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

According to molecular orbital theory, the magnetic behavior of a diatomic species is dictated by the presence of unpaired electrons within its molecular orbitals. While the bond orders for the oxygen series are 2.5, 2, 1.5, and 1 respectively, and the $\mathrm{C}_{2}$ molecule indeed places four electrons in its two degenerate $\pi$ molecular orbitals, the designation of the $\mathrm{O}_{2}^{+}$ ion as diamagnetic is inaccurate. The $\mathrm{O}_{2}^{+}$ ion actually contains one unpaired electron in the $\pi^{*} 2 p$ molecular orbital, which requires it to be characterized as paramagnetic.

In molecular geometry, the magnitude of lone pair-lone pair repulsion is primarily determined by the number of lone pairs residing on the central atom. Among the given molecules, $\mathrm{SF}_{4}$ and $\mathrm{IF}_{5}$ each possess only one lone pair, while $\mathrm{ClF}_{3}$ contains two. Because $\mathrm{XeF}_{2}$ features the highest count with three lone pairs, it inherently experiences the greatest intensity of lone pair-lone pair repulsive interactions.

This problem relies on Boyle's Law, which applies whenever temperature and the amount of gas remain constant, dictating that pressure and volume are inversely proportional. To begin, calculate the number of moles of dioxygen by dividing the mass of $3.2 \text{ g}$ by its molar mass of $32 \text{ g/mol}$, yielding $0.1 \text{ mole}$. Using the molar volume at STP of $22.4 \text{ L/mol}$, the initial volume occupied by the gas is $0.1 \times 22.4 \text{ L} = 2.24 \text{ L}$.

Because the temperature is held constant, the product of pressure and volume remains fixed, described by the equation $P_1 V_1 = P_2 V_2$ where $P_1 = 1 \text{ atm}$ and $V_1 = 2.24 \text{ L}$. When the gas is moved to a vessel where the pressure $P_2$ is reduced to one-third of the original, or $\frac{1}{3} \text{ atm}$, the new volume $V_2$ is determined by substituting these values into the expression $1 \text{ atm} × 2.24 \text{ L} = \frac{1}{3} \text{ atm} × V_2$. Solving this equality leads to $V_2 = 3 × 2.24 \text{ L}$, resulting \in a final volume of $6.72 \text{ L}$.

The geometry of these molecules depends on the arrangement of valence electrons around the central atom, where the repulsion between bonding pairs and lone pairs dictates the final structure. Ammonia ($\mathrm{NH}_3$) features a central nitrogen atom with three bonding pairs and one lone pair, resulting in a trigonal pyramidal geometry. In $\mathrm{ClF}_3$, the chlorine atom is surrounded by three fluorine atoms and two lone pairs, which forces the structure into a T-shape to minimize electron repulsion. The phosphorus atom in $\mathrm{PCl}_5$ forms five bonds with no lone pairs, establishing a symmetric trigonal bipyramidal arrangement. Lastly, $\mathrm{BrF}_5$ consists of a central bromine atom bonded to five fluorine atoms with one lone pair, which distorts the parent octahedral geometry into a square pyramidal shape.

The ease of liquefaction for a gas is directly determined by its critical temperature, $T_c$, following the relationship $\text{Ease of Liquefaction} \propto T_c$. Since a gas can only liquify when the ambient temperature is reduced to or below its $T_c$, the gas with the highest critical temperature will transition to the liquid state first as the system cools. Among the given values, the critical temperature for $\mathrm{NH}_3$ is 405.5 K, which is significantly higher than 304.1 K for $\mathrm{CO}_2$, 126 K for $\mathrm{N}_2$, and 5.3 K for $\mathrm{He}$. Therefore, $\mathrm{NH}_3$ is the first to reach the conditions necessary for liquefaction.

A molecule is considered non-polar if the geometric arrangement of its bonds results in a net dipole moment of zero, meaning the individual bond polarities effectively cancel each other out. Within $SbCl_5$, the central atom adopts a highly symmetrical trigonal bipyramidal structure that allows the bond dipoles to negate one another completely. Because the net vector summation of these bond moments is zero, the molecule demonstrates no overall polarity.

