angle in water is less than that of ammonia.

 

Bond strength represents the extent of bonding between two atoms forming a molecule. The larger the bond energy, the stronger is the bond and the greater is the bond order.

 

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Bond lengths are expressed in terms of Angstrom (10−10m) or picometer (10−12pm) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.

In an ionic compound, the bond length is the sum of the ionic radii of the constituting atoms (d=r++r−).

In a covalent compound, it is the sum of their covalent radii (d=rA+rB) 

 

According to experimental findings, all carbon to oxygen bonds in CO32− are equivalent.

Hence, it is inadequate to represent CO32− ion by a single Lewis structure having two single bonds and one double bond.

Therefore, carbonate ion is described as a resonance hybrid of the following structures

 

The given structures cannot be taken as the canonical forms of the resonance hybrid of

H

3

P

O

3

H3PO3 because the positions of the atoms have changed.

 

1. SO3

Ans: 

 

2. NO2 

Ans:

 

3. NO3− 

Ans: Do it yourself (Practice)

 

1.K and S

Ans: the electronic configuration of K and S are as follows:

K: 2, 8, 8, 1

S: 2, 8, 6

   K⋅:S¨:  

Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:

2. Ca and O

Ans: The electronic configurations of Ca and O are as follows: 

Ca: 2, 8, 8, 2 

O: 2, 6 

Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as:

 

3.Al and N

Ans: The electronic configurations of Al and N are as follows: 

Al: 2, 8, 3 

N: 2, 5 

Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminum has three electrons more than Neon. Hence, the electron transference can be shown as:

 

According to experimental results, the dipole moment of carbon dioxide is zero. This is possible only if the molecule is linear so that the dipole moments of C-O bonds are equal and opposite to nullify each other.

 

Resultant μ=0D 

 

H2O, on the other hand, has a dipole moment value of 1.84 D (though it is a triatomic molecule as CO2 ). The value of the dipole moment suggests that the structure of

H2O molecule is bent where the dipole moment of O-H bonds are unequal.

 

In heteronuclear molecules, polarization arises due to a difference in the electronegativities of the constituents of atoms. As a result, one end of the molecule acquires a positive charge while the other end becomes negative. Hence, a molecule is said to possess a dipole. 

The product of the magnitude of the charge and the distance between the centers of positive-negative charges is called the dipole moment (μ) of the molecule. It is a vector quantity and is represented by an arrow with its tail at the positive center and head pointing towards the negative center.

Dipole moment (μ) = charge (Q) X distance separation (r)

The S I unit of dipole moment is ‘esu’ 1esu=3.335×10−30cm

Dipole moment is the measure of the polarity of a bond. It is used to differentiate between polar and nonpolar bonds since all non-polar molecules (e.g H2,O2) have zero dipole moments. It is also helpful in calculating the percentage ionic character of a molecule.

 

Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself. The electronegativity of any given element is not constant. It varies according to the element to which it is bonded. It is not a measurable quantity. It is only a relative number. 

On the other hand, electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.

 

When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons are not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom. 

As a result, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this type of bond is called a polar covalent bond. 

HCl, for example, contains a polar covalent bond. Chlorine atoms are more electronegative than hydrogen atoms. Hence, the bond pair lies towards chlorine and therefore, it acquires a partial negative charge.

 

The correct Lewis structure for acetic acid is as follows:

 

The electronic configuration of carbon atom is: 6C=1s22s22p2

In the excited state, the orbital picture of carbon can be represented as:

Hence, the carbon atom undergoes Sp3 hybridization in CH4 molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp2. 

However, an atom of carbon does not have d-orbitals to undergo dsp2 hybridization. 

Hence, the structure of CH4 cannot be square planar. Moreover, with a bond angle of900  in square planar, the stability of CH4 will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH4.

 

the Lewis structure for BeH2is as follows:

 There is no lone pair at the central atom (Be) and there are two bond pairs. Hence,

BeH2 is of the type AB2 . It has a linear structure. 

Dipole moments of each H-Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, BeH2 molecule has zero dipole moment.

 

In both molecules, i.e NH3 and NF3, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF3 is greater than NH3. However, the net dipole moment of NH3 (1.46 D) is greater than that of NF3 (0.24 D). 

