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

RBSE Class 11 Chemistry Chapter 4 Multiple Choice Questions

 

Question 1. PCl5 exists but NCl5 does not, why?
(a) Nitrogen do not have vacant d-orbitals
(b) Ionisation energy of nitrogen is very high.
(c) Nitrogen molecule is not similar to chlorine molecule.
(d) None of the options
Answer: (a) Nitrogen do not have vacant d-orbitals
In simple words: Phosphorus can form five bonds because it has empty d-orbitals it can use. Nitrogen, however, does not have these empty d-orbitals, which stops it from making five bonds like in NCl5.

🎯 Exam Tip: Remember that elements in the third period and beyond (like Phosphorus) have available d-orbitals which allow them to expand their octet, unlike second-period elements (like Nitrogen).

 

Question 3. The bond order of N2 is
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (a) 3
In simple words: The bond order for a nitrogen molecule (\( \mathrm{N}_{2} \)) is 3, which means there is a triple bond between the two nitrogen atoms. This strong triple bond makes the \( \mathrm{N}_{2} \) molecule very stable.

🎯 Exam Tip: Bond order indicates the number of bonds between two atoms; higher bond order means stronger and shorter bonds.

 

Question 4. Which of the following has lone pair of electrons?
(a) NO
(b) CO
(c) NH3
(d) O2
Answer: (a) NO
In simple words: Nitric oxide (\( \mathrm{NO} \)) has an odd number of electrons, and due to its structure, it contains a lone pair.

🎯 Exam Tip: When determining lone pairs, always draw the Lewis structure and consider all valence electrons for each atom.

 

Question 5. Which of the following molecule/ion shows \( \mathrm{sp}^{2} \) hybridisation?
(a) \( \mathrm{BF}_{3} \) and \( \mathrm{NH}_{2}^{-} \)
(b) \( \mathrm{NO}_{2} \) and \( \mathrm{NH}_{3} \)
(c) \( \mathrm{BF}_{3} \) and \( \mathrm{NO}_{2}^{-} \)
(d) \( \mathrm{NH}_{2}^{-} \) and \( \mathrm{H}_{2}\mathrm{O} \)
Answer: (c) \( \mathrm{BF}_{3} \) and \( \mathrm{NO}_{2}^{-} \)
In simple words: Both boron trifluoride (\( \mathrm{BF}_{3} \)) and the nitrite ion (\( \mathrm{NO}_{2}^{-} \)) have a central atom that uses \( \mathrm{sp}^{2} \) hybrid orbitals. This leads to a trigonal planar geometry for \( \mathrm{BF}_{3} \) and a bent shape for \( \mathrm{NO}_{2}^{-} \).

🎯 Exam Tip: To find hybridisation, count the number of sigma bonds and lone pairs around the central atom; 3 gives \( \mathrm{sp}^{2} \).

RBSE Class 11 Chemistry Chapter 4 Very Short Answer Type Questions

 

Question 7. Write the electronic configuration of \( \mathrm{Li}_{2} \) molecule.
Answer: The electronic configuration of the \( \mathrm{Li}_{2} \) molecule is \( (\sigma 1s)^2 (\sigma^*1s)^2 (\sigma 2s)^2 \). This configuration shows that the molecule has a bond order of one, indicating a single covalent bond.
In simple words: For a lithium molecule, electrons fill up specific energy levels called molecular orbitals. It has two electrons in the \( \sigma 1s \) orbital, two in the \( \sigma^*1s \) orbital, and two in the \( \sigma 2s \) orbital.

🎯 Exam Tip: Remember to fill molecular orbitals according to the Aufbau principle, Pauli's exclusion principle, and Hund's rule, just like atomic orbitals.

 

Question 8. How do you express the bond strength in terms of bond order?
Answer: Bond strength is directly related to bond order. A higher bond order means there are more bonds between atoms, which makes the bond stronger. Therefore, as bond order increases, bond strength also increases. For instance, a triple bond is stronger than a double bond.
In simple words: If atoms have a higher bond order (more shared electron pairs), their bond is stronger.

🎯 Exam Tip: Recall that bond order also affects bond length; higher bond order means shorter bond length and stronger bond.

 

Question 9. Arrange the following molecules in increasing order of energy: \( \mathrm{N}_{2} \), \( \mathrm{O}_{2} \), \( \mathrm{Cl}_{2} \), \( \mathrm{F}_{2} \).
Answer: The increasing order of energy (specifically, bond dissociation energy) for these molecules is: \( \mathrm{F}_{2} < \mathrm{Cl}_{2} < \mathrm{O}_{2} < \mathrm{N}_{2} \). This order reflects the strength of the chemical bonds, with triple bonds like in \( \mathrm{N}_{2} \) being the strongest, followed by double and then single bonds. The weaker bond in \( \mathrm{F}_{2} \) compared to \( \mathrm{Cl}_{2} \) is due to strong lone-pair repulsion in the small fluorine molecule.
In simple words: From weakest to strongest bonds, the order is fluorine, then chlorine, then oxygen, and finally nitrogen. It takes the most energy to break the bond in nitrogen gas.

🎯 Exam Tip: Remember that bond energy generally increases with bond order and decreases for larger atoms in the same group due to poorer orbital overlap, with exceptions like \( \mathrm{F}_{2} \) due to lone pair repulsion.

 

Question 10. Explain the important aspects of resonance with reference to the \( \mathrm{CO}_{3}^{2-} \) ion.
Answer: Resonance is a way to describe bonding in molecules or ions where a single Lewis structure cannot represent the true bonding. For the carbonate ion (\( \mathrm{CO}_{3}^{2-} \)), there are three possible resonance structures. These structures show that the electrons are delocalized, meaning they are spread out over the entire ion rather than being fixed in one place. The actual structure of the carbonate ion is a hybrid or average of these three resonating structures, which is more stable than any single Lewis structure. Each oxygen atom shares the negative charge, and the bond lengths are all equal and intermediate between single and double bonds.
In simple words: Resonance means the real structure of an ion, like carbonate, is a mix of several possible pictures, not just one. The electrons are shared around the whole ion, making it very stable.

🎯 Exam Tip: When explaining resonance, always mention electron delocalization, increased stability, and the fact that the actual structure is a hybrid, not rapidly shifting between forms.

