GSEB Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure

Question 1. Explain the formation of a chemical bond.
Answer: Atoms from the same or different elements combine, which leads to the creation of a chemical bond between them. Several theories, proposed over time, help explain how these chemical bonds form. Different element atoms combine to complete their octets or duplets, helping them achieve a stable inert gas configuration. They accomplish this by either transferring electrons or sharing electrons. Another way to explain chemical bond formation is that it lowers the energy of the system, allowing the constituent atoms to gain greater stability.
In simple words: A chemical bond forms when atoms join together. They do this to become stable, either by sharing or moving electrons around, which also lowers their energy.

Exam Tip: When explaining chemical bond formation, remember to mention both electron transfer/sharing and the goal of achieving a stable electron configuration (octet/duplet) to attain lower energy and greater stability.

Question 2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Answer:
Mg ⋅
Na ⋅
⋅ B ⋅
: \( \ddot{\text{O}} \):
: \( \dot{\text{N}} \):
⋅ \( \ddot{\text{Br}} \):
In simple words: To draw Lewis dot symbols, first find out how many valence electrons each atom has. Then, put that many dots around the atom's symbol.

Exam Tip: Always remember that Lewis dot symbols only show the valence electrons, which are the outermost electrons involved in bonding. For elements in the same group, the number of valence electrons is the same.

Question 3. Write Lewis symbols for the following atoms and ions: S and S\(^{2-}\), Al and Al\(^{3+}\); H and H\(^-\).
Answer:
Sulfur: :\( \ddot{\text{S}} \):
Sulfide ion: :\( \ddot{\text{S}} \)\(^{2-}\)
Aluminum: ⋅ Al ⋅
Aluminum ion: Al\(^{3+}\)
Hydrogen: H ⋅
Hydride ion: H\(^-\)
In simple words: For atoms, show the valence electrons as dots. For ions, add or remove dots to reflect the charge, and put brackets around the ion with the charge outside.

Exam Tip: Pay close attention to the charge on the ion. A negative charge means more electrons (added dots), and a positive charge means fewer electrons (removed dots).

Question 4. Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, CO\(_{3}\)\(^{2-}\), HCOOH
Answer:
For H2S: H:\( \ddot{\text{S}} \):H or H-S-H

For SiCl4:
         : \( \ddot{\text{Cl}} \):
            |
      :\( \ddot{\text{Cl}} \):Si:\( \ddot{\text{Cl}} \): or :\( \ddot{\text{Cl}} \)-Si-\( \ddot{\text{Cl}} \):
            |
         : \( \ddot{\text{Cl}} \):

For BeF2: :\( \ddot{\text{F}} \):Be::\( \ddot{\text{F}} \): or :\( \ddot{\text{F}} \)-Be-\( \ddot{\text{F}} \):

For CO\(_{3}\)\(^{2-}\):
             : \( \ddot{\text{O}} \):
            //
        [ :\( \ddot{\text{O}} \)-C-\( \ddot{\text{O}} \): ]\(^{2-}\)

For HCOOH:
             : \( \ddot{\text{O}} \):
            //
        H-C-\( \ddot{\text{O}} \)-H
In simple words: Draw each atom, add its valence electrons as dots, then connect atoms with lines for shared electron pairs (bonds) until each atom (except hydrogen) has eight electrons around it. For ions, add or subtract electrons for the charge.

Exam Tip: When drawing Lewis structures, always start by counting the total valence electrons. Identify the central atom, usually the least electronegative one (except hydrogen), and then arrange the other atoms around it. Distribute electrons to satisfy the octet rule for all atoms, forming multiple bonds if necessary.

Question 5. Define the octet rule. Write its significance and limitations.
Answer: The octet rule states that atoms can combine by either transferring valence electrons from one atom to another (gaining or losing) or by sharing valence electrons. This allows them to achieve an octet (eight electrons) in their outermost shells.
**Significance:** Achieving eight electrons (octet formation) in the outermost shells of atoms leads to a stable electronic configuration. This is because all noble gases (except Helium) have eight electrons in their outermost shells, and they generally do not react chemically. All other elements in the periodic table have fewer than eight electrons in their valence shells, which makes them chemically reactive. Thus, atoms of elements combine to acquire eight electrons in their outermost shells to gain inertness or stability, similar to the nearest noble gas.
**Limitations:** (1) **Formations of compounds involving hydrogen:** A hydrogen atom has only one electron in its valence shell, needing just one more electron to gain the nearest noble gas configuration of Helium. Therefore, hydrogen aims to complete its duplet (acquiring 2 electrons in its valence shell) rather than an octet.
(2) **Formation of electron-deficient compounds:** This includes compounds like BeCl\(_{2}\), BF\(_{3}\), AlCl\(_{3}\). In these examples, the central atom (Be in BeCl\(_{2}\), B in BF\(_{3}\), and Al in AlCl\(_{3}\)) has fewer than eight electrons. These compounds are considered electron-deficient. For example:
:\( \ddot{\text{Cl}} \):Be:\( \ddot{\text{Cl}} \): (Beryllium has only 4 electrons around it)
       : \( \ddot{\text{F}} \):
      /
  :\( \ddot{\text{F}} \):B:\( \ddot{\text{F}} \): (Boron has only 6 electrons around it)

(3) **Formation of expanded octet compounds:** This includes compounds like PCl\(_{5}\), SF\(_{6}\), IF\(_{7}\). In all these compounds, the central atom has more than eight electrons. For example:
            :\( \ddot{\text{Cl}} \):
            |
        :\( \ddot{\text{Cl}} \):P:\( \ddot{\text{Cl}} \): (Phosphorus has 10 electrons around it)
            |
            :\( \ddot{\text{Cl}} \):

(4) **Formation of compounds of noble gases:** Noble gases, which the octet rule is based on, already have eight electrons in their valence shells and thus should not form compounds. However, Xenon (Xe) has recently been found to form compounds like XeF\(_{2}\), XeF\(_{4}\), XeF\(_{6}\), etc.
(5) **Odd electron bonds/Odd electron molecules:** There are certain molecules and ions, such as NO and \( \text{O}_{2}^{-} \), where the bonded atoms contain an odd number of electrons between them. For example:
Nitric oxide [NO]: :\( \dot{\text{N}} \):\( \ddot{\text{O}} \): or :\( \ddot{\text{N}} \):\( \dot{\text{O}} \):
Superoxide ion (\( \text{O}_{2}^{-} \)): [ :\( \ddot{\text{O}} \):\( \dot{\text{O}} \): ]\(^{-} \) or [ \( \ddot{\text{O}} \)- \( \dot{\text{O}} \): ]\(^{-} \)

(6) It does not explain the shape of the molecules and their relative stability.
In simple words: The octet rule says atoms try to get eight electrons in their outer shell to be stable, like noble gases. But this rule doesn't always work. For example, hydrogen only needs two, some atoms can have fewer than eight or more than eight, and even some noble gases can form bonds. Also, it doesn't tell us about a molecule's shape.

Exam Tip: When defining the octet rule, clearly state its aim for stable electron configuration. For significance, link it to noble gas stability. For limitations, give a specific example for each point: duplet rule (H), incomplete octet (BeCl\(_{2}\), BF\(_{3}\)), expanded octet (PCl\(_{5}\), SF\(_{6}\)), noble gas compounds (XeF\(_{2}\)), and odd-electron molecules (NO).

Question 6. Write the favorable factors for the formation of an ionic bond.
Answer: The formation of an ionic bond mostly depends on the following factors:
1. The ease with which positive and negative ions form from their neutral atoms.
2. The arrangement of positive and negative ions in the solid state, known as the lattice of the crystalline compound. These arrangements, in turn, depend on:
    (1) **Low Ionization Enthalpy:** The atom that is losing electrons to create a positive ion (cation) must lose them easily. This means the element must have a low ionization enthalpy. For example, for M(g) \( \rightarrow \) M\(^+\)(g) + e\(^-\) [Ionization enthalpy], M should have a low ionization enthalpy, which is common for active metals like Na, K, Mg, and Ca.
    (2) **High Negative Electron Gain Enthalpy:** The atom that is gaining electrons must be able to hold onto them effectively. This means it must have a high negative value of electron gain enthalpy. For example, for X(g) + e\(^-\) \( \rightarrow \) X\(^-\)(g) [Electron gain enthalpy].
    (3) **High Enthalpy of Lattice Formation:** The energy released when the necessary number of positive and negative ions combine to create one mole of the ionic compound is called lattice energy or enthalpy. A higher value of lattice energy for the resulting ionic compound means greater stability of the compound, which makes its formation easier. For example, M\(^+\)(g) + X\(^-\)(g) \( \rightarrow \) MX(s) (1 Mole) [Lattice enthalpy].
In simple words: For an ionic bond to form easily, the first atom should easily lose electrons (low energy needed), the second atom should easily gain electrons (high energy released), and when they join, a lot of energy should be released to make the compound strong and stable.

Exam Tip: Remember the three key factors for ionic bond formation: low ionization enthalpy (for cation formation), high negative electron gain enthalpy (for anion formation), and high lattice energy (for crystal stability). These factors ensure the overall process is energetically favorable.

