Chemical Bonding And Molecular Structure – NEET UG ChemistryPractice Questions & PYQs

The bond order for diatomic species is calculated according to molecular orbital theory using the expression $B.O. = \frac{N_b - N_a}{2}$, where $N_b$ denotes the number of electrons in bonding orbitals and $N_a$ denotes the number of electrons in antibonding orbitals. For the neutral nitrogen molecule $\mathrm{N}_2$, which contains 10 bonding and 4 antibonding electrons, the bond order is $\frac{10-4}{2} = 3$. Adding an electron to form the nitrogen anion $\mathrm{N}_2^-$ increases the antibonding count to 5, resulting in a bond order of $\frac{10-5}{2} = 2.5$, while the addition of a second electron to form the nitrogen dianion $\mathrm{N}_2^{2-}$ further increases the antibonding electrons to 6, giving a bond order of $\frac{10-6}{2} = 2$. By comparing these calculated values, the nitrogen species arranged in terms of increasing bond order is $\mathrm{N}_2^{2-} < \mathrm{N}_2^- < \mathrm{N}_2$.

According to molecular orbital theory, the bond order of a species is determined by calculating half the difference between the number of bonding electrons and antibonding electrons. For $\mathrm{He}_2^{+}$ with 3 electrons, this calculation gives a bond order of $\frac{1}{2}(2-1) = 0.5$. The $\mathrm{O}_2^{-}$ species has 10 bonding and 7 antibonding electrons for a bond order of $\frac{1}{2}(10-7) = 1.5$, whereas $\mathrm{NO}$ possesses 10 bonding and 5 antibonding electrons for a bond order of $\frac{1}{2}(10-5) = 2.5$. Finally, $\mathrm{C}_2^{2-}$ has 10 bonding and 4 antibonding electrons, yielding a bond order of $\frac{1}{2}(10-4) = 3$, which establishes the increasing order as $\mathrm{He}_2^{+} < \mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-}$.

The bond angles in these species are primarily determined by the hybridisation of the central nitrogen atom and the degree of repulsion exerted by lone pairs versus bonding electrons on the surrounding oxygen atoms. Specifically, $NO_2^{+}$ adopts an $sp$ hybridised linear structure with no lone pairs, resulting in a bond angle of $180^{\circ}$. In contrast, both $NO_2$ and $NO_2^{-}$ feature $sp^2$ hybridisation, but the presence of an additional lone pair in $NO_2^{-}$ compared to the single unpaired electron in $NO_2$ exerts greater repulsive force, which further compresses the O-N-O bond angle and establishes it as the smallest of the three species.

Bond order represents the effective number of chemical bonds between atoms and is determined by calculating half the difference between the number of bonding and antibonding electrons. Applying this principle to the given species reveals that $He_2^{+}$ has a bond order of 0.5, $O_2^{-}$ possesses a bond order of 1.5, $NO$ stands at 2.5, and $C_2^{2-}$ reaches a value of 3.0. Sorting these magnitudes from lowest to highest produces the mathematical sequence $He_2^{+} < O_2^{-} < NO < C_2^{2-}$.

Species are isostructural when they share the same hybridization state and molecular geometry, which are defined by the arrangement of bonding pairs and lone pairs around the central atom. Calculating the steric number for $\mathrm{BrO_3^-}$ yields $H = \frac{1}{2}(7 + 0 - 0 + 1) = 4$, indicating $sp^3$ hybridization with one lone pair, resulting in a trigonal pyramidal structure. The central atom in $\mathrm{XeO_3}$ follows the same pattern, where $H = \frac{1}{2}(8 + 0 - 0 + 0) = 4$, which also yields $sp^3$ hybridization and one lone pair, confirming that both species possess an identical trigonal pyramidal geometry.

Bond length is inversely proportional to bond order, meaning that as the number of shared electron pairs between atoms increases, the distance between them effectively decreases. Carbon monoxide ($CO$) contains a triple bond and thus exhibits the shortest bond length, while carbon dioxide ($CO_2$) features double bonds and possesses an intermediate bond distance. The carbonate ion ($CO^{2-}_3$) exists as a resonance hybrid with significant single-bond character, resulting in a fractional bond order that gives it the greatest bond length of the group.

Molecules and ions are considered isostructural when they possess identical hybridization and molecular geometry, both of which are determined by the total number of bonding and lone pairs around the central atom. The species $SiCl_4$, $NH^{+}_4$, $SO^{2-}_4$, and $PO^{3-}_4$ all feature a central atom with $sp^3$ hybridization and four bonding pairs, resulting in a symmetric tetrahedral arrangement. In contrast, $SCl_4$ contains a central sulfur atom with four bonding pairs and one additional lone pair, which causes the structure to adopt a distorted tetrahedral or see-saw geometry.

The molecular geometry of a chemical species is determined by the hybridization of its central atom and the presence of non-bonding lone pair electrons. In the case of $NO_2^{+}$, the nitrogen atom undergoes $sp$ hybridization with no lone pairs, which results in a linear structure. In contrast, $O_3$, $NO_2^{-}$, and $SO_2$ all feature lone pairs on their central atoms, which exert repulsive forces that distort the bond angles and force the molecules into a bent or angular shape.

Molecules with the $AB_5$ configuration can theoretically exist in three distinct geometries: planar pentagonal, square pyramidal, and trigonal bipyramidal. Among these, the trigonal bipyramidal arrangement is the most probable configuration because it effectively minimizes the bond pair-bond pair repulsions between the substituents. Consequently, the assertion that every $AB_5$ molecule inherently possesses a square pyramidal structure is an inaccurate statement, as this geometry is not the universal or most stable outcome for such species.

