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

Bromine, acting as the central atom in $\mathrm{BrF}_3$, begins with seven valence electrons and uses three to form single bonds with the surrounding fluorine atoms, leaving behind four non-bonding electrons that form two lone pairs. This configuration results in a total steric number of 5 with $\mathrm{sp}^3\mathrm{d}$ hybridization, where the two lone pairs occupy equatorial positions in the trigonal bipyramidal electron geometry to minimize electron repulsion. This specific placement of lone pairs distorts the structure into a bent T-shape.

In molecular orbital theory, the linear combination of atomic orbitals results in molecular orbitals formed through interference patterns, where destructive interference represented by $\psi_{MO}=\psi_A-\psi_B$ characterizes an antibonding molecular orbital, while constructive interference represented by $\psi_{MO}=\psi_A+\psi_B$ characterizes a bonding molecular orbital. The bond order provides a measure of bond strength defined by the ratio $\frac{N_b - N_a}{2}$, which accounts for the difference between bonding and antibonding electrons. Additionally, the polarity of a bond is quantified by the dipole moment $\mu = Q \times r$, where the product of the charge magnitude $Q$ and the distance $r$ determines the overall molecular dipole.

Determining the hybridization and molecular geometry relies on assessing the total electron domains around the central Xenon atom according to VSEPR theory. $\mathrm{XeO}_3$ features three bonding pairs and one lone pair, yielding $sp^3$ hybridization and a pyramidal shape, while $\mathrm{XeF}_2$ utilizes two bonding pairs and three lone pairs to achieve $sp^3d$ hybridization with a linear structure. Moving to $\mathrm{XeOF}_4$, the presence of five bonding pairs and one lone pair results in $sp^3d^2$ hybridization and a square pyramidal geometry. Lastly, $\mathrm{XeF}_6$ incorporates six bonding pairs and one lone pair, manifesting as $sp^3d^3$ hybridization with a distorted octahedral shape.

Magnetic behavior is dictated by electronic configuration, where species with unpaired electrons display paramagnetism, while those with fully paired electrons are diamagnetic. $\mathrm{O}_2$, $\mathrm{Cu}^{2+}$, and $\mathrm{Fe}^{3+}$ possess 2, 1, and 5 unpaired electrons respectively, resulting in weak attraction to an external magnetic field and magnetization in the same direction. Conversely, $\mathrm{NaCl}$ and $\mathrm{H}_2 \mathrm{O}$ feature completely paired electrons, identifying them as diamagnetic materials that are weakly repelled and magnetized in the opposite direction to the applied field.

To determine the number of lone pairs on a central atom, we compare the total valence electrons of that atom to the number of electrons it contributes to chemical bonding. For $\mathrm{SCl}_2$, the central sulfur atom has 6 valence electrons and forms two single bonds with chlorine atoms, which leaves 4 electrons organized into 2 lone pairs. The central oxygen atom in $\mathrm{O}_3$ is involved in resonance where it contributes to one double bond and one single bond, retaining 1 lone pair to complete its octet requirements. In the case of $\mathrm{ClF}_3$, the chlorine atom possesses 7 valence electrons and uses 3 to bond with fluorine, resulting in 4 unshared electrons that form 2 lone pairs. Finally, $\mathrm{SF}_6$ features a sulfur atom that utilizes all 6 of its valence electrons to form bonds with six fluorine atoms, leaving the central atom with 0 lone pairs.

To identify the lone pairs on the central atom, subtract the number of bonding electrons from the total valence electrons and divide by two. For $\mathrm{CF}_4$, the carbon atom uses all four valence electrons for bonding, resulting in zero lone pairs, while sulfur in $\mathrm{SF}_4$ leaves one lone pair after forming four bonds with its six valence electrons. Following this same calculation, oxygen in $\mathrm{H}_2\mathrm{O}$ contains two lone pairs, and xenon in $\mathrm{XeF}_4$ likewise retains two lone pairs, bringing the total count of species with two lone pairs to exactly two.

Whether a molecule exhibits paramagnetic behavior depends entirely on the presence of unpaired electrons within its molecular orbital configuration; if all electrons are paired, the species is diamagnetic. We can determine this by examining the electron distribution in the molecular orbitals for each given species.

Starting with $\mathrm{B}_2$, which contains 10 electrons, the molecular orbital filling results in two electrons occupying the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals singly, making it paramagnetic. For $\mathrm{Li}_2$ with 6 electrons, the configuration $\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2}$ shows that all electrons are paired, rendering it diamagnetic. Similarly, $\mathrm{C}_2$ possesses 12 electrons, and with its electrons fully paired in the $\sigma$ and $\pi$ orbitals, it is diamagnetic.

