Chemical Bonding And Molecular Structure – JEE Main ChemistryPractice Questions & PYQs

The central iodine atom in the $\mathrm{I}_3^{-}$ ion is surrounded by two bond pairs and three lone pairs, which corresponds to a steric number of five and $sp^3d$ hybridization. According to VSEPR theory, these three lone pairs prioritize equatorial positions within the trigonal bipyramidal electron geometry to minimize electron repulsion, which forces the two bonding iodine atoms into the axial positions. This symmetric configuration results in a linear molecular shape with an I-I-I bond angle of $180^{\circ}$.

Polarity in a molecule occurs when individual bond dipoles fail to cancel each other out, often due to an asymmetric arrangement of electron density or the presence of a lone pair on the central atom. While geometries such as tetrahedral, square planar, and rectangular planar possess high symmetry that results in a zero vector sum of dipole moments-thereby rendering them non-polar-the square pyramidal shape is distinct. Because a square pyramidal $\text{AB}_4$ molecule contains a lone pair on the central atom 'A' that is not cancelled by the surrounding bonds, the molecule exhibits an asymmetric charge distribution that generates a permanent net dipole moment, making it polar.

According to VSEPR theory, the arrangement of electron pairs around a central atom determines the overall electron geometry of a molecule. With sulfur forming 4 $\sigma$ bonds and retaining 1 lone pair, the molecule possesses a total of 5 electron domains. This specific number of electron regions compels the structure to adopt a trigonal bipyramidal geometry to minimize electrostatic repulsion between the pairs.

The strength of a covalent bond is inversely related to its length, meaning that shorter bonds generally require more energy to break due to superior orbital overlap. As you move down the halogen group from fluorine to iodine, the increasing atomic size causes the carbon-halogen bond length to grow in the sequence $C-F < C-Cl < C-Br < C-I$. Because shorter bonds are inherently more stable and require greater energy to dissociate, the average bond energies follow the descending order $C-F > C-Cl > C-Br > C-I$.

Dipole moment is a vector quantity determined by molecular geometry and bond polarity, where a perfectly symmetrical arrangement leads to a net value of zero as individual bond dipoles cancel out. Both $CH_4$ and $CCl_4$ possess a regular tetrahedral geometry, causing the bond moments to cancel perfectly and resulting in a dipole moment of zero for both molecules. In contrast, $CHCl_3$ exhibits an asymmetric structure because the C-H and C-Cl bonds have differing polarities, preventing the cancellation of bond moments and resulting in a non-zero dipole moment, which establishes the order $CH_4 = CCl_4 < CHCl_3$.

The bond angle in a molecule is primarily dictated by VSEPR theory, which states that lone pairs exert stronger repulsive forces on bond pairs than bond pairs do on each other, thereby reducing the bond angles from the ideal geometry. In $CH_4$, the central carbon atom is $sp^3$ hybridized with four bonding pairs and zero lone pairs, resulting in a stable tetrahedral geometry and a maximum bond angle of 109.5 degrees. Conversely, the presence of lone pairs in $NH_3$, $H_2O$, and $H_2S$ introduces additional repulsion that compresses the bond angles to values significantly smaller than 109.5 degrees.

Benzene consists of a hexagonal ring made of six carbon atoms, each of which is $sp^2$ hybridized to maintain its planar geometry. Since $sp^2$ hybridization involves the combination of one s-orbital and two p-orbitals, every individual carbon atom produces exactly 3 hybrid orbitals. By multiplying the 3 orbitals formed per carbon atom by the 6 carbon atoms present in the molecule, the total count is determined by the equation $6 \times 3 = 18$ $sp^2$ hybrid orbitals.

The hybridization of an atom is determined by the number of bonding pairs and lone pairs surrounding the central center, so any chemical process that alters these electron domains inherently changes the hybridization state. When $\text{XeF}_4$ reacts with $\text{SbF}_5$, the reaction proceeds as follows:
$\text{XeF}_4 + \text{SbF}_5 \rightarrow [\text{XeF}_3]^+ + [\text{SbF}_6]^-$
In this transformation, the xenon atom shifts from $sp^3d^2$ hybridization in the neutral $\text{XeF}_4$ molecule to $sp^3d$ hybridization in the resulting $[\text{XeF}_3]^+$ ion due to the loss of a fluoride ligand.

The central chlorine atom in perchloric acid, $\mathrm{HClO_4}$, exists in a +7 oxidation state, which requires it to establish multiple bonds with the surrounding oxygen atoms to achieve a stable electronic configuration. The chlorine atom forms a single covalent bond with one hydroxyl group, and it utilizes its d-orbitals to form three double bonds with the remaining three oxygen atoms to satisfy its valency and minimize formal charges. Consequently, the structural geometry of the molecule contains 3 $\mathrm{Cl}=\mathrm{O}$ bonds.

The magnetic moment of a molecular species is determined by the number of unpaired electrons present, governed by the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons. A value of 1.73 B.M. indicates the presence of a single unpaired electron, as $\sqrt{1(1+2)} = \sqrt{3}$ evaluates to approximately 1.73. Since both $\mathrm{O}_2^-$ and $\mathrm{O}_2^+$ contain exactly one unpaired electron, they align with this magnetic moment, unlike neutral $\mathrm{O}_2$ which holds two unpaired electrons and exhibits a magnetic moment of 2.83 B.M.

