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

Dipole moment measures the net polarity of a molecule, determined by the magnitude of partial charges and the spatial arrangement of bonds. In the case of $\mathrm{NF}_3$, the lone pair on nitrogen opposes the bond dipoles from the fluorine atoms, resulting in a low dipole moment of 0.24 D. Comparatively, $\mathrm{HBr}$ exhibits a bond polarity resulting in 0.78 D, while the bent geometry of $\mathrm{H}_2\mathrm{S}$ leads to a net separation of charge with a value of 0.95 D. Finally, $\mathrm{CHCl}_3$ possesses the highest dipole moment of 1.01 D because the tetrahedral arrangement of its highly polar C-Cl bonds does not cancel out, creating a stronger overall dipole than the other molecules.

The rate of hydrolysis for carboxylic acid derivatives depends on how easily the leaving group can be displaced, which is governed by the strength of the base being released during the reaction. Because weaker bases serve as better leaving groups, the sequence is dictated by the increasing basicity of the departing species: chloride, carboxylate, alkoxide, and amide. Consequently, the reactivity decreases in the order $p > q > r > s$, moving from the highly reactive acid chloride down to the least reactive amide.

The structure of ozone is best described by resonance, where the electrons are delocalized across the three oxygen atoms, resulting in equivalent bonds with an average bond order of 1.5. Because bond length is inversely proportional to bond order, expressed as $\text{Bond length} \propto \frac{1}{\text{Bond order}}$, these bonds fall directly between the characteristics of a single bond with an order of 1.0 and a double bond with an order of 2.0. Consequently, the observed bond length order is $\mathrm{O}=\mathrm{O} < \mathrm{O}_3 < \mathrm{O}-\mathrm{O}$, which confirms that the first statement is true. In contrast, while lone pair-lone pair repulsion does occur within the molecule, it is not the sole factor dictating bond lengths; the bond order established by resonance plays the dominant role in defining these physical dimensions, rendering the second statement false.

Hybridization is fundamentally determined by the steric number of the central atom, which is defined as the sum of its sigma bonds and lone electron pairs. An $sp^2$ hybridization state corresponds to a steric number of three, which results in a trigonal planar geometry where the central atom is surrounded by three distinct regions of electron density.

Among the provided molecules, $\mathrm{SO}_2$, $\mathrm{C}_2\mathrm{H}_4$, $\mathrm{BCl}_3$, $\mathrm{HCHO}$, $\mathrm{C}_6\mathrm{H}_6$, and $\mathrm{BF}_3$ each feature a central atom with exactly three such regions. Whether by forming three single bonds, as seen in the boron halides, or by engaging in a combination of single and double bonds, as in the organic compounds and sulfur dioxide, these six species all utilize $sp^2$ hybrid orbitals for bonding.

The net dipole moment of a molecule dictates its polarity, as an asymmetric arrangement of atoms ensures that individual bond dipoles do not cancel each other out. Highly symmetrical species such as $\mathrm{CCl}_4$, $\mathrm{CO}_2$, and $\mathrm{CH}_2=\mathrm{CH}_2$ result in a zero net dipole moment because their bond polarities effectively neutralize one another. In contrast, the presence of distinct substituents in the tetrahedral structure of $\mathrm{CHCl}_3$ disrupts this balance, leading to a permanent net dipole moment ($\mu \neq 0$) that renders the molecule polar.

Molecular Orbital Theory dictates that the total number of molecular orbitals created is equivalent to the number of atomic orbitals that combine. Because each of the two atoms contributes one $2s$ atomic orbital, they yield two molecular orbitals: $\sigma 2s$ and $\sigma^* 2s$. Similarly, since each atom contributes three $2p$ atomic orbitals ($2p_x$, $2p_y$, and $2p_z$), the six total atomic orbitals form six molecular orbitals: $\sigma 2p_z$, $\sigma^* 2p_z$, $\pi 2p_x$, $\pi 2p_y$, $\pi^* 2p_x$, and $\pi^* 2p_y$. The sum of these, 2 plus 6, equals 8 total molecular orbitals.

According to Molecular Orbital Theory, the distribution of electrons in $\mathrm{O}_2$ and its ions is determined by filling orbitals in order of increasing energy, with the antibonding $\pi^*$ orbitals playing a critical role. The neutral $\mathrm{O}_2$ molecule has 16 electrons, leaving 2 in the $\pi^*$ orbitals, while the cation $\mathrm{O}_2^{+}$ contains 15 electrons, resulting in 1 in the $\pi^*$ set. Similarly, the anion $\mathrm{O}_2^{-}$ possesses 17 electrons, which places 3 in the $\pi^*$ antibonding orbitals. Aggregating these amounts, the total number of electrons in the $\pi^*$ orbitals for the three species is $2 + 1 + 3 = 6$.