This can be explained on the basis of directions of the dipole moments of each individual bond is NH3 and NF3. These directions can be shown as:

Thus, the resultant moment of the N-H bonds adds up to the bond moment of the lone pair (the two being in the same direction), whereas that of the three N – F bonds partly cancels the moment of the lone pair. 

Hence, the net dipole moment of NF3 is less than that of NH3.

 

Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes. 

For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new

sp2 hybrid orbitals. These hybrid orbitals have minimum repulsion between their electron pairs and thus are more stable. Hybridization helps indicate the geometry of the molecule. 

The shape of sp hybrid orbitals: sp hybrid orbitals have a linear shape. They are formed by the intermixing of s and p orbitals as:

 

Shape of sp2 hybrid orbitals: sp2 hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p-orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement as: 

 

Shape of sp3  hybrid orbitals: 

Four sp3 hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. 

The four sp3 hybrid orbitals are arranged in the form of a tetrahedron as:

 

The valence orbital picture of aluminum in the ground state can be represented as:

The orbital picture of aluminum in the excited state can be represented as:

Hence, it undergoes sp2  hybridization to give a trigonal planar arrangement (in AlCl3 ). To form, the empty 3pz  orbital also gets involved and the hybridization changes from

sp2 to sp3. As a result, the shape gets changed to tetrahedral.

 

There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1[π∗(2py)]1

 

Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = Nb  and the number of antibonding orbitals = 4 = Na .

Bond order = 12(Nb−Na)=12(8−4)=2

Similarly the electronic configuration of O2+ can be written as: KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1

Nb=8

Na = 3

Bond order of O2+ = 12(8−3)=2.5 

Similarly the electronic configuration of O2− can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]1

 Nb=8 Na = 5

Bond order of 12(8−5)=1.5 

Similarly the electronic configuration of

O22−  can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]2

Nb=8  Na = 6

Bond order of

O22− = 12(8−6)=1 

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is

O2+>O2>O2−>O22−

 

Boron atoms in BF3 is sp2 hybridized. The orbital picture of boron in the excited state can be shown as:

 

Nitrogen atom in NH3  is sp3  hybridized. The orbital picture of nitrogen can be represented as: 

 

After the reaction has occurred, an adduct F3B.NH3  is formed as hybridization of ‘B’ changes to sp3 . However, the hybridization of ‘N’ remains intact.

 

C2H4 :

The electronic configuration of C-atom in the excited state is:

6C=1s22s12px12py12pz1

In the formation of an ethane molecule (C2H4), one sp2 hybrid orbital of carbon overlaps 

sp2  hybridized orbital of another carbon atom, thereby forming a C-C sigma bond. The remaining two sp2 orbitals of each carbon atom from a sp2−s sigma bond with two hydrogen atoms. The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak n-bond.

 

C2H2: In the formation of C2H2 molecule, each C-atom is sp hybridized with two 2p-orbitals in an unhybridized state. 

One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a C-C sigma bond. The second sp orbital if each C-atom overlaps a half-filled 1s-orbital to form a σ bond. 

The two unhybridized 2p-orbitals of the first carbon undergo sidewise overlap with the 2p orbital of another carbon atom, thereby forming two pi (n) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two n-bonds.

 

1. C2H2

Ans: A single bond is a result of that axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.

Structure of C2H2 can be represented as:

 

Hence, there are three sigma bonds and two pi-bonds in C2H2.

2. C2H4

The structure of C2H4 can be represented as:

 

Hence, there are five sigma bonds and one pi-bond in C2H4 .

 

2py and 2pz orbitals will not form a sigma bond. Taking x-axis as the internuclear axis, 2py and 2pz orbitals will undergo lateral overlapping, thereby is forming a pi (n) bond. Option (c)

1. CH3−CH3 

Ans: 

Both

C1 and C2 are sp3 hybridized

2. CH3−CH=CH2 

Ans: 

 

C1 is sp3 hybridized, C2and C3 are sp2 hybridized

3. CH3−CH2−OH 

Ans: 

 

Both C1 and C2 are sp3 hybridized

4. CH3CHO 

Ans: C1 is sp3 hybridized, C2 is sp2 hybridized. As the second carbon is present as an aldehyde group which contains double bond with oxygen atom and single bond with hydrogen atom.

