NCERT Exemplar for Class 11 Chemistry - Chemical Bonding and Molecular Structure - Free PDF

NCERT Exemplar for Class 11 Chemistry - Chemical Bonding and Molecular Structure - Free PDF Download

Free PDF download of NCERT Exemplar for Class 11 Chemistry Chapter 4 - Chemical Bonding and Molecular Structure solved by expert Chemistry teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 4 - Chemical Bonding and Molecular Structure Exercise questions with solutions to help you to revise the complete syllabus and score more marks in your Examinations.


Summary of Chemical Bonding and Molecular Structure

Following are the three major exceptions to the rule of octaves:

  1. Incomplete Octet of the Central Atom - In some compounds, the number of Electrons present around the central Atom is less than eight. It occurs mainly in compounds of those elements in which the number of valence Electrons is less than four.

  2. Odd Electron molecule - In those molecules in which the total number of Electrons is odd, such as nitric oxide (NO) and nitrogen dioxide (NO), not all Atoms obey the octet rule.

  3. Expanded octet: In addition to 35 and 3p orbitals, 3d-orbitals are also available for Bonding in the elements of the third and beyond the period of the periodic table. Many compounds of these elements have more than eight Electrons around the central Atom. This is called an expanded octet. The octet rule does not apply to these compounds. Some Examples of such compounds are PFs, SF 6, H2SO4 and many covalent compounds.

Favourable Factors for Ionic Bond Formation Following factors are favourable for forming an Ionic Bond

  1. Ionisation Enthalpy – In the formation of a positive ion or cation, one of the Atoms has to give up Electrons, which requires Ionisation Enthalpy. We know that Ionisation Enthalpy is the amount of energy required to remove the outermost Electron from an isolated gaseous Atom; Therefore, the lower the ionisation enthalpy required, the easier the formation of the cation will be. 5- The alkali metals and alkaline earth metals present in the block generally form cations, because their ionisation enthalpy is relatively low.

  2. Electron Gain Enthalpy - liberated in the formation of cations. The Electrons are taken up by the other Atoms participating in the formation of the ionic bond. The tendency of Atoms to accept Electrons depends on the Electron gain enthalpy. The amount of energy released by an isolated gaseous Atom to acquire an Electron and become an anion is called Electron gain enthalpy. It is thus clear that the formation of an anion will be easier if the Electron gain enthalpy is more negative. The halogens present in group 17 have the highest tendency to form anion because their Electron gain enthalpy is very negative. Members of the oxygen family (group 16) also tend to form anions, but this is not easily possible; Because energy is required to form divalent anions (02-).

  3. Lattice Energy or Enthalpy - Ionic compounds occur in the form of crystalline solids and crystals of ionic compounds, cations and anions are arranged regularly in three-dimensional form. Since ions are charged species; Therefore, the energy released in the attraction of ions is called lattice energy or enthalpy. It can be defined as: The energy released when one mole of a crystalline solid is obtained by combining oppositely charged ions is called lattice energy or enthalpy." It is expressed by 'U'.

It is thus clear that the stability of the Ionic Bond or Ionic Compound will be greater if the magnitude of the lattice energy is fixed. In conclusion, if the magnitude of the lattice energy and the negative Electron gain enthalpy is greater than the required ionisation enthalpy, then a stable Chemical bond will be obtained. If they are less then the bond will not be formed.

Electronegativity - The Electronegativity of an element can be defined in such a way that the measure of the tendency of its Atom to attract the shared Electron pair in a covalent bond is called the Electronegativity of the element.

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FAQs (Frequently Asked Questions)

1. Write the factors favourable for the formation of Ionic Bonds.

Favourable Factors for Ionic Bond Formation Following factors are favourable for forming an ionic bond

(1) Ionization Enthalpy - In the formation of a positive ion or cation, one of the Atoms has to give up Electrons, which requires ionisation enthalpy. We know that ionisation enthalpy is the amount of energy required to remove the outermost Electron from an isolated gaseous Atom; Therefore, the lower the ionisation enthalpy required, the easier the formation of the cation will be. 5- The alkali metals and alkaline earth metals present in the block generally form cations, because their ionisation enthalpy is relatively low.

