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

The dipole moment, denoted as $\mu$, is a vector quantity that represents the polarity of a chemical bond, and scientific convention mandates that the vector points from the atom with the partial positive charge toward the atom with the partial negative charge. Since the first statement incorrectly claims the arrow originates at the negative center and points toward the positive center, it is logically false. In contrast, the crossed arrow is the recognized symbol used to visually represent the net shift of electron density within a molecule toward the more electronegative atom, which validates the second statement.

Molecular Orbital theory determines bond order by calculating half the difference between electrons in bonding and antibonding orbitals, though counting total electrons provides an efficient shortcut for simple diatomic species. Species with 14 total electrons, such as $\mathrm{CO}$ and $\mathrm{NO}^{+}$, occupy only bonding orbitals to achieve a triple bond order of 3. In contrast, $\mathrm{O}_2^{2-}$ possesses 18 total electrons, which forces occupancy into high-energy antibonding orbitals that cancel out bonding contributions, resulting in a significantly lower bond order of 1.

The structure of ice is determined by a rigid, open-cage lattice that forms when water molecules transition into a crystalline state. Within this arrangement, the tetrahedral geometry of each $H_2O$ molecule dictates that it must form hydrogen bonds with exactly four neighboring water molecules. This specific coordination creates the expansive, low-density network that characterizes ice compared to its liquid phase.

The central chlorine atom in $\mathrm{ClF}_5$ possesses seven valence electrons, five of which form bonds with fluorine atoms, leaving one lone pair to create a total of six electron domains. According to VSEPR theory, this configuration forces the molecule into a square pyramidal geometry based on an octahedral electronic arrangement. Consequently, this compound manifests as a colourless liquid under standard room temperature conditions.

Molecules are classified as electron-deficient when the central atom possesses fewer than eight electrons in its valence shell, indicating an incomplete octet. Specifically, $\mathrm{B}_2\mathrm{H}_6$ and $\mathrm{BCl}_3$ feature boron atoms surrounded by only six electrons, identifying both as electron-deficient. Since $\mathrm{PH}_3$, $\mathrm{CCl}_4$, $\mathrm{NH}_3$, and $\mathrm{LiH}$ contain atoms with complete valence shells or stable duet configurations, exactly two molecules from the provided set are electron-deficient.

Calculating the total number of electrons in $\mathrm{CH}_4$, $\mathrm{NH}_4^{+}$, and $\mathrm{BH}_4^{-}$ reveals that each species contains 10 electrons, which establishes that they are isoelectronic. Because every central atom is bonded to four hydrogen atoms without any lone pairs, these species all share the same $sp^3$ hybridized tetrahedral structure.

Odd electron molecules contain a total number of valence electrons that prevents complete pairing, characteristic of species like $\mathrm{NO}$. An expanded octet occurs when the central atom is surrounded by more than eight electrons, a state achieved by $\mathrm{H_2SO_4}$ where the sulfur atom expands its valence shell. While molecules such as $\mathrm{BCl}_3$ are classified as even electron molecules and $\mathrm{SF}_6$ as an expanded octet, only the pair containing $\mathrm{NO}$ and $\mathrm{H_2SO_4}$ satisfies the condition of having one odd electron molecule and one expanded octet molecule.

The net dipole moment of a molecule is determined by its molecular geometry, where symmetric structures allow individual bond dipoles to cancel out, resulting in a net dipole moment of zero. For $\mathrm{BeF}_2$, $\mathrm{BF}_3$, and $\mathrm{CCl}_4$, the highly symmetric linear, trigonal planar, and tetrahedral arrangements cause all bond dipoles to negate one another, yielding a net value of zero. Conversely, $\mathrm{H}_2\mathrm{O}$ adopts a bent geometry, $\mathrm{NH}_3$ forms a trigonal pyramidal structure, and $\mathrm{HCl}$ exists as a polar diatomic molecule, all of which maintain an asymmetric charge distribution that produces a non-zero net dipole moment. Counting these specific cases reveals that 3 of the molecules possess a non-zero net dipole moment.

Paramagnetism in chemical species occurs due to the presence of unpaired electrons, while diamagnetic substances contain only paired electrons. Among the listed compounds, $\mathrm{KO_2}$, $\mathrm{NO_2}$, $\mathrm{ClO_2}$, and $\mathrm{NO}$ all contain unpaired electrons, making them paramagnetic. In contrast, $\mathrm{Na_2O}$, $\mathrm{N_2O}$, $\mathrm{SO_2}$, and $\mathrm{Cl_2O}$ possess completely paired electron configurations, classifying them as diamagnetic. This leaves a total of 4 paramagnetic oxides in the provided list.

