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

The peroxide ion, $\mathrm{O}_{2}^{2-}$, is formed by two oxygen atoms that, together with an additional two electrons, contribute 18 electrons to the molecular orbital system. These electrons occupy the energy levels in the sequence $\sigma1s^2, \sigma^*1s^2, \sigma2s^2, \sigma^*2s^2, \sigma2p_z^2, \pi2p_x^2, \pi2p_y^2, \pi^*2p_x^2, \pi^*2p_y^2$. Because every orbital in this configuration is completely filled with a pair of electrons, the ion contains zero unpaired electrons.

According to molecular orbital theory, the stability of a covalent bond is determined by its bond order, which is calculated using the formula $\text{Bond Order} = \frac{N_b - N_a}{2}$, where $N_b$ represents the number of bonding electrons and $N_a$ represents the number of antibonding electrons. This value provides insight into the strength of the bond and the likelihood of the molecule's existence, with a bond order of zero indicating that the species is unstable or does not exist.

Applying this principle, the nitrogen molecule, $\mathrm{N}_2$, with 14 electrons, fills its orbitals such that there are 10 bonding electrons and 4 antibonding electrons, resulting in a bond order of $\frac{10-4}{2} = 3$. The oxygen molecule, $\mathrm{O}_2$, contains 16 electrons, resulting in 10 bonding and 6 antibonding electrons, which gives a bond order of $\frac{10-6}{2} = 2$. For the fluorine molecule, $\mathrm{F}_2$, which has 18 electrons, there are 10 bonding and 8 antibonding electrons, leading to a bond order of $\frac{10-8}{2} = 1$. Finally, the neon molecule, $\mathrm{Ne}_2$, contains 20 electrons that fill bonding and antibonding orbitals equally with 10 electrons each, yielding a bond order of $\frac{10-10}{2} = 0$. Matching these results confirms that $\mathrm{Ne}_2$ corresponds to 0, $\mathrm{N}_2$ to 3, $\mathrm{F}_2$ to 1, and $\mathrm{O}_2$ to 2.

Whether an oxide exhibits paramagnetism depends on the presence of unpaired electrons within its constituent oxygen species. The oxides $\mathrm{Li}_2\mathrm{O}$, $\mathrm{K}_2\mathrm{O}$, and $\mathrm{HgO}$ contain the oxide ion $\mathrm{O}^{2-}$, which features a stable, fully paired electron configuration of $1s^{2}, 2s^{2}, 2p^{6}$, rendering these compounds diamagnetic. Similarly, $\mathrm{Na}_2\mathrm{O}_2$ contains the peroxide ion $\mathrm{O}_2^{2-}$, which possesses a completely filled set of molecular orbitals, meaning all electrons are paired and the substance is diamagnetic. In contrast, the superoxide ion $\mathrm{O}_2^{-}$ present in $\mathrm{KO}_2$ contains an unpaired electron in its $\pi^* 2p$ molecular orbital, which is the defining characteristic of a paramagnetic substance. Consequently, exactly one paramagnetic oxide is present among these compounds.

Determining the number of lone pairs on a central xenon atom requires subtracting the number of electrons involved in bonding from its eight total valence electrons. In $\mathrm{XeF}_2$, the central atom shares two electrons with fluorine atoms, leaving six non-bonding electrons which constitute three lone pairs. The molecule $\mathrm{XeO}_2\mathrm{F}_2$ utilizes two electrons for fluorine and four electrons for oxygen bonds, leaving two electrons, or one lone pair, remaining. For $\mathrm{XeO}_3\mathrm{F}_2$, all eight valence electrons participate in bonding with the surrounding atoms, resulting in zero lone pairs, whereas $\mathrm{XeF}_4$ consumes four valence electrons in bonding with fluorine, leaving four electrons to form two lone pairs.

The assertion is incorrect because hydrogen bonding is a specific phenomenon distinct from the general class of dipole-dipole interactions, and the partial charges ($\delta^{+}$ and $\delta^{-}$) characterizing polar covalent bonds are always significantly less than the fundamental unit electronic charge of $1.6 \times 10^{-19} \, \text{C}$. Conversely, the provided reason is accurate because fluorine possesses the highest electronegativity of all elements, a property that facilitates the specific, symmetrical nature of the hydrogen bonds observed in HF.

