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

Bond angles are primarily determined by the repulsive interactions between lone pairs and bonding pairs of electrons, which compress the bond angles in molecules containing lone pairs. In $\mathrm{NH}_{3}$, $\mathrm{PCl}_{3}$, and $\mathrm{SCl}_{2}$, the presence of lone pairs exerts significant repulsion that forces the bond angles to narrow. Conversely, the $\mathrm{NH}_{4}^{+}$ ion possesses four bonding pairs and no lone pairs, resulting in a symmetric tetrahedral geometry without additional compressive force. This lack of lone pairs leads to the maximum bond angle of 109 degrees among the molecules provided.

The ability of atoms to form stable multiple bonds is governed by the effectiveness of their orbital overlap, specifically $p \pi - p \pi$ bonding, which relies heavily on atomic size. Because nitrogen has a small atomic radius, it can achieve strong $p \pi - p \pi$ overlap, facilitating the formation of the stable triple bond found in $\mathrm{N}_{2}$. In contrast, phosphorus possesses a larger atomic size, which results in significantly weaker $p \pi - p \pi$ overlap, causing it to favor single covalent bonds and aggregate into the $\mathrm{P}_{4}$ structure instead.

Two chemical species are considered isostructural if they share an identical hybridization state and molecular geometry. Both $XeF_{2}$ and $IF_{2}^{-}$ exhibit $sp^{3}d$ hybridization and adopt a linear shape, where the presence of lone pairs in the equatorial positions results in this specific linear alignment.

In the $X-H \cdot s Y$ hydrogen bonding interaction, the high electronegativity of $X$ polarizes the covalent $X-H$ bond, drawing electron density toward $X$ and leaving $H$ with a partial positive charge. When $Y$ interacts with $H$ to form a hydrogen bond, the further attraction of electrons by $Y$ pulls density away from the hydrogen atom, which enhances the polarization of the $X-H$ bond and results in an increase in electron density on $X$ and a decrease on $H$.

Molecular orbital theory dictates that magnetic behavior depends strictly on the presence of unpaired electrons within a molecule. In carbon monoxide, denoted as ${CO}$, the 14 electrons completely fill the available bonding and antibonding molecular orbitals such that every electron remains paired. This complete pairing excludes paramagnetism, distinguishing it from the other listed species which all contain unpaired electrons and thus exhibit paramagnetic properties.

The formation of a $p\pi-d\pi$ bond requires the central atom to have vacant d-orbitals available to accept electrons from the filled p-orbitals of an oxygen atom. Elements such as carbon and nitrogen, found in $CO_3^{2-}$, $NO_3^-$, and $NO_2^-$, are restricted to the second period of the periodic table and lack d-orbitals in their valence shell, making this type of bonding impossible for them. In contrast, phosphorus in the $PO_4^{3-}$ ion resides in the third period and utilizes its vacant 3d orbitals to accommodate back-bonding from the oxygen atoms, thereby facilitating stabilization.

Molecular Orbital Theory establishes that the dissociation energy of a molecule is directly proportional to its bond order, where a higher bond order signifies a stronger bond that requires more energy to break. For the neutral $\mathrm{N}_{2}$ molecule, the bond order is calculated as $\frac{10-4}{2} = 3$, whereas the removal of an electron to form the $\mathrm{N}_{2}^{+}$ ion decreases the bond order to $\frac{9-4}{2} = 2.5$. Because $\mathrm{N}_{2}$ maintains a greater bond order than its cationic counterpart, it inherently exhibits a higher dissociation energy.