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

The substantial difference in bond dissociation energies arises from the phenomenon of back bonding in $\mathrm{BF}_3$, where electron density is donated from the filled p-orbitals of the fluorine atoms into the vacant p-orbital of the central boron atom, resulting in a strong $\mathrm{p}\pi-\mathrm{p}\pi$ interaction. This additional orbital overlap confers partial double bond character to the $\mathrm{B}-\mathrm{F}$ link, which effectively shortens the bond and elevates the energy required to dissociate it to $646\,\mathrm{kJ}\,\mathrm{mol}^{-1}$. In contrast, the carbon atom in $\mathrm{CF}_4$ lacks a vacant p-orbital to accept electron pairs, preventing such $\pi$-bonding and resulting in the lower observed $\mathrm{C}-\mathrm{F}$ bond dissociation energy of $515\,\mathrm{kJ}\,\mathrm{mol}^{-1}$.

The most efficient way to identify species with the same bond order is to verify their total electron count, as isoelectronic molecules inherently share the same molecular orbital configuration and thus the same bond order. Both $\mathrm{CN}^{-}$ and $\mathrm{NO}^{+}$ contain exactly 14 electrons. Applying the bond order formula $B.O. = \frac{1}{2}(N_b - N_a)$ to their respective electronic structures confirms that both species possess a bond order of 3.

The polarizing power of a cation is directly proportional to its charge density, which is defined by the ratio of its ionic charge to its ionic size. When comparing these species, $\mathrm{K}^{+}$ has the largest radius and the lowest charge, resulting in the weakest polarizing effect, whereas $\mathrm{Be}^{2+}$ possesses the smallest radius and a higher charge, yielding the greatest effect. Consequently, the polarizing power increases as we move through the series, resulting in the order $\mathrm{K}^{+} < \mathrm{Ca}^{2+} < \mathrm{Mg}^{2+} < \mathrm{Be}^{2+}$.

Molecular Orbital Theory provides a method for determining chemical properties based on the distribution of electrons in bonding and antibonding molecular orbitals. The bond order is calculated as $\text{Bond Order} = \frac{N_b - N_a}{2}$, where $N_b$ represents bonding electrons and $N_a$ represents antibonding electrons, while magnetic behavior is determined by the presence or absence of unpaired electrons in these orbitals.

Evaluating the ionization processes requires identifying both the bond order shift and the change in magnetic state. For the $C_2 \rightarrow C_2^+$ transition, the species changes from 12 to 11 electrons, where the bond order decreases from 2 ($N_b=8, N_a=4$) to 1.5 ($N_b=7, N_a=4$) and the magnetic property shifts from diamagnetic to paramagnetic. Similarly, the $N_2 \rightarrow N_2^+$ transition involves 14 electrons changing to 13, causing the bond order to drop from 3 ($N_b=10, N_a=4$) to 2.5 ($N_b=9, N_a=4$) while the magnetism changes from diamagnetic to paramagnetic.

In the case of $NO \rightarrow NO^+$, the ionization involves 15 electrons shifting to 14. Neutral $NO$ features $N_b=10$ and $N_a=5$, resulting in a bond order of 2.5 and paramagnetic behavior, whereas the $NO^+$ ion exhibits $N_b=10$ and $N_a=4$, yielding a higher bond order of 3 and a diamagnetic state, satisfying both criteria of increased bond order and changed magnetic behavior. Finally, the $O_2 \rightarrow O_2^+$ transition involves 16 electrons moving to 15, which increases the bond order from 2 ($N_b=10, N_a=6$) to 2.5 ($N_b=10, N_a=5$), but the species remains paramagnetic in both states due to the persistence of an unpaired electron.

The strength of a hydrogen bond is primarily governed by the magnitude of the partial positive charge residing on the hydrogen atom, which increases significantly as the electronegativity of the atom to which it is covalently bonded increases. Because fluorine is the most electronegative element among the options provided, the $\mathrm{F}-\mathrm{H}$ bond induces the greatest electron deficiency on the hydrogen atom compared to bonds involving oxygen or nitrogen. This highly polarized hydrogen atom is subsequently capable of establishing the most intense electrostatic attraction with the lone pair of a nearby electronegative atom, confirming that $\mathrm{F}-\mathrm{H}\cdots\mathrm{F}$ represents the strongest interaction in this set.

The magnetic behavior of a chemical species is determined by the presence or absence of unpaired electrons within its molecular orbitals, where diamagnetism occurs exclusively when all electrons are paired, and paramagnetism occurs when at least one unpaired electron exists. Evaluating the electronic configuration, the $\mathrm{O}_2^{2-}$ ion possesses 18 electrons that completely fill all molecular orbitals, including the $\pi^*$ levels, resulting in a state where no unpaired electrons remain. In contrast, $\mathrm{NO}$ and $\mathrm{O}_2^{+}$ both contain 15 electrons with one unpaired electron in an antibonding orbital, while $\mathrm{O}_2$ contains 16 electrons with two unpaired electrons, all of which exhibit paramagnetic behavior. Therefore, $\mathrm{O}_2^{2-}$ is the only species in the set that displays diamagnetic properties.

