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

The molecular geometry of these species is predicted using VSEPR theory based on the total number of electron domains around the central atom. In $XeOF_2$, the Xenon atom contains 3 bonding electron pairs and 2 lone pairs, resulting in a T-shape geometry derived from a trigonal bipyramidal electron arrangement. Similarly, $XeOF_4$ has 5 bonding electron pairs and 1 lone pair around the central Xenon atom, which organizes the atoms into a square pyramidal geometry based on an octahedral parent structure.

The bond order of a covalent bond in a molecule with multiple resonance structures is the weighted average of the bond orders present in those structures. In hydrazoic acid ($\mathrm{HN_3}$), the contributing resonance forms include $\mathrm{H-N=N^+=N^-}$ and $\mathrm{H-N^--N^+\equiv N}$, where bond (I) alternates between single and double bond character, yielding an average bond order less than 2, while bond (II) alternates between double and triple bond character, resulting in an average bond order greater than 2. Consequently, bond (I) possesses a bond order less than 2, and bond (II) possesses a bond order greater than 2.

An antibonding $\pi$ orbital, or $\pi^*$, is characterized by the out-of-phase, side-on overlap of atomic p-orbitals, which leads to destructive interference and the formation of a nodal plane between the nuclei. This destructive interaction is visually represented by lobes of opposite phases that do not merge, resulting in a region of zero electron density between the nuclei and a higher energy state compared to the constituent atomic orbitals. Because the diagram exhibits this characteristic phase mismatch and lack of electron density in the internuclear region, it serves as a definitive representation of an antibonding $\pi$ interaction.

The number of lone pairs in a molecular ion is determined by the total valence electrons of each atom minus the electrons involved in bonding. In the $I_3^-$ ion, the central iodine atom has 7 valence electrons plus 1 electron from the negative charge, providing a total of 8 valence electrons; as it forms 2 single covalent bonds, the remaining 6 electrons constitute 3 lone pairs. Each of the two terminal iodine atoms possesses 7 valence electrons and forms 1 single bond with the central iodine, leaving $7 - 1 = 6$ electrons, which equate to 3 lone pairs on each terminal atom. Summing the lone pairs from the central atom and the two terminal atoms, the calculation is $3 + 3 + 3 = 9$.

Molecular magnetism is determined by the electronic configuration within molecular orbitals, where the presence of at least one unpaired electron results in paramagnetism, while a fully paired state indicates diamagnetism. Based on Molecular Orbital Theory, $NO$ contains one unpaired electron in a $\pi^*$ orbital, $O_2$ features two unpaired electrons in $\pi^*$ orbitals, and $B_2$ exhibits two unpaired electrons in its $\pi$ bonding orbitals, making all three species paramagnetic. In contrast, $CO$ is the only molecule in this set where all electrons are completely paired within its molecular structure, establishing it as the only species that is not paramagnetic.

The hybridization of a central atom is dictated by its total number of bonding and lone electron pairs, with $sp^3d^2$ hybridization corresponding to an octahedral arrangement of six electron domains. Molecules such as $SF_6$, $BrF_5$, and $[CrF_6]^{3−}$ accommodate six electron pairs around the central atom, consistently displaying $sp^3d^2$ hybridization. Conversely, $PF_5$ involves phosphorus forming bonds with only five fluorine atoms, which results in a trigonal bipyramidal geometry and necessitates $sp^3d$ hybridization.

The triangular planar geometry is characteristic of molecules where the central atom is $sp^2$ hybridized and devoid of lone pairs of electrons, resulting in a bond angle of $120^{\circ}$. In the species $CO_3^{2-}$, $NO_3^{-}$, and $SO_3$, the central atoms exhibit $sp^2$ hybridization with three surrounding bonded atoms and no lone pairs, establishing the planar structure. Other options fail because species such as $NF_3$, $NH_3$, and $NCl_3$ contain a lone pair on the central nitrogen atom, which distorts the geometry into a trigonal pyramid.