The bond order of a species is a quantitative measure of its chemical bond strength, calculated using the relationship $BO = \frac{1}{2}[N_b - N_a]$, where $N_b$ represents the number of electrons in bonding molecular orbitals and $N_a$ denotes the number of electrons in antibonding molecular orbitals. Establishing the bond order for these ions requires determining their total electron count and their subsequent distribution within molecular orbitals.

For $\mathrm{C}_2^{2-}$, which contains 14 electrons, the molecular orbital filling pattern results in 10 bonding electrons and 4 antibonding electrons, yielding a bond order of $\frac{1}{2}(10 - 4) = 3$. The ion $\mathrm{N}_2^{2-}$ possesses 16 electrons, which occupy orbitals resulting in 10 bonding and 6 antibonding electrons, leading to a bond order of $\frac{1}{2}(10 - 6) = 2$. Finally, $\mathrm{O}_2^{2-}$ contains 18 electrons, which results in a configuration with 10 bonding and 8 antibonding electrons, producing a bond order of $\frac{1}{2}(10 - 8) = 1$. Comparing these calculated values, the sequence of bond orders in increasing order is $\mathrm{O}_2^{2-} < \mathrm{N}_2^{2-} < \mathrm{C}_2^{2-}$.

The hybridization of a central atom is determined by its steric number, calculated using the formula $Z = \frac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons of the central atom, $M$ is the number of monovalent atoms, $C$ is the cationic charge, and $A$ is the anionic charge. A molecule or ion exhibits $sp^3d^2$ hybridization when its steric number equals 6. For $PF_5$, $Z = \frac{1}{2}(5 + 5) = 5$, which corresponds to $sp^3d$ hybridization. For $BrF_5$, $Z = \frac{1}{2}(7 + 5) = 6$, which is $sp^3d^2$. For $PCl_3$, $Z = \frac{1}{2}(5 + 3) = 4$, which corresponds to $sp^3$ hybridization. For $SF_6$, $Z = \frac{1}{2}(6 + 6) = 6$, which is $sp^3d^2$. For $[ICl_4]^-$, $Z = \frac{1}{2}(7 + 4 + 1) = 6$, which is $sp^3d^2$. For $ClF_3$, $Z = \frac{1}{2}(7 + 3) = 5$, which is $sp^3d$. Finally, for $IF_5$, $Z = \frac{1}{2}(7 + 5) = 6$, which is $sp^3d^2$. Consequently, the four species $BrF_5$, $SF_6$, $[ICl_4]^-$, and $IF_5$ possess $sp^3d^2$ hybridization.

Molecular orbital theory allows us to determine the magnetic properties of a molecule by examining the occupancy of its orbitals; if every electron is paired, the species is diamagnetic, whereas the presence of any unpaired electrons indicates paramagnetism. For the nitrogen-based species, $\mathrm{N}_2$ possesses 14 electrons that completely fill all lower-energy molecular orbitals, resulting in a diamagnetic state with no unpaired electrons. Conversely, $\mathrm{N}_2^{+}$ with 13 electrons, $\mathrm{N}_2^{-}$ with 15 electrons, and $\mathrm{N}_2^{2-}$ with 16 electrons all possess unpaired electrons in their respective configurations, making them paramagnetic.

Evaluating the oxygen-based series, $\mathrm{O}_2$ holds 16 electrons, leaving 2 unpaired electrons in the antibonding $\pi^{*}$ orbitals, which makes it paramagnetic. Modifying this structure by removing an electron to form $\mathrm{O}_2^{+}$ results in 15 electrons and 1 unpaired electron, while adding an electron to form $\mathrm{O}_2^{-}$ results in 17 electrons and 1 unpaired electron, meaning both of these ions are also paramagnetic. Finally, $\mathrm{O}_2^{2-}$ contains 18 electrons, which effectively fill the antibonding orbitals completely and leave no unpaired electrons, establishing it as diamagnetic. Comparing all the species listed, only $\mathrm{N}_2$ and $\mathrm{O}_2^{2-}$ exhibit diamagnetism.

