Chemical Bonding And Molecular Structure – JEE Main ChemistryPractice Questions & PYQs

To determine if hybridisation changes during a chemical reaction, one must compare the steric number of the central atom in both the reactant and the product. For $\mathrm{AlH}_3$, the steric number is calculated as $\frac{1}{2}(3+3)=3$, which indicates $sp^2$ hybridisation, whereas in the product $\mathrm{AlH}_4^{-}$, the steric number becomes $\frac{1}{2}(3+4+1)=4$, reflecting a change to $sp^3$ hybridisation. In contrast, consider $\mathrm{H}_2\mathrm{O}$, which has a steric number of $\frac{1}{2}(6+2)=4$ and $sp^3$ hybridisation; forming $\mathrm{H}_3\mathrm{O}^{+}$ with a steric number of $\frac{1}{2}(8+3-1)=4$ does not alter this $sp^3$ state. Similarly, $\mathrm{NH}_3$ begins with a steric number of $\frac{1}{2}(5+3)=4$ corresponding to $sp^3$ hybridisation, and this state persists in $\mathrm{NH}_4^{+}$ as its steric number is $\frac{1}{2}(5+4-1)=4$. Thus, the only case among these examples where the hybridisation state of the central atom changes is the conversion of $\mathrm{AlH}_3$ to $\mathrm{AlH}_4^{-}$.

A bond angle of $109^{\circ} 28'$ is characteristic of a regular tetrahedral geometry, which occurs when a central atom is $\mathrm{sp}^3$ hybridized and bonded to four surrounding atoms without any lone pairs. Both $\mathrm{NH}_4^{+}$ and $\mathrm{BF}_4^{-}$ fulfill these conditions because their central atoms exhibit four bond pairs and zero lone pairs, maintaining ideal geometric symmetry. In contrast, molecules like $\mathrm{NH}_3$ contain one lone pair that constricts the angle to $107^{\circ}$, $\mathrm{NH}_2^{-}$ holds two lone pairs resulting in a $104.5^{\circ}$ angle, and $\mathrm{BF}_3$ adopts an $\mathrm{sp}^2$ hybridization state with a $120^{\circ}$ bond angle.

The formation of a stable chemical bond occurs when a system reaches a state of lower potential energy, which happens precisely when the attractive electrostatic forces between the nuclei and electrons of participating atoms outweigh the repulsive forces between the particles. In contrast to other descriptions, pure water does contain $H^+$ and $OH^-$ ions due to equilibrium self-ionization, covalent bonding involves the sharing of electrons rather than transfer, and HF is actually more polar than HBr because its electronegativity difference of 1.9 significantly exceeds the 0.7 difference found in HBr. Thus, the stability of a bond is contingent on successfully balancing these competing electrostatic forces to achieve a configuration that favors attraction.

The bond strength of a molecule is directly proportional to its bond order, which reflects the net stability provided by shared electrons between atoms. This property is determined using the formula $\text{BO} = \frac{1}{2} [ N_{b} - N_{a} ]$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

To determine the trend, we calculate the bond order for each oxygen species by analyzing their orbital occupancies. For $\mathrm{O}_2^{2-}$ (18 electrons), the configuration results in $N_b = 10$ and $N_a = 8$, yielding a bond order of 1. For $\mathrm{O}_2^{-}$ (17 electrons), the counts are $N_b = 10$ and $N_a = 7$, giving a bond order of 1.5. Neutral $\mathrm{O}_2$ (16 electrons) features $N_b = 10$ and $N_a = 6$, resulting in a bond order of 2, while $\mathrm{O}_2^{+}$ (15 electrons) contains $N_b = 10$ and $N_a = 5$, providing the highest bond order of 2.5. Because bond strength increases as bond order increases, these values establish the sequence $\mathrm{O}_2^{2-} < \mathrm{O}_2^{-} < \mathrm{O}_2 < \mathrm{O}_2^{+}$.

Square planar complexes result from a specific spatial arrangement where the central metal atom and its four ligands reside within the same plane, conventionally defined as the $xy$-plane, which necessitates a $dsp^2$ hybridization scheme. To establish coordinate bonds specifically along the $x$ and $y$ axes, the central atom utilizes the non-directional $s$ orbital, the $p_x$ and $p_y$ orbitals that align with these axes, and the $d_{x^2 - y^2}$ orbital, whose lobes are oriented precisely along the $x$ and $y$ coordinates to facilitate stable planar bonding.

A $\pi^*$ antibonding molecular orbital arises from the destructive interference of two atomic orbitals, such as $p_x$ or $p_y$, oriented perpendicular to the internuclear z-axis, which places an additional nodal plane perpendicular to this axis between the two nuclei. This geometric arrangement necessitates a phase inversion across the internuclear region, where the wave function signs of the orbital lobes on one side of the nodal plane are opposite to those on the other side. Consequently, the correct representation is characterized by four lobes with alternating phase signs across the nodal plane between the atomic centers, effectively creating a region of zero electron density between the nuclei.

According to VSEPR theory, molecular geometry is determined by the total number of electron pairs, including both bond pairs and lone pairs, surrounding the central atom. Applying this principle, $ClF_3$ features 3 bond pairs and 2 lone pairs ($AX_3E_2$) resulting in a T-shaped geometry, whereas $BF_3$ possesses 3 bond pairs and no lone pairs ($AX_3$) adopting a trigonal planar arrangement. Furthermore, $XeF_4$ contains 4 bond pairs and 2 lone pairs ($AX_4E_2$) forming a square planar shape, and $SF_4$ with 4 bond pairs and 1 lone pair ($AX_4E_1$) exhibits a see-saw geometry.