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NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Q4.1: Explain the formation of a chemical bond.
Ans: 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 transference of electrons from one atom to another.

Q4.2: Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Ans: Mg: There are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is: NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Na: There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure is: NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

B: There are 3 valence electrons in Boron atom. Hence, the Lewis dot structure is: NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

O: There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure is: NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

N: There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure is: NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Br: There are seven valence electrons in bromine. Hence, the Lewis dot structure is: NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Q4.3: Write Lewis symbols for the following atoms and ions:
S and S2–; Al and Al3+; H and H
Ans: (i) S and S2–
The number of valence electrons in sulphur is 6.
The Lewis dot symbol of sulphur (S) isNCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
The dinegative charge infers that there will be two electrons more in addition to the six valence electrons. Hence, the Lewis dot symbol of S2– is NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure.

(ii) Al and Al3+
The number of valence electrons in aluminium is 3.
The Lewis dot symbol of aluminium (Al) is NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
The tripositive charge on a species infers that it has donated its three electrons. Hence, the Lewis dot symbol is [Al]3+

(iii) H and H
The number of valence electrons in hydrogen is 1.
The Lewis dot symbol of hydrogen (H) is NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
The uninegative charge infers that there will be one electron more in addition to the one valence electron. Hence, the Lewis dot symbol is NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure


Q4.4: Draw the Lewis structures for the following molecules and ions:
H2S, SiCl4, BeF2,NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure, HCOOH
Ans:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure


Q4.5: Define octet rule. Write its significance and limitations.
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 by having an octet in their valence shell.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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:
(a) The rule failed to predict the shape and relative stability of molecules.
(b) It is based upon the inert nature of noble gases. However, some noble gases like xenon and krypton form compounds such as XeF2, KrF2 etc.
(c) 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.


NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure


(d) The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example, NO and NOdo not satisfy the octet rule.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

(e) This rule cannot be applied to those compounds in which the number of electrons surrounding the central atom is less than eight. For example, LiCl, BeH2, AlCletc. do not obey the octet rule.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure


Q4.6: Write the favourable factors for the formation of ionic bond.
Ans: 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:
(i) 
Low ionization enthalpy of metal atom.
(ii) High electron gain enthalpy (Δeg H) of a non-metal atom.
(iii) High lattice energy of the compound formed.

Q4.7: Discuss the shape of the following molecules using the VESPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Ans:
(i) BeCl2:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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.

(ii) BCl3:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

(iii) SiCl4:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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.

(iv) AsF5:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
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.

(v) H2S:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular StructureThe central atom has one lone pair and there are two bond pairs. Hence, H2S is of the type AB2E. The shape is Bent.

(vi) PH3:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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.

Q4.8: Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Ans: The molecular geometry of NH3 and H2O can be shown as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

The two lone pairs present in the oxygen atom of the H2O molecule repels 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 the H2O molecule are greater than that in NH3, the bond angle in water is less than that of ammonia.

Q4.9: How do you express the bond strength in terms of bond order?
Ans: 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.

Q4.10: Define the bond length.
Ans: 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 (10–10 m) or picometer

(10–12 m) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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).

Q4.11: Explain the important aspects of resonance concerning the NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structureion.
Ans: According to experimental findings, all carbon-to-oxygen bonds in NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure are equivalent. Hence, it is inadequate to represent NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure 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:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

 

Q4.12: H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give reasons for the same.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Ans: The given structures cannot be taken as the canonical forms of the resonance hybrid of H3PO3 because the positions of the atoms have changed.

Q4.13: Write the resonance structures for SO3, NO2 andNCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure.
Ans: The resonance structures are:

(a) SO3:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

 (b) NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

(c) NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Q4.14: Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
Ans:  (a) K and S:

The electronic configurations of  K and S are as follows:
K: 2, 8, 8, 1
S: 2, 8, 6

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:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

(b) Ca and O:
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:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

(c) Al and N:

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 aluminium has three electrons more than Neon. Hence, the electron transference can be shown as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Q4.15: Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
Ans: 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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Resultant μ = 0 D

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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.16: Write the significance/applications of dipole moment.
Ans: 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 centres 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 centre and head pointing towards a negative centre.

Dipole moment (μ) = charge (Q) × distance of separation (r)

The SI unit of a dipole moment is ‘esu’.

1 esu = 3.335 × 10–30 cm

Dipole moment is the measure of the polarity of a bond. It is used to differentiate between polar and non-polar 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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.17: Define electronegativity. How does it differ from electron gain enthalpy?
Ans: Electronegativity is defined as the tendency of an element to attract the shared pair of electrons towards itself in a covalent bond. There is no specific unit for electronegativity. Electron gain enthalpy is the energy released when one mole of electron are added to gaseous atoms of an element. 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.

Q4.18: Explain with the help of suitable example polar covalent bond.
Ans: When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons is 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 a bond is called a polar covalent bond.

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.19: Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 & ClF3
Ans: 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.