 

Question 11. Write the hybridization of central atom of the following molecules: \( \mathrm{CCl}_{4} \), \( \mathrm{H}_{2}\mathrm{O} \), \( \mathrm{CO}_{2} \), \( \mathrm{SO}_{2} \).
Answer: The hybridization of the central atom in each molecule is:
Hybridization of carbon in \( \mathrm{CCl}_{4} \) is \( \mathrm{sp}^{3} \). This makes it tetrahedral.
Hybridization of oxygen in \( \mathrm{H}_{2}\mathrm{O} \) is \( \mathrm{sp}^{3} \). This leads to a bent molecular geometry due to two lone pairs.
Hybridization of carbon in \( \mathrm{CO}_{2} \) is \( \mathrm{sp} \). This results in a linear molecular shape.
Hybridization of sulfur in \( \mathrm{SO}_{2} \) is \( \mathrm{sp}^{2} \). This gives it a bent shape due to one lone pair.
In simple words: Different molecules use different types of orbital mixing, called hybridization, for their central atom. Carbon in \( \mathrm{CCl}_{4} \) and oxygen in \( \mathrm{H}_{2}\mathrm{O} \) use \( \mathrm{sp}^{3} \), carbon in \( \mathrm{CO}_{2} \) uses \( \mathrm{sp} \), and sulfur in \( \mathrm{SO}_{2} \) uses \( \mathrm{sp}^{2} \).

🎯 Exam Tip: To quickly determine hybridization, count the number of sigma bonds and lone pairs around the central atom (steric number). A steric number of 4 is \( \mathrm{sp}^{3} \), 3 is \( \mathrm{sp}^{2} \), and 2 is \( \mathrm{sp} \).

 

Question 12. Water is liquid at room temperature. Why?
Answer: Water is liquid at room temperature primarily because of its strong hydrogen bonding. Water molecules are highly polar due to the large difference in electronegativity between oxygen and hydrogen. This polarity causes hydrogen bonds to form between hydrogen atoms of one molecule and oxygen atoms of another. These strong attractions hold water molecules close together, requiring more energy to separate them, thus keeping it in a liquid state at typical room temperatures. For example, similar-sized molecules like methane are gases.
In simple words: Water stays liquid because its molecules stick together strongly using special connections called hydrogen bonds, which need a lot of energy to break apart.

🎯 Exam Tip: Always emphasize hydrogen bonding as the key factor for water's high boiling point and liquid state at room temperature.

 

Question 13. Melting and Boiling point of ionic compounds are high. Why?
Answer: Ionic compounds have high melting and boiling points because the electrostatic forces holding their ions together are very strong. These strong forces exist between oppositely charged ions in a crystal lattice, like in sodium chloride (NaCl). A large amount of energy is needed to overcome these powerful attractions and break down the lattice structure. This is why ionic compounds typically require high temperatures to melt or boil. For instance, common table salt needs a very high temperature to melt.
In simple words: Ionic compounds have high melting and boiling points because their charged particles are very strongly attracted to each other, needing a lot of heat to break them apart.

🎯 Exam Tip: The strength of electrostatic forces and the lattice energy are the main reasons for the high melting and boiling points of ionic compounds.

 

Question 15. Write the resonance structures of \( \mathrm{SO}_{3} \) and \( \mathrm{NO}_{2} \).
Answer:
(i) For \( \mathrm{SO}_{3} \) (sulfur trioxide), there are three resonance structures. Each structure shows sulfur double-bonded to one oxygen and single-bonded to two other oxygens, with a formal charge of -1 on the single-bonded oxygens and zero on the double-bonded oxygen. The electrons are delocalized, resulting in an average of 1.33 bonds between sulfur and each oxygen.
(ii) For \( \mathrm{NO}_{2} \) (nitrogen dioxide), there are two main resonance structures. Nitrogen is double-bonded to one oxygen and single-bonded to another. There is an unpaired electron on the nitrogen atom. The actual structure is an average of these two forms, with the unpaired electron and the negative charge delocalized.
In simple words: Sulfur trioxide has three ways its electrons can be arranged, and nitrogen dioxide has two. These different arrangements, called resonance structures, show how electrons are spread out in the molecules.

🎯 Exam Tip: When drawing resonance structures, remember that only electrons move, not atoms, and the overall charge must remain the same for each contributing structure.

 

Question 16. 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.
Answer: The combination of \( 2\mathrm{p}_{y} \) and \( 2\mathrm{p}_{y} \) orbitals will not form a sigma bond when the x-axis is the internuclear axis. A sigma bond is formed when atomic orbitals overlap head-on (axially) along the internuclear axis. The \( 2\mathrm{p}_{y} \) orbitals are oriented perpendicular to the x-axis, so they can only undergo sideways overlap, which leads to the formation of a pi bond, not a sigma bond. Overlap between s-orbitals or an s-orbital and an axial p-orbital forms a sigma bond.
In simple words: When atoms line up along the x-axis, two \( \mathrm{p}_{y} \) orbitals cannot make a strong straight-on bond (sigma bond). Instead, they overlap from the side, making a weaker pi bond.

🎯 Exam Tip: Remember that sigma bonds result from head-on overlap along the internuclear axis, while pi bonds result from sideways overlap above and below the internuclear axis.

 

Question 17. Which of the following has maximum dipole moment: \( \mathrm{Na}^{+} \), \( \mathrm{K}^{+} \), \( \mathrm{Li}^{+} \) and \( \mathrm{Cs}^{+} \).
Answer: Among the given ions, \( \mathrm{Li}^{+} \) has the maximum polarizing power, which would lead to a higher induced dipole moment in a bond with an anion. This is because dipole moment (or the ability to induce one, in the case of ions) is related to charge density and electronegativity. As you go down a group, the size of the ion increases, and its electronegativity (and charge density) decreases. \( \mathrm{Li}^{+} \) is the smallest ion with the highest charge density, giving it the strongest polarizing power. Therefore, \( \mathrm{Li}^{+} \) would cause the largest induced dipole moment in a bond. For example, \( \mathrm{LiCl} \) would have a higher bond dipole moment than \( \mathrm{NaCl} \).
In simple words: \( \mathrm{Li}^{+} \) has the strongest ability to pull electrons towards itself when forming a bond, which would create the largest charge difference, or dipole moment, in a resulting molecule. This is because it is the smallest and most concentrated ion.

🎯 Exam Tip: For cations, smaller size and higher charge lead to greater polarizing power, meaning they can induce a larger dipole in an anion they bond with.

