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

Dipole-induced dipole interactions occur when a molecule with a permanent dipole moment creates an electric field that distorts the electron cloud of a neighboring non-polar atom or molecule, inducing a temporary dipole. Among the listed pairs, only $\mathrm{HCl}$ and $\mathrm{He}$ atoms satisfy this requirement, as $\mathrm{HCl}$ is a polar molecule ($\mu \neq 0$) and $\mathrm{He}$ is a non-polar atom ($\mu = 0$). In contrast, pairs like $\mathrm{SiF}_4$ and $\mathrm{He}$ or $\mathrm{Cl}_2$ and $\mathrm{CCl}_4$ involve only non-polar species, and the combination of $\mathrm{H}_2\mathrm{O}$ and alcohol involves two polar substances, neither of which represents the specific interaction between a permanent dipole and an induced dipole.

Molecular Orbital Theory determines the bond order of a species by calculating the difference between the number of bonding and antibonding electrons, divided by two. Species that are isoelectronic possess the same total number of electrons, which leads to identical orbital filling patterns and, consequently, the same bond order. Carbon monoxide ($CO$) contains $6 + 8 = 14$ electrons, and the nitrosonium ion ($NO^{+}$) similarly contains $7 + 8 - 1 = 14$ electrons. Because these species are isoelectronic with 14 electrons each, they both result in a bond order of $\frac{10 - 4}{2} = 3$.

Molecular Orbital Theory provides the simplest way to determine magnetic properties by examining the distribution of electrons within orbitals. A species is considered paramagnetic if it contains at least one unpaired electron, whereas it is diamagnetic when all electrons are paired. Species such as $\mathrm{CN}^{-}$, $\mathrm{NO}^{+}$, and CO each contain 14 total electrons, which fully fill the lower energy molecular orbitals and result in a diamagnetic state. In contrast, $\mathrm{O}_2^{-}$ holds 17 electrons, resulting in the configuration $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*1}$, where the presence of a single unpaired electron in the $\pi_{2p_y}^*$ orbital causes it to be paramagnetic.

The electronic configuration of a molecule is governed by the Aufbau principle, which dictates that incoming electrons occupy the lowest available energy levels in the molecular orbital diagram. For the oxygen molecule, the highest energy valence electrons reside in the degenerate antibonding $\pi^*(2p_x)$ and $\pi^*(2p_y)$ orbitals. When an additional electron is added to convert $\mathrm{O}_2$ into the $\mathrm{O}_2^{-}$ ion, it must occupy the next available space, which corresponds to one of these $\pi^*$ antibonding orbitals.

Molecular orbital theory defines bond order as half the difference between the number of bonding electrons and antibonding electrons. For diatomic oxygen species, there are consistently 10 electrons in bonding molecular orbitals, while the number of antibonding electrons varies based on the ionic charge. This relationship is expressed by the calculation:

$\text{Bond order} = \frac{\text{No. of } e^{-} \text{ in bonding M.O} - \text{No. of } e^{-} \text{ in antibonding M.O}}{2}$

For $\mathrm{O}_2^{-}$, the distribution results in 10 bonding electrons and 7 antibonding electrons, yielding a bond order of $\frac{10-7}{2} = 1.5$. Other species show different values based on their electron configurations: $\mathrm{O}_2^{+}$ contains 5 antibonding electrons with a bond order of 2.5, $\mathrm{O}_2$ has 6 antibonding electrons for a bond order of 2, and $\mathrm{O}_2^{2-}$ possesses 8 antibonding electrons, resulting in a bond order of 1.0.

Molecular orbital theory dictates that bond order is determined by the net difference between bonding and antibonding electrons, effectively indicating the relative stability of a chemical bond. Among the listed species, $\mathrm{O}_2^{2-}$ and $\mathrm{B}_2$ each possess a bond order of 1, while other pairings like $\mathrm{O}_2^+$ at 2.5 and $\mathrm{NO}^+$ at 3, $\mathrm{NO}$ at 2.5 and $\mathrm{CO}$ at 3, or $\mathrm{N}_2$ at 3 and $\mathrm{O}_2$ at 2.0, result in mismatched values.

