Chemical Bonding and Molecular Structure
Chemical Bonding is the attractive force holding atoms together in molecules and compounds. The formation of chemical bonds is driven by the tendency of atoms to attain a stable electronic configuration, often resembling the electron configuration of noble gases.
Key Theories of Chemical Bonding:
Ionic Bonding: Involves the complete transfer of electrons between atoms to form ions with opposite charges, which are then attracted to each other.
Covalent bonding: Covalent bonding occurs when atoms share electrons to attain stable electron configurations. Two atoms share a pair of electrons to create a single covalent bond.
Double covalent bond: Two pairs of electrons shared.
Triple covalent bond: Three pairs of electrons shared.
Coordinate Covalent Bonding: Involves one atom donating both electrons for a shared pair, while the other atom contributes only an empty orbital.
Factors Affecting Bond Strength:
Bonds : Bonds with shorter distances between atoms tend to be more robust.
Bond order: Higher bond orders (single, double, triple) typically indicate stronger bonds.
Electronegativity: The difference in electronegativity between bonded atoms affects bond polarity and strength.
Molecular Structure and VSEPR Theory:
VSEPR (Valence Shell Electron Pair Repulsion) Theory: Predicts the molecular geometry based on the repulsion between electron pairs (bonding and lone pairs) around a central atom.
shapes :The most common molecular shapes are linear, bent, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.
Polarity: A molecule is polar if it has a separation of charge, creating a dipole moment.
Factors affecting polarity: Electronegativity differences between atoms and molecular geometry.
Hydrogen Bonding:
A special type of intermolecular force between hydrogen atoms bonded to highly electronegative atoms (typically oxygen, nitrogen, or fluorine).
Strongest type of intermolecular force.
Responsible for properties like high boiling points of water and ammonia.
Other Intermolecular Forces:
Dipole-dipole interactions: Occur between polar molecules.
London dispersion forces: Weakest type of intermolecular force, present in all molecules.
Molecular Orbital Theory (MO Theory):
Describes bonding and antibonding orbitals formed by the combination of atomic orbitals.
Provides a more accurate picture of bonding in molecules, especially for larger molecules and delocalized systems.
By understanding these concepts, you can predict the structure, shape, polarity, and properties of molecules.
1.Explain the formation of a chemical bond.
Ans : Chemical bonding is the attractive force holding atoms together in molecules and compounds. It arises from the tendency of atoms to attain a stable electronic configuration, often resembling the electron configuration of noble gases.
Types of chemical bonding:
Ionic bonding: Involves the complete transfer of electrons between atoms to form ions with opposite charges.
Covalent bonding: Involves the sharing of electrons between atoms to achieve a stable electron configuration.
Metallic bonding: Occurs in metals and involves the sharing of electrons among all the atoms in the metal lattice.
The formation of a chemical bond is influenced by various factors, including:
Electronegativity measures an atom’s power to pull shared electrons closer to itself compared to other atoms in a bond.
Atomic size: The size of the atoms involved in the bond.
Orbital overlap: The extent to which the orbitals of the bonding atoms overlap.
Understanding chemical bonding is essential for understanding the structure, properties, and reactivity of molecules and compounds.
2. Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.
Ans : Mg: .Mg. (2 valence electrons)
Na: .Na (1 valence electron)
B: .B. (3 valence electrons)
O: :O: (6 valence electrons)
N: :N: (5 valence electrons)
Br: :Br: (7 valence electrons)
In Lewis dot symbols, each dot represents a valence electron. The symbol of the element is placed in the center, and the dots are arranged around it to represent the valence electrons.
3. Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H–
Ans : Lewis symbols provide a shorthand notation for the outer electron arrangement of atoms and ions.Here are the Lewis symbols for the given elements and ions:
S: :S: (6 valence electrons)
S²⁻: [:S:]²⁻ (8 valence electrons)
Al: .Al. (3 valence electrons)
Al³⁺: Al³⁺ (0 valence electrons)
H: .H (1 valence electron)
H⁻: H⁻ (2 valence electrons)
In Lewis dot symbols, each dot represents a valence electron. The symbol of the element or ion is placed in the center, and the dots are arranged around it to represent the valence electrons.