5. CH3COOH 

Ans: 

 

C1 is sp3 hybridized, C2 is sp2 hybridized

 

When two atoms combine by sharing their one or more valence electrons, a covalent bond is formed between them. 

The shared pairs of electrons present between the bonded atoms are called bond pairs. 

All valence electrons may not participate in bonding. The electron pairs that do not participate in bonding are called lone pairs of electrons. 

For example, in C2H6  (ethane), there are seven bond pairs but no lone pair present.

 

In H2O , there are two bond pairs and two lone pairs on the central atom (oxygen).

 

The following are the differences between sigma and pi bonds

 

Let us assume the two hydrogen atoms (A and B) with nuclei (NA  and NB ) and electrons (eA  and eB ) are taken to undergo a reaction to form a hydrogen molecule. 

When A and B are at a large distance, there is no interaction between them. As they begin to approach each other, the attractive and repulsive forces start operating. 

Attractive Force Arises Between:

1. Nucleus of one atom and its own electron i.e.NA – eA and NB-eB. 

2. The nucleus of one atom and electron of another atom i.e.NA – eB and NB – eA . 

Repulsive Force Arises Between: 

1. Electrons of two atoms i.e., eA – eB.

2.  Nuclei of two atoms i.e., NA – NB. 

The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.

 

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

 

The given conditions should be satisfied by atomic orbitals of form molecular orbitals:

1. The combining atomic orbitals must have an equivalent or nearly an equivalent energy. This means that during a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

2. The combining atomic orbitals must have proper orientations to make sure that the overlap is maximum.

3. The extent of overlapping should be large

 

The electronic configuration of Beryllium is 1s22s2. The molecular orbital electronic configuration for Be2 molecule can be written as:

Σ1s2σ1s∗2σ2s2σ2s∗2

 Hence, the bond order for Be2is 12(Nb−Na) 

Where,

Nb = number of electrons in bonding orbitals

Na = number of electrons in non-bonding orbitals

∴ Bond order of Be2=12(4−4)=0 

A negative or zero bond order means that the molecule is unstable. Hence, Be2 molecule does not exist.

 

Molecular orbitals are represented by wave functions. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave function.

 

the ground state and excited state outer electronic configuration of phosphorus (Z=15) are:

 

Phosphorus atom is sp3d  hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:

 

The five sp3d hybrid orbitals are directed towards the five corners of the trigonal bipyramidal. Hence, the geometry of PCl5  can be represented as

 

There are five P-Cl sigma bonds is PCl5. Three P-Cl bonds lie in one plane and make an angle of 1200  with each other. These bonds are called equatorial bonds. 

The remaining two P-Cl bonds lie above and below the equatorial plane and make an angle of 900  with the plane. These bonds are called axial bonds. 

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

 

A hydrogen bond is defined as an attractive force acting between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule (may be of the same kind). 

Due to a difference between electro-negativities, the bond pair between hydrogen and the electronegative atom gets drifted away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

Hδ+−Xδ− …….. Hδ+−Xδ−……. Hδ+−Xδ−

The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state. 

There are two types of H-Bonds: 

1. Intermolecular H-bond e.g.,

2. HF, HF,H2O  etc.

3. Intramolecular H-bond e.g., o-nitrophenol 

 

Hydrogen bonds are stronger than VanDerWaals forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.

 

Bond order is defined as one half of the difference between the number of electrons present in the bonding and antibonding orbitals of a molecule. 

If Na is equal to the number of electrons in an antibonding orbital, then  Nb is equal to the number of electrons in a bonding orbital.

Bond order =12(Nb−Na)

If Nb>Na , then the molecule is said to be stable.

However, Nb≤Na , then the molecule is considered to be unstable. 

Bond order of N2  can be calculated from its electronic configuration as:

[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[π(2px)]2[π(2py)]2[π(2pz)]2

Nb=10

Na = 4

Bond order of nitrogen molecule = 12(10−4)=3

There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1[π∗(2py)]1

 

Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = Nb  and the number of antibonding orbitals = 4 = Na .

Bond order = 12(Nb−Na)=12(8−4)=2

Hence, the bond order of oxygen molecule is 2

Similarly the electronic configuration of O2+ can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1

Nb=8 Na = 3

Bond order of O2+ = 12(8−3)=2.5 

Similarly the electronic configuration of O2− can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]1

 Nb=8  Na = 5