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

According to VSEPR theory, bond angles decrease as the number of lone pairs on the central atom increases due to the greater repulsive force exerted by lone pairs compared to bonding pairs. While $\mathrm{CH}_{4}$ maintains a tetrahedral bond angle of $109.5^{\circ}$ due to the absence of lone pairs, $\mathrm{NH}_{3}$ and $\mathrm{H}_{2}\mathrm{O}$ exhibit progressively smaller bond angles of $107^{\circ}$ and $104.5^{\circ}$ because they contain one and two lone pairs, respectively. Therefore, the statement claiming the $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angle is larger than the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is factually incorrect.

The hybridization of the central Xenon atom is determined by the steric number, calculated as $H = \frac{1}{2}(V + M)$, where $V=8$ represents the valence electrons of Xenon and $M=4$ represents the monovalent Fluorine atoms. This calculation yields a steric number of 6, which indicates $sp^3d^2$ hybridization for the central atom. Because there are four bonding pairs and two lone pairs, the lone pairs occupy opposite axial positions to minimize electron-pair repulsion, establishing a square planar molecular geometry.

Hydrogen bonding is categorized by whether the stabilizing force acts between separate molecular units or occurs within a single, larger structure. Cellulose features a complex chain arrangement that facilitates intramolecular hydrogen bonding, where the interaction is contained within the same molecular framework. Conversely, substances like $H_2O_2$, $HCN$, and concentrated $CH_3COOH$ exhibit intermolecular hydrogen bonding, meaning these interactions bridge distinct and separate molecules.

To determine if two chemical species are isoelectronic and isostructural, one must calculate their total electron counts and compare the hybridization and geometry of their central atoms. Both $\mathrm{ClO}_{3}^{-}$ and $\mathrm{SO}_{3}^{2-}$ contain a total of 42 electrons, as determined by the sums $17 + 24 + 1 = 42$ and $16 + 24 + 2 = 42$, which confirms they are isoelectronic. Because both central atoms utilize $sp^3$ hybridization and retain one lone pair, they each adopt a trigonal pyramidal geometry, rendering them isostructural.

Identifying isoelectronic species requires determining if they possess the same total number of valence electrons, while verifying their isostructural nature involves confirming they share identical molecular geometry. The chlorate ion, $\mathrm{ClO}_3^-$, contains 7 valence electrons from chlorine, 18 from three oxygen atoms, and one additional electron from its charge, summing to 26 valence electrons. Similarly, the sulfite ion, $\mathrm{SO}_3^{2-}$, contributes 6 valence electrons from sulfur, 18 from three oxygen atoms, and two extra electrons for its charge, which also equals 26 valence electrons. Because both ions feature a central atom surrounded by three bonding pairs and one lone pair, they adopt the same trigonal pyramidal structure, classifying the pair as both isoelectronic and isostructural.

The hybridization state of a central nitrogen atom is determined by its steric number ($X$), which accounts for valence electrons ($VE$), attached monovalent atoms ($MA$), and net ionic charge ($c$ or $a$) using the formula
$X=\frac{1}{2}(VE+MA-c+a)$
Applying this to nitrogen, which has a valence electron count of 5, allows us to derive the hybridization for each species. For $\mathrm{NO}_{2}^{+}$, the calculation $X=\frac{1}{2}(5+0-1)=2$ indicates $sp$ hybridization, while $\mathrm{NO}_{3}^{-}$ yields $X=\frac{1}{2}(5+0+1)=3$ due to the added negative charge, signifying $sp^{2}$ hybridization. Including the four monovalent hydrogen atoms for $\mathrm{NH}_{4}^{+}$ results \in $X=\frac{1}{2}(5+4-1)=4$ which corresponds to $sp^{3}$ hybridization.

The Valence Shell Electron Pair Repulsion theory governs molecular geometry by quantifying the electrostatic repulsion between electron pairs surrounding a central atom. Because lone pairs are attracted to only one nucleus instead of two, they occupy more space and exert stronger repulsive forces than bond pairs, which are constrained between two atoms. Consequently, the intensity of these repulsive interactions follows the order $lone \ pair-lone \ pair > lone \ pair-bond \ pair > bond \ pair-bond \ pair$.