For ions, the charge is indicated outside the brackets. If an ion has a negative charge, it has gained electrons, so the number of dots increases. If an ion has a positive charge, it has lost electrons, so the number of dots decreases.
4. Draw the Lewis structures for the following molecules and ions : H2S, SiCl4, BeF2, CO3 2− , HCOOH
Ans :
5. Define octet rule. Write its significance and limitations.
Ans : Octet Rule:
The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a stable configuration with eight electrons in their outermost shell (valence shell). This stable configuration is similar to the electron configuration of noble gases, which are generally unreactive due to their full valence shells.
Significance of the Octet Rule:
Predicting Lewis Structures: The octet rule helps us predict the Lewis structures of molecules, which in turn helps us understand their shape, polarity, and reactivity.
Understanding Chemical Bonding: The octet rule is a fundamental principle in understanding the nature of chemical bonding, particularly ionic and covalent bonding.
Explaining Stability: The octet rule explains why elements tend to react with each other to form stable compounds.
Limitations of the Octet Rule:
Exceptions: There are many exceptions to the octet rule, especially for elements in the third period and beyond. These valence shell.
Expanded Octet: Elements in the third period and beyond can have expanded octets, meaning they can accommodate more than eight electrons around them. Examples include sulfur in SF₆ and phosphorus in PF₅.
Incomplete Octet: Some molecules, such as BH₃ (borane), have fewer than eight electrons around the central atom.
Despite its limitations, the octet rule remains a valuable tool for understanding the bonding and structure of many molecules. It provides a simple framework for predicting the Lewis structures and properties of compounds.
6. Write the favourable factors for the formation of ionic bond.
Ans : Favorable Factors for the Formation of Ionic Bonds:
Low Ionization Enthalpy (IE): Elements with low ionization enthalpies readily lose electrons to form cations.
High Electron Gain Enthalpy (EGE): Elements with high electron gain enthalpies readily gain electrons to form anions.
Large Lattice Enthalpy: The formation of an ionic compound involves the release of energy due to the electrostatic attraction between the oppositely charged ions. A large lattice enthalpy favors the formation of an ionic bond.
Difference in Electronegativity: A large difference in electronegativity between the two elements involved in the bond promotes the complete transfer of electrons, leading to the formation of ionic bonds.
In summary, ionic bonds are favored between elements with low ionization enthalpies, high electron gain enthalpies, large lattice enthalpies, and a significant difference in electronegativity.
7. Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Ans :
8.Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Ans : The difference in bond angles between NH₃ and H₂O, despite their similar distorted tetrahedral geometries, can be attributed to the greater repulsion between lone pair-lone pair electrons in H₂O compared to lone pair-bond pair repulsions in NH₃.
NH₃: Nitrogen has one lone pair and three bond pairs. The lone pair-bond pair repulsion is significant, but not as strong as the lone pair-lone pair repulsion in water. The resulting bond angle is 107.3 degrees.
H₂O: Oxygen has two lone pairs and two bond pairs. The repulsion between the two lone pairs is stronger than the repulsion between a lone pair and a bond pair. This increased repulsion pushes the bond pairs closer together, resulting in a smaller bond angle of 104.5°.
In summary, the presence of more lone pairs in H₂O compared to NH₃ leads to stronger lone pair-lone pair repulsions, which compress the bond angle in H₂O.
9. How do you express the bond strength in terms of bond order ?
Ans : A stronger bond is associated with a higher bond order.
Higher bond order: Stronger bond
Lower bond order: Weaker bond
Bond order reflects the multiplicity of a covalent bond, with higher bond orders indicating stronger bonds due to increased electron sharing.
For example:
A single covalent bond, formed by sharing one pair of electrons, has a bond order of 1.
A double covalent bond, formed by sharing two pairs of electrons, has a bond order of 2.
