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

Valence Shell Electron Pair Repulsion (VSEPR) theory determines molecular geometry by calculating the number of bonding and lone pairs of electrons around the central atom to minimize electrostatic repulsion. Ammonia ($\mathrm{NH}_3$) has one lone pair and three bonding pairs, creating a trigonal pyramidal shape, while bromine pentafluoride ($\mathrm{BrF}_5$) possesses five bonding pairs and one lone pair, resulting in a square pyramidal structure. Phosphorus pentachloride ($\mathrm{PCl}_5$) features five bonding pairs in a trigonal bipyramidal arrangement, and methane ($\mathrm{CH}_4$) comprises four bonding pairs, forming a tetrahedral geometry.

Molecular Orbital Theory enables the determination of magnetic character by predicting the filling of molecular orbitals for these species. In this group, $\mathrm{O}_2$ contains two unpaired electrons, while the $\mathrm{O}_2^{2-}$ and $\mathrm{CN}^{-1}$ ions both possess fully paired electrons, resulting in zero unpaired electrons. Meanwhile, the $\mathrm{O}_2^{-1}$ ion and the $\mathrm{NO}$ molecule each exhibit a single unpaired electron in their respective electronic configurations. Therefore, identifying the species with exactly one unpaired electron yields a total count of two.

The hybridization state of a central atom is determined by its steric number ($SN$), calculated using the formula $SN = \frac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons of the central atom, $M$ is the number of monovalent atoms attached, $C$ is the cationic charge, and $A$ is the anionic charge. Applying this to $\mathrm{NO}_3^{-}$, $SN = \frac{1}{2}(5 + 0 + 1) = 3$, indicating $\mathrm{sp}^2$ hybridization, while $\mathrm{BCl}_3$ yields $SN = \frac{1}{2}(3 + 3) = 3$, which also corresponds to $\mathrm{sp}^2$ hybridization. For the $\mathrm{ClO}_2^{-}$ ion, the steric number is $SN = \frac{1}{2}(7 + 0 + 1) = 4$, indicating $\mathrm{sp}^3$ hybridization. Similarly, the central chlorine atom in the $\mathrm{ClO}_3$ radical possesses three sigma bonds and one lone pair, resulting in $SN = 4$ and $\mathrm{sp}^3$ hybridization. Therefore, exactly two of the listed species, $\mathrm{ClO}_2^{-}$ and $\mathrm{ClO}_3$, involve $\mathrm{sp}^3$ hybridization.

The number of lone pairs on the central atom is calculated by subtracting the number of valence electrons involved in bonding from the total valence electrons of the central atom and dividing the result by two. For a compound to possess exactly one lone pair, the central atom must retain two non-bonding valence electrons after forming all necessary bonds. By evaluating the Lewis structures of the given compounds, it is determined that exactly four of them satisfy this condition, as their central atoms each possess a surplus of two electrons once the bonding requirements are accounted for using the expression $\frac{V - B}{2}$, where $V$ represents the valence electrons and $B$ represents the number of electrons shared in bond formation.

Molecular Orbital Theory dictates the magnetic properties of a molecule based on the presence of unpaired electrons, while the bond order is determined by the electronic distribution between bonding and antibonding orbitals. To satisfy the requirements, a species must simultaneously demonstrate paramagnetic behavior and possess a bond order of exactly 1.

Analyzing the provided list, $\mathrm{H}_2$, $\mathrm{O}_2^{2-}$, and $\mathrm{F}_2$ are all diamagnetic despite having a bond order of 1, which excludes them from the count. The species $\mathrm{O}_2^{+}$, $\mathrm{He}_2^{+}$, $\mathrm{N}_2^{2-}$, and $\mathrm{Ne}_2^{+}$ are paramagnetic, but their respective bond orders of 0.5, 2.5, 2, and 0.5 do not match the required value of 1. Only $\mathrm{B}_2$ fulfills both criteria, as it is paramagnetic and maintains a bond order of 1, resulting in a single species that satisfies the condition.

Molecular geometry and magnetic properties are governed by valence shell electron pair repulsion theory and the distribution of valence electrons. The molecule $SO_2Cl_2$ exhibits a tetrahedral structure because the sulfur atom forms four bonding pairs with no lone pairs, whereas $NO$ is paramagnetic due to its odd number of valence electrons. The $NO_2^-$ ion is diamagnetic as all its electrons are paired, and $I_3^-$ possesses a linear geometry because the central iodine atom accommodates three lone pairs in the equatorial positions of a trigonal bipyramidal arrangement, leaving the two bonded iodine atoms in the axial positions at a $180^\circ$ bond angle.

