Chemistry Class 11 NCERT Solutions: Chapter 4 Chemical Bonding and Molecular Structure Part 7

Q: 30. which hybrid orbitals are used by carbon atoms in the following molecules?

(A) Equation

(B) Equation

(C) Equation

(D) Equation

(E) Equation

Answer

(A)

Both Equation

(B)

Equation

(C)

Both Equation

(D)

Equation

(E)

Equation .

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

Answer:

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

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

Q: 32. Distinguish between a sigma and a pi bond.

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

Sigma and Pi Bonds

Sigma ( Equation ) bond

Pi ( Equation ) Bond

(A) It is formed by the end to end overlap of orbitals.

Ti is formed by the lateral overlap of orbitals.

(B) The orbitals involved in the overlapping are Equation

These bonds are formed by the overlap of Equation orbitals only.

(C) It is a strong bond.

It is weak bond.

(D) The electron cloud is symmetrical about the line joining two nuclei.

The electron cloud is not symmetrical.

(E) It consists of one electron cloud, which is symmetrical about the internuclear axis.

There are two electron clouds lying above and below the plane of the atomic nuclei.

(F) Free rotation about Equation bonds is possible.

Rotation is restricted in case of Equation bonds.

Q: 33. Explain the formation of Equation molecule on the basis of valence bond theory.

Answer:

Let us assume that two hydrogen atom (A and B) with nuclei ( Equation ) and electrons Equation are taken to undergo a reaction to from 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., Equation

(B) Nuclei of two atoms i.e., Equation

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

Image: 1

Circle c Circle c: Circle with center A and radius 0.31 Circle d Circle d: Circle with center B and radius 0.33 Circle e Circle e: Circle with center C and radius 0.31 Circle f Circle f: Circle with center D and radius 0.31 Segment h_1 Segment h_1: Segment [G_1, H_1] Segment g_1 Segment g_1: Segment [E_1, F_1] Segment g_2 Segment g_2: Segment [E_2, F_2] Segment h_2 Segment h_2: Segment [G_2, H_2] Segment g_5 Segment g_5: Segment [E_5, F_5] Segment g_6 Segment g_6: Segment [E_6, F_6] Segment g_7 Segment g_7: Segment [E_7, F_7] Segment g_8 Segment g_8: Segment [E_8, F_8] Point H_1 H_1 = (1.7, 1.98) Point H_1 H_1 = (1.7, 1.98) Point F_1 F_1 = (1.5, 1.7) Point F_1 F_1 = (1.5, 1.7) Point F_2 F_2 = (-0.34, 4.02) Point F_2 F_2 = (-0.34, 4.02) Point H_2 H_2 = (-0.14, 4.28) Point H_2 H_2 = (-0.14, 4.28) Point F_5 F_5 = (1.38, 4.24) Point F_5 F_5 = (1.38, 4.24) Point F_6 F_6 = (1.52, 4.02) Point F_6 F_6 = (1.52, 4.02) Point F_7 F_7 = (-0.24, 1.9) Point F_7 F_7 = (-0.24, 1.9) Point F_8 F_8 = (-0.08, 1.68) Point F_8 F_8 = (-0.08, 1.68) text1 = " " - text2 = "-" text1_1 = " " - text2_1 = "-" H_B text3 = "H_B" H_B text3 = "H_B" e_B text4 = "e_B" e_B text4 = "e_B" H_A text5 = "H_A" H_A text5 = "H_A" e_A text6 = "e_A" e_A text6 = "e_A" Attractive Forces text7 = "Attractive Forces"

Image: 2

Circle c Circle c: Circle with center A and radius 0.31 Circle d Circle d: Circle with center B and radius 0.33 Circle e Circle e: Circle with center C and radius 0.31 Circle f Circle f: Circle with center D and radius 0.31 Segment g Segment g: Segment [E, F] Segment h Segment h: Segment [G, H] Segment i Segment i: Segment [I, J] Segment i_1 Segment i_1: Segment [I_1, J_1] Segment i_4 Segment i_4: Segment [I_4, J_4] Segment i_2 Segment i_2: Segment [I_2, J_2] Point J J = (3.36, 3.08) Point J J = (3.36, 3.08) Point I_1 Point I_1: Point on d_1 Point I_1 Point I_1: Point on d_1 Point I_4 Point I_4: Point on d_4 Point I_4 Point I_4: Point on d_4 Point J_2 J_2 = (1.86, 0.94) Point J_2 J_2 = (1.86, 0.94) text1 = " " - text2 = "-" text1_1 = " " - text2_1 = "-" H_B text3 = "H_B" H_B text3 = "H_B" e_B text4 = "e_B" e_B text4 = "e_B" H_A text5 = "H_A" H_A text5 = "H_A" e_A text6 = "e_A" e_A text6 = "e_A" Repulsive Forces text7 = "Repulsive Forces" B text8 = "B" A text9 = "A"

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.

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