 

Question 18. Arrange the halides of silver in increasing order of solubility in water.
Answer: As we move down the halogen group (F, Cl, Br, I), the size of the anion increases, and its electronegativity decreases. This leads to an increase in the covalent character of the silver halide bond. Covalent compounds are generally less soluble in water than ionic compounds. Silver fluoride (AgF) is the most ionic and therefore the most soluble. Silver iodide (AgI) is the most covalent and least soluble. Thus, the increasing order of solubility of silver halides in water is: \( \mathrm{AgI} < \mathrm{AgBr} < \mathrm{AgCl} < \mathrm{AgF} \). This trend is explained by Fajan's rules, where a larger anion is more easily polarized.
In simple words: Silver fluoride dissolves best in water, then silver chloride, then silver bromide, and silver iodide dissolves the least. This is because the bonds become more "covalent" (less ionic) as the halogen gets bigger.

🎯 Exam Tip: Remember Fajan's rules: smaller cation, larger anion, and higher charges on ions increase the covalent character of an ionic bond, reducing solubility in polar solvents like water.

 

Question 19. \( \mathrm{NaCl} \) is an ionic compound whereas \( \mathrm{CuCl} \) is a covalent compound. Explain the reason.
Answer: The difference in character between \( \mathrm{NaCl} \) and \( \mathrm{CuCl} \) is explained by Fajan's rules, specifically the concept of polarizing power. When a cation approaches an anion, it distorts the anion's electron cloud; this is called polarization. The more an anion is polarized, the more covalent the bond becomes. Factors influencing covalent character include the size of the anion, charge on the ions, and the electronic configuration of the cation. \( \mathrm{Na}^{+} \) has a noble gas configuration (8 electrons in its outermost shell), while \( \mathrm{Cu}^{+} \) has a pseudo noble gas configuration (18 electrons in its outermost shell). Cations with a pseudo noble gas configuration have greater polarizing power than those with a noble gas configuration, even if they have similar size and charge. Although \( \mathrm{Cu}^{+} \) (0.96 Å) and \( \mathrm{Na}^{+} \) (0.95 Å) are similar in size and both have a +1 charge, \( \mathrm{Cu}^{+} \)'s 18-electron shell makes it a stronger polarizer. Therefore, \( \mathrm{CuCl} \) exhibits more covalent character than \( \mathrm{NaCl} \).
In simple words: \( \mathrm{CuCl} \) is more covalent than \( \mathrm{NaCl} \) because the \( \mathrm{Cu}^{+} \) ion, even though similar in size to \( \mathrm{Na}^{+} \), has a special electron setup (18 electrons outside the nucleus) that makes it much better at pulling on the chloride ion's electrons, making the bond less purely ionic.

🎯 Exam Tip: When comparing ionic and covalent character, always consider Fajan's rules, especially the electronic configuration of the cation (noble gas vs. pseudo noble gas).

 

Question 20. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
\( \mathrm{AlCl}_{3} + \mathrm{Cl}^{-} \rightarrow \mathrm{AlCl}_{4}^{-} \)
Answer: In aluminum chloride (\( \mathrm{AlCl}_{3} \)), the aluminum atom is \( \mathrm{sp}^{2} \) hybridized. It forms three bonds with chlorine atoms, resulting in a trigonal planar geometry. Aluminum has empty \( 3\mathrm{p}_{z} \) orbitals. When \( \mathrm{AlCl}_{3} \) reacts with a chloride ion (\( \mathrm{Cl}^{-} \)) to form the tetrachloroaluminate ion (\( \mathrm{AlCl}_{4}^{-} \)), the empty \( 3\mathrm{p}_{z} \) orbital of aluminum gets involved in hybridization. The aluminum atom undergoes a change in hybridization from \( \mathrm{sp}^{2} \) to \( \mathrm{sp}^{3} \). This change in hybridization also transforms the molecular geometry from trigonal planar in \( \mathrm{AlCl}_{3} \) to tetrahedral in \( \mathrm{AlCl}_{4}^{-} \). This allows aluminum to form four bonds.
In simple words: In \( \mathrm{AlCl}_{3} \), aluminum uses \( \mathrm{sp}^{2} \) hybridization, giving it a flat triangular shape. When it gains another chlorine atom to become \( \mathrm{AlCl}_{4}^{-} \), its hybridization changes to \( \mathrm{sp}^{3} \), making its shape a 3D pyramid with four points (tetrahedral).

🎯 Exam Tip: Remember that adding a ligand can change the hybridization and geometry of the central atom if empty orbitals are available for bonding.

RBSE Class 11 Chemistry Chapter 4 Short Answer Type Questions

 

Question 21. Define octet rule. Write its significance and limitations.
Answer: The octet rule states that atoms tend to combine in such a way that each atom has eight electrons in its valence shell, achieving a stable electron configuration similar to that of a noble gas. This stability can be reached by gaining, losing, or sharing valence electrons. The octet rule is significant because it helps us understand the bonding and structures of a wide range of compounds, particularly in organic chemistry. However, it has several limitations:
1. **Electron Deficient Molecules:** Some compounds, like \( \mathrm{LiCl} \), \( \mathrm{BeH}_{2} \), and \( \mathrm{BCl}_{3} \), have central atoms (Li, Be, B) with fewer than eight valence electrons. For example, in \( \mathrm{Li}-\mathrm{Cl} \), lithium only has two valence electrons.
2. **Expanded Octets:** Elements in the third period and beyond, which have vacant d-orbitals, can accommodate more than eight electrons around their central atom. Examples include \( \mathrm{PF}_{6} \), \( \mathrm{SF}_{6} \), and \( \mathrm{H}_{2}\mathrm{SO}_{4} \). In these cases, the central atom uses its d-orbitals for bonding.
3. **Odd-Electron Molecules:** Molecules with an odd number of valence electrons, such as nitric oxide (NO) and nitrogen dioxide (\( \mathrm{NO}_{2} \)), cannot satisfy the octet rule for all their atoms. In NO, the nitrogen atom does not complete its octet.
In simple words: The octet rule says atoms try to get eight outer electrons to be stable, like noble gases. It helps explain many bonds but doesn't work for all molecules, especially those with too few or too many electrons around the central atom, or those with an odd number of electrons.

🎯 Exam Tip: When defining the octet rule, mention its goal (stability) and the means (gaining, losing, sharing electrons). For limitations, cite specific examples for each type of exception (electron-deficient, expanded octet, odd-electron species).

 

Question 22. Which of the following pairs are more covalent and why?
(a) CuO and CuS
(b) AgCl and Agl
(c) PbCl2 and PbCl4
(d) NaCl and CuCl
Answer: To determine which compound in each pair is more covalent, we use Fajan's rules, which describe how factors like ion size and charge affect the covalent character of a bond:
* **Cation Size:** Smaller cations have higher charge density and greater polarizing power, leading to more covalent character.
* **Cationic Charge:** Higher charge on the cation increases its polarizing power and thus covalent character.
* **Anion Size:** Larger anions are more easily polarized, leading to more covalent character.
* **Anionic Charge:** Higher charge on the anion increases its polarizability and covalent character.
* **Electronic Configuration:** Cations with a pseudo noble gas configuration (18 electrons in the outermost shell) have greater polarizing power than those with a noble gas configuration (8 electrons).

