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

The dipole moment of a diatomic molecule is defined as the product of the partial charge on each atom and the bond distance, represented by the formula $\mu = q \times d$. Using the provided dipole moment of $1.2\ \mathrm{D}$ and the bond length of $1\ \text{Å}$, we first express the dipole moment in consistent units, where $1.2\ \mathrm{D}$ is equivalent to $1.2 \times 10^{-10}\ \mathrm{esu}\ \text{Å}$. Substituting these values into the formula yields the equation $1.2 \times 10^{-10}\ \mathrm{esu}\ \text{Å} = q \times 1\ \text{Å}$, which simplifies to determine the charge magnitude as $q = 1.2 \times 10^{-10}\ \mathrm{esu}$.

When atomic orbitals overlap to create molecular orbitals, the total number of molecular orbitals generated must equal the combined number of original atomic orbitals. Considering the 2s orbitals, the overlap produces one bonding and one antibonding molecular orbital, represented as $\sigma^*_{2s}$. For the 2p orbitals, the three atomic orbitals from each atom interact to generate six molecular orbitals, consisting of three bonding and three antibonding orbitals, specifically $\sigma^*_{2p_z}$, $\pi^*_{2p_x}$, and $\pi^*_{2p_y}$. Adding the single antibonding orbital from the 2s level to the three antibonding orbitals from the 2p level results in a total of 4 antibonding molecular orbitals.

The central nitrogen atom in $\mathrm{NO}_2^{-}$ adheres to the octet rule to maintain stability, which means it is surrounded by a total of 8 valence electrons through a combination of bonding and lone pair interactions. When constructing the Lewis structure, the nitrogen atom participates in one double bond and one single bond with the surrounding oxygen atoms. After these bonds are accounted for, one lone pair remains on the nitrogen atom, which consists of 2 non-bonding valence electrons.

Hydrogen bonding arises from the electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative element and another electronegative atom nearby. In compounds like $\mathrm{H}_2\mathrm{O}$, $\mathrm{NH}_3$, and $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$, this interaction acts between separate, distinct molecules, defining them as instances of intermolecular hydrogen bonding. In contrast, the remaining choice exhibits intramolecular hydrogen bonding because its structure allows the hydrogen atom to form a bridge between two electronegative sites within the very same molecule rather than between different ones.

A molecule's net dipole moment arises from the vector sum of individual bond dipoles and any lone pair contributions, which is determined by the molecule's specific geometry. Symmetrical structures such as $\mathrm{CH}_4$ and $\mathrm{PF}_5$ result in a net dipole moment of $\mu = 0$ because the bond vectors cancel out perfectly. While both $\mathrm{NH}_3$ and $\mathrm{NF}_3$ possess lone pairs that prevent complete cancellation, the bond dipoles in $\mathrm{NH}_3$ align with the direction of the lone pair to reinforce the total dipole, whereas the highly electronegative fluorine atoms in $\mathrm{NF}_3$ create bond dipoles that oppose the lone pair, resulting in a significantly lower overall value.

According to VSEPR theory, the geometry of a molecule depends on the total count of electron domains, comprising both bonding pairs and lone pairs surrounding the central atom. In the case of $\mathrm{BrF}_5$, the central bromine atom holds seven valence electrons, five of which are involved in bonding with fluorine atoms, while the remaining electrons form one lone pair; this arrangement forces the molecule into a square pyramidal shape derived from an octahedral electronic distribution. Conversely, both $\mathrm{PCl}_5$ and $\mathrm{PF}_5$ feature five bonding pairs with no lone pairs, resulting in a trigonal bipyramidal structure, while $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ adopts a square planar geometry.

A carbocation features a positively charged carbon atom bonded to three distinct atoms or groups, which results in the central carbon undergoing $sp^2$ hybridization. Because the carbon atom is $sp^2$ hybridized and holds an empty p-orbital, the three sigma bonds arrange themselves in space to minimize electron repulsion, establishing a trigonal planar geometry with bond angles of approximately $120^\circ$.

