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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 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:

Answer:

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

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

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

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

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

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

Image:

Structure of Trigonal Bipyramidal

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

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

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

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

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

Answer:

The molecular geometry of and can be shown as:

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

Title: Structure of nitrogen atom and oxygen atom.

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

The two lone pairs present in the oxygen atom of 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 molecule are greater than that in 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 or picometer 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 In a covalent compound, it is the sum of their covalent radii

Image:

Constituting Atoms

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

Answer:

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

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