According to VSEPR theory, the molecular geometry is determined by the total number of valence shell electron pairs around the central atom, which arrange themselves to minimize repulsion. $PCl_5$ features a steric number of 5 with zero lone pairs, resulting in a trigonal bipyramidal geometry, while $SF_6$ has a steric number of 6 with zero lone pairs, giving it an octahedral structure. $BrF_5$ consists of 5 bond pairs and 1 lone pair, creating a square pyramidal shape, and $BF_3$ has 3 bond pairs resulting in a trigonal planar arrangement.

Bond enthalpy is fundamentally determined by the strength of the bond, which is inversely proportional to bond length. As one moves down the halogen group from fluorine to iodine, the atomic radius of the halogen increases, resulting in a longer and weaker C-X bond. This causes a progressive decrease in bond dissociation energy, with values of 452 kJ/mol for $\text{CH}_3-F$, 351 kJ/mol for $\text{CH}_3-Cl$, 293 kJ/mol for $\text{CH}_3-Br$, and 234 kJ/mol for $\text{CH}_3-I$. Thus, the order of bond enthalpy is $\text{CH}_3-F > \text{CH}_3-Cl > \text{CH}_3-Br > \text{CH}_3-I$.

Boron sits at the center of the $BF_3$ molecule, utilizing its three valence electrons to form single covalent bonds with three separate fluorine atoms. Because each of these three covalent bonds consists of two electrons, the central boron atom ends up with a total of 6 electrons in its valence shell, leaving it electron deficient. Since there are three bonding regions around the central atom with no lone pairs, the boron atom adopts $sp^2$ hybridization to form its trigonal planar structure.

Molecular Orbital Theory predicts that a molecule's stability is dictated by its bond order, where a result of zero implies that the species cannot naturally exist. In the case of $\mathrm{He}_2$, the electrons fill both the bonding and antibonding orbitals according to the electronic configuration $\sigma 1s^2, \sigma^* 1s^2$. By applying the bond order formula with the number of bonding electrons ($N_b = 2$) and antibonding electrons ($N_a = 2$), we find that $\frac{1}{2}[2 - 2] = 0$, which proves that no stable bond holds the atoms together.

A molecule possesses a net zero dipole moment when its geometry is sufficiently symmetric such that the vector sum of individual bond dipole moments cancels out completely. In $BF_3$, the trigonal planar structure allows the three $B-F$ bond dipoles to cancel, while both $BeF_2$ and $CO_2$ are linear molecules with equal and opposite bond dipoles that result in a net zero polarity. Similarly, 1,4-dichlorobenzene is a para-substituted benzene ring where the two equal $C-Cl$ bond dipoles are directed in exactly opposite directions, effectively neutralizing each other to yield a non-polar molecule.

Molecular Orbital Theory allows us to determine the magnetic nature of a molecule by analyzing whether its electrons are paired or unpaired. When we construct the electron configuration for oxygen, we find that it possesses two unpaired electrons located in the antibonding $\pi^*$ orbitals, specifically filling the $\pi^* 2p_x$ and $\pi^* 2p_y$ levels singly according to Hund's rule. While the other molecules listed, such as $\mathrm{N}_2$, $\mathrm{H}_2$, and $\mathrm{Li}_2$, contain only paired electrons, this existence of unpaired electrons grants $\mathrm{O}_2$ its characteristic paramagnetic behavior.

Dipole moment is determined by the vector sum of individual bond moments, which depends on molecular geometry, lone pair contributions, and electronegativity differences. $BF_3$ has a symmetric trigonal planar structure that results in a net dipole moment of zero, while $NF_3$ exhibits a small dipole moment because the bond dipoles oppose the direction of the lone pair dipole. In contrast, $NH_3$ has a significantly higher dipole moment than $NF_3$ because the $N-H$ bond dipoles reinforce the lone pair, and $H_2O$ possesses the highest value among these molecules due to its bent geometry and the strongly additive nature of its bond dipoles.

Molecular Orbital Theory determines the bonding nature of a diatomic species by assigning electrons to orbitals based on increasing energy levels. For $\mathrm{C}_2$, which contains 12 total electrons, the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2 p_x^2 = \pi 2 p_y^2$. Since the electrons in the $\sigma 1s, \sigma^* 1s, \sigma 2s,$ and $\sigma^* 2s$ orbitals effectively cancel each other out, the bond is formed entirely by the four electrons occupying the $\pi 2 p_x$ and $\pi 2 p_y$ orbitals, meaning the molecule contains only $\pi$ bonds.