The ionic character of a bond is directly proportional to the difference in electronegativity between the bonded atoms, as a larger gap in their attraction for electrons leads to more pronounced bond polarity. Because $\mathrm{N}_2$ consists of two identical nitrogen atoms, the electronegativity difference is zero, making it purely non-polar covalent and the least ionic member of the group. As we progress, $\mathrm{SO}_2$ and $\mathrm{ClF}_3$ exhibit increasing polar covalent character due to the variations in electronegativity between their respective non-metal components. Finally, $\mathrm{K}_2\mathrm{O}$ and $\mathrm{LiF}$ represent ionic bonds between metals and non-metals, with the bond in $\mathrm{LiF}$ displaying the greatest ionic character due to the extreme electronegativity difference between lithium and fluorine, resulting in the final arrangement of $\mathrm{N}_2 < \mathrm{SO}_2 < \mathrm{ClF}_3 < \mathrm{K}_2\mathrm{O} < \mathrm{LiF}$.

The bond angle in these molecules is primarily determined by their molecular geometry and the number of lone pairs on the central atom, as lone pair-bond pair repulsion compresses the angles more effectively than bond pair-bond pair interactions. $\mathrm{BF}_3$ possesses a trigonal planar structure with no lone pairs, resulting in the largest bond angle among the group. Conversely, $\mathrm{PF}_3$ adopts a pyramidal geometry with one lone pair, while $\mathrm{C\ell F}_3$ experiences the greatest distortion due to the presence of two lone pairs, establishing the overall increasing order as $\mathrm{C\ell F}_3 < \mathrm{PF}_3 < \mathrm{BF}_3$.

Hydrogen bonding occurs exclusively when a hydrogen atom is covalently bonded to a highly electronegative atom such as Nitrogen, Oxygen, or Fluorine, creating a strong dipole that enables electrostatic interaction with the lone pair of an electronegative atom in an adjacent molecule. This phenomenon is strictly dependent on the presence of $N-H$, $O-H$, or $F-H$ bonds within the molecular structure, which facilitates strong intermolecular forces significantly greater than standard dipole-dipole interactions. By identifying molecules that contain these specific polar bonds from the provided set, one determines that exactly five species possess the necessary electronegativity difference and hydrogen atom arrangement to exhibit hydrogen bonding.

Fajans' Rule dictates that ionic character is inversely related to the polarizing power of the cation, which measures its ability to deform the electron cloud of the anion. Because the silver ion ($\mathrm{Ag}^{+}$) possesses a pseudo-noble gas configuration with 18 electrons in its valence shell, it exerts a significantly high polarizing power that pulls the chloride ion's electron density toward itself, inducing greater covalent character. This results in $\mathrm{AgCl}$ being the least ionic compound, with the ionic character increasing across the sequence from $\mathrm{AgCl}$ to $\mathrm{CoCl}_2$ and $\mathrm{BaCl}_2$, before reaching its maximum in $\mathrm{KCl}$.

Bond order is determined by calculating the difference between the number of bonding electrons and antibonding electrons, then dividing by two, using the formula $\frac{N_b - N_a}{2}$. Applying this principle to the given set of molecules reveals their specific bond orders based on their electron configurations. For $\mathrm{C}_2$, which contains 12 electrons, the calculation is $\frac{8-4}{2} = 2$. Similarly, $\mathrm{O}_2$ with 16 electrons results in a bond order of $\frac{10-6}{2} = 2$.

The remaining molecules yield different values: $\mathrm{Be}_2$ with 8 electrons results in $\frac{4-4}{2} = 0$, $\mathrm{Li}_2$ with 6 electrons gives $\frac{4-2}{2} = 1$, and $\mathrm{Ne}_2$ with 20 electrons results in $\frac{10-10}{2} = 0$. Furthermore, $\mathrm{N}_2$ with 14 electrons results in $\frac{10-4}{2} = 3$, while $\mathrm{He}_2$ with 4 electrons results in $\frac{2-2}{2} = 0$. Comparing these results shows that only two molecules, $\mathrm{C}_2$ and $\mathrm{O}_2$, possess a bond order of 2.

Molecular orbital theory establishes that the bond order and magnetic behavior of a species are primarily determined by its total electron count, meaning isoelectronic entities share these identical properties. The acetylide ion consists of 14 electrons, resulting in a bond order of 3 and a diamagnetic state because all electrons are paired. Since $\mathrm{NO}^{+}$ also contains exactly 14 electrons, it shares the same bond order and lack of unpaired electrons as the acetylide ion.

The geometry of molecular species is determined by VSEPR theory, which states that the presence of lone pairs on a central atom causes electron-pair repulsion, typically distorting a structure into a bent or angular shape, whereas a lack of lone pairs on the central atom generally results in a linear arrangement. Examining the given species, both the azide ion $\mathrm{N}_3^{-}$ and the triiodide ion $\mathrm{I}_3^{-}$ adopt linear geometries due to their specific orbital configurations. In contrast, the nitrite ion $\mathrm{NO}_2^{-}$, ozone $\mathrm{O}_3$, and sulfur dioxide $\mathrm{SO}_2$ each contain a central atom possessing at least one lone pair, which exerts sufficient repulsion to force a bent molecular shape. Accounting for these structural differences confirms that there are exactly 3 bent-shaped molecules in the set.