Question 7. Discuss the shape of the following molecules using the VSEPR model : BeCl\(_{2}\), BCl\(_{3}\), SiCl\(_{4}\), AsF\(_{5}\), H\(_{2}\)S, PH\(_{3}\).
Answer: According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the shape of a molecule depends on the repulsion between electron pairs in the valence shell of the central atom. The order of repulsion between electron pairs is: lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsions. The exact shape of the molecule depends on the total number of electron pairs present around the central atom.
**BeCl\(_{2}\): Linear**
        \( \text{Cl}-\text{Be}-\text{Cl} \)
            \( 180^\circ \)
(No lone pairs of electrons on Be). The central atom has only 2 bond pairs of electrons, leading to a linear shape with a 180° bond angle.
**BCl\(_{3}\): Trigonal Planar**
          \( \text{Cl} \)
          /
      \( \text{Cl}-\text{B}-\text{Cl} \)
          \( 120^\circ \)
(No lone pairs of electrons on B). The central atom has only 3 bond pairs of electrons, resulting in a trigonal planar shape with 120° bond angles.
**SiCl\(_{4}\): Tetrahedral**
          \( \text{Cl} \)
          |
      \( \text{Cl}-\text{Si}-\text{Cl} \)
          |
          \( \text{Cl} \)
          \( 109.5^\circ \)
(No lone pairs of electrons on Si). Silicon has 4 bond pairs of electrons, leading to a tetrahedral shape with approximately 109.5° bond angles.
**AsF\(_{5}\): Trigonal Bipyramidal**
              \( \text{F} \)
              |
        \( \text{F} \)-As-\( \text{F} \)
        /     \ \
    \( \text{F} \)       \( \text{F} \)
      \( 90^\circ, 120^\circ \)
(No lone pairs of electrons on As). Arsenic has 5 bond pairs of electrons, resulting in a trigonal bipyramidal shape with 90° and 120° bond angles.
**H\(_{2}\)S: Bent (Angular)**
        H
      /
   S
    \
    H
If all electron pairs were bond pairs, the shape would have been tetrahedral. However, sulfur has two lone pairs of electrons, which cause greater repulsion, distorting the tetrahedral shape into a bent or angular structure. (2 bond pairs, 2 lone pairs)
**PH\(_{3}\): Pyramidal**
         P
      / | \
    H H H
If there were a bond pair in place of the lone pair, the shape would have been tetrahedral. However, the presence of one lone pair on phosphorus causes greater repulsion, reducing the angle between the bond pairs and resulting in a distorted trigonal bipyramidal/pyramidal shape. (3 bond pairs, 1 lone pair)
In simple words: The VSEPR theory says that electron pairs around a central atom push each other away. Lone pairs push more strongly than bonding pairs. This pushing determines the molecule's shape, like linear for BeCl\(_{2}\), trigonal planar for BCl\(_{3}\), tetrahedral for SiCl\(_{4}\), trigonal bipyramidal for AsF\(_{5}\), bent for H\(_{2}\)S, and pyramidal for PH\(_{3}\).

Exam Tip: To determine molecular shape using VSEPR, first find the central atom. Count its total valence electron pairs (bond pairs + lone pairs). Determine the electron pair geometry (e.g., linear, trigonal planar, tetrahedral). Then, consider how lone pairs affect the molecular geometry, noting that lone pairs cause more repulsion, leading to distorted shapes and smaller bond angles.

Question 8. Although geometries of NH\(_{3}\) and H\(_{2}\)O molecules are distorted tetrahedral, the bond angle in water is less than that of NH\(_{3}\). Discuss.
Answer: Both NH\(_{3}\) and H\(_{2}\)O molecules involve sp\(^{3}\) hybridization, so their expected ideal shape should be tetrahedral. However, their actual geometries are distorted tetrahedral due to the presence of lone pairs of electrons on the central atoms. In NH\(_{3}\), there is one lone pair of electrons on the nitrogen atom, while in H\(_{2}\)O, there are two lone pairs of electrons on the oxygen atom.
According to VSEPR theory, lone pair-lone pair repulsions are stronger than lone pair-bond pair repulsions, which are stronger than bond pair-bond pair repulsions. In H\(_{2}\)O, the two lone pairs on oxygen exert a greater repulsive force on the bonded pairs of electrons compared to the single lone pair in NH\(_{3}\). This stronger repulsion in H\(_{2}\)O causes the bond angle between the bond pairs to shrink more significantly, from 107° in NH\(_{3}\) to 104.5° in H\(_{2}\)O.
Diagram for NH\(_{3}\) (One lone pair):
          \( \dot{\text{N}} \)
        / | \
      H H H
        \( 107^\circ \)

Diagram for H\(_{2}\)O (Two lone pairs):
          \( \ddot{\text{O}} \)
        /   \
      H     H
        \( 104.5^\circ \)
In simple words: Both ammonia (NH\(_{3}\)) and water (H\(_{2}\)O) have a basic tetrahedral shape, but they get squished because of unshared electron pairs. Water has two such pairs, which push harder on the other electrons than the one unshared pair in ammonia. This stronger push makes the H\(_{2}\)O bond angle smaller than the NH\(_{3}\) bond angle.

Exam Tip: When comparing bond angles, remember the key principle of VSEPR theory: lone pair-lone pair repulsions are strongest, followed by lone pair-bond pair, and then bond pair-bond pair. The more lone pairs present, the greater the repulsion and the smaller the bond angle.

Question 9. How do you express bond strength in terms of bond order?
Answer: Bond strength is directly proportional to bond order. A higher bond order indicates a stronger bond.
In simple words: The more bonds there are between two atoms (higher bond order), the stronger those atoms are held together.

Exam Tip: A simple way to remember is: single bond (order 1) is weaker than a double bond (order 2), which is weaker than a triple bond (order 3). So, higher bond order equals stronger bond and shorter bond length.

Question 10. Define the bond length.
Answer: Bond length is defined as the average distance between the centers of the nuclei of two atoms that are chemically bonded together in a molecule.
In simple words: Bond length is how far apart the middle of two bonded atoms are from each other.

Exam Tip: Bond length is typically measured in Ångströms (\( \text{Å} \)) or picometers (pm). It's an average value because atoms vibrate, so the distance isn't absolutely fixed.

Question 11. Explain the important aspects of resonance with reference to the CO\(_{3}\)\(^{2-}\) ion.
Answer: For the carbonate ion (\( \text{CO}_{3}^{2-} \)), a single Lewis structure showing two single bonds and one double bond between carbon and oxygen atoms is not enough to accurately represent the molecule. This is because such a structure would suggest unequal bond lengths, with the double bond being shorter than the single bonds.
However, experimental findings show that all carbon-to-oxygen bonds in \( \text{CO}_{3}^{2-} \) are identical in length. This indicates that the molecule is better described as a resonance hybrid of several contributing structures (canonical forms). In a resonance hybrid, the actual structure is an average of these contributing forms, and electrons are delocalized over all the atoms involved, leading to bonds of intermediate length and strength.
The contributing (canonical) forms for the carbonate ion are:

**Canonical Form I:**
              :\( \ddot{\text{O}} \):\(^{-} \)
              /
        :\( \ddot{\text{O}} \)=C-\( \ddot{\text{O}} \):\(^{-} \)

**Canonical Form II:**
              :\( \ddot{\text{O}} \):\(^{-} \)
              |
        :\( \ddot{\text{O}} \):-C= \( \ddot{\text{O}} \):
                    \(^{-} \)

**Canonical Form III:**
              :\( \ddot{\text{O}} \):
              |
        :\( \ddot{\text{O}} \):\(^{-} \)-C= \( \ddot{\text{O}} \):\(^{-} \)

Therefore, the carbonate ion is best described as a resonance hybrid of these three canonical forms, where the negative charge and the double bond character are spread equally among all three C-O bonds.
In simple words: For the carbonate ion, we can't draw one Lewis picture that shows its real structure because all its bonds are equal. Instead, we use "resonance," which means the real structure is a mix of a few different possible pictures. The electrons are shared over all the bonds, making them all the same length.

Exam Tip: When discussing resonance, emphasize that the actual molecule is *not* rapidly shifting between canonical forms but rather exists as a single, stable resonance hybrid. The key indicators for resonance are identical bond lengths and distributed charge, which a single Lewis structure cannot represent.