A triple covalent bond, formed by sharing three pairs of electrons, has a bond order of 3.
Therefore, a triple bond (higher bond order) is generally stronger than a double bond, which is stronger than a single bond.
10. Define the bond length.
Ans : The bond length represents the average separation between the centers of two bonded atoms. It is an important property of a molecule as it influences its shape, size, and reactivity.
Factors affecting bond length:
Bond order: Higher bond order (more shared electron pairs) generally results in shorter bond lengths.
Atomic radii: Larger atoms have longer bond lengths.
Electronegativity difference: A greater difference in electronegativity between the bonded atoms can lead to shorter bond lengths due to increased ionic character.
Measurement: Bond lengths are typically measured in picometers (pm) or angstroms (Å).
Significance: Bond length is a key factor in determining the stability, reactivity, and properties of a molecule. For example, shorter bond lengths often indicate stronger bonds.
11. Explain the important aspects of resonance with reference to the CO3 2− ion.
Ans :
Resonance in the CO₃²⁻ Ion
Resonance is a concept in chemistry that describes the delocalization of electrons in a molecule or ion. It occurs when multiple valid Lewis structures can be drawn for the same molecule, differing only in the placement of electrons.
In the CO₃²⁻ ion, resonance occurs due to the equivalent positions of the oxygen atoms. All three oxygen atoms are bonded to the central carbon atom, and each oxygen atom shares a double bond with the carbon atom. However, the double bond can be delocalized between any of the three oxygen atoms, leading to three equivalent resonance structures.
Key aspects of resonance in CO₃²⁻:
Equivalent Resonance Structures: All three resonance structures contribute equally to the overall structure of the CO₃²⁻ ion.
Delocalized Electrons: The electrons involved in the double bonds are delocalized over all three oxygen atoms, creating a “cloud” of electron density.
Enhanced Stability: Resonance delocalization increases the stability of the molecule or ion.
Equal Bond Lengths: Due to resonance, all three C-O bonds in the CO₃²⁻ ion are equivalent in length, reflecting the delocalized nature of the electrons.
Reduced Reactivity: Resonance delocalization can reduce the reactivity of a molecule or ion.
12. H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3 ? If not, give reasons for the same.
Ans : The two structures provided cannot be taken as the canonical forms of the resonance hybrid representing H₃PO₃.
Here’s why:
Atom Positions: The positions of the atoms are different in the two structures. In one structure, a hydrogen atom is directly bonded to the phosphorus atom, while in the other, a hydrogen atom is bonded to an oxygen atom. For valid resonance structures, only the positions of electrons can change, not the positions of atoms.
Formal Charges: The formal charges on the atoms are different in the two structures. This violates the principle that resonance structures should have the same formal charges on corresponding atoms.
Therefore, the structures are not equivalent resonance structures and do not accurately represent the resonance hybrid of H₃PO₃.
The correct resonance structures for H₃PO₃ involve the delocalization of the double bond between the phosphorus atom and one of the oxygen atoms.
13.Write the resonance structures for SO3, NO2 and NO3 − .
Ans :
14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N.
Ans :
15.Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
Ans : The difference in the shapes of CO₂ and H₂O molecules, despite both being triatomic, can be explained by their dipole moments.
CO₂:
The CO₂ molecule is linear, with the carbon atom in the center and the oxygen atoms on either side.
The C=O bond exhibits polarity arising from the disparity in electronegativity between carbon and oxygen.
However, the two C=O bonds are identical and point in opposite directions. This results in the dipole moments of the two bonds canceling each other out.
Therefore, CO₂ has a zero dipole moment and is a nonpolar molecule.
H₂O:
The H₂O molecule has a bent shape, with the oxygen atom at the center and the hydrogen atoms at an angle of approximately 104.5°.
The unequal sharing of electrons between oxygen and hydrogen atoms in the O-H bond creates a polar covalent bond.
The two O-H bonds are not identical and do not point in opposite directions. This results in a net dipole moment for the H₂O molecule.