Molecular Orbital Theory dictates that a species exhibits paramagnetic behavior when it contains unpaired electrons in its molecular orbitals. Species such as $CO$ and $NO^{+}$, both having 14 electrons, along with $O_2^{2-}$ with its 18 electrons, possess completely paired electron configurations. Conversely, $B_2$ consists of 10 electrons that fill the bonding orbitals such that two electrons remain unpaired in the degenerate $\pi_{2p}$ molecular orbitals, resulting in paramagnetic properties.

Hybridization is determined by the steric number, which is calculated as the sum of sigma bonds and lone pairs around a central atom. In the linear species $\mathrm{O}=\mathrm{N}^+=\mathrm{O}$, the central nitrogen atom forms two double bonds with oxygen atoms and possesses no lone pairs. Since this configuration results in a steric number of two, the nitrogen atom is in a state of $sp$ hybridization.

Bond angles are determined by the electronic repulsion around the central atom as described by VSEPR theory, where lone pair-bond pair repulsion reduces bond angles more significantly than bond pair-bond pair repulsion. In $CH_4$, the central carbon atom is $sp^3$ hybridized with four bond pairs and zero lone pairs, resulting in a perfect tetrahedral geometry and the largest bond angle of 109.5 degrees. Conversely, the presence of lone pairs in $NH_3$ and $H_2O$ decreases their bond angles, and $PH_3$ exhibits a bond angle near 90 degrees because it does not undergo hybridization according to Drago's rule.

Molecular shape is governed by the hybridization of the central atom and the positioning of its electron lone pairs, which together dictate the final geometry. In the group consisting of $\mathrm{ClF}_3$, $\mathrm{XeOF}_2$, and $\mathrm{XeF}_3^+$, every central atom possesses $\mathrm{sp}^3\mathrm{d}$ hybridization alongside two lone pairs of electrons. Since these lone pairs occupy equatorial sites to minimize repulsion, the atoms are compelled to adopt the same T-shape configuration in each instance.

The strength of an intermolecular interaction depends on the spatial distribution of the charges involved, which dictates the rate at which the force diminishes as separation increases. For an ion-dipole interaction, the force is determined by the variation of the dipole's potential field as the ion moves further away.

Because the dipole potential is proportional to $\frac{1}{r^2}$, the force-being related to the derivative of this potential with respect to the distance-follows an inverse cube law. Expressed mathematically, the interaction force $F$ relates to the dipole moment $\mu$ and distance $r$ such that $F \propto \mu \frac{d}{dr} \left( \frac{1}{r^2} \right)$, which results in the dependency $F \propto \frac{\mu}{r^3}$. This specific inverse cube relationship distinguishes the ion-dipole interaction from other forces that follow different power laws.

Molecular Orbital Theory defines magnetic behavior based on electron pairing, where molecules with no unpaired electrons are classified as diamagnetic. Both $N_2$ and $C_2$ feature completely paired electrons within their molecular orbitals, which accounts for their characteristic diamagnetism. In contrast, the presence of unpaired electrons in the antibonding orbitals of $O_2$ and $S_2$ results in paramagnetic behavior, identifying $C_2$ as a molecule that satisfies the criteria for diamagnetism.

The stability of diatomic species is fundamentally determined by their bond order, where a higher bond order indicates a stronger chemical bond and greater overall stability. When two species share the same bond order, the one with fewer electrons in antibonding orbitals is more stable. To calculate the bond order for these species, we use the formula $\text{Bond Order} = \frac{\text{Bonding Electrons} - \text{Antibonding Electrons}}{2}$. For the neutral $\mathrm{Li}_2$ molecule, which contains 6 electrons, the electronic configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2$, resulting in a bond order of $\frac{4-2}{2} = 1$.