Question 12. H\(_{3}\)PO\(_{3}\) 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 H\(_{3}\)PO\(_{3}\)? If not, give reasons for the same.
Answer: No, these two structures cannot be considered as canonical forms of the resonance hybrid for H\(_{3}\)PO\(_{3}\).
**Structure (1):**
        H
        |
    H:\( \ddot{\text{O}} \):\( \ddot{\text{P}} \):\( \ddot{\text{O}} \):H
            :\( \ddot{\text{O}} \):

**Structure (2):**
        H
        |
    H:\( \ddot{\text{O}} \):\( \ddot{\text{P}} \):\( \ddot{\text{O}} \):H
        ||
        :\( \ddot{\text{O}} \):

The reason is that for structures to be canonical forms, only the positions of electrons (especially pi electrons and lone pairs) can change, while the positions of the atomic nuclei must remain fixed. In the given structures (1) and (2) for H\(_{3}\)PO\(_{3}\), the positions of the hydrogen atoms relative to the phosphorus and oxygen atoms are different. In structure (1), all three hydrogens are bonded to oxygen atoms, while in structure (2), one hydrogen is directly bonded to the phosphorus atom and two hydrogens are bonded to oxygen atoms. Since the positions of atoms (nuclei) have changed, these cannot be resonance structures.
In simple words: For structures to be "resonance forms," only the electrons can move, not the atoms themselves. In the given H\(_{3}\)PO\(_{3}\) examples, the hydrogen atoms are connected differently in each picture. Because atoms have moved, these are not resonance forms, but different molecules or isomers.

Exam Tip: A crucial rule for resonance is that the skeleton of atoms must remain the same in all canonical forms. Only the distribution of pi electrons and lone pairs can vary. If atom positions differ, the structures represent different compounds or isomers, not resonance contributors of the same compound.

Question 13. Write the resonance structures for SO\(_{3}\) NO\(_{2}\) and NO\(_{3}^{-}\)
Answer:
**Resonance structures of SO\(_{3}\):**
              :\( \ddot{\text{O}} \):
              ||
      :\( \ddot{\text{O}} \)=S-\( \ddot{\text{O}} \):  \( \leftrightarrow \)   :\( \ddot{\text{O}} \)-S=\( \ddot{\text{O}} \):  \( \leftrightarrow \)   :\( \ddot{\text{O}} \)=S-\( \ddot{\text{O}} \):
                                                                  ||
                                                                  :\( \ddot{\text{O}} \):

**Resonance structures of NO\(_{2}\):**
              :\( \dot{\text{O}} \):                   :\( \ddot{\text{O}} \):
              /                         /
        [ \( \text{O} \)-\( \dot{\text{N}} \)=\( \text{O} \) ] \(^{+} \)   \( \leftrightarrow \)   [ \( \text{O} \)=\( \dot{\text{N}} \)-\( \text{O} \) ] \(^{+} \)

**Resonance structures of NO\(_{3}^{-}\) ion:**
              :\( \ddot{\text{O}} \):\(^{-} \)
              /
        :\( \ddot{\text{O}} \)=\( \text{N} \)-\( \ddot{\text{O}} \):\(^{-} \)   \( \leftrightarrow \)   :\( \ddot{\text{O}} \)-\( \text{N} \)=\( \ddot{\text{O}} \):   \( \leftrightarrow \)   :\( \ddot{\text{O}} \)=\( \text{N} \)-\( \ddot{\text{O}} \):\(^{-} \)
                                                                        |
                                                                    :\( \ddot{\text{O}} \):\(^{-} \)
In simple words: Resonance happens when a molecule's true structure is a mix of a few different drawings because electrons are shared. For SO\(_{3}\), NO\(_{2}\), and NO\(_{3}^{-}\), you can draw several structures by moving double bonds and lone pairs, and the real molecule is a blend of all these pictures.

Exam Tip: When drawing resonance structures, remember that the total number of valence electrons and formal charges must be conserved across all canonical forms. Use double-headed arrows (\( \leftrightarrow \)) to indicate resonance between structures, and brackets for ions with the charge outside.

Question 14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:
**1. K and S**
**2. Ca and O**
**3. Al and N.**
Answer:
(1) **K and S:**
K\( \cdot \)   \( \rightarrow \)   K\(^{+} \) + e\(^-\)
(2, 8, 8, 1)     (2, 8, 8)
                      or
\( \cdot \)K     \( \rightarrow \)   K\(^{+} \) + e\(^-\)
(2, 8, 8, 1)     (2, 8, 8)

S + 2e\(^-\)   \( \rightarrow \)   S\(^{2-}\)

Combined transfer:
                          \( \text{K} \)
      \( \text{K} \cdot \)     +     \( \ddot{\text{S}} \)     \( \rightarrow \)     2K\(^{+} \) [ :\( \ddot{\text{S}} \): ]\(^{2-}\)

(2) **Ca and O:**
Ca\( \cdot \)   \( \rightarrow \)   Ca\(^{2+} \) + 2e\(^-\)
(2, 8, 8, 2)     (2, 8, 8)

\( \ddot{\text{O}} \): + 2e\(^-\)   \( \rightarrow \)   :\( \ddot{\text{O}} \):\(^{2-}\)
(2, 6)             (2, 8)

Combined transfer:
Ca\( \cdot \)   +   \( \ddot{\text{O}} \):   \( \rightarrow \)   Ca\(^{2+} \) [ :\( \ddot{\text{O}} \): ]\(^{2-}\)

(3) **Al and N:**
Al \( \cdot \)   \( \rightarrow \)   Al\(^{3+} \) + 3e\(^-\)

\( \cdot \)N \( \cdot \) + 3e\(^-\)   \( \rightarrow \)   :\( \ddot{\text{N}} \):\(^{3-}\)

Combined transfer:
Al\( \cdot \)   +   \( \cdot \)N\( \cdot \)   \( \rightarrow \)   Al\(^{3+} \) [ :\( \ddot{\text{N}} \): ]\(^{3-}\)
In simple words: To show electron transfer, draw the Lewis dot symbol for each atom. Then, draw arrows to show electrons moving from the metal (which forms a positive ion) to the non-metal (which forms a negative ion). Finally, write the resulting ions with their correct charges and complete electron shells.

Exam Tip: Always balance the total number of electrons lost by the metal with the total number of electrons gained by the non-metal. The resulting ions should have stable electron configurations, usually resembling noble gases.

Question 15. Although both CO\(_{2}\) and H\(_{2}\)O are triatomic molecules, the shape of the H\(_{2}\)O molecule is bent while that of CO\(_{2}\) is linear. Explain this on the basis of dipole moment.
Answer: Both carbon dioxide (\( \text{CO}_{2} \)) and water (\( \text{H}_{2}\text{O} \)) are triatomic molecules, meaning they consist of three atoms. However, their shapes differ significantly: \( \text{CO}_{2} \) is linear, while \( \text{H}_{2}\text{O} \) has a bent structure. This difference can be explained by their dipole moments.
**Water (H\(_{2}\)O):** The \( \text{H}_{2}\text{O} \) molecule has a dipole moment of 1.84 D. It contains two O-H bonds, both of which are polar due to the electronegativity difference between oxygen and hydrogen. Since the \( \text{H}_{2}\text{O} \) molecule possesses a net dipole moment, it means the two O-H dipoles do not cancel each other out. This immediately rules out a linear structure (H-O-H), as a linear arrangement would cause the bond dipoles to be in opposite directions and cancel, resulting in a zero net dipole moment. Therefore, the two O-H bonds must be inclined to each other at a certain angle, giving \( \text{H}_{2}\text{O} \) an angular or bent shape.
          \( \delta^{-} \)
          \( \text{O} \)
        /   \
      H     H
    \( \delta^{+} \)     \( \delta^{+} \)
        \( 104.5^\circ \)

**Carbon Dioxide (CO\(_{2}\)):** The \( \text{CO}_{2} \) molecule has two C=O polar bonds. Although these polar bonds have the same value of dipole moment, the overall dipole moment of the molecule is found to be zero. This happens because the individual dipole moments in \( \text{CO}_{2} \) are of equal magnitude but their directions are exactly opposite to each other, causing them to cancel out. This cancellation of dipole moments indicates that the \( \text{CO}_{2} \) molecule is linear.
    \( \delta^{-} \) \( \quad \delta^{+} \quad \delta^{-} \)
    \( \text{O} = \text{C} = \text{O} \)
    \( \longleftarrow \longrightarrow \)
    (Dipole moments cancel)

In summary, the presence of a net dipole moment in \( \text{H}_{2}\text{O} \) indicates a bent shape, while the absence of a net dipole moment in \( \text{CO}_{2} \) (due to cancellation of equal and opposite bond dipoles) indicates a linear shape.
In simple words: Even though both have three atoms, water is bent and carbon dioxide is straight. Water has a "dipole moment" because its polar bonds don't cancel out, meaning its shape is not straight. Carbon dioxide's polar bonds pull equally in opposite directions, canceling each other out, so it has a straight shape and no overall dipole moment.

Exam Tip: Remember that a molecule's overall dipole moment depends on both the polarity of its individual bonds and its molecular geometry. If bond dipoles are equal and opposite (e.g., in a linear or perfectly symmetrical molecule), they cancel, resulting in a zero net dipole moment and a nonpolar molecule. If they don't cancel, the molecule is polar and has a non-symmetrical shape.