Molecular Orbital Theory defines the electronic structure of species by assigning electrons to orbitals based on increasing energy levels, where bonding orbitals contribute to molecular stability. The peroxide ion $\mathrm{O}_2^{2-}$ contains 18 total electrons, determined by the 8 electrons from each oxygen atom plus the 2 additional electrons represented by the negative charge. By filling the bonding molecular orbitals according to the Aufbau principle, the electrons occupy the $\sigma 1s^2$, $\sigma 2s^2$, $\sigma 2p_z^2$, $\pi 2p_x^2$, and $\pi 2p_y^2$ orbitals. Since each of these five bonding orbitals accommodates 2 electrons, the total number of electrons residing in all bonding molecular orbitals is 10.

Molecular Orbital Theory provides the framework to determine the electronic structure and magnetism of a molecular ion by filling orbitals according to the Aufbau principle. The superoxide ion, $\mathrm{O}_2^{-}$, contains a total of 17 electrons, which distribute into the molecular orbitals as $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. To find the bond order, we apply the formula $\text{Bond Order} = \frac{n_b - n_a}{2}$ which yields $\frac{10 - 7}{2} = 1.5$ since there are 10 bonding electrons and 7 antibonding electrons. The presence of a single unpaired electron \in the $\pi^*$ orbital confirms that the ion is paramagnetic.

The structure of the $I_3^-$ ion is governed by the central iodine atom, which contributes seven valence electrons plus one additional electron from the negative charge to the system. By forming two single covalent bonds with the flanking iodine atoms, the central iodine utilizes two of its electrons, while the remaining electrons form non-bonding pairs to satisfy its electronic configuration. Consequently, the central iodine atom is left with 3 lone pairs of electrons.

Molecular Orbital Theory provides the framework for determining the bond order of diatomic species by distributing electrons into bonding and antibonding molecular orbitals. The bond order is calculated using the formula $\text{Bond Order} = \frac{N_b - N_a}{2}$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. A bond order of 2.5 is characteristic of Nitric oxide, a molecule composed of nitrogen with 7 electrons and oxygen with 8 electrons. Summing these atomic contributions results in a total of 15 electrons, which distribute into the molecular orbitals to satisfy this specific bond order requirement.

The molecular geometry of these species is governed by the valence shell electron pair repulsion theory, where the presence of lone pairs on the central atom fundamentally alters the bond angles and overall shape. Both $\mathrm{SO}_{3}$, $\mathrm{NO}_{3}^{-}$, and $\mathrm{CO}_{3}^{2-}$ lack lone pairs on their central atoms and adopt a trigonal planar geometry, whereas $\mathrm{PCl}_{3}$ features a lone pair that forces the molecular framework into a trigonal pyramidal shape. Therefore, since three out of the four listed compounds exhibit this flat, non-pyramidal structure, the total number of species satisfying the condition is three.

Molecular Orbital Theory defines bond order by calculating the net difference between bonding and antibonding electrons and dividing that value by two. For carbon monoxide ($\mathrm{CO}$), the arrangement of 10 bonding and 4 antibonding electrons results in a bond order expressed as $\frac{10-4}{2} = 3$. In the case of the nitrosonium ion ($\mathrm{NO}^{\oplus}$), the electronic distribution leads to 8 bonding and 2 antibonding electrons, giving a bond order of $\frac{8-2}{2} = 3$. Comparing these two results shows no difference between them, so setting the relationship $\frac{x}{2} = 0$ confirms that $x = 0$.

To be isostructural, two molecules must share the same hybridization state and molecular geometry, which are dictated by the arrangement of electron pairs around the central atom. For the pairs $\mathrm{SO}_4^{2-}$ and $\mathrm{CrO}_4^{2-}$, as well as $\mathrm{SiCl}_4$ and $\mathrm{TiCl}_4$, the central atoms each undergo $sp^3$ hybridization with four bond pairs and zero lone pairs, resulting in identical tetrahedral structures. Conversely, $\mathrm{NH}_3$ has $sp^3$ hybridization with one lone pair yielding a pyramidal shape, while $\mathrm{NO}_3^{-}$ is $sp^2$ hybridized and trigonal planar, making them structurally distinct. Similarly, $\mathrm{BCl}_3$ adopts $sp^2$ hybridization with a trigonal planar geometry, whereas $\mathrm{BrCl}_3$ features $sp^3 d$ hybridization with two lone pairs and a T-shaped structure, meaning these are also not isostructural.

The hybridization of a central atom is determined by calculating the steric number using the formula $SN = \frac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons, $M$ is the number of monovalent atoms, and $C$ and $A$ represent the cationic and anionic charges, respectively. For $NO_2^-$, the steric number is $\frac{1}{2}(5 + 0 + 1) = 3$, resulting in $sp^2$ hybridization, while for $NO_2^+$, the value is $\frac{1}{2}(5 + 0 - 1) = 2$, indicating $sp$ hybridization. Finally, for $NH_4^+$, the steric number is $\frac{1}{2}(5 + 4 - 1) = 4$, which corresponds to $sp^3$ hybridization.