When we examine $\mathrm{C}_2^{-}$, which has 13 electrons, the additional electron enters the $\sigma_{2p_z}$ orbital, creating a single unpaired electron and thus a paramagnetic species. The ion $\mathrm{O}_2^{2-}$ contains 18 electrons, and its complete orbital filling $\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \sigma_{2p_z^2} \pi_{2p_x^2} = \pi_{2p_y^2} \pi_{2p_x^2}^* = \pi_{2p_y^2}^*$ leaves no room for unpaired electrons, so it is diamagnetic. For $\mathrm{O}_2^{+}$, which has 15 electrons, the configuration leaves one unpaired electron in the antibonding $\pi^*$ orbitals, classifying it as paramagnetic. Finally, $\mathrm{He}_2^{+}$ with 3 electrons follows the configuration $\sigma_{1s^2} \sigma_{1s^1}^*$, leaving one electron unpaired, which makes it paramagnetic. Based on this analysis, $\mathrm{B}_2, \mathrm{C}_2^{-}, \mathrm{O}_2^{+}$, and $\mathrm{He}_2^{+}$ are the four paramagnetic species in the provided list.

Molecular planarity depends on the geometric arrangement of atoms around the central atom, where structures like trigonal planar and square planar sit within a single plane, whereas others exist in three dimensions. $\mathrm{NO}_3^{-}$, $\mathrm{BF}_3$, $\mathrm{SO}_3$, and $\mathrm{XeF}_4$ represent the planar category with their trigonal planar or square planar geometries. The remaining species possess three-dimensional structures: $\mathrm{H}_2\mathrm{O}_2$ has an open-book shape, $\mathrm{PCl}_3$ and $\mathrm{XeO}_3$ are pyramidal, $\mathrm{SF}_4$ is see-saw, and both $\mathrm{PH}_4^{+}$ and $[\mathrm{Al}(\mathrm{OH})_4]^{-}$ are tetrahedral. In total, there are 6 non-planar molecules and ions in the list.

The extent of covalent character in an ionic compound is governed by Fajan's rule, which states that larger, more polarizable anions are more easily distorted by the electrostatic field of a cation. As we move down the halogen group, the ionic radius increases from fluoride to iodide, making the electron cloud of the anion progressively easier to polarize by the fixed calcium ion. This increased polarization leads to greater electron sharing between the ions, resulting in a trend of increasing covalent character from $\mathrm{CaF}_2$ through $\mathrm{CaCl}_2$ and $\mathrm{CaBr}_2$ to $\mathrm{CaI}_2$.

Predicting molecular geometry relies on VSEPR theory, which determines how electron pairs and central atoms arrange themselves to minimize repulsion. For $\mathrm{BrF}_5$, the valence electrons lead to a square pyramidal arrangement, while the $[\mathrm{CrF}_6]^{3-}$ complex assumes an octahedral shape typical of six-coordinated centers. The $\mathrm{O}_3$ molecule exhibits a bent structure due to the lone pair on the central oxygen atom, and $\mathrm{PCl}_5$ adopts a trigonal bipyramidal geometry to accommodate its five chlorine atoms around the phosphorus center.

Applying VSEPR theory to the central xenon atom, which has 8 valence electrons, allows us to determine the lone pair count for each species. In $\text{XeO}_3$, the three double bonds with oxygen atoms utilize 6 electrons, leaving $8 - 6 = 2$ electrons, which corresponds to 1 lone pair. For $\text{XeOF}_4$, the structure involves one double bond with oxygen and four single bonds with fluorine, creating five bonding regions that utilize 6 electrons and leave 1 lone pair. In the case of $\text{XeF}_6$, although the xenon atom has 8 valence electrons and forms six single bonds, the remaining lone pair is often considered stereochemically inactive and does not contribute to the molecular geometry, effectively resulting in a count of 0 for this molecule. Summing these values gives a total of $1 + 1 + 0 = 2$.

The formation of hydrogen bonds requires a hydrogen atom to be covalently bonded to a highly electronegative element, typically nitrogen, oxygen, or fluorine, which creates a significant permanent dipole. Methane ($\mathrm{CH}_4$) lacks these conditions entirely and therefore exhibits no hydrogen bonding, placing it at the lowest end of the strength spectrum. Ammonia ($\mathrm{NH}_3$) features a prominent electronegativity difference between nitrogen and hydrogen, resulting in strong hydrogen bonding, while hydrogen cyanide ($\mathrm{HCN}$) possesses weaker intermolecular interactions that fall between the nonpolar nature of methane and the robust hydrogen bonding observed in ammonia.

Hybridization is fundamentally determined by the steric number of the central atom, which accounts for the sum of sigma bonds and lone pairs around it. Sulfur in $SF_4$ involves five electron domains from four bonds and one lone pair, which aligns with $sp^3d$ hybridization, while the six electron domains in $IF_5$ consist of five bonds and one lone pair, requiring $sp^3d^2$ hybridization. In the $NO_2^{+}$ ion, the central nitrogen atom forms two sigma bonds without any lone pairs to achieve $sp$ hybridization, whereas the $NH_4^{+}$ ion features four sigma bonds with no lone pairs, consistent with $sp^3$ hybridization.