Question 16. Write the significance/applications of dipole moment.
Answer: The dipole moment has several important applications:
1. **Determining the polarity of bonds:** The magnitude of the dipole moment (\( \mu = \text{e} \times \text{d} \)) is directly related to the polarity of the bond. A greater dipole moment indicates higher bond polarity. This principle applies to molecules containing only one polar bond, such as HCl and HBr. In nonpolar molecules like H\(_{2}\), O\(_{2}\), and N\(_{2}\), the dipole moment is zero because there are no charge separations (e = 0). Thus, the dipole moment can distinguish between polar and nonpolar molecules.
2. **Calculating percentage ionic character:** The dipole moment helps to calculate the percentage of ionic character in a bond. For example, if HCl has an observed dipole moment of \( \mu = 1.03 \) D. If HCl were 100% ionic, each end would carry a charge of one unit (4.8 \( \times 10^{-10} \) e.s.u.). Given the bond length (d) in H-Cl is 1.275 Å, the theoretical dipole moment for 100% ionic character (\( \mu_{\text{ionic}} \)) would be:
\( \mu_{\text{ionic}} = \text{e} \times \text{d} = 4.8 \times 10^{-10} \text{ e.s.u.} \times 1.275 \times 10^{-8} \text{ cm} = 6.12 \times 10^{-18} \text{ e.s.u cm} = 6.12 \text{ D} \)

Therefore, the percentage ionic character \( = \frac{\text{Observed dipole moment}}{\text{Calculated dipole moment for 100% ionic character}} \times 100 = \frac{1.03}{6.12} \times 100 = 16.83\% \).
3. **Determining the symmetry (or shape) of molecules:** The dipole moment is an important property for determining the shape of molecules that contain three or more atoms. For instance, if a molecule possesses two or more polar bonds, it will not be symmetrical if it has a net molecular dipole moment, as is the case with water (\( \text{H}_{2}\text{O} \), \( \mu = 1.84 \) D) and ammonia (\( \text{NH}_{3} \), \( \mu = 1.49 \) D). However, if a molecule containing similar atoms around a central atom has a zero overall dipole moment, it implies the molecule is symmetrical, as seen in \( \text{CO}_{2} \), BF\(_{3}\), CH\(_{4}\), and CCl\(_{4}\), etc.
In simple words: Dipole moments are helpful for a few things. They show how polar a bond is, let us calculate how much a bond behaves like an ionic bond, and tell us about the overall shape or symmetry of molecules. If a molecule has a net dipole, it's usually not symmetrical; if it has no net dipole, it's often symmetrical.

Exam Tip: Remember that dipole moment is a vector quantity. Its applications stem from the fact that it indicates the separation of charge within a molecule, which is influenced by both bond polarity and molecular geometry. Always consider how individual bond dipoles add up or cancel out in a molecule.

Question 17. Define electronegativity. How does it differ from electron gain enthalpy?
Answer: Electronegativity of an element is the tendency of its atom to attract a shared pair of electrons towards itself when forming a covalent bond.
**Difference from Electron Gain Enthalpy:** Electronegativity is a property that applies to an atom within a molecule (i.e., when it's already bonded). In contrast, electron gain enthalpy is applicable to atoms in their isolated, gaseous states, describing the energy change when an isolated atom gains an electron.
In simple words: Electronegativity is an atom's ability to pull shared electrons towards itself in a bond. It's different from electron gain enthalpy because electronegativity applies to atoms *in* a molecule, while electron gain enthalpy applies to individual, unbonded atoms gaining an electron.

Exam Tip: Clearly distinguish that electronegativity is about *attraction* of *shared* electrons in a *bond*, while electron gain enthalpy is about *gaining* an electron by an *isolated* atom. Both are related to an atom's electron affinity but are measured under different conditions and describe different phenomena.

Question 18. Explain with the help of a suitable example polar covalent bond.
Answer: A polar covalent bond occurs when atoms in a covalently bonded molecule have different electronegativities. This difference causes the shared electron pair to be attracted more strongly to the more electronegative atom, resulting in a partial negative charge (\( \delta^{-} \)) on that atom and a partial positive charge (\( \delta^{+} \)) on the less electronegative atom.
**Example: Hydrogen Fluoride (HF)** In the hydrogen fluoride (HF) molecule, fluorine is much more electronegative than hydrogen. Therefore, the shared electron pair in the H-F covalent bond is pulled more towards the fluorine atom. This makes the fluorine atom slightly negative (\( \delta^{-} \)) and the hydrogen atom slightly positive (\( \delta^{+} \)).
The HF molecule can be represented as: \( \text{H}:\ddot{\text{F}}: \)   or   \( \overset{\delta^{+}}{\text{H}} \)-\( \overset{\delta^{-}}{\text{F}} \)
Here, \( +\delta \) refers to a small amount of positive charge and \( -\delta \) refers to a small amount of negative charge. This separation of charge creates a dipole, making the H-F bond a polar covalent bond.
In simple words: A polar covalent bond forms when two atoms share electrons unevenly because one atom pulls the electrons more strongly. For example, in HF, fluorine pulls the shared electrons closer to itself than hydrogen, making fluorine slightly negative and hydrogen slightly positive.

Exam Tip: For explaining polar covalent bonds, always use an example that clearly shows electronegativity difference and label the partial charges (\( \delta^{+} \) and \( \delta^{-} \)). The key is unequal sharing of electrons due to differing pulls from the bonded atoms.

Question 19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K\(_{2}\)O, N\(_{2}\), SO\(_{2}\), and CIF\(_{3}\).
Answer: The ionic character of a bond increases with a greater difference in electronegativity between the bonded atoms.
Let's consider the electronegativity differences (approximate values): * N\(_{2}\) (N-N): Electronegativity difference = 0 (pure covalent) * SO\(_{2}\) (S-O): Electronegativity difference (\( \text{O} \approx 3.44, \text{S} \approx 2.58 \)) \( \approx 0.86 \) * ClF\(_{3}\) (Cl-F): Electronegativity difference (\( \text{F} \approx 3.98, \text{Cl} \approx 3.16 \)) \( \approx 0.82 \) (Note: The actual difference for Cl-F is similar to S-O, but the overall geometry of ClF3 and its bond polarization might be slightly different. Based on typical ranking for these common compounds, Cl-F is often considered slightly less polar than S-O in this context, or very similar.) * K\(_{2}\)O (K-O): Electronegativity difference (\( \text{O} \approx 3.44, \text{K} \approx 0.82 \)) \( \approx 2.62 \) * LiF (Li-F): Electronegativity difference (\( \text{F} \approx 3.98, \text{Li} \approx 0.98 \)) \( \approx 3.00 \)
Therefore, the order of increasing ionic character is:
N\(_{2}\) \( < \) SO\(_{2}\) \( < \) ClF\(_{3}\) \( < \) K\(_{2}\)O \( < \) LiF
In simple words: The more two atoms in a bond differ in how strongly they pull electrons, the more "ionic" that bond becomes. So, we list them from the smallest difference to the largest, which gives us the order: N\(_{2}\) (least ionic) then SO\(_{2}\), ClF\(_{3}\), K\(_{2}\)O, and finally LiF (most ionic).

Exam Tip: To determine the order of ionic character, estimate the electronegativity difference for each bond. A larger difference indicates greater ionic character. Remember that pure covalent bonds (like N\(_{2}\)) have zero electronegativity difference.

Question 20. The skeletal structure of CH\(_{3}\)COOH as shown below is correct but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid
        H :O:
        H=C-C-O-H
            H
Answer: The provided skeletal structure incorrectly shows a triple bond between carbon and hydrogen and an incorrect oxygen bonding. The correct Lewis structure for acetic acid (\( \text{CH}_{3}\text{COOH} \)) is:
              :\( \ddot{\text{O}} \):
              ||
    H         C
    |           |
  H-C-C-\( \ddot{\text{O}} \)-H
    |
    H

*Alternatively, showing all lone pairs:*               :\( \ddot{\text{O}} \):
              ||
    H         C
    |           |
  H-C-\( \ddot{\text{C}} \)-\( \ddot{\text{O}} \)-H
    |
    H
In simple words: The given drawing of acetic acid has some wrong bonds, like hydrogen having too many bonds. The correct Lewis structure shows single bonds for all hydrogens, a double bond between carbon and one oxygen, and a single bond between carbon and the other oxygen (which is also bonded to a hydrogen). All atoms should have the right number of bonds and lone pairs to satisfy the octet rule.

Exam Tip: When drawing Lewis structures, remember the octet rule for most atoms (two for hydrogen) and the typical valencies: Carbon forms 4 bonds, Oxygen forms 2 bonds and has 2 lone pairs, Hydrogen forms 1 bond. Always verify that each atom has a stable electron configuration.