In summary, the difference in shape between CO₂ and H₂O is due to the presence of a net dipole moment in H₂O, while CO₂ has a zero dipole moment due to the cancellation of its bond dipoles.
16. Write the significance/applications of dipole moment.
Ans : Significance and Applications of Dipole Moment:
Dipole moment is a measure of the polarity of a molecule, reflecting the separation of charges within it. It has various significant applications in chemistry and related fields:
Molecular Polarity:
Classifies molecules based on their polarity.
Polar molecules have a dipole moment, while nonpolar molecules have a zero dipole moment.
Polarity affects properties like solubility, boiling point, and intermolecular forces.
Molecular Structure:
Can be used to infer the molecular geometry and the arrangement of atoms.
For example, a molecule with a zero dipole moment is likely to be linear or symmetrical.
Spectroscopy:
Dipole moment influences the intensity of absorption bands in infrared and Raman spectroscopy.
This helps in identifying and studying molecules based on their vibrational and rotational properties.
Drug Design:
Understanding the dipole moments of molecules is crucial in drug design, as it affects interactions with biological targets.
Polar molecules tend to be more soluble in water, which is important for drug absorption and distribution.
Materials Science:
Dipole moment plays a role in the properties of materials like polymers and liquid crystals.
For example, polar polymers tend to have higher melting points and stronger intermolecular forces.
Environmental Science:
Dipole moment influences the solubility of pollutants in water and their transport in the environment.
In conclusion, dipole moment is a valuable property that provides insights into molecular structure, reactivity, and interactions. It has applications in various fields, including chemistry, biology, materials science, and environmental science.
17. Define electronegativity. How does it differ from electron gain enthalpy ?
Ans : Electronegativity reflects an atom’s power to pull shared electrons closer to itself in a chemical bond.
Distinguish between electronegativity and electron gain enthalpy:
Context: Electronegativity refers to atoms in a chemical bond, while electron gain enthalpy refers to an isolated atom.
Focus: Electronegativity focuses on the attraction of shared electrons in a bond, while electron gain enthalpy focuses on the energy change when an atom gains an electron.
Measurement: Electronegativity is a qualitative property, often estimated using empirical scales like the Pauling scale. Electron gain enthalpy is a quantitative property that can be measured experimentally.
In summary, electronegativity reflects an atom’s tendency to attract electrons in a molecule, while electron gain enthalpy measures the energy change when an atom gains an electron.
18.Explain with the help of suitable example polar covalent bond.
Ans :
When two atoms with varying electronegativities share electrons, a polar covalent bond is formed.
Electronegativity is a relative property that quantifies an atom’s electron-pulling power in a molecular context.
Example:
HCl (hydrogen chloride) is a polar covalent molecule. Due to chlorine’s greater electronegativity, the electron density in the H-Cl bond is skewed towards the chlorine atom.This creates a partial positive charge on the hydrogen atom (δ+) and a partial negative charge on the chlorine atom (δ-).
The polarity of a molecule can affect its physical and chemical properties, such as solubility, boiling point, and reactivity.
19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
Ans : The provided explanation is correct and comprehensive. Here’s a simplified breakdown:
Key points:
Ionic character: The degree to which a bond is ionic, rather than covalent.
Electronegativity difference: The larger the difference in electronegativity between two bonded atoms, the more ionic the bond is.
Calculation: To determine ionic character, calculate the absolute difference in electronegativity values for the bonded atoms.
Comparison: A higher electronegativity difference indicates a more ionic bond.
Steps:
Find electronegativity values: Look up the electronegativity values of the elements in the bond.
Calculate difference: Subtract the smaller electronegativity value from the larger one.
Compare: The larger the difference, the more ionic the bond.
Example:
LiF: Electronegativity difference = |4.0 – 1.0| = 3.0 (highly ionic)
N₂: Electronegativity difference = |3.0 – 3.0| = 0.0 (covalent)
Conclusion:
The order of increasing ionic character is N₂ < SO₂ < ClF₃ < K₂O < LiF, based on the calculated electronegativity differences.
