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

Bond length in carbon compounds is inversely related to the bond order, meaning that as the number of bonds between two atoms increases, the distance between their nuclei decreases. While ethane features a single bond, ethene involves a double bond, and benzene possesses resonance-stabilized character, the ethyne molecule is defined by a strong triple bond, represented as $\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$. This triple bond exerts the greatest attractive force between the carbon atoms, pulling them closer together than in any of the other structures provided.

Determining the electronic configuration of the neon dimer involves applying the Aufbau principle to distribute its total of 20 electrons across molecular orbitals of increasing energy. Following the sequence $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \sigma 2p_z^2 < (\pi 2p_x^2 = \pi 2p_y^2) < (\pi^* 2p_x^2 = \pi^* 2p_y^2) < \sigma^* 2p_z^2$, we find that the final electrons occupy the antibonding orbital at the top of the energy diagram. This terminal level, denoted as $\sigma_{2p}^*$, represents the highest occupied molecular orbital for the molecule.

Paramagnetism occurs exclusively in chemical species that possess one or more unpaired electrons, as these electrons interact with external magnetic fields. In the case of nitric oxide, denoted as $\overset{\text{..}}{\mathrm{N}}=\mathrm{O}$, the molecular configuration results in the presence of an unpaired electron, which is the defining characteristic behind its paramagnetic behavior. Conversely, the other listed species achieve complete electron pairing within their structures, which renders them diamagnetic.

The bond order of a diatomic molecule is defined as half the difference between the total number of electrons in bonding molecular orbitals, denoted as $N_{b}$, and the number of electrons in antibonding molecular orbitals, denoted as $N_{a}$. According to the electronic configuration provided for the nitrogen molecule, there are 10 bonding electrons and 4 antibonding electrons. Using the bond order formula
$\text{Bond order} = \frac{1}{2}(N_{b} - N_{a})$
we find that
$\text{Bond order} = \frac{1}{2}(10 - 4) = 3$
which characterizes the stable triple bond present in the nitrogen molecule.

The central chlorine atom in $\mathrm{ClF}_3$ possesses seven valence electrons, which combine with the three electrons from the bonded fluorine atoms to create five total electron pairs. According to VSEPR theory, these pairs arrange themselves in a trigonal bipyramidal electronic geometry, where the two lone pairs on the chlorine atom occupy equatorial positions to minimize repulsion. This specific configuration forces the three fluorine atoms into a characteristic T-shaped molecular structure.

Molecular Orbital Theory offers the clearest insight into the electronic configuration of diatomic molecules by tracking how electron removal affects magnetic character and bond stability. In ${N}_{2},$ all electrons are paired, resulting in a diamagnetic species, but ionizing it to ${N}_{2}^{+}$ leaves a single unpaired electron in the $\sigma 2p_z$ orbital, which means it must be paramagnetic rather than diamagnetic. Conversely, neutral ${O}_{2}$ possesses two unpaired electrons in its antibonding $\pi^*$ orbitals, meaning that forming ${O}_{2}^{+}$ removes one of these, thereby decreasing the total paramagnetism. Furthermore, since the electron removed for ${N}_{2}^{+}$ comes from a bonding orbital, the bond order drops from 3 to 2.5 and the bond weakens, while for ${O}_{2}^{+},$ removing an electron from an antibonding orbital increases the bond order from 2 to 2.5.

Formal charge represents the bookkeeping of valence electrons assigned to an atom in a molecule, calculated as $FC = \text{Valence electrons} - (\text{Lone pair electrons} + \frac{1}{2} \times \text{Bonding electrons})$. For the terminal oxygen 2, which participates in a double bond and retains 2 lone pairs, the formal charge is $6 - (4 + \frac{1}{2} \times 4) = 0$. The central oxygen 1 engages in 3 total bonds and holds 1 lone pair, resulting in a formal charge of $6 - (2 + \frac{1}{2} \times 6) = +1$. The final terminal oxygen 3 possesses a single bond and 3 lone pairs, leading to a formal charge of $6 - (6 + \frac{1}{2} \times 2) = -1$. Following this calculation, the formal charges for oxygen atoms 2, 1, and 3 are 0, +1, and -1, respectively.

Molecular geometry is determined by the hybridization of the central atom and the specific arrangement of its bonding and non-bonding electron pairs as predicted by VSEPR theory. $\mathrm{PCl}_5$ utilizes $sp^3d$ hybridization to adopt a trigonal bipyramidal structure, while $\mathrm{BF}_4^{-}$ exhibits a tetrahedral geometry consistent with $sp^3$ hybridization and an absence of lone pairs. $\mathrm{BrF}_5$ undergoes $sp^3d^2$ hybridization with one lone pair, resulting in a square pyramidal shape, and $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ assumes a square planar configuration driven by $dsp^2$ hybridization under the influence of strong cyanide ligands.

The number of anti-bonding electron pairs in the $\mathrm{O}_2^{2-}$ molecular ion is determined by calculating the total electron count as 18, derived from two oxygen atoms and the ionic charge. Distributing these electrons into molecular orbitals following the Aufbau principle leads to the configuration $(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\sigma 2p_z)^{2} (\pi 2p_x)^{2} (\pi 2p_y)^{2} (\pi^{*} 2p_x)^{2} (\pi^{*} 2p_y)^{2}$. Identifying the anti-bonding orbitals, which are designated with an asterisk as $\sigma^{*} 1s$, $\sigma^{*} 2s$, $\pi^{*} 2p_x$, and $\pi^{*} 2p_y$, reveals a total of 8 anti-bonding electrons. Because these electrons occupy orbitals in pairs, the molecule contains 4 anti-bonding electron pairs.

