Chemical bonding & molecular structure practice · Medium

Option C (The electronegativity of chlorine is greater than that of hydrogen) is correct because the difference in electronegativity creates a dipole moment in the HCl molecule.

Option C is correct because both $H_2^{2+}$ and $He_2$ have a bond order of zero, meaning they are not likely to exist.

Option A ($1.98 \mathop {\text{A}}\limits^ \circ$) is correct because the experimentally determined bond distance for Cl-Cl is 1.98 Angstroms.

The correct answer is $s p, s p^2, s p^2$ because carbon 2 is sp hybridized, carbon 4 is sp2 hybridized, and carbon 6 is sp2 hybridized.

Option A (Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1) is correct because larger ions are more polarizable, leading to increased covalent character and decreased solubility.

Option C (3.0, 1.5, 3, 1.33) is correct because it accurately lists the bond orders calculated using Molecular Orbital Theory (MOT) and resonance concepts.

The correct answer is 3 – Linear, 2 – Square planar because ClF2- has 3 lone pairs and a linear geometry, while ClF4- has 2 lone pairs and a square planar geometry.

The correct answer is B ($\pi$-bond is formed from hybrid orbitals) because pi bonds are formed from unhybridized p-orbitals.

Option A ($\mathrm{BeCl}_2, \mathrm{XeF}_2, \mathrm{CO}_2$) is correct because all three species are linear in shape.

Option A ($O_2^ + > {O_2} > O_2^ - > O_2^{2 - }$) is correct because the stability of a species is directly proportional to its bond order, and the bond orders decrease in the order $O_2^+ > O_2 > O_2^- > O_2^{2-}$.

Option B ($\mathrm{H}_2 \mathrm{O} > \mathrm{HF} > \mathrm{NF}_3 > \mathrm{BeF}_2$) is correct because it correctly orders the molecules based on their dipole moments, considering geometry and electronegativity differences.

Option D is correct because the bond length of $O_2$ is shorter than $O_2^-$, as $O_2^-$ has a lower bond order.

Option A ($C{H_3} - C{H_2} - C{H_2} - N{H_2}$) is correct because primary amines have two hydrogen atoms bonded to the nitrogen atom, allowing for more hydrogen bonding and a higher boiling point compared to secondary and tertiary amines.

Option B ($CO$) is correct because it has 14 electrons, resulting in all molecular orbitals being fully occupied, making it diamagnetic (not paramagnetic).

Option B (Molecular orbital) is correct because molecular orbitals describe the electron probability distribution in a molecule.

The correct answer is C because in solid NaCl, the ions are held in fixed positions and cannot move freely to conduct electricity.

The correct answer is D (both Assertion & Reason are False) because adding sodium acetate to acetic acid creates a buffer solution, which resists pH changes, and the ionization of acetic acid is suppressed due to the common ion effect.

If both Assertion & Reason True but Reason not a correct explanation of the assertion because NO is paramagnetic due to the presence of an unpaired electron, and its bond order is indeed 2.5, but the bond order doesn't directly explain the paramagnetism.

Option D ($\mathrm{O}_2^ + > {\mathrm{O}_2} > \mathrm{O}_2^ -$) is correct because the bond order of O₂⁺ is 2.5, O₂ is 2, and O₂⁻ is 1.5, and stability increases with bond order.

Option C (Trigonal bipyramidal) is correct because PF5 has sp3d hybridization, resulting in a trigonal bipyramidal geometry.

Option B (If both Assertion & Reason True but Reason not a correct explanation of the assertion) is correct because while both statements are true, the reason doesn't explain the assertion.

Option D (Statement I is false but Statement II is true) is correct because water is more polar than H2S, and both molecules have a bent structure.

Option B (${NH}_3$) is correct because the dipole moments of the N-H bonds add up constructively, resulting in a larger net dipole moment compared to the other options.

Option D (HF) is correct because dipole moment depends on the electronegativity difference between atoms, and HF has the largest electronegativity difference.

Option C (6) is correct because in $\mathrm{S}{\mathrm{F}_6}$, sulfur forms six single bonds with six fluorine atoms.

Option B ($\mathrm{N}^{3-}$) is correct because it has a complete octet (8 valence electrons) in its outermost shell.

Option B ($OF_2$) has the minimum bond angle because the electronegativity of the side atom (F) is greater than that of the central atom (O), leading to a smaller bond angle.

Option C is correct because both ${ICl}_2^{-}$ and ${XeF}_2$ have a linear shape.

Option A (Bond order) is correct because bond order is defined as half the difference between the number of electrons in bonding and antibonding molecular orbitals.

Option B (A metallic solid) is correct because Cesium (Cs) is a metal and exists as a solid at room temperature due to metallic bonding.