Applying these rules to the pairs:
(a) **CuO and CuS:** \( \mathrm{CuS} \) is more covalent than \( \mathrm{CuO} \). The cation (\( \mathrm{Cu}^{2+} \)) is the same. Sulfur (\( \mathrm{S}^{2-} \)) is a larger anion than oxygen (\( \mathrm{O}^{2-} \)), so \( \mathrm{S}^{2-} \) is more polarizable, increasing the covalent character of \( \mathrm{CuS} \).
(b) **AgCl and AgI:** \( \mathrm{AgI} \) is more covalent than \( \mathrm{AgCl} \). The cation (\( \mathrm{Ag}^{+} \)) is the same. Iodine (\( \mathrm{I}^{-} \)) is a much larger anion than chlorine (\( \mathrm{Cl}^{-} \)), making \( \mathrm{I}^{-} \) more polarizable and thus \( \mathrm{AgI} \) more covalent.
(c) **PbCl2 and PbCl4:** \( \mathrm{PbCl}_{4} \) is more covalent than \( \mathrm{PbCl}_{2} \). The lead cation has a higher charge in \( \mathrm{PbCl}_{4} \) (\( \mathrm{Pb}^{4+} \)) than in \( \mathrm{PbCl}_{2} \) (\( \mathrm{Pb}^{2+} \)). A higher cationic charge leads to greater polarizing power, making \( \mathrm{PbCl}_{4} \) more covalent.
(d) **NaCl and CuCl:** \( \mathrm{CuCl} \) is more covalent than \( \mathrm{NaCl} \). Both \( \mathrm{Na}^{+} \) and \( \mathrm{Cu}^{+} \) have a +1 charge and similar sizes (\( \mathrm{Na}^{+} \) 0.95 Å, \( \mathrm{Cu}^{+} \) 0.96 Å). However, \( \mathrm{Na}^{+} \) has a noble gas configuration (2, 8), while \( \mathrm{Cu}^{+} \) has a pseudo noble gas configuration (2, 8, 18). Cations with a pseudo noble gas configuration have much greater polarizing power. Therefore, \( \mathrm{CuCl} \) is more covalent.
In simple words: Bonds become more "covalent" when the positive ion can strongly pull on the electron cloud of the negative ion. This pulling power is stronger if the positive ion is small and highly charged, or if the negative ion is large and easily stretched. Also, certain electron arrangements in the positive ion (like in copper) make it pull even harder than simpler ions (like in sodium).

🎯 Exam Tip: For comparative questions on covalent character, always state Fajan's rules first, then apply them systematically to each pair, clearly identifying the differing factor (anion size, cation charge, or cation electronic configuration).

 

Question 23. Which out of \( \mathrm{NH}_{3} \) and \( \mathrm{NF}_{3} \) has higher dipole moment and why?
Answer: Ammonia (\( \mathrm{NH}_{3} \)) has a higher dipole moment than nitrogen trifluoride (\( \mathrm{NF}_{3} \)). Both molecules have a pyramidal shape with a lone pair of electrons on the central nitrogen atom. In \( \mathrm{NH}_{3} \), the individual N-H bond dipoles point towards nitrogen because nitrogen is more electronegative than hydrogen. The dipole moment of the lone pair on nitrogen also points in the same direction as the resultant of the N-H bond dipoles (away from the hydrogens and towards the nitrogen). These effects add up, resulting in a large net dipole moment for \( \mathrm{NH}_{3} \) (approx. \( 4.90 \times 10^{-30} \mathrm{Cm} \)). In contrast, in \( \mathrm{NF}_{3} \), fluorine is more electronegative than nitrogen, so the N-F bond dipoles point towards fluorine (away from nitrogen). The dipole moment of the lone pair on nitrogen points opposite to the resultant of the N-F bond dipoles. These opposing vectors partially cancel each other out, leading to a much smaller net dipole moment for \( \mathrm{NF}_{3} \) (approx. \( 0.8 \times 10^{-30} \mathrm{Cm} \)). This illustrates how molecular geometry and vector addition determine the overall dipole moment.
In simple words: \( \mathrm{NH}_{3} \) has a bigger dipole moment because all the small electrical pulls from its bonds and lone pair add up in the same direction. For \( \mathrm{NF}_{3} \), the pulls from its bonds and lone pair go in opposite directions, partly cancelling each other out.

🎯 Exam Tip: To compare dipole moments, consider both the polarity of individual bonds and the molecular geometry, visualizing how bond dipoles and lone pair dipoles add vectorially.

 

Question 24. Distinguish between a sigma and a pi bond.
Answer: A sigma bond (\( \sigma \)) and a pi bond (\( \pi \)) are different types of covalent bonds formed by the overlap of atomic orbitals. Here are their key distinctions:

In simple words: Sigma bonds are formed by a direct, head-on merge of electron clouds, making them strong and allowing free rotation. Pi bonds are formed by a side-by-side merge, are weaker, and prevent free rotation. Pi bonds always occur alongside a sigma bond.

🎯 Exam Tip: Focus on the direction of overlap (axial for sigma, sideways for pi), bond strength, ability to rotate, and their occurrence (sigma can be alone, pi always with sigma).

 

Question 25. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer: For atomic orbitals to combine effectively and form molecular orbitals, three main conditions must be met:
1. **Similar Energy Levels:** The combining atomic orbitals must have the same or very similar energy levels. For example, in a homonuclear diatomic molecule, a 1s orbital can combine with another 1s orbital, but not with a 2s orbital because their energy difference is too large. This energy similarity is crucial for effective mixing.
2. **Maximum Overlap:** The atomic orbitals must overlap to the greatest possible extent. A greater degree of overlap leads to a higher electron density between the nuclei in the resulting molecular orbital, which translates to a stronger bond. The extent of overlap directly influences the strength of the molecular bond formed.
3. **Correct Symmetry:** The combining atomic orbitals must possess the same symmetry with respect to the molecular axis. By convention, the z-axis is often chosen as the molecular axis. Orbitals with different symmetries cannot combine effectively. For instance, a \( 2\mathrm{p}_{z} \) orbital on one atom cannot combine with a \( 2\mathrm{p}_{x} \) orbital on another if the z-axis is the internuclear axis, as their orientations are incompatible for effective bonding.
In simple words: Atomic orbitals can only join to make molecular orbitals if they have almost the same energy, overlap a lot, and match up in shape (symmetry) around the bond line.

