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

Q: 6. Write the favourable factors for the formation of ionic bond.

Answer:

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 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 Equation of a non-metal atom

(iii) High lattice energy of the compound formed.

Q: 7. Discuss the shape of the following molecules using the VSEPR model: Equation

Answer:

Equation

Equation

The central atom has no lone pair and there are two bond pairs. i.e., Equation Hence, it has a linear shape.

Equation

Image showing BCl3 Lewis Structure.

Image Showing BCl3 Lewis Structure.

Image showing BCl3 Lewis Structure.

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

Image showing BCL3 structure.

Image Showing BCL3 Structure.

Image showing BCL3 structure.

Equation

Image showing SiCl4 lewis structure.

Image Showing SiCl4 Lewis Structure.

Image showing SiCl4 lewis structure.

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

Equation

Image:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Triangle poly2 Triangle poly2: Polygon D, B, E Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment e Segment e: Segment [D, B] of Triangle poly2 Segment d_1 Segment d_1: Segment [B, E] of Triangle poly2 Segment b_1 Segment b_1: Segment [E, D] of Triangle poly2 Segment b_1 Segment b_1: Segment [E, D] of Triangle poly2 Segment g Segment g: Segment [D, F] Segment f Segment f: Segment [G, B] Segment h Segment h: Segment [H, E] Segment i Segment i: Segment [A, I] Segment j Segment j: Segment [J, C] Segment k Segment k: Segment [K, L] Segment l Segment l: Segment [E, C] AS text1 = "AS" AS text1 = "AS" F text2 = "F" F text3 = "F" F text2_1 = "F" F text2_2 = "F" F text3_1 = "F"

Structure of Trigonal Bipyramidal

Trigonal bipyramidal geometry is characterized by 5 electron pairs.

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

Equation

Equation

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

Equation

Image showing PH3 lewis structure.

Image Showing PH3 Lewis Structure.

Image showing PH3 lewis structure.

The central atom has one lone pair and there are three bond pairs. Hence, Equation is of the Equation type. Therefore, the shape is trigonal bipyramidal.

Q: 8. although geometries of Equation and Equation molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Answer:

The molecular geometry of Equation and Equation can be shown as:

Image showing NH3 and H2Omolecular geometry.

Image Showing NH3 and H2O Molecular Geometry.

Image showing NH3 and H2Omolecular geometry.

Title: Structure of nitrogen atom and oxygen atom.

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

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

Q: 9 How do you express the bond strength in terms of bond order?

Answer:

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.

Q: 10 Define the bond length.

Answer:

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 Angstron Equation or picometer Equation 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 Equation In a covalent compound, it is the sum of their covalent radii Equation

Image:

Polygon poly1 Polygon poly1: Polygon[E, F, 4] Polygon poly1 Polygon poly1: Polygon[E, F, 4] Circle c Circle c: Circle through B with center A Circle d Circle d: Circle through D with center C Segment f Segment f: Segment [E, F] of Polygon poly1 Segment g Segment g: Segment [F, G] of Polygon poly1 Segment i Segment i: Segment [H, E] of Polygon poly1 Segment j Segment j: Segment [I, J] Segment k Segment k: Segment [K, L] Segment l Segment l: Segment [M, N] Point I Point I: Point on h Point I Point I: Point on h Point K K = (-3.88, 7.28) Point K K = (-3.88, 7.28) Point L L = (-2.18, 7.3) Point L L = (-2.18, 7.3) Point M M = (-5, 7.28) Point M M = (-5, 7.28) A text1 = "A" B text2 = "B" r_A text3 = "r_A" r_A text3 = "r_A" r_B text4 = "r_B" r_B text4 = "r_B"

Constituting Atoms

Constituting atoms with bond lengths.

Q: 11. Explain the important aspects of resonance with reference to the Equation ion.

Answer:

According to experimental findings, all carbon to oxygen bonds in Equation are equivalent. Hence, it is inadequate to represent Equation 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:

Image showing Lewis Structures.

Image Showing Lewis Structures.

Image showing Lewis Structures.

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