Chemistry Class 11 NCERT Solutions: Chapter 4 Chemical Bonding and Molecular Structure Part 5

Q: 22. Explain why Equation molecule has a zero dipole moment although the Equation bonds are polar.

Answer:

The Lewis structure for Equation is as follows:

Equation

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

Image showing BeH2 linear structure.

Image Showing BeH2 Linear Structure.

Image showing BeH2 linear structure.

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

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

Answer:

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

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

Image showing NF3 and NH3 bonds.

Image Showing NF3 and NH3 Bonds.

Image showing NF3 and NH3 bonds.

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 Equation is less than that of Equation

Q: 24. What is meant by hybridization of atomic orbitals? Describe the shapes of Equation hybrid orbitals.

Answer:

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 Equation 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:

Image showing Shape of sp hybrid orbitals.

Image Showing Shape of Sp Hybrid Orbitals.

Image showing Shape of sp hybrid orbitals.

Shape of Equation hybrid orbitals:

Equation 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:

figure:

Image showing Shape of sp2 hybrid orbitals.

Image Showing Shape of sp2 Hybrid Orbitals.

Image showing Shape of sp2 hybrid orbitals.

Circle c Circle c: Circle through A, B, C Circle d Circle d: Circle through E with center D Circle e Circle e: Circle through G with center F Circle e_1 Circle e_1: Circle through G_1 with center F_1 Circle e_2 Circle e_2: Circle through G_2 with center F_2 Segment f Segment f: Segment [K, L] Segment g Segment g: Segment [M, N] Vector u Vector u: Vector[H, I] Vector u Vector u: Vector[H, I] Vector v Vector v: Vector[G, J] Vector v Vector v: Vector[G, J] text1 = " " - text2 = "-" text1_1 = " " - text2_1 = "-" X text3 = "X" Z text4 = "Z"

Structure of sp2hybrid Orbital

Structure of sp2 hybrid orbitals formed by intermixing one s-orbital with three p-orbitals

Four Equation hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. The four Equation hybrid orbitals are arranged in the form of a tetrahedron as:

Figure:

Circle c Circle c: Circle through A, B, C Circle d Circle d: Circle through E with center D Circle e Circle e: Circle through G with center F Circle e_1 Circle e_1: Circle through G_1 with center F_1 Circle e_2 Circle e_2: Circle through G_2 with center F_2 Circle p Circle p: Circle through T with center S Circle k Circle k: Circle through V with center U Segment h Segment h: Segment [O, P] Segment f Segment f: Segment [W, Z] Segment g Segment g: Segment [A_1, B_1] Segment i Segment i: Segment [C_1, D_1] Vector u Vector u: Vector[H, I] Vector u Vector u: Vector[H, I] Vector v Vector v: Vector[G, J] Vector v Vector v: Vector[G, J] Point P P = (4.57, 3.19) Point P P = (4.57, 3.19) Point D_1 D_1 = (8.8, 0.18) Point D_1 D_1 = (8.8, 0.18) text1 = " " - text2 = "-" text1_1 = " " X text3 = "X" Z text4 = "Z" text1_2 = " " text1_3 = " " - text2_3 = "-" - text2_4 = "-" y text5 = "y"

Structure of sp3 Hybrid Orbitals

Structure of sp3 hybrid orbitals formed by intermixing one s-orbital with three p-orbitals

Q: 25. Describe the change in hybridization (if any) of the Al atom in the following reaction. Equation

Answer:

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

Orbital picture of Al ground state

Orbital Picture of Al Ground State

Orbital picture of Al ground state

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

Orbital picture of Al excited state

Orbital Picture of Al Excited State

Orbital picture of Al excited state

Hence, it undergoes Equation hybridization to give a trigonal planar arrangement Equation To form Equation the empty Equation orbital also gets involved and the hybridization changes from Equation As a result, the shape gets changed to tetrahedral.

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

Equation

Answer:

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

Image showing Boron sp2 hybridized.

Image Showing Boron sp2 Hybridized.

Image showing Boron sp2 hybridized.

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

Image showing NH3 sp3 hybridized.

Image Showing NH3 sp3 Hybridized.

Image showing NH3 sp3 hybridized.

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

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