🎯 Exam Tip: Clearly list and explain each of the three conditions (energy, overlap, symmetry) with simple examples or reasons for full marks on this concept.

 

Question 27. Among \( \mathrm{HCl} \), \( \mathrm{H}_{2}\mathrm{O} \) and \( \mathrm{NH}_{3} \), which has higher boiling point and why?
Answer: The order of boiling points is \( \mathrm{H}_{2}\mathrm{O} > \mathrm{NH}_{3} > \mathrm{HCl} \). Water (\( \mathrm{H}_{2}\mathrm{O} \)) has the highest boiling point because it can form the most extensive network of hydrogen bonds. Each water molecule can form up to four hydrogen bonds with surrounding water molecules due to having two hydrogen atoms and two lone pairs. Ammonia (\( \mathrm{NH}_{3} \)) can form fewer hydrogen bonds (up to three per molecule) because it has three hydrogen atoms but only one lone pair. Hydrogen chloride (\( \mathrm{HCl} \)) forms the weakest hydrogen bonds, leading to the lowest boiling point among the three. The strength of the hydrogen bonding directly affects the amount of energy needed to overcome these intermolecular forces and boil the substance. Water's ability to form more hydrogen bonds makes it unique.
In simple words: Water boils at the highest temperature because its molecules can make more strong "hydrogen bonds" with each other compared to ammonia and hydrogen chloride.

🎯 Exam Tip: To explain boiling point differences due to hydrogen bonding, compare the number of potential hydrogen bonds each molecule can form, linking it to the energy required to overcome intermolecular forces.

 

Question 29. Dipole moment of \( \mathrm{CO}_{2} \) molecule is zero where as \( \mathrm{SO}_{2} \) has some dipole moment. Explain the reason.
Answer: The difference in dipole moment between \( \mathrm{CO}_{2} \) and \( \mathrm{SO}_{2} \) is due to their molecular geometries. For \( \mathrm{CO}_{2} \), the Lewis structure shows a central carbon atom double-bonded to two oxygen atoms (\( \mathrm{O}=\mathrm{C}=\mathrm{O} \)). It has a linear geometry with a bond angle of 180°. Although each C=O bond is polar, the two bond dipoles are equal in magnitude and point in opposite directions, thus they cancel each other out. This results in a net dipole moment of zero, making \( \mathrm{CO}_{2} \) a nonpolar molecule. For \( \mathrm{SO}_{2} \), the central sulfur atom is bonded to two oxygen atoms and also has one lone pair of electrons. This leads to a bent molecular geometry, not linear. Even though the S=O bonds are polar, their dipoles do not cancel each other out because of the bent shape and the presence of the lone pair. The resultant of the bond dipoles and the lone pair dipole creates a net nonzero dipole moment, making \( \mathrm{SO}_{2} \) a polar molecule. Sulfur dioxide's bent shape is crucial for its polarity.
In simple words: \( \mathrm{CO}_{2} \) has no dipole moment because its straight shape causes the two bond pushes to cancel each other out. \( \mathrm{SO}_{2} \) has a dipole moment because its bent shape means the bond pushes don't cancel, creating a net electrical imbalance.

🎯 Exam Tip: Always consider both bond polarity and molecular geometry (including lone pairs) when determining if a molecule has a net dipole moment.

 

Question 30. \( \mathrm{BaSO}_{4} \) is an ionic compound, yet it is insoluble in water. Why?
Answer: Barium sulfate (\( \mathrm{BaSO}_{4} \)) is an ionic compound, but it is insoluble in water because of its very high lattice enthalpy. Lattice enthalpy is the energy required to break apart one mole of an ionic compound into its gaseous ions. Both \( \mathrm{Ba}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) ions are large. The strong electrostatic attraction between these large, oppositely charged ions results in a very stable crystal lattice that is difficult to break apart. For an ionic compound to dissolve, the energy released during hydration (when water molecules surround the ions) must be greater than or equal to the lattice enthalpy. In the case of \( \mathrm{BaSO}_{4} \), the lattice enthalpy is significantly higher than its hydration enthalpy, making it energetically unfavorable for the compound to dissolve in water. This makes \( \mathrm{BaSO}_{4} \) very stable as a solid.
In simple words: \( \mathrm{BaSO}_{4} \) does not dissolve in water because its positive and negative ions are very strongly glued together. The energy needed to break them apart is much more than the energy released when water tries to pull them into solution.

🎯 Exam Tip: When discussing solubility of ionic compounds, always compare lattice enthalpy with hydration enthalpy. Insoluble compounds typically have a lattice enthalpy much greater than their hydration enthalpy.

 

Question 31. Explain any four factors which affect the solubility of ionic compounds.
Answer: The solubility of ionic compounds is influenced by several factors:
1. **Solute-Solvent Attractions (Nature of Solvent):** Ionic compounds are generally most soluble in polar solvents like water. This is because polar solvent molecules have partial charges that can strongly attract and surround the charged ions of the solute, pulling them away from the crystal lattice. The stronger these attractions, the higher the solubility. For example, nonpolar solvents cannot dissolve ionic compounds well.
2. **Common Ion Effect:** The presence of a common ion in the solvent significantly reduces the solubility of an ionic compound. If an ionic compound is sparingly soluble, like \( \mathrm{CaSO}_{4} \), adding a source of either \( \mathrm{Ca}^{2+} \) or \( \mathrm{SO}_{4}^{2-} \) ions to the water will shift the dissolution equilibrium back towards the solid \( \mathrm{CaSO}_{4} \), causing less of it to dissolve.
\( \mathrm{CaSO}_{4} (\mathrm{s}) \rightleftharpoons \mathrm{Ca}^{2+} (\mathrm{aq}) + \mathrm{SO}_{4}^{2-} (\mathrm{aq}) \)
3. **Temperature:** For most ionic compounds, increasing the temperature increases their solubility. This is because the dissolution process for many ionic compounds is endothermic (absorbs heat). According to Le Chatelier's principle, adding heat favors the endothermic direction, thus increasing solubility.
\( \mathrm{CaSO}_{4} (\mathrm{s}) + \text{Heat} \rightleftharpoons \mathrm{Ca}^{2+} (\mathrm{aq}) + \mathrm{SO}_{4}^{2-} (\mathrm{aq}) \)
4. **Lattice Energy and Hydration Energy:** For an ionic compound to dissolve, its crystal lattice must break apart, which requires energy (lattice energy). This energy is then compensated by the energy released when ions are surrounded by solvent molecules (hydration energy). If hydration energy is greater than lattice energy, the compound is soluble. If lattice energy is much higher, it is insoluble. This is a fundamental thermodynamic balance.
In simple words: How much an ionic compound dissolves depends on: how well water molecules pull its ions apart, if there are already similar ions in the water (which stops more from dissolving), the temperature, and the balance between the energy needed to break the solid apart and the energy released when water surrounds the ions.