Molecular Orbital Theory determines the magnetic behavior of species by analyzing the occupancy of their energy levels, where the presence of unpaired electrons classifies a molecule as paramagnetic, while their absence indicates it is diamagnetic. In the case of the peroxide ion $\mathrm{O}_2^{2-}$, the total of 18 electrons fills every bonding and antibonding orbital completely, ensuring no unpaired electrons exist. In contrast, $\mathrm{B}_2$ with 10 electrons contains two unpaired electrons in its $\pi_{2p}$ orbitals, $\mathrm{N}_2^{+}$ with 13 electrons possesses a single unpaired electron in its $\sigma_{2p_z}$ orbital, and $\mathrm{O}_2$ with 16 electrons exhibits two unpaired electrons within its antibonding $\pi_{2p}^*$ orbitals.

Bond lengths remain uniform in molecules with high symmetry, where all surrounding atoms occupy equivalent positions relative to the central atom. Both $\mathrm{BF}_4^{-}$ and $\mathrm{SiF}_4$ feature $\mathrm{sp}^3$ hybridization and adopt a regular tetrahedral geometry with bond angles of $109^{\circ} 28'$, ensuring that each fluorine atom is equidistant from the center. Similarly, $\mathrm{XeF}_4$ utilizes $\mathrm{sp}^3 \mathrm{d}^2$ hybridization to form a square planar structure, which dictates equal bond lengths for all its interactions.

Conversely, $\mathrm{SF}_4$ is defined by $\mathrm{sp}^3 \mathrm{d}$ hybridization with four bond pairs and one lone pair, leading to a distorted trigonal bipyramidal geometry often described as a see-saw shape. This structural distortion, caused by the lone pair, results in bond angles of $< 120^{\circ}$ and $179^{\circ}$ instead of the expected ideal values. Within this arrangement, the molecule contains both axial and equatorial bonds, and because axial bonds are inherently longer and weaker than equatorial bonds, the overall bond lengths in $\mathrm{SF}_4$ are unequal.

The variation in bond angles across group 15 hydrides is primarily governed by the magnitude of lone pair-bond pair (lp-bp) repulsion, which changes as the central atom increases in size. In $\mathrm{NH}_3$, the nitrogen atom is small and highly electronegative, causing the lone pair to experience significant repulsion with the bonding electron pairs, which pushes the bonds apart to an angle of $106^{\circ}$. Conversely, as we descend the group to $\mathrm{SbH}_3$, the central atom becomes much larger, leading to a more stable lone pair that exerts significantly less lp-bp repulsion on the bonding pairs. This reduction in repulsive force allows the bond angles to contract towards $91^{\circ}$.

Lattice energy represents the strength of the electrostatic attraction holding ions together in a crystal lattice, which is governed by the relationship $F = K \frac{q_1 q_2}{r^2}$. Because the force scales directly with the product of the ionic charges $q_1$ and $q_2$ and decreases as the internuclear distance $r = r_+ + r_-$ increases, both the magnitude of the charges and the physical dimensions of the ions are the critical factors influencing lattice energy.

Molecular Orbital Theory determines the magnetic nature of a species based on the pairing of electrons within its orbitals, where a substance is classified as diamagnetic only if it contains no unpaired electrons. The hydrogen molecule, $\mathrm{H}_2$, contains two electrons that completely occupy the bonding orbital with a configuration of $\sigma_b^2$, meaning all electrons are paired. Other species such as $\mathrm{H}_2^+$, $\mathrm{H}_2^-$, and $\mathrm{He}_2^+$ possess unpaired electrons, rendering them paramagnetic; specifically, $\mathrm{H}_2^+$ has a configuration of $\sigma_b^1$, while both $\mathrm{H}_2^-$ and $\mathrm{He}_2^+$ feature a $\sigma_b^2 \sigma_a^{*1}$ arrangement.

Bond length is inversely related to bond order, meaning that as the bond order increases, the electrostatic attraction between atoms strengthens and the distance between them decreases. Because $\mathrm{NO}$ has a bond order of 2.5 and $\mathrm{NO}^{+}$ has a higher bond order of 3, the bond in $\mathrm{NO}^{+}$ is shorter than the bond in $\mathrm{NO}$. Consequently, the bond length in $\mathrm{NO}$ is greater than in $\mathrm{NO}^{+}$.

Hybridization is determined by calculating the steric number of an atom, which is the sum of its sigma bonds and lone electron pairs. In $\mathrm{H}_3 \mathrm{BO}_3$, the central boron atom forms three sigma bonds with oxygen atoms and possesses no lone pairs, resulting in a steric number of three, which confirms its $\mathrm{sp}^2$ hybridization. Conversely, each oxygen atom is bonded to one boron atom and one hydrogen atom while retaining two lone pairs, creating four electron domains that necessitate $\mathrm{sp}^3$ hybridization.