The hybridization of a central atom is determined by the sum of its bonding electron pairs and lone pairs, where three electron domains indicate $\mathrm{sp}^2$ hybridization and four electron domains indicate $\mathrm{sp}^3$ hybridization. In the $\mathrm{BF}_3$ molecule, the boron atom is bonded to three fluorine atoms with no lone pairs, and in the $\mathrm{NO}_2^{-}$ ion, the central nitrogen atom is surrounded by two oxygen atoms and one lone pair, meaning both species possess three electron domains and consequently exhibit $\mathrm{sp}^2$ hybridization. In contrast, molecules such as $\mathrm{H}_2\mathrm{O}$ and $\mathrm{NH}_2^{-}$ feature four electron domains-consisting of bonding pairs and lone pairs-which dictates an $\mathrm{sp}^3$ hybridization state.

The molecular geometry of interhalogen compounds is determined using the VSEPR theory, where the steric number ($SN$) calculated as half the sum of the central atom's valence electrons and the number of monovalent surrounding atoms dictates the arrangement of bond pairs and lone pairs. For $\mathrm{ICl}$, the $SN$ is 4 with three lone pairs resulting in a linear geometry, whereas $\mathrm{ICl}_3$ has an $SN$ of 5 with two lone pairs, producing a T-shape. $\mathrm{ClF}_5$ corresponds to an $SN$ of 6 with one lone pair resulting in a square pyramidal structure, and $\mathrm{IF}_7$ has an $SN$ of 7 with zero lone pairs, leading to a pentagonal bipyramidal geometry.

A molecule possesses a non-zero dipole moment ($\mu \neq 0$) if its three-dimensional geometry is asymmetric, preventing the individual bond dipoles from canceling one another. Highly symmetric molecules such as $\mathrm{BeCl}_2$, $\mathrm{BCl}_3$, $\mathrm{XeF}_4$, $\mathrm{CCl}_4$, $\mathrm{CO}_2$, and $\mathrm{H}_2$ possess balanced bond vectors that result in a net dipole moment of $\mu = 0$. Conversely, species containing lone pairs or distinct bonding environments, specifically $\mathrm{NF}_3$, $\mathrm{H}_2\mathrm{O}$, $\mathrm{H}_2\mathrm{S}$, $\mathrm{HBr}$, and $\mathrm{HCl}$, exhibit a permanent polarity, resulting in a total of 5 such species.

Molecular orbital theory describes the formation of new orbitals through the linear combination of atomic wave functions. When wave functions combine out of phase, they undergo destructive interference, which creates an antibonding molecular orbital, denoted as $\sigma^{*}$. This physical cancellation of wave amplitudes is represented mathematically by the subtraction of the participating wave functions, resulting in the expression $\psi_A - \psi_B$.

The octet rule dictates that atoms typically gain or share electrons to achieve a stable configuration with 8 valence electrons. In the compound $\mathrm{A}_2\mathrm{B}$, the atom B exhibits an oxidation state of -2, which signifies that the atom has accepted 2 additional electrons to reach this full valence shell. Subtracting these 2 acquired electrons from the 8 required for a complete octet reveals that the neutral atom must possess 6 electrons in its valence shell.

When aluminium chloride dissolves in an acidified aqueous solution, the aluminium ion behaves as a strong Lewis acid and coordinates with six water molecules to form the stable complex $[\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6]^{3+}$. Because the central metal atom is surrounded by six ligands in this hexacoordinated structure, the electron pairs are arranged to minimize repulsion. This specific spatial arrangement results in an octahedral geometry.

The $\pi$ bonding molecular orbital is formed through the constructive sideways overlap of atomic orbitals, which accumulates higher electron density in the regions directly above and below the internuclear axis, rendering the first statement incorrect. In contrast, the $\pi^*$ antibonding molecular orbital arises from destructive interference, creating a nodal plane between the nuclei where the probability of finding an electron is zero, which confirms that the second statement is accurate.

A molecule possesses a net zero dipole moment when its structural symmetry allows all individual bond dipoles to perfectly cancel each other out. This condition is satisfied by $\mathrm{CH}_4$ with its tetrahedral geometry, $\mathrm{BF}_3$ due to its trigonal planar arrangement, and $\mathrm{CO}_2$ because of its linear shape. In contrast, $\mathrm{H_2O}$, $\mathrm{HF}$, $\mathrm{NH}_3$, and $\mathrm{SO}_2$ contain lone pairs or adopt shapes that result in an asymmetric charge distribution, yielding a non-zero dipole moment. Consequently, there are 3 molecules in the provided set that exhibit a zero dipole moment.