Question 21. Apart from tetrahedral geometry, another possible geometry for CH\(_{4}\) is the square planer with the four H atoms at the corners of the square and the C atom at the center. Explain why CH\(_{4}\) is not square planar.
Answer: Methane (\( \text{CH}_{4} \)) is not square planar because carbon undergoes sp\(^{3}\) hybridization, which leads to a tetrahedral geometry, and carbon does not possess the d orbitals necessary for dsp\(^{2}\) hybridization, which would result in a square planar arrangement.
Let's consider the electronic configuration of carbon (atomic number 6): * **Ground state:** \( 1s^{2} \, 2s^{2} \, 2p_{x}^{1} \, 2p_{y}^{1} \) * **Excited state:** To form four bonds, carbon promotes one electron from the 2s orbital to the empty 2p\(_{z}\) orbital. \( 1s^{2} \, 2s^{1} \, 2p_{x}^{1} \, 2p_{y}^{1} \, 2p_{z}^{1} \)
In this excited state, carbon can undergo sp\(^{3}\) hybridization. This involves one 2s orbital and three 2p orbitals mixing to form four equivalent sp\(^{3}\) hybrid orbitals. These four sp\(^{3}\) hybrid orbitals are directed towards the corners of a regular tetrahedron, resulting in a tetrahedral geometry with bond angles of 109.5°.
For a square planar arrangement with the carbon atom at the center and four hydrogen atoms at the corners, dsp\(^{2}\) hybridization would be required. This type of hybridization involves one d orbital, one s orbital, and two p orbitals. However, carbon is a second-period element and does not have any available d orbitals in its valence shell (it only has 2s and 2p orbitals). Since d orbitals are not available, dsp\(^{2}\) hybridization is not possible for carbon.
Therefore, \( \text{CH}_{4} \) exclusively adopts a tetrahedral geometry, not a square planar one.
In simple words: Methane isn't flat and square-shaped because carbon's electrons mix in a way that creates a 3D, tetrahedral shape. Carbon doesn't have the special "d" orbitals needed to make a flat, square shape; it only has "s" and "p" orbitals for bonding.

Exam Tip: When explaining molecular geometry, always connect it to the central atom's hybridization. For carbon, its ability to only use s and p orbitals for hybridization (sp, sp\(^{2}\), sp\(^{3}\)) limits its geometries to linear, trigonal planar, and tetrahedral, respectively. The absence of d orbitals prevents it from forming square planar structures via dsp\(^{2}\) hybridization.

Question 22. Explain why the BeH\(_{2}\) molecule has a zero dipole moment although the Be-H bonds are polar.
Answer: The BeH\(_{2}\) molecule has a zero dipole moment despite having polar Be-H bonds because of its linear geometry and the cancellation of bond dipoles.
Beryllium in BeH\(_{2}\) undergoes sp hybridization. This hybridization results in two sp hybrid orbitals that are oriented at 180° to each other, leading to a linear shape for the BeH\(_{2}\) molecule.
The individual Be-H bonds are polar because there is a difference in electronegativity between beryllium (approx. 1.57) and hydrogen (approx. 2.20). This means there is a small positive charge on beryllium and a small negative charge on hydrogen, creating a bond dipole for each Be-H bond.
However, because the molecule is linear, these two individual Be-H bond dipoles are equal in magnitude and point in opposite directions along the same axis. As dipole moment is a vector quantity (having both magnitude and direction), these two opposing bond dipoles cancel each other out completely.
The cancellation can be visualized as:
        \( \longleftarrow \)     \( \longrightarrow \)
      \( \overset{\delta^{-}}{\text{H}} \)- \( \overset{\delta^{+}}{\text{Be}} \)- \( \overset{\delta^{-}}{\text{H}} \)
          \( 180^\circ \)
Resultant \( \mu = 0 \).
Therefore, the net molecular dipole moment of BeH\(_{2}\) is zero, making the molecule nonpolar overall, even though its individual bonds are polar.
In simple words: BeH\(_{2}\) has no overall "dipole moment" even though each Be-H bond is "polar." This is because the molecule is straight, like a line. The two pulls of electrons in opposite directions are equal, so they cancel each other out, leaving no net pull in any single direction.

Exam Tip: When explaining zero dipole moment for molecules with polar bonds, always refer to two key aspects: the molecular geometry (e.g., linear, trigonal planar, tetrahedral) and the symmetrical arrangement of equal bond dipoles that leads to their cancellation. This is common for symmetrical molecules like \( \text{CO}_{2} \), CCl\(_{4} \), and BeH\(_{2} \).

Question 23. Which out of NH\(_{3}\) and NF\(_{3}\) has a higher dipole moment and why?
Answer: Ammonia (NH\(_{3}\)) has a significantly higher dipole moment (1.46 D) compared to nitrogen trifluoride (NF\(_{3}\)) (0.24 D).
Both NH\(_{3}\) and NF\(_{3}\) have a pyramidal shape with a lone pair of electrons on the nitrogen atom. The difference in their dipole moments arises from the different directions of the bond moments for N-H and N-F bonds.
* **Electronegativity:** * In NH\(_{3}\), nitrogen (\( \text{EN} \approx 3.0 \)) is more electronegative than hydrogen (\( \text{EN} \approx 2.1 \)). So, the N-H bond moments point from hydrogen towards nitrogen. * In NF\(_{3}\), fluorine (\( \text{EN} \approx 4.0 \)) is more electronegative than nitrogen (\( \text{EN} \approx 3.0 \)). So, the N-F bond moments point from nitrogen towards fluorine.
* **Direction of Bond Moments and Lone Pair Moment:** * **For NH\(_{3}\):** The lone pair on nitrogen contributes a dipole moment pointing away from the nitrogen nucleus. The three N-H bond moments also point towards the nitrogen atom. All these bond moments and the lone pair moment are in the same general direction, adding up to a larger net dipole moment.
Lone pair of electrons \( \uparrow \)
        N
      / | \
    H H H
(Resultant of 3 N-H bonds points upwards, along with lone pair moment)

* **For NF\(_{3}\):** The lone pair on nitrogen contributes a dipole moment pointing away from the nitrogen nucleus. However, the three N-F bond moments point *away* from the nitrogen atom (towards the more electronegative fluorine atoms). Therefore, the bond moments and the lone pair moment are in opposite directions, leading to a partial cancellation. This cancellation results in a much smaller net dipole moment for NF\(_{3}\).
Lone pair of electrons \( \uparrow \)
        N
      / | \
    F F F
(Resultant of 3 N-F bonds points downwards, opposing the lone pair moment)

Because the bond dipoles and lone pair dipole reinforce each other in NH\(_{3}\) but partially oppose each other in NF\(_{3}\), NH\(_{3}\) has a much higher dipole moment.
In simple words: Ammonia (NH\(_{3}\)) has a much stronger electric pull (dipole moment) than nitrogen trifluoride (NF\(_{3}\)). This is because in ammonia, all the small pulls from the bonds and the unshared electron pair add up in the same direction. But in nitrogen trifluoride, the pulls from the bonds go in the opposite direction to the pull from the unshared electron pair, making them partly cancel out and resulting in a weaker overall pull.

Exam Tip: When comparing dipole moments of molecules with similar geometry, always analyze the direction of both bond dipoles (based on electronegativity) and the lone pair dipole. If they reinforce each other, the net dipole moment will be higher; if they oppose, it will be lower due to cancellation.

Question 24. What is meant by the hybridization of atomic orbitals? Describe the shapes of sp, sp\(^{2}\), sp\(^{3}\) hybrid orbitals.
Answer: **Hybridization** is defined as the mixing of atomic orbitals that belong to the same atom but have slightly different energies. This mixing leads to a redistribution of energy, resulting in the formation of new orbitals that have equal energies and identical shapes. These newly formed orbitals are known as hybrid orbitals.
Let's describe the shapes of common hybrid orbitals:
1. **sp Hybridization:** * Here, one s orbital and one p orbital from the same atom combine. * They form two sp hybrid orbitals. * Each sp hybrid orbital has 1/2 s character and 1/2 p character. * The two sp hybrid orbitals are oriented at 180° to each other, giving a linear shape. *
\[ \text{s-orbital} \quad + \quad \text{p-orbital} \quad \xrightarrow{\text{Hybridisation}} \quad \text{Two sp hybrid orbitals (linear, } 180^\circ \text{)} \]
2. **sp\(^{2}\) Hybridization:** * Here, one s orbital and two p orbitals from the same atom combine. * They form three sp\(^{2}\) hybrid orbitals. * Each sp\(^{2}\) hybrid orbital has 1/3 s character and 2/3 p character. * The three sp\(^{2}\) hybrid orbitals arrange themselves in a trigonal planar shape with bond angles of 120°. *
\[ \text{s-orbital} \quad + \quad \text{Two p-orbitals} \quad \xrightarrow{\text{Hybridisation}} \quad \text{Three sp}^{2} \text{ hybrid orbitals (trigonal planar, } 120^\circ \text{)} \]
3. **sp\(^{3}\) Hybridization:** * Here, one s orbital and three p orbitals from the same atom combine. * They form four sp\(^{3}\) hybrid orbitals. * Each sp\(^{3}\) hybrid orbital has 1/4 s character and 3/4 p character. * The four sp\(^{3}\) hybrid orbitals orient themselves towards the corners of a regular tetrahedron, giving a tetrahedral shape with bond angles of 109.5°. *
\[ \text{s-orbital} \quad + \quad \text{Three p-orbitals} \quad \xrightarrow{\text{Hybridisation}} \quad \text{Four sp}^{3} \text{ hybrid orbitals (tetrahedral, } 109.5^\circ \text{)} \]In simple words: Hybridization is when different types of electron clouds (orbitals) in an atom mix together to create new, equal-energy electron clouds. These new clouds, called hybrid orbitals, then determine the molecule's shape. For example, 'sp' makes a straight line, 'sp\(^{2}\)' makes a flat triangle, and 'sp\(^{3}\)' makes a pyramid-like shape with four points.