Molecular Orbital Theory assesses the strength of chemical bonds by examining the electron distribution across bonding and antibonding orbitals. Based on this, $\mathrm{O}_2^{2+}$ contains 14 electrons and exhibits a bond order of 3.0, while $\mathrm{O}_2^+$ holds 15 electrons with a bond order of 2.5, and $\mathrm{O}_2^-$ possesses 17 electrons with a bond order of 1.5. Consequently, the increasing sequence of bond orders for these species is $\mathrm{O}_2^- < \mathrm{O}_2^+ < \mathrm{O}_2^{2+}$.

According to molecular orbital theory, the stability of a molecular species is directly proportional to its bond order, which is calculated using the formula $\text{Bond order} = \frac{1}{2}(N_b - N_a)$. By applying this calculation to each species, we determine that $\mathrm{O}_2^{+}$ has a bond order of $\frac{1}{2}(10-5) = 2.5$, while the neutral $\mathrm{O}_2$ molecule has a bond order of $\frac{1}{2}(10-6) = 2$. Continuing with the addition of electrons, the species $\mathrm{O}_2^{-}$ yields a bond order of $\frac{1}{2}(10-7) = 1.5$, and $\mathrm{O}_2^{2-}$ results in the lowest bond order of $\frac{1}{2}(10-8) = 1.0$. Because stability decreases as the bond order decreases, the species follow the stability order of $\mathrm{O}_2^{+} > \mathrm{O}_2 > \mathrm{O}_2^{-} > \mathrm{O}_2^{2-}$.

Two chemical species are isostructural if they share the same hybridization of the central atom and possess identical molecular geometry. In $\text{XeF}_4$, the xenon atom undergoes $sp^3d^2$ hybridization and contains two lone pairs, resulting in a square planar geometry, while $\text{XeO}_4$ features $sp^3$ hybridization with four double bonds and zero lone pairs, creating a tetrahedral geometry. Conversely, diamond and silicon carbide both exhibit a tetrahedral $sp^3$ network, $\text{NH}_3$ and $\text{PH}_3$ share a trigonal pyramidal structure, and both $\text{SiCl}_4$ and $\text{PCl}_4^+$ possess a tetrahedral geometry, confirming those pairs are indeed isostructural.

Molecular Orbital Theory determines the strength and stability of chemical species by calculating bond order, which is the half-difference between the number of bonding and antibonding electrons. For the species provided, the total electron count dictates the bond order directly: $\mathrm{O}_2^-$ has 17 electrons resulting in a bond order of 1.5, $\mathrm{O}_2$ has 16 electrons resulting in a bond order of 2, and $\mathrm{O}_2^+$ has 15 electrons resulting in a bond order of 2.5. Comparing these values demonstrates the progression $\mathrm{O}_2^{-} < \mathrm{O}_2 < \mathrm{O}_2^{+}$, as the bond order increases with the removal of electrons from antibonding orbitals.

The bond angle around a central nitrogen atom is primarily determined by its hybridization state and the electron density surrounding it, as these factors dictate the geometry of the molecule. Specifically, in the nitronium ion $\text{NO}^{+}_2$, the nitrogen atom undergoes $sp$ hybridization, which results in a linear molecular shape with a bond angle of 180 degrees. Other species such as $\text{NO}_2^-$, $\text{NO}_3^-$, and $\text{NO}_2$ exhibit different hybridization or electronic environments that force smaller bond angles, following the increasing order $\text{NO}_2^- < \text{NO}_3^- < \text{NO}_2 < \text{NO}_2^+$. This structural arrangement confirms that the linear geometry of the nitronium ion provides the largest possible separation between the terminal oxygen atoms.

Determining the dipole moment relies on molecular geometry and symmetry, as molecules with a perfectly symmetrical distribution of charge, such as linear $\text{CO}_2$ and tetrahedral $\text{CH}_4$, possess a net dipole moment of zero because the individual bond polarities cancel each other out. In contrast, $\text{NH}_3$ adopts a trigonal pyramidal structure where the lone pair on the nitrogen atom and the three $\text{N-H}$ bond dipoles reinforce one another, creating a substantial cumulative dipole. Although $\text{NF}_3$ also features a pyramidal geometry, the fluorine atoms pull electron density away from the nitrogen, effectively opposing the direction of the lone pair's dipole and resulting in a much smaller net value than the significant dipole found in $\text{NH}_3$.