A regular tetrahedral geometry requires $\mathrm{sp}^3$ hybridization, where the central atom is bonded to four surrounding atoms with no lone pairs, yielding a bond angle of $109^{\circ} 28'$. Among the choices provided, $\mathrm{XeF}_4$ exhibits $\mathrm{sp}^3 d^2$ hybridization and a square planar structure, $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ features $\mathrm{dsp}^2$ hybridization resulting in a square planar geometry because the $\mathrm{CN}^{-}$ ion acts as a strong field ligand, and $\mathrm{SF}_4$ adopts a see-saw structure via $\mathrm{sp}^3 d$ hybridization. This confirms that the $\mathrm{BF}_4^{-}$ ion is the only structure among the options that satisfies the requirements for a regular tetrahedral arrangement, as it possesses four bonding pairs around the central boron atom with no lone pairs.

Molecular geometry determines the spatial distribution of ligands, and the octahedral arrangement inherent to $sp^3 d^2$ hybridization maximizes orthogonal interactions between bond pairs compared to other common coordination geometries. In an octahedral structure, the central atom is surrounded by six substituents that create twelve unique $90^{\circ}$ angles between adjacent bonds. Other hybridization states yield fewer of these interactions: square planar $dsp^2$ configurations feature only four $90^{\circ}$ angles, while trigonal bipyramidal arrangements found in $dsp^3$ and $sp^3 d$ hybridization contribute six and eight $90^{\circ}$ angles, respectively.

To determine the bond angles, one must consider hybridization and the influence of lone pairs on molecular geometry. $\mathrm{H}_2\mathrm{S}$ exhibits a bond angle of approximately $90^{\circ}$ because the central sulfur atom follows Drago's rule, resulting in bonding through pure p-orbitals rather than hybrid orbitals. In $\mathrm{NH}_3$, the $sp^3$ hybridized nitrogen atom features one lone pair that compresses the bond angle to $107^{\circ}$, whereas $\mathrm{SiH}_4$ adopts a regular tetrahedral geometry with a bond angle of $109^{\circ} 28'$. Finally, $\mathrm{BF}_3$ assumes a trigonal planar structure due to $sp^2$ hybridization, yielding the largest angle of $120^{\circ}$.

A molecule possesses a permanent dipole moment when its geometry is sufficiently asymmetric that the vector sum of its individual bond dipoles does not equal zero. Both $\mathrm{NO}_2$ and $\mathrm{O}_3$ adopt bent, asymmetric shapes that prevent these bond dipoles from canceling out, resulting in a net dipole moment where $\mu \neq 0$. In contrast, $\mathrm{CO}_2$ and $\mathrm{SiF}_4$ exhibit highly symmetric linear and tetrahedral structures, respectively, which cause the individual bond dipoles to perfectly negate one another, yielding a total dipole moment of $\mu = 0$.

According to Drago's rule, central atoms from the third period or higher bonded to elements of low electronegativity do not undergo hybridization, causing the bonds to form using pure p-orbitals with a bond angle of approximately 90 degrees. This applies specifically to $\mathrm{SH}_2$, distinguishing it from $\mathrm{OH}_2$ and $\mathrm{NH}_3$, which undergo $\text{sp}^3$ hybridization leading to bond angles of 104.5 degrees and 107 degrees respectively. Because $\mathrm{SO}_2$ exhibits $\text{sp}^2$ hybridization with a larger bond angle of 117 degrees, $\mathrm{SH}_2$ possesses the smallest bond angle among these molecules.

According to VSEPR theory, molecular geometry is determined by the arrangement of bonding and lone electron pairs around the central atom. In $\mathrm{XeF}_2$, the central xenon atom possesses two bonding pairs and three lone pairs, leading to an $sp^3d$ hybridized state and a linear molecular geometry where the three equatorial lone pairs cancel each other's repulsive effects. Similarly, $\mathrm{CO}_2$ features a central carbon atom bonded to two oxygen atoms via double bonds with zero lone pairs, which results in $sp$ hybridization and a linear geometry to maximize the separation between the bonding electron domains.

The hybridization of a carbon atom is determined by the number of sigma ($\sigma$) bonds it forms, where a carbon atom bonded to four other atoms through single bonds adopts $sp^3$ hybridization, while the presence of double or triple bonds results in $sp^2$ or $sp$ hybridization, respectively. In the molecule $\text{CH}_3\text{CH}_2\text{OH}$, the central methylene carbon atom is linked to two hydrogen atoms, one methyl group, and one hydroxyl group, creating four distinct $\sigma$ bonds. Because this specific carbon is saturated and lacks any pi bonds, it exists in an $sp^3$ hybridized state, unlike the carbonyl carbons in the other provided species which are involved in double bonds and therefore demonstrate $sp^2$ hybridization.