Chlorine undergoes a disproportionation reaction when treated with hot and concentrated sodium hydroxide, yielding sodium chloride and sodium chlorate. Sodium chloride corresponds to (X) because it generates a white precipitate in the presence of silver nitrate, identifying (Y) as sodium chlorate. The structural chemistry of the chlorate ion, $ClO_3^-$, determines the bond properties in compound (Y).

The chlorate ion exists as a resonance hybrid where five total bonds are distributed equally among three chlorine-oxygen linkages. Calculating the average bond order requires dividing the total number of bonds by the number of bonding positions, which is mathematically represented as $1 + \frac{2}{3}$. This division results in a value of $\frac{5}{3}$, yielding an average bond order of 1.66 or 1.67.

The magnitude of electrostatic forces is determined by Coulomb's Law, $F \propto \frac{q_1q_2}{r^2}$, which dictates that interaction strength relies primarily on the product of the charges involved. Ion-ion interactions are the strongest because they involve the attraction between two full formal charges, whereas ion-dipole interactions are weaker due to involving only one full charge interacting with a partial charge. Finally, dipole-dipole interactions are the weakest of the three as they arise from the attraction between two permanent partial charges, leading to an overall decreasing strength of ion-ion > ion-dipole > dipole-dipole.

Intermolecular and inter-ionic forces are defined by how their potential energy changes relative to the spatial separation $r$ between particles. Ion-ion interactions result from the direct electrostatic attraction between charged species, giving an energy dependence of $E \propto \frac{1}{r}$, which corresponds to matching (a). In contrast, dipole-dipole interactions rely on the alignment of permanent dipoles and follow the relationship $E \propto \frac{1}{r^{3}}$, correlating to matching (c), while London dispersion forces, which arise from transient fluctuations in electron density, exhibit the steepest decay as $E \propto \frac{1}{r^{6}}$, matching (d).

The potential energy of an $H_2$ molecule is determined by the balance between attractive electronic interactions and repulsive nuclear forces as the atoms approach each other. At large separations, the potential energy is zero, but as the atoms move closer, attractive forces lower the energy, creating a well that reaches a minimum at the equilibrium bond length. If the internuclear distance is reduced further, dominant electrostatic repulsion between the two nuclei causes the potential energy to rise sharply, resulting in the characteristic asymmetric curve of a stable chemical bond.

While phosphorus pentachloride exhibits a trigonal bipyramidal geometry in its gaseous and liquid forms, it undergoes self-ionization upon solidification to maximize lattice stability. This transition results in the formation of an ionic solid consisting of tetrahedral $[PCl_4]^+$ cations and octahedral $[PCl_6]^-$ anions. Consequently, the molecular form is replaced by an ionic arrangement that optimizes the crystal structure.

The strength of a chemical bond is directly proportional to its bond order, such that a lower bond order equates to a weaker bond. The species $\mathrm{NO}^{+}$ has the highest bond order of 3, while $\mathrm{NO}$ and $\mathrm{NO}^{2+}$ possess an intermediate bond order of 2.5. Because $\mathrm{NO}^{-}$ has the lowest bond order of 2, it exhibits the minimum bond strength of the group.

The stiffness and stability of a chemical bond are determined by the depth of its potential energy curve, where a deeper minimum in $U$ indicates a more stable and robust interaction. Because the A-B molecule sits at the lowest potential energy point among the options provided, it possesses the largest bond enthalpy and shortest bond length, identifying it as the stiffest bond in the set.

To determine the bond order and magnetic behavior of the cyanide ion, $\mathrm{CN}^{-}$, we first calculate the total number of electrons available for bonding. Carbon contributes 6 electrons, nitrogen contributes 7, and the additional negative charge provides one more, resulting in a total of 14 electrons to distribute into molecular orbitals. Filling these orbitals in order of increasing energy gives the configuration $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2$. The bond order is calculated by taking the difference between bonding and antibonding electrons and dividing by two, which yields $\frac{1}{2}(10 - 4) = 3$. Because every electron occupies a paired state within these orbitals, the species lacks unpaired electrons and is therefore diamagnetic.

Determining the molecular geometry of these xenon compounds requires analyzing their hybridization states based on electron pair distribution around the central atom. The $[\mathrm{XeF}_5]^{-}$ ion undergoes $sp^3 d^3$ hybridization to accommodate its electron pairs, resulting in a pentagonal planar structure. In contrast, $\mathrm{XeO}_3 \mathrm{F}_2$ adopts $sp^3 d$ hybridization, where the arrangement of the surrounding atoms dictates a trigonal bipyramidal geometry.

According to Molecular Orbital Theory, bond length is inversely proportional to bond order, meaning a higher bond order results in a shorter bond distance. Assessing the electronic configurations reveals that the bond order for $C_2^{2-}$ is 3, for $N_2^{2-}$ is 2, for $O_2^{2-}$ is 1, and for $O_2$ is 2. Because $C_2^{2-}$ exhibits the highest bond order of 3 and has all electrons paired, it is classified as diamagnetic and represents the species with the shortest bond length among the options provided.