For atomic orbitals to undergo effective linear combination to form molecular orbitals, they must possess comparable energy levels. Furthermore, these orbitals need to align with the same symmetry relative to the molecular axis and achieve the maximum possible degree of overlap. Because disparate symmetry and minimal overlap prevent proper interaction, only the criteria involving matched energy and consistent symmetry act as the fundamental requirements for the process.

The resultant dipole moment of a molecule is determined by the vector sum of individual bond dipoles and the lone pair orbital dipole on the central atom. In $\mathrm{NH}_3$, the electronegativity of nitrogen is greater than that of hydrogen, causing the $\mathrm{N}-\mathrm{H}$ bond moments to point toward the nitrogen atom, which aligns with the direction of the lone pair's orbital dipole and results in a large net dipole moment. Conversely, because fluorine is the most electronegative element, the $\mathrm{N}-\mathrm{F}$ bond moments point away from the nitrogen atom, directly opposing the direction of the lone pair's orbital dipole and causing significant cancellation that results in a smaller net dipole moment for $\mathrm{NF}_3$.

Hybridization is determined by the steric number of the central atom, which is calculated as the sum of its sigma bonds and lone pairs, where a value of 4 consistently signifies $\text{sp}^3$ hybridization. Applying this to the provided species, $\text{NH}_3$ possesses three bonds and one lone pair, the network structure of $\text{SiO}_2$ features silicon bonded to four oxygens, $\text{H}_2\text{O}$ contains two sigma bonds and two lone pairs, and $\text{CH}_4$ exhibits four sigma bonds, all resulting in $\text{sp}^3$ hybridization. Meanwhile, $\text{SO}_2$ and $\text{BF}_3$ are $\text{sp}^2$ hybridized due to their steric number of 3, and $\text{BeCl}_2$ and $\text{CO}_2$ are $\text{sp}$ hybridized with a steric number of 2. Summing these instances confirms that 4 species in the list utilize $\text{sp}^3$ hybrid orbitals.

The bonding profile of a molecule is determined by its structure, where all single covalent bonds are $\sigma$ bonds and the auxiliary components of multiple bonds are identified as $\pi$ bonds. In the ethylene molecule, $C_2H_4$, there are four carbon-hydrogen single bonds and a single carbon-carbon double bond. Because every single bond contributes one $\sigma$ bond and a double bond contains both a $\sigma$ and a $\pi$ bond, the total count yields 5 $\sigma$ bonds and 1 $\pi$ bond.

The reaction of $\mathrm{NiS}$ with aqua regia is described by the equation $3\mathrm{NiS} + 2\mathrm{HNO}_3 + 6\mathrm{HCl} \rightarrow 3\mathrm{NiCl}_2 + 2\mathrm{NO} + 3\mathrm{S} + 4\mathrm{H}_2\mathrm{O}$, which identifies product $\mathrm{A}$ as $\mathrm{NiCl}_2$. Product $\mathrm{B}$ corresponds to the stable hexahydrate complex $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2$ that forms upon crystallization from the aqueous solution. This complex contains 6 water molecules coordinated to the central nickel ion, contributing a total of $6 \times 2 = 12$ protons. These protons are directly coordinated to the metal center and, within this specific structural context, are considered not to participate in the intermolecular hydrogen bonding network of the crystal lattice, resulting in a count of 12.

Resonance energy characterizes the thermodynamic stabilization achieved when electrons are delocalized throughout a molecule, placing the actual hybrid structure at a lower potential energy than the lowest energy contributing Lewis structure. If we denote the energy of this most stable canonical form as $E_{\text{canonical}}$ and the energy of the actual hybrid as $E_{\text{hybrid}}$, the resonance energy is defined by the difference $E_{\text{resonance}} = E_{\text{canonical}} - E_{\text{hybrid}}$. This reduction in total energy serves as a quantitative measure of the increased stability inherent in resonant systems compared to static bonding representations.

A molecule exhibits a net dipole moment of zero when its molecular geometry is sufficiently symmetric to ensure that individual bond dipoles cancel one another out. Among the molecules listed, $\mathrm{H}_2$ is nonpolar due to its homonuclear nature, while $\mathrm{CO}_2$ and $\mathrm{BeF}_2$ utilize linear structures to align opposing bond moments. Furthermore, the trigonal planar symmetry of $\mathrm{BF}_3$ and the tetrahedral geometry of both $\mathrm{CH}_4$ and $\mathrm{SiF}_4$ also result in perfect cancellation, bringing the final count of these nonpolar compounds to 6.