Isostructural species are those that share both an identical hybridization state and a matching molecular geometry, as their central atoms are surrounded by an equal number of bonding and lone pairs. For $\text{BF}_4^{-}$ and $\text{NH}_4^{+}$, the central atoms are each bonded to four atoms with no lone pairs, resulting in $\text{sp}^3$ hybridization and a tetrahedral shape for both. Conversely, pairs like $\text{NF}_3$ and $\text{BF}_3$ or $\text{NH}_3$ and $\text{NO}_3^{-}$ differ in their structural properties due to disparities in their hybridization states or molecular geometries.

According to VSEPR theory, the distribution of valence electrons around a central atom determines its bonding and non-bonding electron pairs. In the molecule $PCl_3$, the central phosphorus atom possesses five valence electrons, three of which participate in single covalent bonds with chlorine atoms, while the remaining two constitute one lone pair, resulting in a configuration of three bond pairs and one lone pair.

Stability in Lewis structures is determined by minimizing formal charges and maximizing the total number of covalent bonds within the molecule. For a $SO_3$ molecule, the arrangement that achieves the lowest energy state is one where the sulfur atom forms double bonds with each of the three surrounding oxygen atoms. By maximizing the number of covalent bonds in this manner, the formal charges are minimized, resulting in the most preferred and stable configuration.

Molecular orbital theory establishes an inverse relationship between bond order and bond length, as higher bond order implies stronger attraction between atoms that pulls them closer together. By determining the electron configurations, the bond orders are calculated to be 2.5 for $\mathrm{O}_2^+$, 2 for $\mathrm{O}_2$, 1.5 for $\mathrm{O}_2^-$, and 1 for $\mathrm{O}_2^{2-}$. Since $\mathrm{O}_2^+$ possesses the highest bond order among these species, it exhibits the minimum bond length.

To determine the orbital geometry of these ions, we must calculate the hybridization state of the central atom by analyzing its bonding pairs and lone pairs. The nitrite ion ($NO_2^-$) and the nitrate ion ($NO_3^-$) each possess a central nitrogen atom that achieves $sp^2$ hybridization, thereby sharing the same geometric arrangement. In contrast, $NH_2^-$ and $NH_4^+$ both exhibit $sp^3$ hybridization, whereas the $SCN^-$ ion is characterized by $sp$ hybridization.

Bond length is fundamentally determined by the size of the bonded atoms and the multiplicity of the bond, where smaller atomic radii and higher bond orders yield shorter distances between nuclei. The C-H bond emerges as the shortest at 107 pm because hydrogen's small size allows for a very close approach to carbon, followed by the C=C double bond at 134 pm, which is shortened by the increased electron density inherent in multiple bonding. Among the single bonds, the C-O bond measures 141 pm, making it shorter than the C-C bond at 154 pm because the smaller oxygen atom allows for a closer inter-nuclear distance than the larger carbon atom.

Molecular Orbital Theory dictates that magnetic behavior is determined by the presence or absence of unpaired electrons within a molecule's energy levels. The oxygen molecule ($O_2$) is paramagnetic because it contains two unpaired electrons in its antibonding molecular orbitals, while the dioxygen cation ($O_2^{+}$) also exhibits paramagnetism due to the presence of one unpaired electron. Unlike other combinations that involve diamagnetic species such as the peroxide ion ($O_2^{2-}$), this specific pair maintains unpaired electrons in both of its components.

The stability of a diatomic species is determined by its bond order, which is calculated as half the difference between the number of bonding electrons and antibonding electrons. In the case of $\mathrm{Be}_2$, the molecular orbital configuration is $\sigma 1s^2 \sigma^* 1s^2 \sigma^* 2s^2$, resulting in a bond order of $\frac{4-4}{2}=0$. Because a bond order of zero signifies that no stable net bond is formed, this species cannot exist under normal conditions.