The $\mathrm{Li}_2^{+}$ ion, containing 5 electrons, has the configuration $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^1$, which gives it a bond order of $\frac{3-2}{2} = 0.5$. Similarly, the $\mathrm{Li}_2^{-}$ ion, with 7 electrons, features the configuration $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^1$, resulting in a bond order of $\frac{4-3}{2} = 0.5$. Since both ions have an identical bond order, their relative stability is governed by the number of antibonding electrons. The $\mathrm{Li}_2^{-}$ ion contains more antibonding electrons than $\mathrm{Li}_2^{+}$, making $\mathrm{Li}_2^{+}$ the more stable species. Consequently, the stability increases in the order $\mathrm{Li}_2^{-} < \mathrm{Li}_2^{+} < \mathrm{Li}_2$.

Molecular orbital theory predicts the stability of a chemical species based on its bond order, which is calculated as the net difference between electrons in bonding and antibonding orbitals. When the calculated bond order is zero, it indicates that the antibonding interactions completely negate the bonding interactions, making the species energetically unstable and essentially non-existent. Since both $H_2^{2+}$ and $He_2$ result in a bond order of zero, they represent the pair that is not likely to exist.

Two molecules are considered isostructural when they share identical hybridization states and molecular geometries. While $\mathrm{PF}_5$ involves $\mathrm{sp}^3 d$ hybridization to arrange five bond pairs into a trigonal bipyramidal structure, $\mathrm{BrF}_5$ contains five bond pairs and one lone pair around the central atom. This electron configuration requires $\mathrm{sp}^3 d^2$ hybridization, which forces the molecule into a square pyramidal shape. Since their hybridization states and structural geometries do not align, these two species fail to meet the criteria for being isostructural.

The covalent character of an ionic compound is determined by the polarizing power of its cation, which increases as the positive charge density on the ion rises. This power is directly proportional to the magnitude of the cation's charge and inversely proportional to its ionic radius. Because $\mathrm{Al}^{3+}$ carries a higher positive charge and is smaller in size than the $\mathrm{Mg}^{2+}$, $\mathrm{Sn}^{2+}$, and $\mathrm{Fe}^{2+}$ cations, it is significantly more effective at polarizing the electron cloud of the chloride ion. Consequently, $\mathrm{AlCl}_3$ exhibits the most pronounced covalent character among these compounds.

Iodine acts as the central atom in $\mathrm{IF}_7$, where it engages in seven single bonds with fluorine atoms, leading to $\text{sp}^3 d^3$ hybridization to accommodate all seven bonding electron pairs. This specific steric number of seven forces the molecule into a pentagonal bipyramid shape, as this unique arrangement allows the electron pairs to be spaced as far apart as possible to minimize repulsive forces.

The hybridization of the central nitrogen atom is determined by the steric number, calculated as $SN = \frac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons of the central atom, $M$ is the number of monovalent atoms bonded to it, $C$ is the cationic charge, and $A$ is the anionic charge. Applying this formula yields $SN = 3$ for $\mathrm{NO}_3^{-}$ ($V=5$, oxygen is divalent, $A=1$), $SN = 2$ for $\mathrm{NO}_2^{+}$ ($V=5$, oxygen is divalent, $C=1$), and $SN = 4$ for $\mathrm{NH}_4^{+}$ ($V=5$, four monovalent hydrogen atoms, $C=1$), corresponding to $\mathrm{sp}^2$, $\mathrm{sp}$, and $\mathrm{sp}^3$ hybridization respectively.

Bond length is inversely proportional to bond order, meaning that as the bond order increases, the bond becomes stronger and shorter. The bond order for a diatomic species is calculated using the formula $\text{Bond order} = \frac{1}{2}(N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. For the given species, the calculated bond orders are 3.0 for $\mathrm{O}_2^{2+}$ (14 electrons), 2.5 for $\mathrm{O}_2^{+}$ (15 electrons), 1.5 for $\mathrm{O}_2^{-}$ (17 electrons), and 1.0 for $\mathrm{O}_2^{2-}$ (18 electrons). Consequently, $\mathrm{O}_2^{2+}$ has the highest bond order, which corresponds to the shortest bond length.