Surface tension is fundamentally governed by the strength of intermolecular attractive forces, with hydrogen bonding acting as a key contributor to these cohesive interactions. Among the given substances, ${H}_{2} {O}$ possesses the most extensive and potent network of hydrogen bonds compared to ${C}_{2} {H}_{5} {OH}$ and ${CH}_{3} {OH}$, whereas ${C}_{6} {H}_{6}$ does not exhibit such strong bonding capabilities. This superior intermolecular attraction results in ${H}_{2} {O}$ demonstrating the maximum surface tension of the group.

Trigonal planar geometry is characterized by a central atom bonded to three surrounding atoms with a steric number of three and zero lone pairs, which corresponds to $sp^2$ hybridization and bond angles of 120 degrees. The molecule $BF_3$ satisfies these conditions because the central boron atom possesses three valence electrons, all of which are utilized to form three covalent bonds with fluorine atoms, leaving no lone pairs. Conversely, $NH_3$ and $PCl_3$ have a lone pair that results in a trigonal pyramidal shape, while $IF_3$ contains two lone pairs on the central iodine atom, leading to a T-shaped geometry.

Bond length is fundamentally determined by bond order, as higher bond orders involve more shared electrons that pull the bonded nuclei closer together. In molecular oxygen ($O_2$), the presence of a strong double bond results in the shortest separation of 1.21 Å, while the resonance structure of ozone ($O_3$) leads to an intermediate bond length of 1.28 Å. Hydrogen peroxide ($H_2O_2$) contains a single covalent bond, which results in the longest O-O bond length of 1.48 Å. Consequently, the sequence of increasing bond length is $O_2 < O_3 < H_2O_2$.

Isostructural species are defined by having the same hybridization and molecular geometry, which depends on the arrangement of electron pairs around the central atom. While $SiF_4$ utilizes $sp^3$ hybridization to adopt a symmetric tetrahedral structure, $SF_4$ involves $sp^3d$ hybridization and includes a lone pair, leading to a distorted tetrahedral or see-saw geometry that clearly differentiates its structural configuration from that of $SiF_4$.

The net dipole moment of a molecule relies on its spatial geometry, which determines whether the individual bond dipoles reinforce one another or cancel out. $\mathrm{H}_2\mathrm{O}$ features an angular geometry with a bond angle of 105 degrees, preventing the bond dipoles from canceling and rendering the molecule polar. Conversely, $\mathrm{BeF}_2$ adopts a linear configuration that aligns the bond dipoles in opposite directions, causing them to neutralize each other entirely and resulting in no net dipole moment.

A regular octahedral molecule features a central atom $M$ surrounded by six ligands $X$ oriented symmetrically along the positive and negative directions of the $x$, $y$, and $z$ axes. Since each axis connects two ligands directly opposite one another through the central atom, these pairings form linear $X-M-X$ angles of $180^{\circ}$. Counting these collinear pairs across the three spatial axes reveals exactly three such bond angles.

VSEPR theory explains that molecules adopt specific shapes to minimize electron repulsion, prioritizing the elimination of the most intense interactions. In the $BrF_{3}$ molecule, the central bromine atom possesses two lone pairs that are placed in the equatorial positions of the trigonal bipyramid. This specific configuration is adopted because it results in $(lp-lp)$ repulsion $= 0$, which is the primary factor in determining the molecular structure.

When placed in these equatorial positions, the remaining interactions consist of $(lp-bp)$ repulsion $= 4$ and $(bp-bp)$ repulsion $= 2$. The equatorial placement is critical because it ensures the strongest interaction, $(lp-lp)$ repulsion, is completely eliminated, thereby stabilizing the bent T-shaped geometry.

Sigma bonds are formed by the direct head-on overlap of orbitals, which provides greater stability compared to the side-by-side overlap that characterizes a pi bond. To determine the energy of a pi bond, we subtract the energy of a single $\mathrm{C}-\mathrm{C}$ bond, which is 347 kJ/mol, from the total energy of a $\mathrm{C}=\mathrm{C}$ bond, given as 619 kJ/mol. This calculation reveals that the pi bond energy is actually 272 kJ/mol, making the statement identifying the energies as 264 kJ/mol and 347 kJ/mol factually incorrect.

Chemical structures containing central atoms with accessible vacant $d$ orbitals can participate in $\pi$ bonding through the overlap of these orbitals with the filled $p$ orbitals of peripheral atoms, a process known as $p\pi-d\pi$ bonding. In the sulphite ion, $\mathrm{SO}_{3}^{2-}$, the central sulphur atom is $sp^3$ hybridized, with three hybrid orbitals forming $\sigma$ bonds and one lone pair occupying the fourth tetrahedral position. Because sulphur possesses vacant $3d$ orbitals that remain excluded from hybridization, it can facilitate $\pi$ interactions with oxygen, a feature impossible for the second-period elements in the other ions that lack such $d$ orbitals.

Isoelectronic species are defined as chemical entities that possess an identical total count of electrons. In a molecule of $\mathrm{CO}$, carbon provides 6 electrons and oxygen provides 8, resulting in a total of 14 electrons. Correspondingly, the $\mathrm{CN}^{-}$ ion includes 6 electrons from carbon, 7 from nitrogen, and 1 additional electron due to the negative charge, which also yields a sum of 14 electrons. Since both species contain the same number of electrons and share a similar molecular orbital configuration, they are considered isoelectronic.

The central nitrogen atom typically possesses 5 valence electrons, but the negative charge on the $\mathrm{NO}_{3}^{-}$ ion contributes one additional electron, bringing the total involved electrons to 6. In the structure of the nitrate ion, the nitrogen atom forms one double bond and two single bonds to coordinate with the three oxygen atoms. This arrangement results in a total of 4 shared bond pairs around the nitrogen center, leaving 0 lone pairs remaining on the atom.