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

VSEPR theory determines molecular geometry based on the steric number of the central atom, where a linear shape is achieved by $sp$ hybridized species with no lone pairs or $sp^3d$ hybridized species with three equatorial lone pairs. Among the listed species, $\mathrm{XeF}_2$ and $\mathrm{I}_3^-$ are linear because they possess $sp^3d$ hybridization with three lone pairs occupying equatorial positions. Additionally, $\mathrm{CO}_2$, $\mathrm{C}_3 \mathrm{O}_2$, and $\mathrm{BeCl}_2$ adopt linear geometries as they contain a central atom with $sp$ hybridization and zero lone pairs. The remaining species, including $\mathrm{I}_3^+$, $\mathrm{SO}_2$, and $\mathrm{BCl}_2^-$, exhibit bent geometries due to the presence of lone pairs on their central atoms, which exert repulsive forces that distort the bond angles.

The extent of hydrogen bonding in water is dictated by the level of molecular order and the presence of external solutes. Ice possesses the highest degree of hydrogen bonding because its molecules are locked in a fixed, highly ordered crystalline lattice, ensuring maximum network formation. Transitioning to liquid water introduces kinetic energy that breaks and reforms these connections dynamically, resulting in fewer stable bonds than in the solid state. Furthermore, the presence of impurities significantly interferes with the cohesive interactions between water molecules, further reducing the overall extent of hydrogen bonding in the mixture compared to both ice and pure liquid water.

The number of lone pairs on a central atom is determined using the formula $LP = \frac{1}{2}(V - B)$, where $V$ represents the valence shell electrons of the central atom and $B$ denotes the number of electrons shared through bonding. By constructing the Lewis structure for each molecule in the provided list, one subtracts the electrons involved in bonding ($B$) from the central atom's valence electrons ($V$) and halves the difference to find the lone pair count. Applying this systematic analysis to every candidate confirms that exactly four of the listed molecules possess two lone pairs on their central atom, thereby satisfying the specified condition.

To determine the magnetic behavior of these compounds, one must evaluate the electronic configuration of the oxygen species present, as the existence of unpaired electrons leads to paramagnetism while paired electrons result in diamagnetism. Lithium oxide involves the $\mathrm{O}^{2-}$ ion, which has a fully occupied valence shell, rendering it diamagnetic. Sodium peroxide incorporates the $\mathrm{O}_2^{2-}$ ion with all electrons paired in its molecular orbitals, while potassium superoxide features the $\mathrm{O}_2^{-}$ ion, which possesses an unpaired electron in an antibonding molecular orbital, making the former diamagnetic and the latter paramagnetic.

The number of $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bonds in these phosphorus compounds is governed by their specific molecular architecture, defined by the connectivity of phosphorus atoms through oxygen bridges. $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7$ consists of two phosphoric acid groups joined by a single central oxygen atom in the structure $(HO)_2P(O)-\mathrm{O}-P(O)(OH)_2$, which establishes 1 $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bond.

As a cyclic trimer, $(\mathrm{HPO}_3)_3$ contains a six-membered ring of alternating phosphorus and oxygen atoms that requires 3 $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bonds to maintain the closed cycle. The $\mathrm{P}_4\mathrm{O}_{10}$ molecule adopts an adamantane-like cage arrangement where four phosphorus atoms located at the vertices of a tetrahedron are linked by oxygen atoms along the six edges, resulting in 6 $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bonds in total.

Fajan's Rule serves as the primary guideline for evaluating the covalent character within an ionic lattice. It dictates that covalent character intensifies when cations possess a small ionic radius or high charge, and when anions are large and highly polarizable, as these factors facilitate the distortion of the anion's electron cloud by the cation's positive field. Consequently, comparing halides like $\mathrm{KF}$ and $\mathrm{KI}$ shows that $\mathrm{KI}$ exhibits higher covalent character because the iodide ion is larger and more easily polarized than the fluoride ion. Between $\mathrm{LiF}$ and $\mathrm{KF}$, the smaller $\mathrm{Li}^+$ cation exerts a stronger polarizing force than the larger $\mathrm{K}^+$ cation, making $\mathrm{LiF}$ the more covalent species.