Q4.20: The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular StructureAns: The correct Lewis structure for acetic acid is as follows:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.21: Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?
Ans: Electronic configuration of carbon atom:
6C: 1s2 2s2 2p2
In the excited state, the orbital picture of carbon can be represented as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Hence, carbon atom undergoes sp3 hybridization in CH4 molecule and takes a tetrahedral shape.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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 of 90° 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.

Q4.22: Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
Ans: The Lewis structure for BeH2 is as follows:
NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular 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.

Q4.23: Which out of NH3 and NF3 has a higher dipole moment and why?
Ans: 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 the directions of the dipole moments of each individual bond in NF3 and NH3. These directions can be shown as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Thus, the resultant moment of the N–H bonds add 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.

Q4.24: What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Ans: 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.

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Shape of sp2 hybrid orbitals:

sphybrid 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:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.25: Describe the change in hybridisation (if any) of the Al atom in the following reaction.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular StructureAns: The valence orbital picture of aluminium in the ground state can be represented as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Hence, it undergoes sp2 hybridization to give a trigonal planar arrangement (in AlCl3).

To form AlCl4, the empty 3pz orbital also gets involved and the hybridization changes from sp2 to sp3. As a result, the shape gets changed to tetrahedral.

Q4.26: Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3  NH3 → F3B.NH3
Ans: Boron atom in BF3 is sp2 hybridized. The orbital picture of boron in the excited state can be shown as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

After the reaction has occurred, an adduct F3B⋅NH3 is formed and hybridization of ‘B’ changes to sp3. However, the hybridization of ‘N’ remains unchanged because N shares its lone pair with electron deficient B.

Q4.27: Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
Ans: C2H:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

In the formation of an ethane molecule (C2H4), one sp2 hybrid orbital of carbon overlaps a sp2 hybridized orbital of another carbon atom, thereby forming a C-C sigma bond.

The remaining two sp2 orbitals of each carbon atom form 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 π-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 of 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 (π) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two π-bonds.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.28: What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2
(b) C2H4
Ans: A single bond is a result of the 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:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

The structure of C2H4 can be represented as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

Q4.29: Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px (c) 2py and 2py (d) 1s and 2s.
Ans: (c)
Axial overlapping results in the formation of sigma bond. and sideways overlapping results in the formation of pi bond . Sigma bond involves overlapping of s - s , s - p and p - p atomic orbitals .
Only (c ) won't form sigma bond because on taking x - axis as the internuclear axis, 2py and 2py orbitals overlap sideways ( lateral overlapping ) resulting in the formation of pi bond .
Hence, option ( c ) is answer .

Q4.30: Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3–CH3 
(b) CH3–CH=CH2 
(c) CH3-CH2-OH 
(d) CH3-CHO 
(e) CH3COOH

Ans:- (a)

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Both C1 and C2 are sp3 hybridized.

(b)

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

C1 is sp3 hybridized, while C2 and C3 are sp2 hybridized.

(c)

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Both C1 and C2 are sp3 hybridized.

(d)

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

C1 is sp3 hybridized and C2 is sp2 hybridized.

(e)

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

C1 is sp3 hybridized and C2 is sp2 hybridized.

Q4.31: What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Ans: 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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
Q4.32: Distinguish between a sigma and a pi bond.
Ans: The following are the differences between sigma and pi-bonds:
NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

 

Q4.33: Explain the formation of H2 molecule on the basis of valence bond theory.
Ans: Let us assume that two hydrogen atoms (A and B) with nuclei (NA and NB) and electrons (eand 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:
(a) Nucleus of one atom and its own electron i.e., NA – eA and NB – eB.
(b) Nucleus of one atom and electron of another atom i.e., NA – eB and NB – eA.

Repulsive force arises between:
(a) Electrons of two atoms i.e., eA – eB.
(b) 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.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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.

Q4.34: Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Ans: The given conditions should be satisfied by atomic orbitals to form molecular orbitals:

(a) The combining atomic orbitals must have the same or nearly the same energy. This means that in 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.

(b) The combining atomic orbitals must have proper orientations to ensure that the overlap is maximum.

(c) The extent of overlapping should be large.

Q4.35: Use molecular orbital theory to explain why the Be2 molecule does not exist.
Ans: The electronic configuration of Beryllium isNCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure.

The molecular orbital electronic configuration for Be2 molecule can be written as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Hence, the bond order for Beis  NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Where,

Nb = Number of electrons in bonding orbitals

Na = Number of electrons in anti-bonding orbitals

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular StructureBond order of Be2 = (4-4) / 2 = 0

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

Q4.36: Compare the relative stability of the following species and indicate their magnetic properties;
O2,NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure,NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure (superoxide), NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure(peroxide)
Ans: There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = Nand the number of anti-bonding orbitals = 4 = Na.

