Chemical bonding & molecular structure practice · Mixed difficulty

Option D (${M}_3\left({PO}_4\right)_2$) is correct because if MO has a +2 and -2 charge, then M has a +2 charge. Phosphate has a -3 charge, so the formula will be M3(PO4)2.

Option B (Kössel-Lewis) is correct because Kössel and Lewis independently developed theories explaining valence based on the inertness of noble gases and the tendency of atoms to achieve a noble gas configuration.

Option B (${M}_3\left({PO}_4\right)_2$) is correct because if the metal chloride is ${MCl}_2$, then the metal has a +2 charge ($M^{2+}$). To balance the -3 charge of two phosphate ions ($2PO_4^{3-}$), three metal ions are required.

Option A ($BH_3$) is correct because Boron in $BH_3$ has only 6 electrons in its valence shell, resulting in $sp^2$ hybridization, not $sp^3$.

Option A (Polarization of an anion is maximum by high charged cation) is correct because according to Fajan's rule, a highly charged cation will cause greater polarization of the anion.

Option C is correct because the dipole moments of N-F bonds partially cancel each other due to the direction of the bond dipoles and the lone pair on nitrogen.

Option A (RCH2NHCH3) is correct because it contains an N-H bond, which is necessary for hydrogen bonding.

Option B (One sigma-bond and one pi-bond) is correct because a double bond consists of one sigma bond formed by head-on overlap and one pi bond formed by sideways overlap.

Option A (More volatile than p – nitro phenol) is correct because o-nitrophenol exhibits intramolecular hydrogen bonding, which reduces intermolecular forces and increases volatility compared to p-nitrophenol, which exhibits intermolecular hydrogen bonding.

Option D (${sp}^3 {~d}$) is correct because $ClF_3$ has 3 bond pairs and 2 lone pairs, resulting in a steric number of 5, which corresponds to $sp^3d$ hybridization.

Option A (Statement-1 is true, statement-2 is true and statement- 2 is correct explanation for statement-1) is correct because both statements are true and statement 2 correctly explains why $NO^+$ and $CO^+$ are more stable than $NO$ and $CO$ respectively.

Option D (Lattice enthalpy) is correct because lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous ions.

Option D ($ds{p^3}$-hybrid orbitals are directed towards the corners of a regular tetrahedron) is wrong because $sp^3$ hybrid orbitals are directed towards the corners of a regular tetrahedron, not $dsp^3$ which would have a different geometry.

Option D ($180^\circ$) is correct because the dipole moments of the two Y atoms cancel each other out, resulting in a minimum (zero) net dipole moment.

Option C ($C{F_4}$) is correct because carbon tetrafluoride has polar C-F bonds, but due to its symmetrical tetrahedral geometry, the bond dipoles cancel each other out, resulting in a zero net dipole moment.

Option A is correct because a $\pi$-bond always exists in conjunction with a $\sigma$-bond; it cannot exist independently.

Option D ($25 \%$) is correct because it is calculated by dividing the actual dipole moment by the theoretical dipole moment assuming full charge separation.

Option A (VB theory) is correct because Valence Bond (VB) theory introduced the concept of directional nature of covalent bonds based on the overlap of atomic orbitals.

Option B (Covalent) is correct because an electronegativity difference between 0.4 and 1.7 typically indicates a polar covalent bond.

Option D (NaI) is correct because, according to Fajan's rule, larger anions increase covalent character, and iodide (I-) is the largest anion among the given options.

Option D (All are correct) is correct because resonance leads to decreased bond length, decreased energy, and increased stability of the molecule.

Option C (2.5) is correct because the bond order of NO is calculated using Molecular Orbital Theory, resulting in a bond order of 2.5.

Option C ($1.33{A^0}$) is the closest to the actual C=C bond length in a $sp^2$ hybridized molecule, which is approximately 1.34 Å. Option B ($1.39{A^0}$) is the C-C bond length in benzene.

Option B (${O}_2^{+}$) is correct because it has the highest bond order (2.5) among the given species, which directly correlates with bond energy.

Option C (Van der Waal's force) is correct because Van der Waals forces are weak intermolecular forces arising from temporary dipoles.

Option B (III > II > I > IV) is correct because the strength of a sigma bond depends on the extent of overlapping, which is greater for 2p-2p than 2s-2p, 2s-2s, and 3s-3s.

Option D (These molecules have exceptionally high attractive forces in their lattice) is correct because the high lattice energy (strong attractive forces) prevents the ions from being easily separated and solvated by water molecules.

Option D (2) is correct because propyne has one triple bond, which contains two pi bonds.

Option C ($\mathrm{O}_2^{2 - }$) is correct because peroxides contain the peroxide ion, which is $\mathrm{O}_2^{2 - }$.

Option C (Small cation & Large anion) is correct because according to Fajan's rule, small, highly charged cations and large, highly charged anions favor covalent bond formation due to increased polarization.