20. The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
Ans :
21.Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ?
Ans : The square planar geometry is not possible for CH₄ because of the VSEPR (Valence Shell Electron Pair Repulsion) theory. This theory states that electron pairs (both bonding and lone pairs) around a central atom will arrange themselves in a way that minimizes repulsion between them.
In CH₄, there are four bonding pairs (one between the carbon atom and each hydrogen atom). According to VSEPR theory, these bonding pairs will arrange themselves in a way that maximizes the distance between them. A square planar arrangement would result in the hydrogen atoms being closer together than in a tetrahedral arrangement. This would lead to greater repulsion between the electron pairs, making the square planar geometry less stable.
Therefore, the tetrahedral geometry, where the hydrogen atoms are arranged at the corners of a tetrahedron with the carbon atom at its center, is the most stable arrangement for CH₄, as it minimizes electron pair repulsion.
22. Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
Ans : BeH₂ is a linear molecule with a zero dipole moment despite the polarity of its Be-H bonds.
This is due to the symmetry of the molecule. In a linear arrangement, the two Be-H bonds are oriented directly opposite each other. This means that the dipole moments of the two bonds are equal in magnitude but point in opposite directions. As a result, the dipole moments cancel each other out, resulting in a net dipole moment of zero.
Therefore, even though the Be-H bonds are polar, the symmetrical arrangement of these bonds in BeH₂ leads to a nonpolar molecule with a zero dipole moment.
23. Which out of NH3 and NF3 has higher dipole moment and why ?
Ans : NH₃ has a higher dipole moment than NF₃.
Here’s why:
Electronegativity Difference: Fluorine (F) is more electronegative than hydrogen (H). This means that in the N-F bond, the electrons are pulled closer to the fluorine atom, creating a larger partial negative charge on fluorine and a larger partial positive charge on nitrogen.
Lone Pair Orientation: In NH₃, the lone pair on the nitrogen atom is oriented in the same direction as the bond dipoles, reinforcing the overall dipole moment. In NF₃, the lone pair is oriented in the opposite direction to the bond dipoles, partially canceling out their effects.
Therefore, due to the greater electronegativity difference in N-H bonds and the favorable orientation of the lone pair in NH₃, it has a higher dipole moment than NF₃.
24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Ans : Hybridisation of Atomic Orbitals:
Hybridisation is the process of mixing atomic orbitals of similar energy to form new hybrid orbitals with different shapes and energies. This mixing allows for the formation of stronger bonds and more stable molecular structures.
Shapes of Hybrid Orbitals:
sp:
Results from the combination of one s and one p orbital.
Linear arrangement.
Bond angle of 180 degrees.
sp²:
Arises from the mixing of an s orbital and two p orbitals.
Trigonal planar arrangement.
Bond angle of 120 degrees.
sp³:
Is generated by the blending of one s orbital and three p orbitals.
Tetrahedral arrangement.
Bond angle of 109.5 degrees.
The hybridization of atomic orbitals in a molecule can be determined by counting the number of sigma bonds and lone pairs around the central atom. For example, a central atom with two sigma bonds and no lone pairs would be sp hybridized.
25. Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl3 + Cl– ——> AlCl4- .
Ans : In the reaction AlCl₃ + Cl⁻ → AlCl₄⁻, the hybridization of the Al atom changes from sp² to sp³.
Here’s a breakdown:
AlCl₃:
Al has 3 valence electrons.
It undergoes sp² hybridization to form 3 hybrid orbitals that bond with the 3 Cl atoms.
This gives AlCl₃ a trigonal planar geometry.
AlCl₄⁻:
When Cl⁻ is added, Al now has 4 electron pairs to bond with.
To accommodate these 4 pairs, Al undergoes sp³ hybridization, forming 4 hybrid orbitals.
This change in hybridization results in a tetrahedral geometry for AlCl₄⁻.
Therefore, the hybridization of the Al atom changes from sp² in AlCl₃ to sp³ in AlCl₄⁻.