🎯 Exam Tip: When explaining factors affecting solubility, define each factor and provide a clear example or chemical principle (like Le Chatelier's principle for temperature and common ion effect).

 

Question 32. What do you understand by polar covalent bond. Explain with example.
Answer: A polar covalent bond is a type of chemical bond where electrons are shared unequally between two atoms. This unequal sharing happens when two atoms with different electronegativities (i.e., different abilities to attract electrons) form a covalent bond. The electrons are pulled more strongly towards the atom with higher electronegativity, creating a partial negative charge on that atom (\( \delta^{-} \)) and a partial positive charge on the less electronegative atom (\( \delta^{+} \)). This separation of charges creates a dipole within the bond. For example, in a hydrogen chloride (\( \mathrm{HCl} \)) molecule, chlorine is more electronegative than hydrogen. As a result, the shared electrons are drawn closer to the chlorine atom. This gives chlorine a slight negative charge and hydrogen a slight positive charge, making the H-Cl bond a polar covalent bond. This partial charge is not a full ionic charge but a slight unevenness in electron distribution.
In simple words: A polar covalent bond is like sharing a toy unevenly; one atom pulls the shared electrons more strongly, becoming slightly negative, while the other becomes slightly positive. Hydrogen chloride is a good example.

🎯 Exam Tip: Define electronegativity when explaining polar covalent bonds, and clearly indicate the partial positive (\( \delta^{+} \)) and partial negative (\( \delta^{-} \)) charges in your example.

 

Question 33. Which of the following compound has weakest and strongest hydrogen bonding: \( \mathrm{NH}_{3} \), \( \mathrm{PH}_{3} \), \( \mathrm{H}_{2}\mathrm{O} \) and \( \mathrm{H}_{2}\mathrm{S} \).
Answer: Hydrogen bonding occurs when a hydrogen atom bonded to a highly electronegative atom (like nitrogen, oxygen, or fluorine) is attracted to a lone pair of electrons on another electronegative atom. The conditions for strong hydrogen bonding are:
1. Hydrogen must be bonded to a highly electronegative atom (N, O, or F).
2. The electronegative atom should also be small.

Based on these conditions:
* **\( \mathrm{H}_{2}\mathrm{O} \):** Water has the **strongest** hydrogen bonding. Oxygen is highly electronegative and small, and each water molecule can form extensive hydrogen bond networks.
* **\( \mathrm{NH}_{3} \):** Ammonia forms strong hydrogen bonds, but weaker than water. Nitrogen is highly electronegative and small, but \( \mathrm{NH}_{3} \) forms fewer hydrogen bonds per molecule compared to \( \mathrm{H}_{2}\mathrm{O} \).
* **\( \mathrm{H}_{2}\mathrm{S} \):** Hydrogen sulfide forms very weak hydrogen bonds. Sulfur is less electronegative than oxygen and larger, making its hydrogen bonds significantly weaker than in water.
* **\( \mathrm{PH}_{3} \):** Phosphine (\( \mathrm{PH}_{3} \)) exhibits the **weakest** (practically negligible) hydrogen bonding. Phosphorus has similar electronegativity to hydrogen (around 2.1), so the P-H bond is essentially nonpolar. Therefore, it cannot form effective hydrogen bonds.

Thus, \( \mathrm{H}_{2}\mathrm{O} \) has the strongest hydrogen bonding, and \( \mathrm{PH}_{3} \) has the weakest (or no) hydrogen bonding among the given compounds. This is clearly shown by their boiling points: \( \mathrm{H}_{2}\mathrm{O} \) (100°C) > \( \mathrm{NH}_{3} \) (-33°C) > \( \mathrm{H}_{2}\mathrm{S} \) (-60°C) > \( \mathrm{PH}_{3} \) (-87°C).
In simple words: Water has the strongest hydrogen bonds because oxygen is very good at attracting hydrogen in this way. Phosphine (\( \mathrm{PH}_{3} \)) has almost no hydrogen bonds at all because phosphorus is not strong enough to pull hydrogen electrons effectively.

🎯 Exam Tip: To identify hydrogen bonding strength, look for H bonded to N, O, or F. The smaller and more electronegative the atom, the stronger the hydrogen bond. If H is bonded to other elements, hydrogen bonding is usually negligible.

 

RBSE Class 11 Chemistry Chapter 4 Long Answer Type Questions

 

Question 35. Draw diagrams showing the formation of a triple bond between carbon atoms in C2H2 molecule.
Answer: The C2H2 molecule, also known as acetylene, forms a triple bond between its two carbon atoms. This triple bond consists of one strong sigma (\( \sigma \)) bond and two weaker pi (\( \pi \)) bonds. Each carbon atom is also single-bonded to one hydrogen atom via a sigma bond. Both carbon atoms are sp-hybridized, leading to a linear geometry with a bond angle of 180 degrees. The two pi bonds are formed by the sidewise overlap of the unhybridized 2py and 2pz atomic orbitals on each carbon atom.

In simple words: The C2H2 molecule has a strong triple bond between its two carbon atoms, made of one sigma and two pi bonds. Each carbon also has a sigma bond with a hydrogen atom. This gives the molecule a straight-line shape.

🎯 Exam Tip: Remember that sp hybridization leads to linear geometry and triple bonds have one sigma and two pi bonds. Clearly label all bonds and orbitals in your diagram.

 

Question 36. Explain the formation of H2 molecule on the basis of Valence Bond Theory.
Answer: According to Valence Bond Theory (VBT), a hydrogen molecule (\( \text{H}_2 \)) forms when two hydrogen atoms approach each other. Each hydrogen atom has one electron in its 1s orbital. When these atoms come close, their atomic orbitals overlap. This overlap allows the electrons to be shared between both nuclei, creating a covalent bond. The formation of this bond releases energy, making the \( \text{H}_2 \) molecule more stable than two separate hydrogen atoms.
As the atoms get closer, new attractive forces (between the nucleus of one atom and the electron of the other) and repulsive forces (between the two nuclei and between the two electrons) arise. A bond forms when the attractive forces are stronger than the repulsive forces. The system reaches maximum stability at a specific distance called the bond length, where the potential energy is at its minimum. For \( \text{H}_2 \), this bond length is approximately 74 pm, and the energy released is called bond energy.
In simple words: Two hydrogen atoms join to make an H2 molecule. Each atom shares its electron, and their orbitals overlap. This makes a strong, stable bond and releases energy, which is why H2 is more stable than single hydrogen atoms.