When looking at tin compounds, $\mathrm{SnCl}_4$ features a $\mathrm{Sn}^{4+}$ cation, which carries a higher charge than the $\mathrm{Sn}^{2+}$ cation in $\mathrm{SnCl}_2$, thereby increasing the overall covalent character. Furthermore, comparing $\mathrm{CuCl}$ and $\mathrm{NaCl}$ reveals that $\mathrm{CuCl}$ is more covalent; even though both cations have a +1 charge, the $\mathrm{Cu}^+$ ion possesses a pseudo-noble gas $d^{10}$ electron configuration that is more effective at polarizing the chloride anion compared to the stable noble gas configuration of the $\mathrm{Na}^+$ ion. These applications of Fajan's Rule confirm the validity of the relationships described in statements B, C, and E.

The total count of valence electrons in a molecule or ion determines whether it possesses an even or odd number of electrons, with ionic charges requiring an adjustment to the valence sum of the constituent atoms. Calculating these values for the provided list yields: $\mathrm{NO}_2$ as $5 + 12 = 17$, $\mathrm{ICl}_4^{-}$ as $7 + 28 + 1 = 36$, $\mathrm{BrF}_3$ as $7 + 21 = 28$, $\mathrm{ClO}_2$ as $7 + 12 = 19$, $\mathrm{NO}_2^{+}$ as $5 + 12 - 1 = 16$, and $\mathrm{NO}$ as $5 + 6 = 11$. Consequently, $\mathrm{ICl}_4^{-}$, $\mathrm{BrF}_3$, and $\mathrm{NO}_2^{+}$ are the species with an even total number of electrons, meaning there are exactly three such items in the set.

The identification of square planar geometry relies on understanding the central atom's coordination and electronic configuration, where four ligands around a central atom can result in this shape if specific lone pair distributions or electronic $d$-configurations are present. Within the given set, $\mathrm{XeF}_4$ and $\mathrm{BrF}_4^{-1}$ demonstrate this geometry through their lone pair arrangements, while the transition metal complexes $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{+2}$ and $[\mathrm{PtCl}_4]^{-2}$ adopt the shape due to their specific orbital occupancy. Since $\mathrm{SF}_4$ is seesaw-shaped and the remaining species, specifically $\mathrm{SiF}_4$, $\mathrm{BF}_4^{-}$, and $[\mathrm{FeCl}_4]^{2-}$, are tetrahedral, exactly four of these species possess a square planar structure.

Predicting molecular geometry relies on VSEPR theory, where the specific arrangement of bonding and lone electron pairs around a central atom dictates the spatial shape of the molecule. The hydronium ion, $\mathrm{H}_3\mathrm{O}^{+}$, adopts a trigonal pyramidal shape due to the presence of one lone pair and three bonding pairs on the oxygen atom, whereas the acetylide anion is linear as a result of the triple bond between its carbon atoms. The ammonium ion, $\mathrm{NH}_4^{+}$, features four bonding pairs and zero lone pairs around the central nitrogen, establishing a tetrahedral geometry. Conversely, the chlorite ion, $\mathrm{ClO}_2^{-}$, is bent because the central chlorine atom accommodates two lone pairs alongside two bonding pairs, creating significant repulsion that forces the molecule into a V-shape.

The count of lone pairs on a central atom is determined by considering the total number of valence electrons, adjusting for any ionic charge, and subtracting the electrons engaged in covalent bonding. Iodine in $\mathrm{IF}_7$ employs all seven of its valence electrons to form bonds, resulting in zero lone pairs, while $\mathrm{ICl}_4^-$ retains four non-bonding electrons after forming four bonds, amounting to two lone pairs. In the case of xenon, the central atom uses six electrons for bonding in $\mathrm{XeF}_6$ to leave one lone pair remaining, whereas in $\mathrm{XeF}_2$, it utilizes only two electrons for bonding, which leaves three lone pairs behind.

