NCERT Solutions for Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

NCERT Solutions for Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2 Class 11 Chemistry book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 11 Chemistry chapter wise NCERT solution for Class 11 Chemistry Some Basic Concepts of Chemistry Part 1 and all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

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Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

NCERT Class 11 Chemistry Chapter wise Solutions

  • Chapter 1 – Some Basic Concepts of Chemistry
  • Chapter 2 – Structure of Atom
  • Chapter 3 – Classification of Elements and Periodicity in Properties
  • Chapter 4 – Chemical Bonding and Molecular Structure
  • Chapter 5 – States of Matter
  • Chapter 6 – Thermodynamics
  • Chapter 7 – Equilibrium
  • Chapter 8 – Redox Reactions
  • Chapter 9 – Hydrogen
  • Chapter 10 – The s-Block Elements
  • Chapter 11 – The p-Block Elements
  • Chapter 12 – Organic Chemistry – Some Basic Principles and Techniques
  • Chapter 13 – Hydrocarbons
  • Chapter 14 – Environmental Chemistry

NCERT Solutions for Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

21. Apart from tetrahedral geometry, another possible geometry for is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why is not square planar?

Ans. Electronic configuration of carbon atom:

6C:

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

14160575681318.jpg

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

14160575733521.jpg

For a square planar shape, the hybridization of the central atom has to be . However, an atom of carbon does not have d-orbitalsto undergo dsp2 hybridization. Hence, the structure of cannot be square planar.

Moreover, with a bond angle of 90° in square planar, the stability of will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for .


22. Explain why molecule has a zero dipole moment although the Be-H bonds are polar.

Ans. The Lewis structure for BeH2 is as follows:

14160575790139.jpg

There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, is of the type . It has a linear structure.

14160575832789.jpg

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


23. Which out of has higher dipole moment and why?

Ans. In both molecules i.e., , 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 is greater than . However, the net dipole moment of (1.46 D) is greater than that of (0.24 D).

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

14160575902523.jpg

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.


24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, 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 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:

14160576006439.jpg

Shape of hybrid orbitals:

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:

14160576044202.jpg

Shape of hybrid orbitals:

Four 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:

14160576100649.jpg


25. Describe the change in hybridisation (if any) of the Al atom in the followingreaction.

Ans. The valence orbital picture of aluminium in the ground state can be represented as:

14160576169169.jpg

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

14160576199408.jpg

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

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


26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?

Ans. Boron atom in hybridized. The orbital picture of boron in the excited state can be shown as:

14160576253206.jpg

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

14160576291888.jpg

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


27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in molecules.

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

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

The remaining two 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.

14160576373401.jpg

: In the formation of molecule, each 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 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.

14160576392933.jpg


28. What is the total number of sigma and pi bonds in the following molecules?

(a)

(b)

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 can be represented as:

14160576428396.jpg

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

The structure of can be represented as:

1416057646484.jpg

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


29. Considering x-axis as the internuclear axis which out of the following will notform a sigma bond and why? (a) 1s and 1s (b) 1s and (c) (d) 1s and 2s.

Ans. orbitals will not a form a sigma bond. Taking x-axis as the internuclear axis, orbitals will undergo lateral overlapping, thereby forming a pi (π) bond.


30. Which hybrid orbitals are used by carbon atoms in the following molecules?

Ans. (a)

14160576617124.jpg

Both hybridized.

(b)

14160576624118.jpg

hybridized, while hybridized.

(c)

14160576655682.jpg

Both hybridized.

(d)

14160576662363.jpg

hybridized and hybridized.

(e)

14160576672209.jpg

hybridized and hybridized.


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 (ethane), there are seven bond pairs but no lone pair present.

14160576707906.jpg

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

14160576713677.jpg


32. Distinguish between a sigma and a pi bond.

Ans. The following are the differences between sigma and pi-bonds:


Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

33. Explain the formation of 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 (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:

(a) Nucleus of one atom and its own electron i.e., 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., .

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

14160576923531.jpg

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.


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.


Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

35. Use molecular orbital theory to explain why the molecule does not exist.

Ans. The electronic configuration of Beryllium is .

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

1416057711372.jpg

Hence, the bond order for is

Where,

= Number of electrons in bonding orbitals

= Number of electrons in anti-bonding orbitals

Bond order of = 0

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


36. Compare the relative stability of the following species and indicate their magnetic properties;

, , (superoxide), (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:

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

Bond order

= 2

Similarly, the electronic configuration of can be written as:

Nb = 8

Na = 3

Bond order of

= 2.5

Electronic configuration of ion will be:

= 8

= 5

Bond order of =

= 1.5

Electronic configuration of ion will be:

= 8

= 6

Bond order of =

= 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 14160577289148.gif.


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.


Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

.38. Describe the hybridisation in case of. 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:

Ground state:

14160577410028.jpg

Excited state:

14160577415846.jpg

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

1416057744011.jpg

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

14160577446574.jpg

There are five P-Cl sigma bonds in . 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.


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.

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,etc.

(ii) Intramolecular H-bond e.g., o-nitrophenol

14160577550861.jpg

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


Class 11 Chemistry Chemical Bonding and Molecular Structure Part 2

40. What is meant by the term bond order? Calculate the bond order of: , and .

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 is equal to the number of electrons in a bonding orbital.

Bond order =

If , then the molecule is said be stable. However, if , then the molecule is considered to be unstable.

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

14160577674076.gif

Number of bonding electrons, = 10

Number of anti-bonding electrons, = 4

Bond order of nitrogen molecule

= 3

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

14160577693633.gif

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

Bond order

= 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of can be written as:

14160577719446.gif

= 8

= 3

Bond order of

= 2.5

Thus, the bond order of is 2.5.

The electronic configuration of ion will be:

14160577764411.gif

= 8

= 5

Bond order of =

= 1.5

Thus, the bond order of ion is 1.5.

NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure Part 2

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