🎯 Exam Tip: When explaining bond formation using VBT, always mention the overlap of atomic orbitals, sharing of electrons, and the balance between attractive and repulsive forces that leads to a stable molecule and bond energy.

 

Question 37. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Answer: In phosphorus pentachloride (\( \text{PCl}_5 \)), the central phosphorus atom undergoes \( \text{sp}^3\text{d} \) hybridization. This means one s, three p, and one d orbital mix to form five equivalent \( \text{sp}^3\text{d} \) hybrid orbitals. These five hybrid orbitals arrange themselves in a trigonal bipyramidal geometry around the phosphorus atom. Three of these orbitals lie in a plane at 120-degree angles to each other (equatorial positions), while the other two lie above and below this plane along the vertical axis (axial positions).

The axial bonds in \( \text{PCl}_5 \) are longer than the equatorial bonds due to greater repulsion experienced by the axial bond pairs. The three equatorial bond pairs repel each other at 120 degrees, while the two axial bond pairs experience more intense 90-degree repulsions from the equatorial bond pairs. To minimize these stronger repulsions, the axial bonds stretch out and become slightly longer, resulting in weaker bonds compared to the equatorial ones.
In simple words: The central phosphorus atom in \( \text{PCl}_5 \) uses a special mix of orbitals called \( \text{sp}^3\text{d} \) hybridization. This gives it a shape like a triangular pyramid with two points. The two bonds that point straight up and down (axial bonds) are a bit longer than the three bonds that lie flat in the middle (equatorial bonds). This happens because the up-and-down bonds get pushed more by the flat bonds, so they stretch out to make more space.

🎯 Exam Tip: When describing \( \text{PCl}_5 \) hybridization, explicitly mention \( \text{sp}^3\text{d} \) hybridization, trigonal bipyramidal geometry, and the reasons for the different bond lengths based on bond pair-bond pair repulsion (axial bonds experience more 90° repulsion).

 

Question 38. What is meant by hybridisation of atomic orbitals? Describe sp, sp² and sp³ hybridization with examples.
Answer: Hybridization is the process where atomic orbitals of slightly different energies mix together to form a new set of equivalent hybrid orbitals. These new hybrid orbitals have the same energy and shape, and they arrange themselves in space to minimize repulsion, which makes the molecule more stable. Only orbitals from the same atom or ion and with similar energies take part in hybridization. The number of hybrid orbitals formed is equal to the number of atomic orbitals that mixed.

(a) **sp Hybridization:** This involves the mixing of one s orbital and one p orbital. It results in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbital has 50% s-character and 50% p-character. Molecules with sp hybridization have a linear geometry, and the bond angle is 180°. For example, in ethyne (\( \text{C}_2\text{H}_2 \)), each carbon atom is sp-hybridized.

(b) **sp² Hybridization:** This involves the mixing of one s orbital and two p orbitals. It results in the formation of three equivalent sp² hybrid orbitals. These three hybrid orbitals are arranged in a trigonal planar arrangement with bond angles of 120°. Each sp² hybrid orbital has 33.3% s-character and 66.7% p-character. For example, in ethene (\( \text{C}_2\text{H}_4 \)) and boron trifluoride (\( \text{BF}_3 \)), the carbon and boron atoms are sp²-hybridized, respectively.

(c) **sp³ Hybridization:** This involves the mixing of one s orbital and three p orbitals. It results in the formation of four equivalent sp³ hybrid orbitals. These four hybrid orbitals are arranged in a tetrahedral geometry with bond angles of 109.5°. Each sp³ hybrid orbital has 25% s-character and 75% p-character. For example, in methane (\( \text{CH}_4 \)) and ammonia (\( \text{NH}_3 \)), the carbon and nitrogen atoms are sp³-hybridized, respectively.
In simple words: Hybridization is when different types of atomic orbitals mix to make new, equal orbitals. These new orbitals help atoms bond better and make molecules stable. The type of hybridization (like sp, sp², sp³) tells us the shape of the molecule. For example, sp makes a straight line, sp² makes a flat triangle, and sp³ makes a pyramid shape.

🎯 Exam Tip: For hybridization questions, define the process, state the orbitals involved, the number of hybrid orbitals formed, the percentage of s and p character, the resulting geometry, and provide a common example for each type (sp, sp², sp³).

 

Question 39. What is Molecular Orbital Theory. With the help of energy levels in homonuclear diatomic orbitals, arrange the following species in increasing order of stability \( \mathrm{0}_{2}^{2-}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+} \) Answer:
Answer: Molecular Orbital Theory (MOT) explains how atomic orbitals combine to form molecular orbitals when atoms come together to form a molecule. In MOT, electrons are not seen as belonging to individual atoms but are spread over the entire molecule in molecular orbitals. The main ideas are:
1. Atomic orbitals with similar energies and correct symmetry combine to form new molecular orbitals.
2. Electrons in molecular orbitals are influenced by all the nuclei in the molecule.
3. The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, they form one bonding molecular orbital (lower energy, more stable) and one antibonding molecular orbital (higher energy, less stable).
4. Electrons fill molecular orbitals according to the Aufbau principle, Pauli's exclusion principle, and Hund's rule.

To arrange the given oxygen species by stability, we calculate their bond orders:
**Molecular Orbital Configurations and Bond Orders:**
\( \text{O}_2^{2-} \): Total electrons = 16 + 2 = 18
\( \sigma 1\text{s}^2 \sigma^* 1\text{s}^2 \sigma 2\text{s}^2 \sigma^* 2\text{s}^2 \sigma 2\text{p}_z^2 \pi 2\text{p}_x^2 \pi 2\text{p}_y^2 \pi^* 2\text{p}_x^2 \pi^* 2\text{p}_y^2 \)
Bond order \( = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 8) = \frac{2}{2} = 1.0 \)

\( \text{O}_2^{-} \): Total electrons = 16 + 1 = 17
\( \sigma 1\text{s}^2 \sigma^* 1\text{s}^2 \sigma 2\text{s}^2 \sigma^* 2\text{s}^2 \sigma 2\text{p}_z^2 \pi 2\text{p}_x^2 \pi 2\text{p}_y^2 \pi^* 2\text{p}_x^2 \pi^* 2\text{p}_y^1 \)
Bond order \( = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 7) = \frac{3}{2} = 1.5 \)