In the crystalline structure of $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$, the five water molecules serve distinct structural roles rather than behaving as equivalent units. Four of these water molecules are directly coordinated to the central $\mathrm{Cu}^{2+}$ ion, while the single remaining water molecule is held within the crystal lattice through hydrogen bonds that bridge the sulfate group to the coordinated water ligands.

Molecular Orbital Theory describes the stability of these species by focusing on the occupancy of bonding versus antibonding orbitals, where bond order is calculated as half the difference between bonding and antibonding electrons. While the neutral $O_2$ molecule possesses a bond order of 2, removing an electron to form $O_2^+$ leaves more bonding electrons relative to antibonding ones, raising the bond order to 2.5. Conversely, adding electrons to antibonding orbitals reduces the bond order to 1.5 for $O_2^-$ and 1 for $O_2^{2-}$. Consequently, the overall bond order sequence follows the trend $O_2^+ > O_2 > O_2^- > O_2^{2-}$.

Molecular geometry is primarily determined by the hybridization state of the central atom, where the presence of lone pairs typically distorts bond angles and results in angular or bent structures. For ${NO}_{2}$, ${O}_{3}$, and ${Cl}_{2}{O}$, the central atoms possess lone pairs that result in $sp^{2}$ or $sp^{3}$ hybridization, causing these molecules to adopt a bent or V-shape. In contrast, the azide ion ${N}_{3}^{-}$ features a central nitrogen atom that is $sp$ hybridized with no lone pairs, which dictates a strictly linear geometry.

The VSEPR theory allows us to predict molecular geometry by calculating the total steric number, which is the sum of the number of bonded atoms and lone pairs attached to the central atom. With three single bonds and two lone pairs, the central atom exhibits a steric number of 5, which corresponds to $sp^3d$ hybridization and an underlying trigonal bipyramidal electron geometry. To minimize electron-electron repulsion, the two lone pairs occupy the equatorial positions of this structure, forcing the three remaining bonding atoms to align themselves into a T-shaped molecular arrangement.

The spin-only magnetic moment of a molecular species is determined by the number of unpaired electrons in its molecular orbitals, calculated using the formula $\mu_S = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. For the $\mathrm{B}_{2}^{+}$ ion, which contains a total of nine electrons, the molecular orbital filling order follows the sequence $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_y^1$. Because only a single electron occupies the $\pi 2p_y$ orbital, the species contains exactly one unpaired electron, meaning $n = 1$. Substituting this value into the formula yields $\mu_S = \sqrt{1(1+2)} = \sqrt{3}$ BM. Given that $\sqrt{3} = 1.73$, the magnetic moment is $1.73$ BM, which can be expressed as $173 \times 10^{-2}$ BM. Therefore, the value needed to satisfy the expression is 173.

According to the Valence Shell Electron Pair Repulsion theory, the oxygen atom in an $\mathrm{H_2O}$ molecule is $sp^3$ hybridized due to the presence of two lone pairs and two sigma bonds, which normally implies a theoretical tetrahedral geometry with a bond angle of 109.5°. However, the central oxygen atom experiences distinct electrostatic forces because the repulsion between lone pairs is inherently stronger than the repulsion between bonding pairs. This enhanced repulsion pushes the $\mathrm{H-O-H}$ bond angle down to 104.5°, confirming that the difference in repulsion magnitudes is the correct explanation for this observed geometric distortion.

Determining the number of lone pairs on a central atom involves subtracting the number of electrons involved in bonding from the total number of valence electrons associated with that atom. In the case of ${SF}_{4}$, the central sulfur atom possesses six valence electrons and shares four of them with fluorine atoms, which results in one lone pair. The ${BF}_{4}^{-}$ ion features a central boron atom that utilizes all four available valence electrons for bonding, leaving zero lone pairs, whereas ${ClF}_{3}$ utilizes only three of chlorine's seven valence electrons, leaving two lone pairs behind. Arsenic in ${AsF}_{3}$ uses three of its five valence electrons for bonding, leaving one lone pair, while phosphorus in ${PCl}_{5}$ utilizes all five valence electrons for bonding, resulting in zero lone pairs.

For ${BrF}_{5}$, the central bromine atom uses five of its seven valence electrons to bond with fluorine, which leaves one lone pair, while xenon in ${XeF}_{4}$ utilizes only four of its eight valence electrons, creating two lone pairs. Finally, in ${SF}_{6}$, sulfur uses all six of its valence electrons for bonding, resulting in zero lone pairs. Counting the species that possess exactly two lone pairs reveals that only ${ClF}_{3}$ and ${XeF}_{4}$ satisfy this condition, leading to a final count of 2.