Chlorine is a halogen belonging to group 17, which provides it with seven valence electrons available for bonding. In the $ClF_3$ molecule, the chlorine atom utilizes three of these valence electrons to form single covalent bonds with three separate fluorine atoms. This leaves four non-bonding electrons, which pair up to form two lone pairs positioned around the central chlorine atom.

Molecular Orbital Theory identifies bond strength by calculating bond order as half the difference between the number of bonding and antibonding electrons. For the species $\mathrm{CN}^{-}$, the electronic configuration results in $10$ bonding and $4$ antibonding electrons, yielding a bond order of $\frac{10-4}{2} = 3$. Applying this calculation to the remaining species reveals bond orders of $\frac{10-5}{2} = 2.5$ for $\mathrm{NO}$, $\frac{9-4}{2} = 2.5$ for $\mathrm{CN}$, and $\frac{8-4}{2} = 2$ for $\mathrm{CN}^{+}$. Consequently, $\mathrm{CN}^{-}$ possesses the highest bond order among these species.

Two chemical species are classified as isoelectronic if they possess an identical total number of electrons, and they are defined as isostructural if they share the same geometric arrangement of atoms. Evaluating the reference compound $\mathrm{XeF}_{2}$, we observe that it contains 72 total electrons and adopts a linear geometry. In comparison, $\mathrm{TeI}_{2}$ possesses 158 electrons with a bent structure, $\mathrm{IBr}_{2}^{-}$ holds 124 electrons while maintaining a linear shape, $\mathrm{IF}_{3}$ is composed of 80 electrons in a T-shaped configuration, and $\mathrm{BeCl}_{2}$ contains 38 electrons with a linear arrangement. Since none of these alternatives match the total electron count of 72 found in $\mathrm{XeF}_{2}$, no pair satisfies the criteria for being both isoelectronic and isostructural.

Molecular Orbital Theory determines bond strength via bond order, which is calculated as half the difference between the number of bonding electrons and antibonding electrons: $B.O. = \frac{1}{2}(N_b - N_a)$. Evaluating the given pairs requires calculating these values for each species. While $\mathrm{O}_2$ features a bond order of $2$ due to its $10$ bonding and $6$ antibonding electrons, $\mathrm{NO}^{+}$ exhibits a bond order of $3$ because it has $10$ bonding and $4$ antibonding electrons. Similarly, comparing $\mathrm{N}_2$ with a bond order of $3$ against $\mathrm{O}_2^{-}$ with a bond order of $1.5$ (derived from $10$ bonding and $7$ antibonding electrons) confirms they are unequal, just as $\mathrm{CO}$ at $3$ and $\mathrm{NO}$ at $2.5$ do not match. Conversely, both $\mathrm{CN}^{-}$ and $\mathrm{CO}$ contain $10$ bonding and $4$ antibonding electrons, resulting in an identical bond order of $3$ for both species.

Molecular geometry is determined by the hybridization of the central atom and the spatial arrangement of electron pairs around it. In the molecule $\mathrm{BCl}_{3}$, the central boron atom undergoes $sp^{2}$ hybridization to accommodate three bonding pairs without any lone pairs. This results in a symmetric trigonal planar structure where the atoms are positioned to minimize repulsion, leading to uniform bond angles of 120 degrees.

Molecular geometry is dictated by the hybridization of the central atom and the number of lone pair electrons present as defined by VSEPR theory. $\mathrm{SeF}_{4}$ exhibits a seesaw shape because it undergoes $sp^{3}d$ hybridization with one lone pair on the central selenium atom, whereas $\mathrm{CH}_{4}$ possesses a tetrahedral geometry due to $sp^{3}$ hybridization without any lone pairs. Consequently, the statement claiming they have the same shape is incorrect, unlike the others where $\mathrm{PH}_{5}$ and $\mathrm{BiCl}_{5}$ are chemically unstable, $\mathrm{SO}_{2}$ utilizes sulfur $d$-orbitals for $p \pi-d \pi$ bonding, and $\mathrm{I}_{3}^{+}$ adopts a bent geometry due to two lone pairs on the central iodine atom.