To determine the total number of lone pairs in an ozone ($O_3$) molecule, we analyze the distribution of valence electrons across its three oxygen atoms based on its resonance-stabilized structure. In this arrangement, the central oxygen atom is double-bonded to one terminal oxygen atom and single-bonded to the other. The terminal oxygen atom involved in the double bond retains two lone pairs, the terminal oxygen atom involved in the single bond carries three lone pairs, and the central oxygen atom contributes one lone pair. Summing these individual contributions results in a total of 6 lone pairs for the molecule.

Molecular Orbital Theory provides the framework to determine how changing the electron count in a molecule alters its magnetic behavior and bond strength. As the neutral molecule $\mathrm{NO}$ converts to the positive ion $\mathrm{NO}^{+}$, its bond order increases from 2.5 to 3, causing the paramagnetic nature of the original molecule to become diamagnetic.

The molecular geometry of these compounds is determined using Valence Shell Electron Pair Repulsion (VSEPR) theory, which accounts for the arrangement of valence shell electron pairs around the central atom to minimize repulsion. $XeF_4$ features four bond pairs and two lone pairs, resulting in a square planar geometry (A-II), while $SF_4$ contains four bond pairs and one lone pair, which adopts a see-saw structure (B-I). Consequently, $NH_4^+$ exhibits a tetrahedral geometry due to four bond pairs and no lone pairs (C-IV), whereas $BrF_3$ has three bond pairs and two lone pairs, characterizing it as bent T-shaped (D-III).

Bent's rule establishes that more electronegative atoms prefer to be attached to hybrid orbitals with lower s-character, which conversely forces the remaining bonds to incorporate higher s-character. Because fluorine is more electronegative than hydrogen, the oxygen atoms in $\mathrm{F}_2\mathrm{O}_2$ direct more s-character into the $\mathrm{O}-\mathrm{O}$ bond compared to the oxygen atoms in $\mathrm{H}_2\mathrm{O}_2$; since higher s-character leads to shorter, stronger bonds, the $\mathrm{O}-\mathrm{O}$ bond in $\mathrm{F}_2\mathrm{O}_2$ is shorter, meaning the $\mathrm{O}-\mathrm{O}$ bond in $\mathrm{H}_2\mathrm{O}_2$ is longer. Furthermore, the $\mathrm{O}-\mathrm{H}$ bond is physically shorter than the $\mathrm{O}-\mathrm{F}$ bond because the hydrogen atom has a significantly smaller atomic radius than the fluorine atom.

The number of lone pairs on the central Xenon atom is determined by the formula $\frac{1}{2}(V - B - C)$, where $V$ is the valence electrons of the central atom ($8$ for Xenon), $B$ is the number of bonding electrons, and $C$ is the net ionic charge. For $\mathrm{XeF}_6$, the central atom forms $6$ bonds, leaving $8 - 6 = 2$ electrons, which constitutes $1$ lone pair. In $\mathrm{XeOF}_4$, Xenon forms $4$ bonds with fluorine and $2$ bonds with oxygen, leaving $8 - (4 + 2) = 2$ electrons, which corresponds to $1$ lone pair. For $\mathrm{XeO}_2\mathrm{F}_2$, Xenon forms $2$ bonds with fluorine and $4$ bonds with oxygen, leaving $8 - (2 + 4) = 2$ electrons, which also constitutes $1$ lone pair. Finally, for $\mathrm{XeF}_5^{+}$, the positive charge reduces the available valence electrons to $7$, and forming $5$ bonds leaves $7 - 5 = 2$ electrons, resulting in $1$ lone pair.

Although bond dissociation energy typically decreases as atomic size increases down the halogen group, fluorine behaves anomalously due to its extremely small atomic radius. The close proximity of the atoms in the fluorine molecule leads to strong interelectronic repulsions between the lone pairs of electrons, which significantly weakens the $\mathrm{F}-\mathrm{F}$ bond relative to the $\mathrm{Cl}-\mathrm{Cl}$ bond. As a result, the trend for bond dissociation energy is $\mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{F}_2 > \mathrm{I}_2$, making the chlorine molecule the one with the highest energy.