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

The reaction sequence commences with aniline, which serves as the starting material for electrophilic aromatic substitution. Upon treatment with $\text{Br}_2/\text{water}$, the highly activating amino group directs bromine atoms to the two ortho positions and the para position, yielding 2,4,6-tribromoaniline as product A. Subsequently, diazotization with $\text{NaNO}_2/\text{HCl}$ at $273-278\,\text{K}$ converts the amino group into a diazonium salt, forming 2,4,6-tribromobenzenediazonium chloride, which serves as intermediate B.

The final step involves the reaction of the diazonium salt with $\text{HBF}_4$ followed by $\text{NaNO}_2/\text{Cu}$. This specific sequence facilitates the replacement of the diazonium group ($-\text{N}_2^+$) with a nitro group ($-\text{NO}_2$), a transformation known as the Sandmeyer-type nitration. Consequently, the diazonium group is substituted by a nitro group, producing 1-nitro-2,4,6-tribromobenzene, which corresponds to the structure of 1,3,5-tribromo-2-nitrobenzene.

To identify $sp^3d$ hybridization, one calculates the steric number by summing the number of sigma bonds and lone pairs, where a value of 5 signifies this specific state. Evaluating the species, $\mathrm{BrF}_5$ has a steric number of 6 ($sp^3d^2$), $\mathrm{XeF}_5^-$ has a steric number of 7 ($sp^3d^3$), and $\mathrm{BF}_4^-$ has a steric number of 4 ($sp^3$). Similarly, $\mathrm{ICl}_4^-$ and $\mathrm{XeF}_4$ are both $sp^3d^2$ due to a steric number of 6, while $\mathrm{NH}_4^+$ results in a steric number of 4 ($sp^3$).

The species $\mathrm{SF}_4$ ($4\sigma + 1\,\text{lp} = 5$), $\mathrm{ClF}_3$ ($3\sigma + 2\,\text{lp} = 5$), $\mathrm{XeF}_2$ ($2\sigma + 3\,\text{lp} = 5$), and $\mathrm{ICl}_2^-$ ($2\sigma + 3\,\text{lp} = 5$) all feature a steric number of 5, confirming their $sp^3d$ hybridization. Counting these instances confirms that exactly 4 species in the provided list possess $sp^3d$ hybridization.

Two molecules are considered isostructural if they share the same hybridization and molecular geometry, which are determined by the distribution of bonding pairs and lone pairs around the central atom. In $\mathrm{SF}_4$, the sulfur central atom is surrounded by four bonding pairs and one lone pair, resulting in a seesaw geometry associated with $sp^3d$ hybridization. Both $\mathrm{IF}^\oplus_4$ and $\mathrm{XeO}_2\mathrm{F}_2$ possess this identical arrangement of electron pairs, leading to the same seesaw molecular shape. In contrast, $\mathrm{CH}_4$ adopts a tetrahedral geometry, while $\mathrm{BrF}^\ominus_4$ and $\mathrm{XeF}_4$ are square planar, making these species structurally distinct from $\mathrm{SF}_4$.

Hybridization is determined by calculating the steric number of the central atom, which is the sum of its $\sigma$ bonds and lone pairs. In $\mathrm{SO}_2$, the sulfur atom is surrounded by 2 $\sigma$ bonds and 1 lone pair, resulting in a total of 3 and thus $sp^2$ hybridization. Similarly, the central nitrogen in $\mathrm{NO}_2^{-}$ holds 2 $\sigma$ bonds and 1 lone pair, which also corresponds to $sp^2$ hybridization. Finally, the central nitrogen in $\mathrm{N}_3^{-}$ connects via 2 $\sigma$ bonds with no lone pairs, yielding a steric number of 2 and identifying the hybridization as $sp$.

Molecular geometry is dictated by VSEPR theory, which predicts a linear arrangement of $180^{\circ}$ for species with two bonding pairs and no lone pairs on the central atom, or for hypervalent molecules where lone pairs distribute symmetrically in equatorial positions to minimize repulsion. From the provided list, $\mathrm{BeCl}_2$, $\mathrm{CO}_2$, $\mathrm{N}_3^{-}$, $\mathrm{XeF}_2$, $\mathrm{NO}_2^{+}$, and $\mathrm{I}_3^{-}$ fulfill these criteria, whereas $\mathrm{SO}_2$, $\mathrm{NO}_2$, $\mathrm{F}_2\mathrm{O}$, and $\mathrm{O}_3$ deviate into bent structures due to the presence of non-bonding electrons. This results in exactly six chemical species that exhibit a linear geometry.

According to VSEPR theory and hybridization concepts, the $\mathrm{ClF}_3$ molecule, which exhibits $sp^3d$ hybridization, can theoretically form three distinct structural isomers by arranging its two lone pairs and three bonding pairs across the axial and equatorial positions of a trigonal bipyramidal geometry. Statement I is accurate in noting the existence of these three potential isomers, but Statement II is incorrect because lone pairs exert greater repulsive forces and are most stable when occupying equatorial positions to minimize repulsion with bonding pairs at 90 degrees; placing lone pairs in axial positions significantly increases repulsion, making such configurations the least stable.