Molecular Orbital theory defines bond order as half the difference between the number of bonding and antibonding electrons, allowing us to characterize the strength of chemical bonds. Calculating this for the provided species, $\mathrm{CN}^{-}$, $\mathrm{NO}^{+}$, and $\mathrm{O}_2^{2+}$ each contain 14 electrons, which consistently results in a bond order of 3. Meanwhile, $\mathrm{O}_2$ has 16 electrons, corresponding to a bond order of 2, and $\mathrm{O}_2^{+}$ contains 15 electrons, yielding a bond order of 2.5. By comparing these values, we determine that three of the five species share an identical bond order.

A molecule exhibits a net dipole moment if its geometric arrangement is unsymmetrical, preventing the individual bond dipoles from canceling one another out. Among the listed substances, the highly symmetrical structures of $\mathrm{BF}_3$, $\mathrm{BeF}_2$, $\mathrm{SiF}_4$, $\mathrm{CCl}_4$, and $\mathrm{PF}_5$ result in a zero net dipole moment due to the perfect cancellation of bond polarities. In contrast, $\mathrm{NF}_3$, $\mathrm{CHCl}_3$, and $\mathrm{H}_2 \mathrm{S}$ lack this structural symmetry, ensuring that their bond dipoles do not cancel, thus yielding a non-zero net resultant dipole moment for these 3 molecules.

The hybridization of the central atom depends on its steric number, defined as the total count of sigma bonds and lone pairs. In $\mathrm{PF}_5$, the phosphorus atom is surrounded by five sigma bonds with no lone pairs, resulting in a steric number of five and $\mathrm{sp}^3 \mathrm{d}$ hybridization. By equating this configuration to the general form $\mathrm{sp}^x \mathrm{d}^y$, it becomes evident that the exponent of the $d$ orbital is 1, meaning $y=1$.

The square pyramidal structure is characteristic of interhalogen molecules with an $AX_5E$ configuration, specifically those containing five $\sigma$ bonds and one lone pair around the central halogen atom. Among the provided compounds, $\mathrm{BrF}_5$, $\mathrm{IF}_5$, and $\mathrm{ClF}_5$ possess this electronic arrangement and adopt the square pyramidal geometry. Since the remaining molecules, such as $\mathrm{ClF}_3$, $\mathrm{BrF}_3$, $\mathrm{I}_2 \mathrm{Cl}_6$, $\mathrm{IF}_7$, and $\mathrm{ClF}$, exhibit alternative structures like T-shaped, dimeric, pentagonal bipyramidal, or linear forms, there are exactly three instances of the square pyramidal arrangement.

In the $\mathrm{SF}_4$ molecule, the sulfur atom exhibits $\mathrm{sp}^3 \mathrm{d}$ hybridization and adopts a trigonal bipyramidal electron geometry where the lone pair occupies an equatorial position to minimize electronic repulsions. Because this arrangement places the lone pair at $90^{\circ}$ relative to the two axial bond pairs, the structure results in exactly two lone pair-bond pair repulsions at this angle.

The phosphorus pentachloride molecule adopts a trigonal bipyramidal geometry resulting from $\mathrm{sp}^3 d$ hybridization of the central phosphorus atom. Within this arrangement, the three equatorial chlorine atoms reside in a single plane, while the remaining two occupy axial positions. Due to the repulsion from the equatorial bonding pairs, the axial bonds experience greater bond length and are consequently weaker than the equatorial bonds, making the claim that they are stronger inherently false.

Bond order is calculated as $\text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons})$, meaning it represents the net stability provided by the electronic configuration. Electrons located in antibonding molecular orbitals destabilize the molecule, so removing one from such an orbital reduces the total count of antibonding electrons, which mathematically increases the bond order and strengthens the bond. Among the given species, both NO and $O_2$ have their highest energy electrons occupying antibonding molecular orbitals. Therefore, the removal of an electron from either NO or $O_2$ leads to an increased bond order and a stronger resulting chemical bond.

Molecular orbital theory determines bond order using the expression $BO = \frac{1}{2}(N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. Because electrons added beyond the $\mathrm{O}_2$ configuration occupy antibonding orbitals, the bond order systematically decreases as the electron count increases. Calculating these values for the given species yields bond orders of 2.5 for $\mathrm{O}_2^+$, 2.0 for $\mathrm{O}_2$, 1.5 for $\mathrm{O}_2^-$, and 1.0 for $\mathrm{O}_2^{2-}$. Based on these values, the progression for increasing bond order is $\mathrm{O}_2^{2-} < \mathrm{O}_2^{-} < \mathrm{O}_2 < \mathrm{O}_2^{+}$.