26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
Ans : Yes, there is a change in hybridization for both B and N atoms in the reaction.
Reaction: BF₃ + NH₃ → F₃B-NH₃
Analysis:
BF₃:
Boron (B) has 3 valence electrons.
To form bonds with the three fluorine atoms, boron undergoes sp² hybridization. This results in a trigonal planar geometry for BF₃.
NH₃:
Nitrogen (N) has 5 valence electrons.
To form bonds with the three hydrogen atoms, nitrogen undergoes sp³ hybridization. This results in a trigonal pyramidal geometry for NH₃.
F₃B-NH₃:
When BF₃ and NH₃ react, the boron atom forms a coordinate covalent bond with the lone pair on the nitrogen atom.
In this new complex, the boron atom now has 4 electron pairs to bond with (3 from the original BF₃ and 1 from the lone pair on NH₃).
To accommodate these 4 pairs, boron undergoes sp³ hybridization.
Nitrogen, in this complex, still has 4 electron pairs (3 from the original NH₃ and 1 from the shared pair with boron). So, its hybridization remains sp³.
Conclusion:
Boron (B): Changes from sp² hybridization in BF₃ to sp³ hybridization in the complex.
Nitrogen (N): Maintains its sp³ hybridization.
Therefore, there is a change in hybridization for boron, but nitrogen retains its sp³ hybridization in the reaction.
27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
Ans :
28. What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2
(b) C2H4
Ans : (a) H—C = C—H
Sigma bond = 3 Π bonds = 2
29. Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s
(b) 1s and 2px ;
(c) 2py and 2py
(d) 1s and 2s.
Ans : The incorrect option is (c) 2py and 2py.
Reasoning:
Sigma Bonds
The axial overlap of atomic orbitals forms sigma bonds.
Sigma bonds are characterized by end-to-end overlap along the internuclear axis.
2py Orbitals
The 2py orbitals are oriented perpendicular to the internuclear axis, forming a dumbbell shape.
The 2py orbitals have a side-to-side orientation.
Lateral Overlap of 2py Orbitals
The lateral overlap of 2py orbitals does not lead to the formation of a sigma bond.
The side-to-side overlap of 2py orbitals results in a pi bond, not a sigma bond.
A pi bond is formed when the parallel lobes of two p orbitals overlap side by side.
The other options (a), (b), and (d) all involve head-to-head overlap of atomic orbitals and can form sigma bonds.
30.Which hybrid orbitals are used by carbon atoms in the following molecules? (a)CH3–CH3;
(b) CH3–CH=CH2;
(c) CH3-CH2-OH;
(d) CH3-CHO
(e) CH3COOH
Ans :
31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.
Ans : Bond Pairs and Lone Pairs of Electrons
In a molecule, electrons are either involved in forming bonds between atoms (bonding pairs) or are not involved in bonding (lone pairs).
Bond Pairs:
Definition: Electron pairs that are shared between two atoms to form a covalent bond.
Example: In the water molecule (H₂O), the two O-H bonds involve two bond pairs of electrons.
Lone Pairs:
Definition: Electron pairs that are not involved in bonding and are localized on a single atom.
Example: In the water molecule (H₂O), the oxygen atom has two lone pairs of electrons.
These bond pairs and lone pairs influence the molecular geometry and polarity of a molecule, as explained by VSEPR theory.
32. Distinguish between a sigma and a pi bond.
Ans : Sigma (σ) Bond:
Formation: Formed by the head-to-head overlap of atomic orbitals along the internuclear axis.
Shape: Cylindrical shape around the internuclear axis.
Strength: Generally stronger than pi bonds due to greater overlap.
Examples: Single bonds in molecules like H₂, HCl, and CH₄.
Pi (π) Bond:
Formation: Formed by the side-by-side overlap of atomic orbitals.
Shape: Dumbbell shape above and below the internuclear axis.
Strength: Generally weaker than sigma bonds due to less overlap.
Examples: Double bonds (e.g., in O₂) and triple bonds (e.g., in N₂) contain pi bonds in addition to sigma bonds.