Fajans' rules establish the foundation for evaluating the covalent character of ionic bonds, which occurs when the electron cloud of an anion is distorted by the electrical field of a cation. This covalent interaction is governed by the polarising power of the cation, which exerts the attractive force, and the polarisability of the anion, which indicates the ease with which its electron cloud can be deformed, leading to an observable extent of distortion of the anion. Although these three aspects directly dictate the degree of covalent character, the polarising power of the anion is not a valid factor because anions are inherently polarisable rather than polarising. Therefore, among the variables listed, there are exactly three factors that determine the percent covalent character of an ionic bond.

Both sulfur dioxide ($\mathrm{SO}_2$) and water ($\mathrm{H}_2\mathrm{O}$) adopt a bent or V-shaped molecular geometry due to the presence of lone pairs on the central atom, which makes the first statement accurate. In terms of bond angles, $\mathrm{SO}_2$ is $sp^2$ hybridized with a bond angle of approximately $120^\circ$, while $\mathrm{H}_2\mathrm{O}$ is $sp^3$ hybridized with a bond angle of roughly $104.5^\circ$. Because the $sp^2$ hybridization and the repulsion from multiple bonds lead to a wider angle in $\mathrm{SO}_2$, the claim in the second statement that its bond angle is smaller is incorrect.

The VSEPR theory dictates the structure of $\mathrm{OF}_2$, where the central oxygen atom forms two covalent bonds with fluorine while holding two lone pairs of electrons. These lone pairs exert strong repulsive forces that compress the F-O-F bond angle to approximately 102 degrees, which is indeed less than 104.5 degrees, and they force the molecule into a bent or V-shaped geometry. Because oxygen is less electronegative than the attached fluorine atoms, it assumes a positive oxidation state of +2 rather than a negative one. Given these chemical properties, the accurate characterizations of the molecule are that it contains two lone pairs on the oxygen, adopts a V-shape, and features a bond angle smaller than 104.5 degrees.

The number of lone pairs on a central atom is calculated by subtracting the electrons involved in bonding from the total valence electrons of the central atom, accounting for any ionic charge. For $\mathrm{ClO}_3^{-}$, the central chlorine atom has 7 valence electrons plus 1 from the negative charge, and after forming bonds with three oxygen atoms, it retains 1 lone pair. In $\mathrm{XeF}_4$, the xenon atom possesses 8 valence electrons and forms 4 bonds, leaving 4 electrons or 2 lone pairs.

For $\mathrm{SF}_4$, the sulfur atom has 6 valence electrons and forms 4 bonds, leaving 2 electrons or 1 lone pair. Finally, the central iodine atom in $\mathrm{I}_3^{-}$ has 7 valence electrons plus 1 from the negative charge; it forms 2 bonds, leaving 6 electrons, which corresponds to 3 lone pairs. Comparing these values, the maximum number of lone pairs on the central atom among the given species is 3.

Understanding the Lewis structures of nitrogen oxides reveals the specific bonding patterns that define their chemical stability. The molecule $\mathrm{N}_2\mathrm{O}_4$ exists as a symmetric dimer characterized by a direct $\mathrm{N}-\mathrm{N}$ single bond, whereas the paramagnetic radical $\mathrm{NO}_2$ features a characteristic $\mathrm{N}=\mathrm{O}$ bond. In the structure of $\mathrm{N}_2\mathrm{O}_5$, an oxygen atom bridges two nitro groups to create an $\mathrm{N}-\mathrm{O}-\mathrm{N}$ linkage, while the linear $\mathrm{N}_2\mathrm{O}$ molecule is distinguished by resonance structures containing an $\mathrm{N}=\mathrm{N}$ or $\mathrm{N} \equiv \mathrm{N}$ bond between the nitrogen atoms.