Bond order NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

(8 - 4) / 2 = 2

Similarly, the electronic configuration of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structurecan be written as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Nb = 8
Na = 3

Bond order of  NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

= 2.5

Electronic configuration of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structureion will be:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Nb = 8

Na = 5

Bond order of  NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

= 1.5

Electronic configuration of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structurethe ion will be:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Nb = 8
Na = 6
Bond order of  NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure = 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 NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Q4.37: Write the significance of a plus and a minus sign shown in representing the orbitals.
Ans: 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.

Q4.38: Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Ans: The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

There are five P–Cl sigma bonds in PCl5. Three P–Cl bonds lie in one plane and make an angle of 120° 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 90° 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.

Q4.39: Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Ans: 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 electronegativities, the bond pair between hydrogen and the electronegative atom gets drifted far away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

There are two types of H-bonds:
(i) Intermolecular H-bond e.g., HF, H2O etc.
(ii) Intramolecular H-bond e.g., o-nitrophenol

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

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

Q4.40: What is meant by the term bond order? Calculate the bond order of: N2, O2, NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular StructureandNCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure.
Ans: Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

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

Bond order = NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

If Nb > Na, then the molecule is said be stable. However, if Nb ≤ Na, then the molecule is considered to be unstable.

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Number of bonding electrons, Nb = 10

Number of anti-bonding electrons, Na = 4

Bond order of nitrogen molecule NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure = 3

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

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = Nand the number of anti-bonding electrons = 4 = Na.

Bond order NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

= 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structurecan be written as:

NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

Nb = 8

Na = 3

Bond order of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

= 2.5
Thus, the bond order of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structureis 2.5.
The electronic configuration of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structureion will be:
NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular StructureNb = 8
Na = 5
Bond order of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure= NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure
= 1.5

Thus, the bond order of NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structureion is 1.5.

The document NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure is a part of the NEET Course Chemistry Class 11.
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FAQs on NCERT Solutions Class 11 Chemistry Chapter 4 - Chemical Bonding & Molecular Structure

1. What is chemical bonding?
Ans. Chemical bonding refers to the process by which atoms combine to form molecules or compounds. It involves the interaction between the outermost electrons of atoms, which are known as valence electrons. These electrons can be shared, transferred, or redistributed between atoms to achieve a stable configuration, resulting in the formation of chemical bonds.
2. What are the different types of chemical bonding?
Ans. There are three main types of chemical bonding: 1. Ionic bonding: In this type of bonding, electrons are transferred from one atom to another, forming positively charged ions (cations) and negatively charged ions (anions). The electrostatic attraction between these oppositely charged ions holds them together in an ionic compound. 2. Covalent bonding: Covalent bonding occurs when atoms share electrons to achieve a stable electron configuration. This type of bonding is commonly observed in nonmetallic elements and results in the formation of molecules. 3. Metallic bonding: Metallic bonding occurs in metals, where the valence electrons are delocalized and shared among all the atoms in a metallic lattice. This leads to the characteristic properties of metals, such as high electrical and thermal conductivity.
3. How is a chemical bond formed?
Ans. A chemical bond is formed through the interaction of valence electrons between atoms. The most common ways in which chemical bonds are formed are: 1. Electron transfer: In ionic bonding, one atom donates electrons to another atom, resulting in the formation of positive and negative ions. The electrostatic attraction between these ions holds them together in an ionic bond. 2. Electron sharing: In covalent bonding, atoms share one or more pairs of electrons to achieve a stable electron configuration. This sharing of electrons leads to the formation of covalent bonds and the creation of molecules. 3. Electron pooling: In metallic bonding, valence electrons are pooled or delocalized among all the atoms in a metallic lattice. This pooling of electrons allows metals to conduct electricity and heat efficiently.
4. What is the importance of chemical bonding in determining the properties of substances?
Ans. Chemical bonding plays a crucial role in determining the properties of substances. Here are a few reasons why: 1. Stability: Chemical bonding allows atoms to achieve a stable electron configuration, which is the most energetically favorable state. This stability influences the physical and chemical properties of substances. 2. Bond strength: The type and strength of chemical bonds determine the physical properties of substances, such as melting point, boiling point, hardness, and conductivity. For example, substances with strong covalent bonds tend to have high melting and boiling points. 3. Reactivity: The type of chemical bonding affects the reactivity of substances. For instance, substances with ionic bonds are more likely to undergo chemical reactions, as the electrostatic attraction between ions can be easily disrupted. 4. Solubility: Chemical bonding influences the solubility of substances in different solvents. Polar substances with polar covalent or ionic bonds tend to dissolve in polar solvents, while nonpolar substances with nonpolar covalent bonds are soluble in nonpolar solvents.
5. How do we represent chemical bonding?
Ans. Chemical bonding is represented using Lewis dot structures or electron dot structures. In these structures, the valence electrons of atoms are represented as dots around the atomic symbol. The number of dots corresponds to the number of valence electrons. The electrons are paired up and shared between atoms to form bonds. The Lewis dot structures provide a visual representation of how atoms are bonded together in a molecule or compound.
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