In summary, sigma bonds are stronger due to greater overlap and are formed by head-to-head overlap, while pi bonds are weaker due to less overlap and are formed by side-by-side overlap.
33. Explain the formation of H2 molecule on the basis of valence bond theory.
Ans : Formation of the H₂ Molecule According to Valence Bond Theory
Valence Bond Theory (VBT) explains the formation of a chemical bond as the overlapping of half-filled atomic orbitals of two atoms. In the case of the H₂ molecule, the 1s orbitals of two hydrogen atoms overlap to form a sigma bond.
Steps:
Approach of Hydrogen Atoms: Two hydrogen atoms, each with a single electron in their 1s orbitals, approach each other.
Orbital Overlap: As the atoms come closer, their 1s orbitals begin to overlap. This overlap is head-to-head, meaning the two orbitals overlap directly along the internuclear axis.
Electron Pairing: The two unpaired electrons in the 1s orbitals pair up to form a shared electron pair.
Sigma Bond Formation: This shared electron pair constitutes a sigma bond, which holds the two hydrogen atoms together in the H₂ molecule.
Key Points:
The combination of atomic orbitals results in the creation of molecular orbitals.
The shared electrons in the molecular orbital are attracted to both nuclei, creating a stable bond.
The H₂ molecule is a simple example of covalent bonding, where two atoms share electrons to achieve a stable electron configuration.
34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Ans : Important Conditions for the Linear Combination of Atomic Orbitals to Form
Molecular Orbitals:
Similar Energies: The atomic orbitals involved in the combination should have similar energies. This is because orbitals with significantly different energies do not overlap effectively.
Proper Overlap: The atomic orbitals must overlap with each other. The type of overlap determines the type of molecular orbital formed (sigma or pi).
Symmetry: The atomic orbitals must have the same symmetry with respect to the internuclear axis. For example, two s orbitals or two p orbitals with the same orientation can overlap.
Same Spin: The electrons in the atomic orbitals must have opposite spins for effective bonding. This is due to Pauli’s exclusion principle, which states that no two electrons in an atom can have the same set of quantum numbers.
When these conditions are met, the atomic orbitals can combine to form molecular orbitals, which represent the distribution of electrons in the molecule. The resulting molecular orbitals can be bonding orbitals (lower in energy) or antibonding orbitals (higher in energy).
35.Use molecular orbital theory to explain why the Be2 molecule does not exist.
Ans : Why Be₂ Molecule Does Not Exist: A Molecular Orbital Theory Explanation
Molecular Orbital Theory (MO theory) provides a framework for understanding the bonding and stability of molecules. It involves the combination of atomic orbitals to form molecular orbitals, which represent the distribution of electrons in the molecule.
For Be₂:
Atomic Orbital Configuration: Each beryllium atom has the electron configuration 1s² 2s².
Molecular Orbital Formation: When two beryllium atoms combine, their 2s orbitals interact to form two molecular orbitals: a bonding σ₂s orbital and an antibonding σ₂s* orbital.
Electron Filling: Each beryllium atom contributes two valence electrons, for a total of four electrons. These electrons fill the σ₂s orbital.
Bond Order Calculation:
The bond order is calculated as the difference between the bonding and antibonding electrons, divided by two. For Be₂, the bond order is zero, indicating no net bonding.
A bond order of 0 indicates that there are no net bonding interactions between the two beryllium atoms. This means that the Be₂ molecule is unstable and does not exist.
In summary, the Be₂ molecule does not exist because its molecular orbital configuration results in a bond order of 0, indicating no net bonding interaction between the two beryllium atoms.
36. Compare the relative stability of the following species and indicate their magnetic properties; O2, O2, O2– (Superoxide),O22- (peroxide)
Ans : Comparison of O₂, O₂⁻, and O₂²⁻
Stability:
O₂: The neutral oxygen molecule has a bond order of 2, indicating a stable double bond.
O₂⁻ (Superoxide): The superoxide ion has a bond order of 1.5, indicating a weaker bond than O₂. It is less stable due to the presence of an unpaired electron.