The boiling point of a molecular compound is dictated by the strength of its intermolecular forces, which is why $\mathrm{NH}_3$ exhibits a higher boiling point than $\mathrm{PH}_3$ due to the presence of strong hydrogen bonding between ammonia molecules. The statement provided as the reason is factually incorrect because it erroneously reverses the nature of these forces by suggesting that $\mathrm{PH}_3$ is associated through hydrogen bonding while $\mathrm{NH}_3$ is not. In reality, $\mathrm{PH}_3$ lacks this strong hydrogen bonding interaction, which is the exact reason it has a lower boiling point than ammonia.

The bond order of a molecule corresponds to the number of chemical bonds between a pair of atoms, and for carbon monoxide, the stable $\mathrm{C} \equiv \mathrm{O}$ structure results in a bond order of 3. Because the nitrosonium ion $\mathrm{NO}^{+}$ is isoelectronic with carbon monoxide, possessing the same total number of valence electrons, it also features a triple bond arrangement with a bond order of 3. Combining these two individual bond order values results in a total sum of $3 + 3 = 6$.

Determining molecular geometry requires analyzing the arrangement of electron pairs around the central atom, where a trigonal bipyramidal shape specifically corresponds to a steric number of five with no lone pairs. Among the listed species, $\mathrm{PF}_5$, $\mathrm{PCl}_5$, and $\mathrm{Fe}(\mathrm{CO})_5$ all feature a central atom bonded to five surrounding ligands without residual lone pairs, resulting in the characteristic trigonal bipyramidal arrangement. Conversely, $\mathrm{BrF}_5$ adopts a square pyramidal geometry, $[\mathrm{PtCl}_4]^{2-}$ maintains a square planar structure, and $\mathrm{BF}_3$ aligns with a trigonal planar configuration due to having only three bond pairs.

The net dipole moment of a molecule is determined by the vector sum of individual bond dipoles and the lone pair orbital dipole. In $NH_3$, the nitrogen atom is more electronegative than hydrogen, so the $N-H$ bond dipoles point towards nitrogen and align in the same direction as the lone pair dipole, resulting in a large net dipole moment of approximately $1.47\,D$. In $NF_3$, fluorine is more electronegative than nitrogen, causing the $N-F$ bond dipoles to point away from nitrogen towards the fluorine atoms, which opposes the direction of the lone pair dipole and results in a significantly smaller net dipole moment of approximately $0.24\,D$. Consequently, statement I is false because $NH_3$ has a greater dipole moment than $NF_3$, and statement II is false because it mischaracterizes the relative alignment of the dipoles in both molecules.

The atomic size of elements in group 15 generally increases as we move down the group due to the addition of successive electron shells. While this trend leads to a notable increase in the covalent radius from nitrogen to phosphorus, the subsequent increase from arsenic to bismuth is significantly smaller. This disparity exists because the heavier elements contain completely filled $d$ and $f$-orbitals, which provide poor shielding for the valence electrons against the increasing nuclear charge, thus limiting the expected expansion in atomic size. Therefore, although the general increase in size down the group is true, it does not inherently account for the diminished growth observed in the heavier elements of the group.

The octet rule states that atoms tend to bond in such a way that they attain eight electrons in their valence shells, though exceptions occur when central atoms possess incomplete octets, expanded octets, or an odd total number of valence electrons. Incomplete octets are found in $\mathrm{BF}_3$, where the central boron atom has only six valence electrons, and $\mathrm{BeF}_2$, where the central beryllium atom has only four valence electrons. Expanded octets, which occur in elements capable of utilizing $d$-orbitals to accommodate more than eight electrons, are present in $\mathrm{H}_2\mathrm{SO}_4$, where sulfur shares 12 valence electrons, and $\mathrm{PCl}_5$, where phosphorus shares 10 valence electrons. Odd-electron species that violate the rule due to the impossibility of pairing every electron include $\mathrm{NO}_2$ and $\mathrm{ClO}_2$, as both contain an unpaired electron and do not complete an octet.

Conversely, the molecules $\mathrm{CO}_2, \mathrm{CH}_4, \mathrm{SiF}_4, \mathrm{C}_2\mathrm{H}_6, \mathrm{CHCl}_3,$ and $\mathrm{CBr}_4$ strictly adhere to the octet rule because their central carbon or silicon atoms achieve a stable configuration of eight valence electrons through shared bonding pairs. By evaluating each structure, it is determined that $\mathrm{NO}_2, \mathrm{H}_2\mathrm{SO}_4, \mathrm{BF}_3, \mathrm{ClO}_2, \mathrm{PCl}_5,$ and $\mathrm{BeF}_2$ are the specific compounds from the provided list that function as exceptions to the octet rule.