Drying speed depends on the ability of a material to attract and hold onto water molecules through intermolecular forces. Cotton fibers are composed of cellulose, which contains numerous hydroxyl groups that establish a dense network of hydrogen bonds with water, resulting in longer retention times compared to synthetic nylon. Because nylon possesses fewer sites for these specific interactions, it interacts less strongly with moisture and dries rapidly, which contradicts the claim that nylon has greater hydrogen bonding with water than cotton.

Determining the target molecule requires isolating which species possesses a non-zero dipole moment and the maximum number of lone pairs on its central atom. Although $\mathrm{XeF}_2$ features three lone pairs, it lacks a dipole moment due to its linear geometry, whereas $\mathrm{ClF}_3$ retains a net dipole and contains two lone pairs, which is the highest amount among the polar candidates in this set. This specific arrangement of five electron regions around the central chlorine atom necessitates an $sp^{3}d$ hybridization.

According to Molecular Orbital Theory, a molecule is paramagnetic if it possesses one or more unpaired electrons in its molecular orbitals. For $\mathrm{O}_2$ and $\mathrm{S}_2$, the electrons occupy degenerate $\pi^*$ antibonding orbitals singly, resulting in two unpaired electrons. Conversely, $\mathrm{N}_2$, $\mathrm{F}_2$, and $\mathrm{Cl}_2$ feature completely filled molecular orbitals with all electrons paired, identifying them as diamagnetic.

Lewis theory defines non-bonded electrons as the valence electrons remaining on atoms after satisfying the octet rule through covalent bonding. In the $\mathrm{NO}_2^{-}$ ion, the central nitrogen atom forms a double bond with one oxygen atom and a single bond with another, requiring the oxygen in the double bond to retain 4 non-bonded electrons, the oxygen in the single bond to retain 6, and the nitrogen to retain 2. Summing these non-bonded electrons across the entire structure results in 12.

$\mathrm{ClF}_3$ adopts a trigonal bipyramidal electron geometry where the two lone pairs reside in the equatorial positions to minimize inter-electronic repulsion, establishing that $n = 2$. Identifying which ions possess this same number of unpaired electrons requires examining the $d$-orbital occupancy of each metal cation.

Vanadium, with an atomic number of 23, forms the $\mathrm{V}^{3+}$ ion with an electronic configuration of $[\mathrm{Ar}] 3d^2$, which results in two unpaired electrons. Nickel ($Z=28$) in the $\mathrm{Ni}^{2+}$ state has a $[\mathrm{Ar}] 3d^8$ configuration, where the arrangement of the eight $d$-electrons leaves two unpaired, and titanium ($Z=22$) in the $\mathrm{Ti}^{2+}$ state also has a $[\mathrm{Ar}] 3d^2$ configuration, possessing two unpaired electrons. In contrast, both $\mathrm{Ti}^{3+}$ and $\mathrm{Cu}^{2+}$ have electronic configurations of $[\mathrm{Ar}] 3d^1$ and $[\mathrm{Ar}] 3d^9$ respectively, each resulting in only one unpaired electron.

The polarity of a molecule depends on the electronegativity difference between the bonded atoms; because the $C-Cl$ bond possesses a higher dipole moment than the $C-Br$ bond, $CH_3Cl$ is indeed more polar than $CH_3Br$. While molecular polarity contributes to intermolecular attraction, the boiling point is primarily governed by the total strength of intermolecular forces, including van der Waals dispersion forces which increase significantly with molecular mass and size. Consequently, even though $CH_3Cl$ is more polar, $CH_3Br$ exhibits a higher boiling point due to the stronger dispersion forces associated with the larger bromine atom, which outweigh the dipole-dipole differences between the two molecules.

The octet rule dictates that atoms are generally stable with eight valence electrons, yet many molecules exhibit structural deviations based on their specific bonding arrangements. Compounds such as $\mathrm{CCl}_4$ and $\mathrm{CO}_2$ align with the octet rule by maintaining 8 electrons, whereas $\mathrm{BCl}_3$ and $\mathrm{AlCl}_3$ represent the incomplete octet category with only 6 electrons. Species like $\mathrm{NO}$ and $\mathrm{NO}_2$ contain 7 electrons, identifying them as molecules with an incomplete octet featuring an odd electron. Conversely, $\mathrm{H}_2\mathrm{SO}_4$ with 12 electrons and $\mathrm{PCl}_5$ with 10 electrons showcase an expanded octet, as their central atoms effectively accommodate more than the standard eight electrons.