Exam Tip: When defining hybridization, emphasize the mixing of orbitals *on the same atom* to form *new, equivalent orbitals*. For describing sp, sp\(^{2}\), and sp\(^{3}\) hybridization, always mention the number of orbitals mixed, the number of hybrid orbitals formed, their s/p character, and the resulting geometry and bond angle.

Question 25. Describe the change in hybridization (if any) of the Al atom in the following reaction. AlCl\(_{2}\) + Cl\(^{-}\) \( \rightarrow \) AlCl\(_{4}^{-}\)
Answer: Let's assume the question meant AlCl\(_{3}\) as a common starting material for AlCl\(_{4}^{-}\) formation.
For the Al atom in the reaction where it changes from \( \text{AlCl}_{3} \) to \( \text{AlCl}_{4}^{-} \), there is a change in hybridization.
**Initial state: Al in AlCl\(_{3}\)** * The electronic configuration of \( _{13}\text{Al} \) in its ground state is \( 1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{2} \, 3p_{x}^{1} \). * In its excited state (for bonding), it would be \( 1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{1} \, 3p_{x}^{1} \, 3p_{y}^{1} \). * Aluminum undergoes sp\(^{2}\) hybridization, mixing one 3s orbital and two 3p orbitals to form three sp\(^{2}\) hybrid orbitals. * This leads to a trigonal planar geometry for \( \text{AlCl}_{3} \), with three Al-Cl bonds and an empty 3p\(_{z}\) orbital.
**Final state: Al in AlCl\(_{4}^{-}\)** * When \( \text{AlCl}_{3} \) reacts with a chloride ion (Cl\(^{-}\)), the empty 3p\(_{z}\) orbital on aluminum accepts a lone pair of electrons from the Cl\(^{-}\) ion to form a coordinate covalent bond. * To accommodate this new bond, the aluminum atom changes its hybridization. It now mixes one 3s orbital, three 3p orbitals, and this empty 3p\(_{z}\) orbital effectively participates, leading to sp\(^{3}\) hybridization. * This sp\(^{3}\) hybridization results in a tetrahedral geometry for the \( \text{AlCl}_{4}^{-} \) ion.
Thus, the hybridization of the Al atom changes from sp\(^{2}\) in \( \text{AlCl}_{3} \) to sp\(^{3}\) in \( \text{AlCl}_{4}^{-} \).
In simple words: When aluminum goes from being in AlCl\(_{3}\) to AlCl\(_{4}^{-}\), its hybridization changes. In AlCl\(_{3}\), it's sp\(^{2}\) (flat triangle shape). But when it adds an extra chloride ion, it changes to sp\(^{3}\) (3D pyramid shape) to make space for the new bond.

Exam Tip: For reactions involving the formation of complex ions, look for a change in the number of electron domains around the central atom. An increase in the number of bonds (especially from accepting a lone pair) often indicates a shift to a higher hybridization state (e.g., sp\(^{2}\) to sp\(^{3}\)), resulting in a change in geometry.

Question 26. Is there any change in the hybridization of B and N atoms as a result of the following reaction?
Answer: The implicit reaction is the formation of an adduct between \( \text{BF}_{3} \) and \( \text{NH}_{3} \), such as \( \text{BF}_{3} + \text{NH}_{3} \rightarrow \text{H}_{3}\text{N:BF}_{3} \).
Let's analyze the hybridization of B and N before and after the reaction:
**Before the reaction:** * **In BF\(_{3}\):** Boron (B) is surrounded by three bond pairs and no lone pairs. It undergoes sp\(^{2}\) hybridization, resulting in a trigonal planar geometry. * **In NH\(_{3}\):** Nitrogen (N) is surrounded by three bond pairs and one lone pair. It undergoes sp\(^{3}\) hybridization, resulting in a pyramidal geometry.
**After the reaction (in the adduct H\(_{3}\)N:BF\(_{3}\)):** * **Boron in H\(_{3}\)N:BF\(_{3}\):** Boron, which was sp\(^{2}\) hybridized in \( \text{BF}_{3} \), acts as a Lewis acid and accepts a lone pair of electrons from the nitrogen of \( \text{NH}_{3} \). This forms a new coordinate covalent bond. Now, boron is surrounded by four bond pairs. To accommodate these four electron pairs, its hybridization changes from sp\(^{2}\) to sp\(^{3}\). This results in a tetrahedral arrangement around boron. * **Nitrogen in H\(_{3}\)N:BF\(_{3}\):** Nitrogen, which was sp\(^{3}\) hybridized in \( \text{NH}_{3} \), donates its lone pair to boron. Even after donating the lone pair, nitrogen is still surrounded by four electron domains (three N-H bond pairs and one N-B coordinate bond). Therefore, its hybridization remains sp\(^{3}\), maintaining a tetrahedral arrangement of electron pairs (though the molecular geometry around N remains pyramidal if considering the groups, or tetrahedral for the total electron domains).
So, as a result of the reaction, the hybridization of **B changes from sp\(^{2}\) to sp\(^{3}\)**, while the hybridization of **N remains unchanged as sp\(^{3}\)**.
In simple words: When \( \text{BF}_{3} \) and \( \text{NH}_{3} \) combine, the boron atom in \( \text{BF}_{3} \) changes its electron mixing from sp\(^{2}\) to sp\(^{3}\) because it gains a new bond. But the nitrogen atom in \( \text{NH}_{3} \) stays sp\(^{3}\) hybridized, even though it forms a new bond, because it just uses its existing lone pair for that bond.

Exam Tip: For adduct formation reactions, pay close attention to the central atom that *accepts* the electron pair (Lewis acid). Its hybridization will likely change to accommodate the new bond. The atom that *donates* the electron pair (Lewis base) often retains its original hybridization if its electron domain count doesn't fundamentally change.

Question 27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in (a) C\(_{2}\)H\(_{2}\) (b) C\(_{2}\)H\(_{4}\) molecules.
Answer:
**(a) Formation of a triple bond (one sigma and two pi bonds) in C\(_{2}\)H\(_{2}\) (Ethyne)**
In the formation of ethyne (\( \text{C}_{2}\text{H}_{2} \)) molecule, both carbon atoms undergo sp hybridization. Each carbon atom has two unhybridized p orbitals (2p\(_{x}\) and 2p\(_{y}\)) remaining after sp hybridization.
* **Sigma (\( \sigma \)) bonds:** * One sp hybrid orbital from each carbon atom overlaps axially (end-on) to form a C-C sigma bond. * The other sp hybrid orbital of each carbon atom overlaps axially with the half-filled s orbital of a hydrogen atom, forming two C-H sigma bonds. *
\[ \text{H} \underset{\sigma}{---} \text{C} \underset{\sigma}{---} \text{C} \underset{\sigma}{---} \text{H} \]
* **Pi (\( \pi \)) bonds:** * Each of the two unhybridized p orbitals (2p\(_{x}\) and 2p\(_{y}\)) of both carbon atoms overlap sidewise (lateral overlap) to form two \( \pi \) bonds between the carbon atoms. One \( \pi \) bond forms from the overlap of 2p\(_{x}\) orbitals, and another \( \pi \) bond forms from the overlap of 2p\(_{y}\) orbitals.
Therefore, the triple bond between the two carbon atoms in C\(_{2}\)H\(_{2}\) is composed of one sigma (\( \sigma \)) bond and two pi (\( \pi \)) bonds.
(Diagram showing axial overlap for sigma and lateral overlap for pi bonds for C\(_{2}\)H\(_{2}\) would be here, similar to the source PDF)
**(b) Formation of a double bond (one sigma and one pi bond) in C\(_{2}\)H\(_{4}\) (Ethene)**
In the formation of ethene (\( \text{C}_{2}\text{H}_{4} \)) molecule, both carbon atoms undergo sp\(^{2}\) hybridization. Each carbon atom has one unhybridized p orbital (2p\(_{z}\)) remaining after sp\(^{2}\) hybridization.
* **Sigma (\( \sigma \)) bonds:** * One sp\(^{2}\) hybrid orbital from each carbon atom overlaps axially to form a C-C sigma bond. * The other two sp\(^{2}\) hybrid orbitals of each carbon atom overlap axially with the s orbitals of two hydrogen atoms, forming two C-H sigma bonds per carbon (a total of four C-H sigma bonds). *
\[ \underset{H}{\overset{H}{|}}\text{C} \underset{\sigma}{==} \underset{H}{\overset{H}{|}}\text{C} \]
* **Pi (\( \pi \)) bond:** * The unhybridized p orbital (2p\(_{z}\)) of one carbon atom overlaps sidewise with the similar p orbital of the other carbon atom to form a weak \( \pi \) bond. This \( \pi \) bond consists of two equal electron clouds, one distributed above and one below the plane of the carbon and hydrogen atoms.
Thus, in the ethene (\( \text{C}_{2}\text{H}_{4} \)) molecule, the carbon-carbon bond consists of one sp\(^{2}\)-sp\(^{2}\) sigma bond and one pi (\( \pi \)) bond between the p orbitals which are not used in the hybridization and are perpendicular to the plane of the molecule. The C-C bond length is 134 pm. The C-H bond is sp\(^{2}\)-s sigma with a bond length of 108 pm. The H-C-H bond angle is 117.6°, while the H-C-C angle is 121°.
(Diagram showing axial overlap for sigma and lateral overlap for pi bonds for C\(_{2}\)H\(_{4}\) would be here, similar to the source PDF)
In simple words: For a triple bond (like in C\(_{2}\)H\(_{2}\)), the carbons use 'sp' mixing. They make one strong 'sigma' bond by head-on overlap and two weaker 'pi' bonds by side-to-side overlap. For a double bond (like in C\(_{2}\)H\(_{4}\)), the carbons use 'sp\(^{2}\)' mixing. They make one 'sigma' bond and one 'pi' bond. Sigma bonds are head-on, while pi bonds are side-to-side overlaps of remaining electron clouds.