The hybridization of a central atom depends on its steric number, which represents the total count of bonding electron pairs and lone electron pairs surrounding that atom. In $\mathrm{CH}_4$, $\mathrm{BF}_4^{-}$, and $\mathrm{NH}_4^{+}$, the central atom in each case is surrounded by four bonding electron pairs and zero lone pairs, resulting in a steric number of 4 and $sp^3$ hybridization. Conversely, the sulfur atom in $\mathrm{SF}_4$ is bonded to four fluorine atoms and retains one lone pair, leading to a steric number of 5, which necessitates $sp^3d$ hybridization.

The hybridisation of a central atom is determined by the steric number formula $H = \frac{1}{2}(V + M - C + A)$, where $V$ denotes valence electrons, $M$ represents monovalent atoms, and $C$ and $A$ adjust for ionic charges. For $\mathrm{BF}_3$, substituting the values yields $H = \frac{1}{2}(3 + 3 - 0 + 0) = 3$, confirming $sp^2$ hybridisation. Similarly, for $\mathrm{NO}_2^{-}$, the calculation gives $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$, which also corresponds to $sp^2$ hybridisation.

Hybridization of a central atom is determined by calculating the steric number using the formula $H = \frac{1}{2} (V + M - C + A)$, where $V$ represents the valence electrons of the central atom, $M$ is the number of monovalent atoms surrounding it, and $C$ and $A$ account for the cationic and anionic charges, respectively. For the species $SF_4$, $I_3^{-}$, and $PCl_5$, this calculation consistently results in a value of 5, indicating $sp^3d$ hybridization for each. Specifically, sulfur in $SF_4$ provides $\frac{1}{2}(6 + 4) = 5$, the central iodine in $I_3^{-}$ calculates to $\frac{1}{2}(7 + 2 + 1) = 5$, and phosphorus in $PCl_5$ results in $\frac{1}{2}(5 + 5) = 5$. The $[\mathrm{SbCl}_5]^{2-}$ ion differs because the antimony central atom, having 5 valence electrons, is bonded to 5 monovalent chlorine atoms with an additional anionic charge of 2, leading to $\frac{1}{2}(5 + 5 + 2) = 6$, which corresponds to $sp^3d^2$ hybridization.

The hybridization and geometry of a molecule are determined by the number of electron pairs surrounding the central atom. For $\mathrm{NO}_3^{-}$, the calculation of electron pairs results in a value of three, indicating $sp^2$ hybridization and a trigonal planar structure. In contrast, the calculation for $\mathrm{H}_3\mathrm{O}^{+}$ yields four electron pairs, which signifies $sp^3$ hybridization and a distinct pyramidal geometry. Because both the hybridization state and the structural arrangement differ between these two species, they are characterized by different properties.

Methanol, $\mathrm{CH}_3\mathrm{OH}$, features a hydroxyl group where an electronegative oxygen atom is directly bonded to a hydrogen atom, inducing strong polarity across the molecule. This polarity facilitates the formation of hydrogen bonds, which are significantly stronger than the other types of intermolecular forces present in the liquid. As a result, the primary energy requirement for converting liquid $\mathrm{CH}_3\mathrm{OH}$ into a gas is the disruption of these specific, strong hydrogen bonds.

Determining the hybridization of a central atom involves calculating its steric number, which is the total count of sigma bonds and lone pairs, where a value of 3 indicates $sp^2$ hybridization. Boron in $\mathrm{BF}_3$ contains 3 sigma bonds with 0 lone pairs, and nitrogen in $\mathrm{NO}_2^{-}$ exhibits 2 sigma bonds with 1 lone pair, both resulting in a steric number of 3. Conversely, nitrogen in $\mathrm{NH}_2^{-}$ and oxygen in $\mathrm{H}_2\mathrm{O}$ both possess 2 sigma bonds and 2 lone pairs, creating a steric number of 4 that points to $sp^3$ hybridization.