The dipole moment of $\mathrm{CH_3-CH=CH-CH=O}$ is greater than that of $\mathrm{CH_3-CH_2-CH_2-CH=O}$ because the conjugated $\pi$-system in the former facilitates the delocalization of electrons, leading to increased resonance-induced charge separation and higher overall polarity. In contrast, the $\mathrm{C_1}-\mathrm{C_2}$ bond in the conjugated system acquires significant partial double bond character due to resonance stabilization, which makes it shorter than the standard, pure single $\mathrm{C_1}-\mathrm{C_2}$ bond found in the saturated molecule $\mathrm{CH_3-CH_2-CH_2-CH=O}$. Consequently, Statement-I is true while Statement-II is false, as the conjugated bond length is shorter, not longer, than its saturated counterpart.

VSEPR theory enables the calculation of bond pairs and lone pairs by evaluating the valence electrons of the central atom and the number of surrounding bonding atoms. The $\mathrm{ICl}_2^{-}$ ion contains an iodine central atom with 8 valence electrons, forming 2 bond pairs and retaining 3 lone pairs, which corresponds to a ratio of 2:3. The $\mathrm{H}_2\mathrm{O}$ molecule involves an oxygen atom with 6 valence electrons, creating 2 bond pairs and 2 lone pairs, resulting in a ratio of 2:2. In $\mathrm{SO}_2$, the sulfur atom has 6 valence electrons and shares them to form 2 bond pairs with oxygen atoms, leaving 1 lone pair on the sulfur atom, which gives a ratio of 2:1. Finally, the $\mathrm{XeF}_4$ molecule utilizes the xenon central atom with 8 valence electrons to form 4 bond pairs and 2 lone pairs, establishing a ratio of 4:2.

To determine which molecule meets these criteria, we must evaluate the central atom's hybridization and geometry based on its electron distribution. For $\mathrm{SF}_4$, the sulfur atom contributes six valence electrons, forming four bonds with fluorine atoms while keeping one lone pair. With four bond pairs and one lone pair, the steric number is 5, which confirms that the hybridization of the central sulfur atom is $\mathrm{sp}^3 d$. Because the lone pair occupies an equatorial position, it exerts unequal repulsion on the surrounding bond pairs, causing the molecule to adopt a see-saw shape where the axial and equatorial bond lengths differ, specifically satisfying the relationship $\mathrm{(S-F)}_a > \mathrm{(S-F)}_e$.

Identifying the elements requires recognizing their periodic properties: element $X$ with the highest electronegativity is fluorine, element $Y$ with the second highest is oxygen, and element $A$ is the non-radioactive noble gas xenon, which possesses a lower ionization enthalpy than the others. This composition yields the molecule $\mathrm{XeOF}_4$, where the central xenon atom uses four valence electrons to bond with fluorine, two to form a double bond with oxygen, and retains one lone pair. Because the total of five bonding domains and one lone pair creates a steric number of six, the central xenon atom adopts an octahedral electron geometry, resulting in a square pyramidal molecular shape.

The molecules $\mathrm{CH}_4$, $\mathrm{NH}_3$, and $\mathrm{H}_2\mathrm{O}$ all contain a central atom surrounded by four electron domains, meaning they are all $\mathrm{sp}^3$ hybridized. As the number of lone pairs increases from $\mathrm{CH}_4$ to $\mathrm{NH}_3$ to $\mathrm{H}_2\mathrm{O}$, the increased electron-pair repulsion progressively reduces the bond angles from the ideal tetrahedral 109.5 degrees to 107.5 degrees and 104.5 degrees, respectively. Furthermore, the dipole moments follow an increasing order, where the nonpolar $\mathrm{CH}_4$ has a value of 0, whereas the polar molecules $\mathrm{NH}_3$ and $\mathrm{H}_2\mathrm{O}$ possess values of 1.47 and 1.85, establishing the relation $\mathrm{CH}_4 < \mathrm{NH}_3 < \mathrm{H}_2\mathrm{O}$.

Successful formation of molecular orbitals necessitates that atomic orbitals possess compatible symmetry and sufficiently close energy levels to enable effective overlap. The $2s$ and $2p_z$ orbitals constitute the only pair among the choices provided that meets these criteria, as their relatively small energy difference facilitates efficient linear combination. Other orbital combinations are precluded from forming molecular orbitals in this context primarily because they lack this necessary energy proximity.

Square pyramidal geometry arises in molecules with an $AX_5E$ configuration, where the central atom is surrounded by five bonding pairs and one lone pair, resulting in $sp^3d^2$ hybridization. In $\mathrm{BrF}_5$, the bromine atom utilizes five valence electrons to form bonds with fluorine atoms, leaving one lone pair, while in $\mathrm{XeOF}_4$, the xenon atom uses four electrons for single bonds with fluorine and two for a double bond with oxygen, also resulting in one lone pair. Molecules like $\mathrm{PCl}_5$ and $\mathrm{SbF}_5$ possess five bonding pairs with no lone pairs, leading to a trigonal bipyramidal structure instead.