Exam Tip: When illustrating multiple bond formation, clearly differentiate between sigma and pi bonds. Sigma bonds result from head-on overlap of hybrid or atomic orbitals and form the single bond framework. Pi bonds result from lateral (sideways) overlap of unhybridized p orbitals and are added to a pre-existing sigma bond.

Question 28. What is the total number of sigma and pi bonds in the following molecules?
Answer:
1. There are a total of three sigma bonds (one C-C sigma and two C-H sigma) and two pi bonds (between the two C atoms) in C2H2.
2. There are a total of five sigma bonds (one C-C sigma and four C-H sigma) and only one pi bond (between the two C atoms) in C2H4.
In simple words: For C2H2, count 3 sigma bonds (1 C-C, 2 C-H) and 2 pi bonds. For C2H4, count 5 sigma bonds (1 C-C, 4 C-H) and 1 pi bond.

Exam Tip: Remember that single bonds are always sigma bonds, double bonds consist of one sigma and one pi bond, and triple bonds have one sigma and two pi bonds.

Question 29. Considering the x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
1. 1s and 1s
2. 1s and 2px
3. 2py and 2py
4. 1s and 2s.
Answer:
3. Since the x-axis is considered the internuclear axis, and the y-axis is perpendicular to the x-axis, therefore, 2py and 2py atomic orbitals can only overlap side-wise (lying on the y-axis). Because overlapping of y-orbitals side-wise results in the formation of a pi \( (\pi) \) bond, 2py and 2py will not form a sigma bond.
In simple words: When the x-axis is the central line for bonding, two 2py orbitals can only overlap sideways, forming a pi bond, not a sigma bond. Sigma bonds need a direct end-to-end overlap along the internuclear axis.

Exam Tip: Recall that sigma bonds form from head-on (axial) overlap, while pi bonds form from sideways (lateral) overlap of orbitals. The chosen internuclear axis is crucial for determining overlap type.

Question 30. Which hybrid orbitals are used by carbon atoms in the following molecules?
1. CH3 - CH3;
2. CH3-CH=CH2;
3. CH3-CH2-OH;
4. CH3-CHO
5. CH3COOH
Answer:
1. In CH3 - CH3, both carbon atoms use sp³ hybrid orbitals.
2. In CH3-CH=CH2, the carbon atom in CH3 uses sp³ hybrid orbitals, and the carbon atoms in CH=CH2 use sp² hybrid orbitals.
3. In CH3-CH2-OH, both carbon atoms use sp³ hybrid orbitals.
4. In CH3-CHO, the carbon atom in CH3 uses sp³ hybrid orbitals, and the carbon atom in CHO uses sp² hybrid orbitals.
5. In CH3COOH, the carbon atom in CH3 uses sp³ hybrid orbitals, and the carbon atom in COOH uses sp² hybrid orbitals.
In simple words: The type of hybrid orbital (sp³, sp², sp) depends on the number of single, double, or triple bonds and lone pairs around each carbon. For example, carbons with all single bonds are sp³, those with one double bond are sp², and those with a triple bond are sp.

Exam Tip: To determine hybridization, count the number of sigma bonds and lone pairs around the central atom. 4 groups = sp³, 3 groups = sp², 2 groups = sp.

Question 31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer: The electrons that exist between two atoms and are shared by them are known as bonded pairs of electrons. For instance, an H2 molecule possesses only a bond pair of electrons. It can be represented as \( H-H \). Conversely, lone pairs of electrons are those that are not shared by another atom. For example, in NH3, the nitrogen atom has three bond pairs with hydrogen atoms and one lone pair of electrons. In an H2O molecule, the oxygen atom has two lone pairs of electrons and two bond pairs with hydrogen atoms.
In simple words: Bond pairs are shared electrons between atoms, like in H2. Lone pairs are unshared electrons on an atom, like on nitrogen in NH3 or oxygen in H2O.

Exam Tip: Clearly differentiate between shared (bonding) and unshared (non-bonding) electron pairs in your definitions and examples, as they both influence molecular geometry.

Question 32. Distinguish between sigma and a pi bond.
Answer:

Exam Tip: Always remember that a single bond is a sigma bond, a double bond contains one sigma and one pi bond, and a triple bond contains one sigma and two pi bonds.

Question 33. Explain the formation of H2 molecule on the basis of valence bond theory.
Answer: In an H2 molecule, the 1s atomic orbitals of both hydrogen atoms overlap end-on (s-s overlap). This overlap results in the formation of a sigma bond between them. Each hydrogen atom contributes one electron to the shared pair, forming a stable covalent bond. The electron density is concentrated between the nuclei, attracting them and holding the molecule together. The following diagram shows the overlap:

In simple words: Two hydrogen atoms come together, and their electron clouds (1s orbitals) overlap directly end-to-end. This overlap creates a shared electron bond, called a sigma bond, which holds the H2 molecule strongly together.

Exam Tip: When explaining bond formation, always mention the type of orbitals involved (e.g., s-s overlap), the resulting bond type (e.g., sigma bond), and how electron sharing leads to stability.

Question 34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer: The linear combination of atomic orbitals to form molecular orbitals happens only if the following conditions are met:
1. The combining atomic orbitals must possess the same or very similar energy levels. For instance, a 1s orbital and a 2s orbital cannot combine effectively because their energy difference is too great.
2. The combining atomic orbitals must show the same symmetry around the molecular axis. Typically, the z-axis is considered the molecular axis. Orbitals with different symmetries (e.g., a 2pz orbital with a 2px or 2py orbital) will not combine, even if their energies are similar.
3. The combining atomic orbitals must overlap as much as possible. A greater extent of overlap leads to higher electron density between the nuclei of the molecular orbital, making the bond stronger.
In simple words: For atomic orbitals to combine and make new molecular orbitals, they must have similar energy, the right shape to match up, and they need to overlap a lot.

Exam Tip: Remember these three key rules for LCAO. Energy similarity ensures effective mixing, symmetry dictates which orbitals can interact, and maximum overlap leads to stable molecular orbitals.

Question 35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
Answer: The electronic configuration of Beryllium \( (\text{Be}) \) is 4, which means \( 1s^2 2s^2 \). According to molecular orbital theory, the hypothetical molecular orbital configuration for Be2 would be \( \text{KK}[\sigma_{2s}]^2[\sigma^*_{2s}]^2 \).
To find the bond order, we use the formula: Bond order \( = \frac {1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \).
For Be2, the number of electrons in bonding orbitals is 2 (from \( [\sigma_{2s}]^2 \)), and the number of electrons in antibonding orbitals is 2 (from \( [\sigma^*_{2s}]^2 \)).
So, Bond order \( = \frac {1}{2} (2 - 2) = \frac {1}{2} (0) = 0 \).
Since the bond order is zero, it means there is no net attractive force between the two beryllium atoms. Therefore, the Be2 molecule does not exist as a stable entity.
In simple words: Beryllium atoms do not form a stable Be2 molecule because, when their orbitals combine, they end up with an equal number of bonding and antibonding electrons. This results in a bond order of zero, meaning no actual bond holds the atoms together.

Exam Tip: Always calculate the bond order using the molecular orbital configuration. A bond order of zero (or a negative number) indicates that the molecule will not exist in a stable form.

Question 36. Compare the relative stability of the following species and indicate their magnetic properties O2, \( \mathrm{O}_{2}^{+} \), \( \mathrm{O}_{2}^{-} \), \( \mathrm{O}_{2}^{2-} \) (peroxide)?
Answer: The molecular orbital configurations for O2, \( \mathrm{O}_{2}^{+} \), \( \mathrm{O}_{2}^{-} \), and \( \mathrm{O}_{2}^{2-} \) are as follows:
O2: \( (\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1 \)
Bond order \( = \frac {1}{2} (10 - 6) = \frac {4}{2} = 2 \). It has two unpaired electrons (one in \( \pi^*_{2p_x} \) and one in \( \pi^*_{2p_y} \) molecular orbitals). Therefore, O2 is paramagnetic.