(2) Electron Gain Enthalpy - liberated in the formation of cations. The Electrons are taken up by the other Atoms participating in the formation of the ionic bond. The tendency of Atoms to accept Electrons depends on the Electron gain enthalpy. The amount of energy released by an isolated gaseous Atom to acquire an Electron and become an anion is called Electron gain enthalpy. It is thus clear that the formation of an anion will be easier if the Electron gain enthalpy is more negative. The halogens present in group 17 have the highest tendency to form anion because their Electron gain enthalpy is very negative. Members of the oxygen family (group 16) also tend to form anions, but this is not easily possible; Because energy is required to form a divalent anion.

(3) Lattice energy or enthalpy- Ionic compounds occur in the form of crystalline solids and crystals of ionic compounds, the cations and anions are arranged regularly in three-dimensional form. Since ions are charged species; Therefore, the energy released in the attraction of ions is called lattice energy or enthalpy. It can be defined as: The energy released when one mole of a crystalline solid is obtained by combining oppositely charged ions is called lattice energy or enthalpy." It is expressed by 'U'.

It is thus clear that the stability of the ionic bond or ionic compound will be greater if the magnitude of the lattice energy is fixed. In conclusion, if the magnitude of the lattice energy and the negative Electron gain enthalpy is greater than the required ionisation enthalpy, then a stable Chemical bond will be obtained. If they are less then the bond will not be formed.

2. How to articulate bond strength in terms of bond order?

If the bond dissociation enthalpy is high then the bond will be stronger and the bond enthalpy increases as the bond order increases. This is clear from the fact that the bond strength and the bond order are proportional to each other. For Example, the bond order of N2 is 3 and its bond enthalpy is 945kJ mol-1. Similarly, the bond order of 02 is 2 and its bond enthalpy is 498 jmol-1. Among these, the N2 bond will be stronger. 

3. Define bond-length.

The equilibrium distance between the nuclei of the bonded atoms in a molecule is called the bond length. The bond length values ​​are usually expressed in picometers (1 pm = 10-12 m).

In the ionic compounds, the bond length between two bonded Atoms is obtained by adding their ionic radii. Similarly, in the covalent compounds, the bond length between two bonded Atoms is obtained by adding up their covalent (Atomic) radii. As the bond order increases, the bond strength will also increase.

4. State the important applications of dipole moment.

Important Applications of Dipole Moment Following are some important applications of dipole moment

(1) Predicting the nature of the molecules - Molecules with a fixed dipole moment are polar, whereas molecules with zero dipole moment are nonpolar. Thus, BeF2 (u = 0 D) is nonpolar, while H2O (u = 1.84D) is polar.

(2) Predicting the Molecular Structure of the molecules - we know that Atomic gases; For Example, the dipole moment of inert gases etc. is zero, that is, they are non-polar, but diAtomic molecules are polar and nonpolar; Like - H2O2. Adi is nonpolar (u = 0) and CO is polar. The Structure of these molecules is also linear.

TriAtomic molecules are also polar and non-polar. CO2, CS2, etc. are nonpolar; Because for them = 0; Therefore, the Structure of these molecules is linear.

Water molecule is polar because = 184D; Hence its Structure cannot be linear. It has an angular Structure and each O-H bond has an angle of 104°5'. Similarly, H25 and SO2 also have angular Structures; Because the values ​​for these are 0.90 D and 1.71 D respectively. Molecules having four Atomicities are also polar and nonpolar. 4 = 0 for the BCl3 molecule i.e. it is nonpolar. Therefore, its Structure is similar to that of an isosceles triangle.

(3) Determining the polarity of the bonds - The ionic property of polarity in a covalently bonded compound depends on the Electronegativity of the Atoms of the elements used in the formation of that bond. Thus, the difference in Electronegativity and dipole moment of the Atoms of the bond.

The difference in the Electronegativity of the Atoms of the bond.

(4) Determining the ionic percentage of the bonds - Dipole moment values ​​help in finding the ionic percentage of polar bonds. It is the ratio of the observed dipole moment or the experimentally determined dipole moment to the total Electron transfer dipole moment (theoretical).

5. What do you understand about polar covalent bonds?

Polar covalent compound - In many molecules, one Atom is more negatively Electronegative than the other Atom, so it tends to attract the Electron pair of the covalent bond towards itself so that the Electron pair is correctly located at the centre of the molecule. does not remain in the Atom, but is attracted to the Atom of the more negatively Electronegative element. Because of this one. A positive charge (which has low Electronegativity) on one Atom and a negative charge (which has a high negative Electronegativity) on another Atom is produced. The molecule thus obtained is called a polar covalent compound and the bond formed in it is called a polar covalent bond.

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