Molecular Orbital Theory describes the filling of orbitals based on an energy order that shifts for diatomic molecules with more than 14 electrons because of reduced $s-p$ mixing. For $\mathrm{N}_2$ with 14 electrons, the electronic configuration is $\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 = \pi 2 p_y^2 \sigma 2 p_z^2$, meaning the highest occupied molecular orbital is $\sigma 2 p_z^2$. Since this orbital is completely filled, there are 0 unpaired electrons. In the case of $\mathrm{N}_2^{+}$, which has 13 electrons, the removal of one electron from the highest level leaves a $\sigma 2 p_z^1$ configuration, resulting in 1 unpaired electron.

For $\mathrm{O}_2$, which contains 16 electrons, the energy ordering places the antibonding orbitals as the highest occupied level, specifically $\pi^* 2 p_x^1 = \pi^* 2 p_y^1$. Because these are degenerate orbitals, Hund's rule mandates that the electrons occupy them singly to minimize repulsion, resulting in 2 unpaired electrons. Finally, $\mathrm{O}_2^{+}$ possesses 15 electrons, which results in a configuration ending in $\pi^* 2 p_x^1 = \pi^* 2 p_y^0$. This leaves only 1 unpaired electron in the highest occupied molecular orbital.

The molecular geometry of a species is determined by VSEPR theory, where a central atom with a steric number of 6, comprising 5 bond pairs and 1 lone pair, adopts a square pyramidal structure. Among the given species, $\mathrm{PF}_5$ features a steric number of 5 resulting in a trigonal bipyramidal structure, while $\mathrm{BrF}_4^{-}$ and $\mathrm{ICl}_4^{-}$ possess a steric number of 6 with 4 bonding pairs and 2 lone pairs, leading to a square planar geometry. In contrast, $\mathrm{IF}_5$, $\mathrm{BrF}_5$, and $\mathrm{XeOF}_4$ each exhibit a steric number of 6 with 5 bonding pairs and 1 lone pair, which generates a square pyramidal geometry. This results in a total of 3 species having a square pyramidal structure.

Valence bond theory determines lone pairs on a central atom by subtracting the number of shared bonding electrons from the total valence electrons available, which for iodine is 7. For $\mathrm{IF}_5$, the central iodine utilizes 5 electrons to bond with fluorine, leaving 2 non-bonding electrons that constitute 1 lone pair. In the $\mathrm{IF}_7$ molecule, all 7 valence electrons are exhausted in bonds with the seven fluorine atoms, resulting in 0 lone pairs. Combining these values, the total sum is $1 + 0 = 1$.

The stability of alkali metal superoxides relies heavily on the size of the cation, as larger alkali metal ions are necessary to provide sufficient lattice energy to stabilize the bulky superoxide anion. While heavier metals such as potassium, rubidium, and cesium readily form stable crystalline superoxides, sodium superoxide $\mathrm{NaO}_2$ is inherently unstable and does not exist as a stable isolated compound under standard conditions. Because the relatively small sodium ion cannot effectively balance the lattice, the formation of the superoxide is thermodynamically unfavorable, leading the element to favor the formation of the more stable peroxide, $\mathrm{Na}_2\mathrm{O}_2$.

$\mathrm{NF}_3$ adopts a trigonal pyramidal geometry due to the presence of a lone pair on the nitrogen atom that repels the bonded fluorine atoms, refuting the claim of a planar structure. Regarding dipole moments, the water molecule possesses a higher polarity than $\mathrm{H}_2\mathrm{S}$ because oxygen is substantially more electronegative than sulfur, generating a more significant net dipole. The bond length comparison holds true because $\mathrm{N}_2$ contains a triple bond with a bond order of 3, whereas $\mathrm{O}_2$ features a double bond with a bond order of 2; since higher bond order leads to a shorter distance between atoms, $\mathrm{N}_2$ is indeed shorter. Additionally, the principle that isoelectronic molecules or ions share identical bond orders is a valid rule in chemical bonding theory, as these species possess the same total number of electrons and equivalent molecular orbital occupancy.