Class 11 Chemistry

Chemical Bonding

NCERT Solution

Part 1

Question 1: Explain the formation of a chemical bond.

Answer: Formation of chemical bond can be explained with the help of Lewis theory. Lewis said that an atom contains a positively charged kernel and a cube surrounding the kernel. The kernel contains the nucleus and inner electrons. The cube represents the outer shell. Each corner of the cube can accommodate 1 electron. Thus, the outer shell can have a maximum of 8 electrons. When the outer shell is filled with 8 electrons, it provides stability to atom, as in case of noble gases. All other elements tend to attain noble gas configuration in outer shell. Formation of bond results in outer shell getting fully filled. Chemical bond can be formed either by transfer of electrons or by sharing of electrons.

Question 2: Write Lewis dot symbols of atoms of the following elements: Mg, Na, B, O, N, Br

Answer: Number of dots indicates number of electrons in outer shell.

Lewis Dot Structure of Mg

Question 3: Write Lewis symbols for the following atoms and ions: S and S2-, Al and Al3+, H and H-

Answer: Lewis symbols of elements and their respective ions are given below.

Lewis Dot Structure of Al

Question 4: Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, CO32-, HCOOH


Lewis Dot Structure of H2S Lewis Dot Structure of SiCl4 Lewis Dot Structure of BeF2 Lewis Dot Structure of Carbonate Lewis Dot Structure of Formic Acid

Question 5: Define octet rule. Write its significance and limitations.

Answer: Octet Rule: (Kossel and Lewis 1916): Atoms can combine either by transfer of valence electrons or by sharing of valence electrons in order to attain an octet in their valence shells.

Significance of octet rule: This rule make it easy to understand the sharing or transfer of electrons during formation of a molecule.

Limitations of Octet Rule

Incomplete octet of the central atom: In some compounds, the number of electrons surrounding the central atom is less than eight. It is especially the case with elements having less than four valence electrons. Examples: LiCl, BeH2 and BCl3

Odd-electron Molecules: Octet rule is not satisfied in molecules with odd number of electrons. Exmaples: NO and NO2

Expanded Octet: Elements in and beyond the third period have 3d orbitals also available for bonding (apart from 3s and 3p orbitals). In a number of compounds of these elements, there are more thatn eight valence electrons around the central atom. Such a condition is called expanded octet. Examples: PF5, SF6, H2SO4, etc.

Question 6: Write the favorable factors for the formation of ionic bond.

Answer: Following are the factors favorable for the formation of ionic bond:

Ionic bonds are formed more easily between elements with comparatively low ionization enthalpies and elements with comparatively high negative value of electron gain enthalpy.

Lattice enthalpy gives a qualitative measure of relative stability of an ionic compound. Crystal structure gets stabilized due to lattice enthalpy.

Question 7: Discuss the shape of the following molecules using VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3

Answer: The central atom in BeCl2 has no lone pair and two bond pairs, so it is of type AB2. Hence its shape is linear.

The central atom in BCl3 has no lone pair and three bond pairs, so it is of type AB3. Hence its shape is trigonal planar.

The central atom in SiCl4 has no lone pair and four bond pairs, so it is of type AB4. Hence its shape is tetrahedral.

The central atom in H2S has two lones pairs and two bond pairs, so it is of type AB3E2. Hence its shape is bent.

The central atom in PH3 has one lone pair and three bond pairs, so it is of type AB3E. Hence its shape is trigonal pyramidal.

Question 8: Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Answer: The central atom in NH3 has three bond pairs and one lone pair, while the central atom in H2O has two bond pairs and one lone pair. As a result, there is greater repulsion between two bond pairs and two lone pairs in water molecule. Due to this, bond angle in water is less than that of ammonia.

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

Answer: Bond strength increases with bond order. The bond order for single, double and triple bonds is 1, 2 and 3 in that order. So, bond strength of single bond is less than that of double or triple bond.

Question 10: Define the bond length.

Answer: The equilibrium distance between the nuclei of two bonded atoms in a molecule is called bond length.

Question 11: Explain the important aspects of resonance with reference to the CO32- ion.

Answer: There are three canonical forms of carbonate ion, as shown in this figure. The resonance hybrid of these three structures gives the more accurate structure of carbonate ion.

Canonical forms of carbonate ion