\( \mathrm{O}_{2}^{+} \): \( (\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 \)
Bond order \( = \frac {1}{2} (10 - 7) = \frac {3}{2} = 1.5 \). There is one unpaired electron present in \( \pi^*_{2p_x} \) molecular orbital. Hence, \( \mathrm{O}_{2}^{+} \) is paramagnetic.

\( \mathrm{O}_{2}^{-} \): \( (\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^1 \)
Bond order \( = \frac {1}{2} (10 - 7) = \frac {3}{2} = 1.5 \). It has one unpaired electron present in \( \pi^*_{2p_y} \) molecular orbital. Hence, \( \mathrm{O}_{2}^{-} \) is also paramagnetic, but less paramagnetic than \( \mathrm{O}_{2}^{+} \).

\( \mathrm{O}_{2}^{2-} \): \( (\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^2 \)
Bond order \( = \frac {1}{2} (10 - 8) = \frac {2}{2} = 1 \). There are no unpaired electrons present in \( \mathrm{O}_{2}^{2-} \). Therefore, \( \mathrm{O}_{2}^{2-} \) is diamagnetic.

The stability of these species varies directly with the bond order. Thus, stability increases from \( \mathrm{O}_{2}^{2-} \) to \( \mathrm{O}_{2}^{+} \) in the following order:
Bond order: O2 (2) \( > \) \( \mathrm{O}_{2}^{+} \) (1.5) \( = \) \( \mathrm{O}_{2}^{-} \) (1.5) \( > \) \( \mathrm{O}_{2}^{2-} \) (1)
However, between \( \mathrm{O}_{2}^{+} \) and \( \mathrm{O}_{2}^{-} \), \( \mathrm{O}_{2}^{+} \) is generally considered more stable due to fewer electrons in antibonding orbitals.
The final order of stability and bond order is: \( \mathrm{O}_{2}^{+} (2.5) > \mathrm{O}_{2} (2) > \mathrm{O}_{2}^{-} (1.5) > \mathrm{O}_{2}^{2-} (1) \).
In simple words: To compare stability, we calculate the bond order for each oxygen species. Higher bond order means more stable. For magnetic properties, if there are any single electrons in orbitals, it's paramagnetic; if all electrons are paired, it's diamagnetic.

Exam Tip: For molecular orbital theory problems, always draw the energy level diagram to correctly fill electrons and then calculate bond order and determine magnetic properties. Remember that bond order directly correlates with stability.

Question 37. Write the significance of plus and a minus sign shown in representing the orbitals.
Answer: Crests of the electron wave are usually marked with a '+' sign, and troughs are marked with a '-' sign. Therefore, a bonding molecular orbital (MO) forms when a '+' part combines with a '+' part, or a '-' part combines with a '-' part of the electron wave. Conversely, antibonding MOs form from the overlap of a '+' part with a '-' part. These '+' or '-' signs do not relate to the electrical charges on the orbitals; they represent the phase of the wave function.
In simple words: The plus and minus signs on orbitals show the wave-like nature of electrons, like the high and low points of a wave. When similar signs (phases) overlap, they form strong bonds, and when different signs overlap, they create weaker, antibonding interactions.

Exam Tip: Understand that the signs (+/-) on orbitals refer to the phase of the wave function, not electrical charge. Proper phase matching is critical for constructive (bonding) overlap, while opposite phases lead to destructive (antibonding) overlap.

Question 38. Describe the hybridization in the case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Answer: For phosphorus (P) in PCl5, its atomic number (Z) is 15. The ground state electronic configuration is \( 1s^2 2s^2 2p^6 3s^2 3p^3 \).
In PCl5, the phosphorus atom undergoes sp³d hybridization. This involves one 3s orbital, three 3p orbitals, and one 3d orbital (specifically, the \( 3d_{z^2} \) orbital) mixing to form five new sp³d hybrid orbitals. These hybrid orbitals adopt a trigonal bipyramidal geometry.

The shape of PCl5 is trigonal bipyramidal, with three equatorial bonds and two axial bonds.
The P-Cl equatorial bond length is 2.04 Å, while the P-Cl axial bond length is 2.10 Å. The reason for the longer axial bond length is that the three P-Cl equatorial bonds are repelled by only two electron pairs (the two axial bond pairs). Conversely, each axial bond pair experiences stronger repulsion from three equatorial bond pairs (at 90° angles). This greater repulsion exerted on the axial bond pairs causes them to be slightly longer and weaker than the equatorial bonds.
In simple words: In PCl5, phosphorus uses sp³d hybridization, forming a shape like two pyramids joined at their bases. The bonds sticking straight up and down (axial) are slightly longer than the bonds around the middle (equatorial) because they get pushed away more by the other electrons.

Exam Tip: For PCl5, remember the sp³d hybridization and the trigonal bipyramidal geometry. Critically, understand that the axial bonds are longer and weaker due to greater 90° repulsions from the equatorial bonds.

Question 39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer: A hydrogen bond forms when a hydrogen atom in one molecule is bonded to a highly electronegative atom (like nitrogen, oxygen, or fluorine), and this hydrogen atom is then attracted to another highly electronegative atom in the same or a different molecule. The electronegative atom attracts the shared pair of electrons, making the hydrogen atom slightly positive. This slightly positive hydrogen atom then forms a weak attractive interaction with another electronegative atom. This bond is weaker than a covalent bond but stronger than van der Waals forces. It is shown by dotted lines:
\( \ldots H - X \ldots H - X \ldots H - X \ldots \)
As a result of hydrogen bonding, an H-atom simultaneously connects two electronegative atoms: one via a covalent bond and the other via a hydrogen bond. This is often called forming a "hydrogen bridge." A hydrogen bond is a weak bond, but it is stronger than van der Waals forces.
In simple words: A hydrogen bond is a special weak attraction when a hydrogen atom (attached to N, O, or F) gets attracted to another N, O, or F atom. It's weaker than normal bonds but stronger than other weak attractions called van der Waals forces.

Exam Tip: Remember the three key electronegative atoms involved in hydrogen bonding (N, O, F) and that the hydrogen atom must be directly bonded to one of them. Also, distinguish its strength relative to covalent and van der Waals interactions.

Question 40. What is meant by the term bond order ? Calculate the bond order of : N2, O2, \( \mathrm{O}_{2}^{+} \) and \( \mathrm{O}_{2}^{-} \).
Answer: Bond order is defined as half the difference between the number of electrons in bonding orbitals \( (N_b) \) and the number of electrons in antibonding orbitals \( (N_a) \).
Bond order \( = \frac {1}{2} (N_b - N_a) \), where \( N_b \) is the number of electrons occupying bonding orbitals and \( N_a \) is the number of electrons occupying antibonding orbitals.

Molecular orbitals of N2: KK \( (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2 \)
For N2: \( N_b = 8 \) (2 in \( \sigma_{2s} \), 4 in \( \pi_{2p} \), 2 in \( \sigma_{2p_z} \)), \( N_a = 2 \) (in \( \sigma^*_{2s} \)).
Bond order of N2 \( = \frac {1}{2} (8 - 2) = \frac {6}{2} = 3 \).

Molecular orbitals of O2: KK \( (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1 \)
For O2: \( N_b = 8 \) (2 in \( \sigma_{2s} \), 2 in \( \sigma_{2p_z} \), 4 in \( \pi_{2p} \)), \( N_a = 4 \) (2 in \( \sigma^*_{2s} \), 2 in \( \pi^*_{2p} \)).
Bond order of O2 \( = \frac {1}{2} (8 - 4) = \frac {4}{2} = 2 \).

Molecular orbitals of \( \mathrm{O}_{2}^{+} \): KK \( (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 \)
For \( \mathrm{O}_{2}^{+} \): \( N_b = 8 \), \( N_a = 3 \) (2 in \( \sigma^*_{2s} \), 1 in \( \pi^*_{2p_x} \)).
Bond order of \( \mathrm{O}_{2}^{+} = \frac {1}{2} (8 - 3) = \frac {5}{2} = 2.5 \).

Molecular orbitals of \( \mathrm{O}_{2}^{-} \): KK \( (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^1 \)
For \( \mathrm{O}_{2}^{-} \): \( N_b = 8 \), \( N_a = 5 \) (2 in \( \sigma^*_{2s} \), 3 in \( \pi^*_{2p} \)).
Bond order of \( \mathrm{O}_{2}^{-} = \frac {1}{2} (8 - 5) = \frac {3}{2} = 1.5 \).
In simple words: Bond order tells us how many bonds are effectively holding atoms together. You find it by subtracting electrons in "anti-bonds" from electrons in "bonds" and then dividing by two. A higher bond order means a stronger, more stable connection.

Exam Tip: To accurately calculate bond order, always write out the full molecular orbital configuration, ensuring the correct filling order of orbitals (especially for B2, C2, N2 vs. O2, F2). Remember to count electrons in both bonding and antibonding orbitals.