O₂²⁻ (Peroxide): The peroxide ion has a bond order of 1, indicating an even weaker bond than the superoxide ion. It is the least stable among the given species due to the presence of two unpaired electrons.
Magnetic Properties:
O₂: The neutral oxygen molecule has two unpaired electrons, making it paramagnetic. This means it is attracted to a magnetic field.
O₂⁻ (Superoxide): The superoxide ion has one unpaired electron, making it paramagnetic as well.
O₂²⁻ (Peroxide): The peroxide ion has all its electrons paired, making it diamagnetic. This means it is not attracted to a magnetic field.
In summary:
Stability: O₂ > O₂⁻ > O₂²⁻
Magnetic Properties: O₂ and O₂⁻ are paramagnetic, while O₂²⁻ is diamagnetic.
The stability and magnetic properties of these species are directly related to their bond orders and the presence or absence of unpaired electrons.
37. Write the significance of a plus and a minus sign shown in representing the orbitals.
Ans : The plus and minus signs shown in representing orbitals signify the phase of the wave function.
In molecular orbital theory, orbitals are represented by wave functions, which are mathematical functions that describe the probability of finding an electron in a particular region of space. The plus and minus signs indicate the relative phase of the wave function at different points in space.
Positive sign: Represents a region of the orbital where the wave function is positive. This can be thought of as a “crest” or a “peak” of the wave.
Negative sign: Represents a region of the orbital where the wave function is negative. This can be thought of as a “trough” or a “valley” of the wave.
The phase of the wave function is important in determining whether two orbitals can combine to form a bonding or antibonding molecular orbital. Bonding orbitals are formed when orbitals with the same phase overlap, while antibonding orbitals are formed when orbitals with opposite phases overlap.
38. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Ans :
Hybridization in PCl₅:
Phosphorus (P) in PCl₅ has five valence electrons. To accommodate the five chlorine atoms, it undergoes sp³d hybridization. This involves mixing one s orbital, three p orbitals, and one d orbital to form five equivalent hybrid orbitals.
Bond Lengths:
The five P-Cl bonds in PCl₅ are not equivalent in length. The axial bonds (those lying along the principal axis) are longer than the equatorial bonds (those lying in the equatorial plane).
Reason:
This difference in bond length is due to lone pair-bond pair repulsion. In PCl₅, the lone pair on the phosphorus atom occupies a hybrid orbital that is more localized along the principal axis. This lone pair exerts a greater repulsion on the axial bonds compared to the equatorial bonds, causing them to be slightly longer.
In summary, the hybridization in PCl₅ is sp³d, and the axial bonds are longer than the equatorial bonds due to the repulsion from the lone pair on the phosphorus atom.
39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Ans : A hydrogen bond is a specific type of intermolecular interaction involving a hydrogen atom bonded to a highly electronegative atom and a lone pair on another electronegative atom.
Strength: Hydrogen bonds are significantly stronger than van der Waals forces. While van der Waals forces are caused by temporary fluctuations in electron density, hydrogen bonds involve a more permanent dipole-dipole interaction between the hydrogen atom and the lone pair. This results in a stronger attraction between the molecules.
Significance of Hydrogen Bonding:
High boiling points: Hydrogen bonding is responsible for the unusually high boiling points of substances like water and ammonia compared to other molecules of similar molecular weight.
Structure and properties: Hydrogen bonding plays a crucial role in determining the structure and properties of many biological molecules, such as proteins and DNA.
Solubility: Hydrogen bonding affects the solubility of substances in water. Substances that can form hydrogen bonds with water are more likely to be soluble in it.
In conclusion, hydrogen bonding is a stronger type of intermolecular force compared to van der Waals forces, and it has significant implications for the properties of many substances.
40. What is meant by the term bond order? Calculate the bond order of : N2, O2, O2 + and O2 – .
Ans : Bond order is defined as the half of the difference between the number of electrons present in bonding and antibonding molecular orbitals.