\( \text{O}_2 \): Total electrons = 16
\( \sigma 1\text{s}^2 \sigma^* 1\text{s}^2 \sigma 2\text{s}^2 \sigma^* 2\text{s}^2 \sigma 2\text{p}_z^2 \pi 2\text{p}_x^2 \pi 2\text{p}_y^2 \pi^* 2\text{p}_x^1 \pi^* 2\text{p}_y^1 \)
Bond order \( = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 6) = \frac{4}{2} = 2.0 \)

\( \text{O}_2^{+} \): Total electrons = 16 - 1 = 15
\( \sigma 1\text{s}^2 \sigma^* 1\text{s}^2 \sigma 2\text{s}^2 \sigma^* 2\text{s}^2 \sigma 2\text{p}_z^2 \pi 2\text{p}_x^2 \pi 2\text{p}_y^2 \pi^* 2\text{p}_x^1 \)
Bond order \( = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 5) = \frac{5}{2} = 2.5 \)

**Order of Bond Order:** \( \text{O}_2^{+} > \text{O}_2 > \text{O}_2^{-} > \text{O}_2^{2-} \)
Since stability is directly proportional to bond order, the increasing order of stability is:
\( \text{O}_2^{2-} < \text{O}_2^{-} < \text{O}_2 < \text{O}_2^{+} \)
In simple words: Molecular Orbital Theory tells us how electrons fill combined atomic orbitals to form molecular orbitals in a molecule. To find which oxygen molecule is most stable, we calculate its 'bond order'. A higher bond order means a stronger and more stable bond. Based on calculations, the oxygen molecule with a positive charge (\( \text{O}_2^{+} \)) is the most stable, and the one with two negative charges (\( \text{O}_2^{2-} \)) is the least stable among these options.

🎯 Exam Tip: When using MOT, clearly state its postulates. For stability comparisons, calculate the bond order for each species. Remember that stability is directly proportional to bond order (and inversely proportional to the number of antibonding electrons). Also, make sure to write out the full molecular orbital configuration to avoid errors in counting bonding and antibonding electrons.

 

Question 40. What do you mean by term bond order? Calculate the bond order of: N2, C2, H2-, N2-.
Answer: Bond order is a measure of the number of chemical bonds between a pair of atoms. It is defined as one-half the difference between the number of electrons in bonding molecular orbitals (\( \text{N}_b \)) and the number of electrons in antibonding molecular orbitals (\( \text{N}_a \)). A higher bond order generally indicates a stronger and more stable bond.
Bond Order (\( \text{B.O.} \)) \( = \frac{1}{2} (\text{N}_b - \text{N}_a) \)
Integral bond order values (1, 2, or 3) correspond to single, double, or triple bonds, respectively. A positive bond order means a stable molecule, while a zero or negative bond order indicates an unstable molecule.

(a) **Bond order of \( \text{N}_2 \):**
Total electrons = 7 + 7 = 14
Molecular orbital electronic configuration:
\( (\sigma 1\text{s})^2 (\sigma^* 1\text{s})^2 (\sigma 2\text{s})^2 (\sigma^* 2\text{s})^2 (\pi 2\text{p}_x)^2 (\pi 2\text{p}_y)^2 (\sigma 2\text{p}_z)^2 \)
Number of bonding electrons (\( \text{N}_b \)) = 2 + 2 + 2 + 2 + 2 = 10
Number of anti-bonding electrons (\( \text{N}_a \)) = 2 + 2 = 4
Bond order \( = \frac{1}{2} (10 - 4) = \frac{6}{2} = 3 \)
In simple words: Nitrogen gas (\( \text{N}_2 \)) has 14 electrons. When these electrons fill the molecular orbitals, there are 10 bonding electrons and 4 antibonding electrons. So, its bond order is 3, meaning it has a very strong triple bond.

(b) **Bond order of \( \text{C}_2 \):**
Total number of electrons = 6 + 6 = 12
Molecular orbital electronic configuration:
\( (\sigma 1\text{s})^2 (\sigma^* 1\text{s})^2 (\sigma 2\text{s})^2 (\sigma^* 2\text{s})^2 (\pi 2\text{p}_x)^2 (\pi 2\text{p}_y)^2 \)
Number of bonding electrons (\( \text{N}_b \)) = 2 + 2 + 2 = 8
Number of anti-bonding electrons (\( \text{N}_a \)) = 2 + 2 = 4
Bond order \( = \frac{1}{2} (8 - 4) = \frac{4}{2} = 2 \)
In simple words: The \( \text{C}_2 \) molecule has 12 electrons. These fill orbitals with 8 bonding electrons and 4 antibonding electrons. This gives a bond order of 2, which means it has a double bond.

(c) **Bond order of \( \text{H}_2^{-} \):**
Total electrons = 1 + 1 + 1 (for the negative charge) = 3
Molecular orbital electronic configuration:
\( (\sigma 1\text{s})^2 (\sigma^* 1\text{s})^1 \)
Number of bonding electrons (\( \text{N}_b \)) = 2
Number of anti-bonding electrons (\( \text{N}_a \)) = 1
Bond order \( = \frac{1}{2} (2 - 1) = \frac{1}{2} = 0.5 \)
In simple words: The \( \text{H}_2^{-} \) ion has 3 electrons. Two go into bonding orbitals, and one goes into an antibonding orbital. This results in a bond order of 0.5, meaning it has a half-bond, which is less stable than a full single bond.

(d) **Bond order of \( \text{N}_2^{-} \):**
Total electrons = 14 + 1 (for the negative charge) = 15
Molecular orbital electronic configuration:
\( (\sigma 1\text{s})^2 (\sigma^* 1\text{s})^2 (\sigma 2\text{s})^2 (\sigma^* 2\text{s})^2 (\pi 2\text{p}_x)^2 (\pi 2\text{p}_y)^2 (\sigma 2\text{p}_z)^2 (\pi^* 2\text{p}_x)^1 \)
Number of bonding electrons (\( \text{N}_b \)) = 10
Number of anti-bonding electrons (\( \text{N}_a \)) = 4 + 1 = 5
Bond order \( = \frac{1}{2} (10 - 5) = \frac{5}{2} = 2.5 \)
In simple words: The \( \text{N}_2^{-} \) ion has 15 electrons. Out of these, 10 are in bonding orbitals and 5 are in antibonding orbitals. This gives it a bond order of 2.5, which is stronger than a double bond but weaker than a triple bond.

🎯 Exam Tip: When calculating bond order, always write down the full molecular orbital configuration for each species. Carefully count bonding and antibonding electrons, and remember that a higher bond order correlates with greater stability and shorter bond length.