Resonance in the $\mathrm{PO}_{4}^{3-}$ ion causes the negative charge to be delocalized across all four oxygen atoms while also averaging the bond character across the structure. Because the total charge of -3 is distributed equally among the four oxygen atoms, the formal charge on each is calculated as $-\frac{3}{4}$, resulting in -0.75. To determine the bond order, we recognize that within the 13 valence electrons, accounting for the -3 charge leaves 10 electrons involved in bond formation, which represents 5 total bonds shared between the central phosphorus and the four oxygen atoms. By distributing these 5 bonds across the 4 possible P-O sites, the ratio $\frac{5}{4}$ yields an average bond order of 1.25.

Hydrogen bond strength is directly proportional to the electronegativity of the atom attached to the hydrogen, as this determines the magnitude of the electrostatic attraction between dipole ends. Fluorine is the most electronegative element, which results in the $HF$ bond having the highest polarity among the choices provided. This extreme dipole character facilitates a much stronger interaction between the molecules compared to the hydrogen bonding found in $H_2O$, $NH_3$, or $H_2S$.

Hybridization is determined by counting the sigma bonds and lone pairs around a central atom, where a sum of three indicates $sp^{2}$ hybridization. In the $C_2H_4$ molecule, each carbon atom forms three sigma bonds-two with hydrogen atoms and one with the adjacent carbon atom-which characterizes the $sp^{2}$ state. By contrast, $BeCl_2$ and $C_2H_2$ possess only two electron domains leading to $sp$ hybridization, while the single bonds in $C_2H_6$ create four electron domains around each carbon atom, signifying $sp^{3}$ hybridization.

A perfectly symmetrical tetrahedral geometry with bond angles of exactly 109° 28′ requires a central atom to be bonded to four identical substituents. Within $CCl_4$, the central carbon atom is surrounded by four equivalent chlorine atoms, forming a regular tetrahedron that distributes the bond angles uniformly. Because the other listed compounds, such as $CH_3Cl$, $CHI_3$, and $CHCl_3$, possess a mixture of hydrogen and halogen atoms, their structures undergo distortion that forces their bond angles to deviate from the ideal tetrahedral value.

Chemical bond strength is determined by the atomic size of the bonded atoms, as smaller atoms can approach each other more closely, leading to a shorter and stronger bond that requires a greater $E_{\text{diss}}$ to overcome. Therefore, the magnitude of the dissociation energy serves as an indirect measure of atomic radius, where a higher energy value indicates a smaller atom. Among the given bonds, C-D has the highest energy of 485 kJ/mol, indicating that element D possesses the smallest atomic size.

To identify the bonding characteristics of these molecules, we evaluate the electron structure of each compound by counting the sigma ($\sigma$) bonds, pi ($\pi$) bonds, and lone pairs present. Ethene ($\mathrm{C}_2\mathrm{H}_4$) features a carbon-carbon double bond, which contributes four $\mathrm{C-H}$ sigma bonds and one $\mathrm{C-C}$ sigma bond plus one pi bond, resulting in five sigma bonds and one pi bond that matches description IV. Ethyne ($\mathrm{C}_2\mathrm{H}_2$) contains a carbon-carbon triple bond, creating two $\mathrm{C-H}$ sigma bonds and one $\mathrm{C-C}$ sigma bond along with two pi bonds, matching description I. Methane ($\mathrm{CH}_4$) consists of four carbon-hydrogen single bonds, establishing a total of four sigma bonds as described in III. Lastly, ammonia ($\mathrm{NH}_3$) has three nitrogen-hydrogen single bonds and one lone pair on the nitrogen atom, fitting description II.

The strength of intermolecular and intramolecular interactions is determined by the nature of the attractive forces involved, which range from weak physical associations to robust chemical bonds. Among these, van der Waals' forces are the weakest, followed by hydrogen bonding, which provides a more significant attraction. Dipole-dipole interactions exhibit higher strength than hydrogen bonding in this specific classification, while the covalent bond remains the strongest because it involves the direct sharing of electron pairs between atoms.

Carbon bond length is inversely related to bond order, meaning that as the number of shared electron pairs increases, the attraction between atoms strengthens and the bond distance shortens. Alkanes such as propane and butane consist of single bonds measuring 1.54 Å, and the alkene propene features a double bond at 1.34 Å. Propyne contains a carbon-carbon triple bond, which reaches the minimum length of 1.20 Å among the choices because the higher bond order pulls the atoms closest together.

Molecular polarity is dictated by the spatial arrangement of atoms, where a symmetric geometry can result in the cancellation of bond dipoles. In the case of $CS_2$, the atoms are arranged in a linear configuration, represented as $S=C=S$, which causes the bond dipoles to point in exactly opposite directions. Since these equal and opposing dipoles neutralize each other perfectly, the molecule lacks a permanent dipole moment.

Molecular Orbital Theory determines the bond order of a molecule by calculating half the difference between the number of bonding and antibonding electrons, which indicates the relative strength of the bond. When we examine the oxygen series, removing an electron from the antibonding orbital of ${O}_{2}$ increases the bond order to 2.5 in ${O}_{2}^{+}$, while adding electrons to these orbitals reduces the bond order to 1.5 in ${O}_{2}^{-}$ and 1.0 in ${O}_{2}^{2-}$. Because the neutral ${O}_{2}$ molecule has a bond order of 2, the cation ${O}_{2}^{+}$ possesses the highest bond order of the group.

The bond length between atoms generally decreases as the bond order increases, as stronger interactions pull the nuclei closer together. In $O_2$, the oxygen atoms are linked by a double bond, resulting in the shortest separation of 1.21 Å. The ozone molecule $O_3$ exhibits a resonance structure with an effective bond order between one and two, which leads to an intermediate length of 1.28 Å. By contrast, $H_2O_2$ features a single covalent bond, producing the longest bond length of 1.48 Å.