Chemical bonding & molecular structure practice · Mixed difficulty

Option B is correct because the equilibrium constant for the reaction D ⇌ 4A is found by manipulating the given equilibrium constants: K = Z / (X² * Y).

Option B (16) is correct because in the structure of $P_4O_{10}$, each phosphorus atom is bonded to four oxygen atoms, resulting in a total of 16 P-O bonds.

The correct answer is C ($Cl > I > Br$) because the stability of oxides decreases down the group due to decreasing electronegativity difference and weaker bonding.

Option A is correct because the lattice energy of ionic compounds depends on both the charge and size of the ions, as described by Coulomb's law.

Option B (Two molecular orbitals) is correct because the combination of two atomic orbitals always results in the formation of two molecular orbitals (one bonding and one antibonding).

Option B is correct because HCl is a covalent compound with weak intermolecular forces, resulting in a lower melting point compared to the ionic compounds CsF, HF, and LiF.

Option C (${N_2}$) is correct because in $N_2$, due to less intermixing, the ${\sigma _{2{p_z}}}$ molecular orbital is filled after ${\pi _{2{p_x}}}$ and ${\pi _{2{p_y}}}$ molecular orbitals.

Option B ($\mathrm{PO}_4^{3 - }$) is correct because phosphorus has available d-orbitals, allowing for $p\pi - d\pi$ bonding with oxygen.

Option B is correct because a low ionization energy for A and a high electron affinity for B favor the formation of ions ${A^ + }$ and ${B^ - }$ respectively, leading to an ionic bond.

Option C (Mulliken) is correct because Robert S. Mulliken is credited with the development of the Molecular Orbital Theory.

Option B ($\mathrm{O}_2^{-} > \mathrm{O}_2 > \mathrm{O}_2^+$) is correct because bond length is inversely proportional to bond order, and the bond orders are O2- (1.5), O2 (2), and O2+ (2.5).

Option C ($I - I$) is correct because bond energy decreases as the size of the atom increases due to reduced overlap of atomic orbitals.

Option C is correct because Hydrogen Fluoride (HF) exhibits the strongest hydrogen bonding due to the high electronegativity of Fluorine.

Option A (The energy of the resonance hybrid is more than the energy of any single canonical structure) is correct because the resonance hybrid is more stable and thus has lower energy than any individual contributing canonical structure.