The shape of a polyatomic species is determined by the hybridization state of its central atom and the number of lone pair electrons influencing electron-pair repulsion. Both $\mathrm{N}_3^{-}$ and $\mathrm{CO}_2$ undergo sp hybridization, which results in a linear geometry for both molecules. While $\mathrm{NO}_2^{-}$ also features sp$^2$ hybridization, the presence of a lone pair on the central nitrogen atom distorts the structure into a bent shape. In contrast, $\mathrm{NO}_3^{-}$ is sp$^2$ hybridized with no lone pairs on the nitrogen, creating the necessary conditions for a plane triangular arrangement.

Hybridization is determined by calculating the steric number of a central atom, which is the sum of its sigma bonds and lone pairs. For the molecule ${NF}_{3}$, the nitrogen atom is bonded to three fluorine atoms and holds one lone pair, yielding a total of four electron domains, which indicates ${sp}^{3}$ hybridization. Likewise, the central oxygen atom in ${H}_{2} {O}$ participates in two sigma bonds with hydrogen atoms while retaining two lone pairs, resulting in four electron domains that also establish ${sp}^{3}$ hybridization.

The ethene molecule, represented by the chemical formula $C_2H_4$, features a central double bond between two carbon atoms, with each carbon atom covalently bonded to two hydrogen atoms. A single covalent bond constitutes one sigma ($\sigma$) bond, whereas a double bond comprises one $\sigma$ bond and one pi ($\pi$) bond. Summing the bonds in the structure yields four $C-H$ sigma bonds plus one $C-C$ sigma bond, and one $C-C$ pi bond, totaling 5 $\sigma$ and 1 $\pi$ bonds.

Isostructural species possess identical hybridization and molecular geometry, which are determined by the arrangement of bonding pairs and lone pairs around the central atom according to the valence shell electron pair repulsion theory. The $\mathrm{XeF}_2$ molecule contains a central xenon atom with two bonding pairs and three lone pairs, resulting in a steric number of 5 and a linear molecular geometry. The $\mathrm{ICl}_2^{-}$ ion is isostructural because it also utilizes 8 valence electrons to generate a steric number of 5 with three lone pairs, forcing the iodine and chlorine atoms into the same linear configuration.

A molecule is polar when its net dipole moment, $\mu \neq 0$, arises from an asymmetric geometry that prevents individual bond dipoles from cancelling. Highly symmetrical molecules such as $\mathrm{SiF}_4$, $\mathrm{XeF}_4$, and $\mathrm{BF}_3$ exhibit a net dipole moment of $\mu = 0$ because their structural symmetry forces the bond dipoles to effectively neutralize. In contrast, $\mathrm{SF}_4$ features an asymmetric see-saw shape resulting from its $\mathrm{sp}^3\mathrm{d}$ hybridisation with 4 bond pairs and 1 lone pair. This configuration ensures that the bond dipoles do not cancel out, leading to a permanent polarity where $\mu \neq 0$.

Electron deficiency occurs when a central atom lacks sufficient valence electrons to complete its octet through standard two-center, two-electron bonds. Boron, as a group 13 element, enters bonding with only three valence electrons, which results in the boron atom in $BH_3$ being surrounded by just six electrons rather than the stable eight. This characteristic incomplete octet persists in its dimeric form, $(BH_3)_2$, which employs specialized three-center-two-electron bonds to partially stabilize the deficiency, making it a textbook example of an electron-deficient compound compared to the stable, octet-complete structures of the other listed hydrides.

Molecular Orbital Theory provides the necessary tools to track shifts in electronic stability and magnetism during ionization. Bond energy is directly correlated with bond order, so increasing this value strengthens the bond, while the presence of unpaired electrons dictates paramagnetic behavior. By examining the electron configuration of each species, we can see how the removal of a single electron impacts both these properties simultaneously.

The molecule $\mathrm{NO}$ possesses an odd number of electrons with a bond order of 2.5, exhibiting paramagnetism because of the single electron in its antibonding $\pi^{\star} 2 \mathrm{p}_x$ orbital. Upon ionization to $\mathrm{NO}^{+}$, this lone electron is removed, which raises the bond order to 3 and ensures all remaining electrons are paired, effectively changing the molecule's magnetic nature from paramagnetic to diamagnetic. In contrast, the ionization of $\mathrm{O}_2$ increases the bond order to 2.5 but retains paramagnetism, while the ionization of both $\mathrm{C}_2$ and $\mathrm{N}_2$ results in a decrease in bond order, making $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$ the only process that fulfills both criteria.