According to VSEPR theory, molecular shape is determined by the steric number ($SN$), calculated as $\frac{1}{2}(\text{valence electrons} + \text{monovalent atoms} \pm \text{charge})$. For $\mathrm{SF}_4$, $SN = 5$ with one lone pair yields a see-saw shape, while $\mathrm{BrF}_3$ has $SN = 5$ and two lone pairs, resulting in a bent T-shape. $\mathrm{BrO}_3^{-}$ has $SN = 4$ with one lone pair, leading to a pyramidal geometry, and $\mathrm{NH}_4^{+}$ has $SN = 4$ with no lone pairs, maintaining a perfect tetrahedral shape.

The molecular geometry is predicted using VSEPR theory based on the total number of valence shell electron pairs, where lone pairs significantly influence the final shape. $\mathrm{BrF}_5$ has 5 bond pairs and 1 lone pair on the central atom resulting in a square pyramidal shape (IV), while $\mathrm{H_2O}$ has 2 bond pairs and 2 lone pairs forming a bent geometry (III). $\mathrm{ClF}_3$ possesses 3 bond pairs and 2 lone pairs arranged in a T-shape (I), and $\mathrm{SF}_4$ has 4 bond pairs and 1 lone pair exhibiting a see-saw structure (II).

A molecule is considered non-polar when its net dipole moment is zero, a condition typically resulting from a symmetric structure where the bond dipoles cancel each other out perfectly. In this list, $\mathrm{H}_2$ is non-polar as a homonuclear diatomic molecule, while $\mathrm{CO}_2$, $\mathrm{CH}_4$, and $\mathrm{BF}_3$ exhibit non-polarity due to the perfect cancellation of dipoles in their linear, tetrahedral, and trigonal planar geometries, respectively. All other molecules listed, such as $\mathrm{HF}$, $\mathrm{H}_2\mathrm{O}$, $\mathrm{SO}_2$, $\mathrm{NH}_3$, $\mathrm{HCl}$, and $\mathrm{CHCl}_3$, possess permanent dipole moments resulting from their asymmetric shapes, which prevents the neutralization of individual bond polarities.

Hydrogen bonding arises when a hydrogen atom is covalently bonded to a highly electronegative atom such as $\mathrm{F}$, $\mathrm{O}$, or $\mathrm{N}$, resulting in a strong dipole-dipole attraction. The intensity of this interaction is heavily influenced by the physical state of the substance, as the magnitude of hydrogen bonding is greater in the solid state than in the liquid state due to closer molecular packing. While o-nitrophenol is characterized by intramolecular hydrogen bonding, substances like $\mathrm{HF}$ exhibit intermolecular hydrogen bonding, highlighting the structural distinctions between the two. These interactions exert a powerful influence on the chemical properties of a compound, directly affecting macroscopic traits such as density, melting point, and boiling point.

According to VSEPR theory, molecular geometry is determined by the total number of bonding and non-bonding electron pairs around the central atom, where a trigonal pyramidal shape specifically arises from a central atom surrounded by three bonding pairs and one lone pair. Among the given species, only $\mathrm{SO}_3^{2-}$ possesses this geometry because the central sulfur atom has one lone pair and three oxygen bonds, while $\mathrm{S}_2\mathrm{O}_3^{2-}$, $\mathrm{SO}_4^{2-}$, and $\mathrm{S}_2\mathrm{O}_7^{2-}$ maintain tetrahedral geometry because their central sulfur atoms are bonded to four atoms with no remaining lone pairs.

Molecular Orbital Theory determines the electronic configuration of species by populating bonding and antibonding orbitals according to total electron count, the Aufbau principle, and Hund's rule. Identifying species with exactly one unpaired electron requires examining the occupancy of their highest energy molecular orbitals. In $\mathrm{N}_2$ and $\mathrm{CN}^{-}$, which contain 14 electrons, all electrons are paired, whereas the 16-electron $\mathrm{O}_2$ molecule holds two unpaired electrons in its antibonding $\pi^*$ orbitals. The 18-electron $\mathrm{O}_2^{2-}$ ion similarly results in a fully paired configuration.

Calculating the configurations for the remaining species reveals that $\mathrm{C}_2^{-}$ with 13 electrons and $\mathrm{O}_2^{-}$ with 17 electrons both leave a single unpaired electron in their highest energy orbitals. Furthermore, the simple $\mathrm{H}_2^{+}$ ion containing only one electron and the $\mathrm{He}_2^{+}$ ion with three electrons follow the same pattern of having exactly one unpaired electron. Evaluating the set as a whole confirms that these 4 species satisfy the condition of containing a single unpaired electron.