To identify the species with unequal bond lengths, one must examine the symmetry of each molecular geometry. Molecules like ${XeF}_{4}$, ${SiF}_{4}$, and ${BF}_{4}^{-}$ exhibit high degrees of symmetry-specifically square planar and tetrahedral shapes-where all bonding positions are equivalent, leading to identical bond lengths for all substituents. Conversely, ${SF}_{4}$ undergoes $sp^3d$ hybridization with a trigonal bipyramidal electron geometry that includes a lone pair, forcing the molecule into an asymmetrical see-saw shape. Within this structure, the axial ${S-F}$ bonds experience different electronic environments than the equatorial bonds, causing the axial bonds to be longer and thus unequal to the others.

Molecular geometry is determined by the total number of electron domains surrounding the central atom, as predicted by VSEPR theory. For an $AB_3$ interhalogen species to exhibit a T-shaped structure, the central atom must form three sigma bonds with the surrounding atoms and retain two lone pairs. These two lone pairs occupy the equatorial positions to minimize electronic repulsion, resulting in the characteristic T-shaped arrangement.

Melting point in ionic compounds is primarily determined by lattice energy, which increases as the ionic charges increase and decreases as the interionic distance becomes larger. In the case of $\mathrm{LiF}$ and $\mathrm{LiCl}$, the ions each possess a charge of $+1$ and $-1$, meaning the smaller size of the fluoride ion compared to the chloride ion allows for a stronger lattice and a higher melting point for $\mathrm{LiF}$. For $\mathrm{MgO}$ and $\mathrm{NaCl}$, the presence of $2+$ and $2-$ ions in $\mathrm{MgO}$ creates a much stronger electrostatic attraction than the $1+$ and $1-$ ions in $\mathrm{NaCl}$, resulting in a significantly higher melting point for $\mathrm{MgO}$.

The ability of a central atom to undergo hydrolysis largely depends on its accessibility to water molecules and the availability of vacant coordination sites. While halides like $\mathrm{BF}_3$, $\mathrm{SiCl}_4$, and $\mathrm{PCl}_5$ readily react with water, $\mathrm{SF}_6$ remains chemically inert. This resistance is primarily attributed to significant steric crowding, as the six fluorine atoms surrounding the sulfur atom effectively shield it from any nucleophilic attack by water.

Resonance stability arises when a molecule distributes $\pi$-electron density across multiple equivalent positions, a condition requiring specific structural characteristics. Among the listed species, only $\mathrm{O}_2$ and $\mathrm{CO}_3^{2-}$ feature exactly one $\pi$-bond, which serves as the limiting factor for selection. The $\mathrm{CO}_3^{2-}$ ion uniquely satisfies the requirement for the maximum number of canonical forms by delocalizing this $\pi$-bond across three equivalent oxygen atoms, unlike $\mathrm{O}_2$ which does not exhibit resonance.

Magnetic behavior of a solid is determined by the configuration and orientation of its electrons. $\mathrm{NaCl}$ is diamagnetic because all of its electrons are paired, while $\mathrm{O}_{2}$ acts as a paramagnetic substance due to the presence of two unpaired electrons. Furthermore, $\mathrm{Fe}_{3}\mathrm{O}_{4}$ shows ferrimagnetism because its magnetic moments are aligned in opposite directions in unequal numbers, whereas $\mathrm{MnO}$ is antiferromagnetic as the magnetic moments of its ions align to cancel each other out completely.

According to Molecular Orbital Theory, the stability of a chemical species is determined by its bond order, which is calculated as half the difference between the number of bonding and antibonding electrons. A positive bond order indicates a stable species, while a bond order of zero implies that the attractive and repulsive forces are balanced, meaning the species does not exist.

Applying this to the given species, $\mathrm{O}_2^{2-}$ results in a bond order of $\frac{10-8}{2} = 1$, while $\mathrm{He}_2^{-}$ and $\mathrm{He}_2^{+}$ each yield a bond order of $\frac{3-2}{2} = 0.5$ and $\frac{2-1}{2} = 0.5$ respectively, indicating they can all exist. In contrast, the electronic configuration of $\mathrm{Be}_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$, which results in an equal number of bonding and antibonding electrons. Because its calculated bond order is $\frac{4-4}{2} = 0$, $\mathrm{Be}_2$ cannot form a stable bond and therefore does not exist.