Chemical bonding & molecular structure practice · Mixed difficulty

The correct answer is C ($LiCl < BeCl_2 < BCl_3 < CCl_4$) because covalent character increases with the polarizing power of the cation, which increases with decreasing size and increasing charge.

The correct answer is C (only $\mathrm{K}{\mathrm{O}_2}$) because $\mathrm{K}{\mathrm{O}_2}$ contains the superoxide ion which has one unpaired electron.

Option C (${H}_2 {O}$) is correct because it has the highest dipole moment among the given molecules due to its bent geometry and electronegativity difference between oxygen and hydrogen.

Option D is correct because $ClF_3$ has $sp^3d$ hybridization, while $BrF_5$, $ICl_4^-$, and $XeF_4$ all have $sp^3d^2$ hybridization.

The correct answer is D ($ds{p^3}$-hybrid orbitals are directed towards the corners of a regular tetrahedron) because $ds{p^3}$ hybridization leads to a square planar or other geometries, not a tetrahedral geometry. Tetrahedral geometry is associated with $sp^3$ hybridization.

Option C (2) is correct because the molecular orbital configuration of O2 has two unpaired electrons in the antibonding π* orbitals, making it paramagnetic.

Option A (Kossel's postulations) is correct because Kossel's theory explains ion formation through electron transfer and the formation of ionic compounds.

Option D (${AB}$ has lesser ionic bond character than ${CD}$) is correct because the magnitude of charge calculated for CD is greater than that for AB, indicating a higher ionic character.

Option B (5) is correct because after the excitation of two electrons, the chlorine atom can form 5 covalent bonds.

Option C (Magnesium sulphide) is correct because MgS has the highest lattice energy due to higher charges and smaller sizes, making it the most stable salt in water (least soluble).

Option B is correct because $\mathrm{N(SiH_3)_3}$ has $\mathrm{sp^2}$ hybridization and a planar shape due to back bonding from N to Si.

Option C ($NO \rightarrow NO^+$) is correct because the bond order increases from 2.5 to 3, and the paramagnetic NO becomes diamagnetic $NO^+$ by losing an electron from an antibonding orbital.

Option A is correct because both statements (a) and (b) accurately describe the magnetic properties of oxygen and boron molecules based on their electronic configurations.

Option B (II < III < IV < I) is correct because it accurately reflects the increasing order of bond angles based on VSEPR theory and hybridization.

Option B is correct because Mg and F will form the most stable ionic bond due to the higher charge on Mg (+2) leading to a greater lattice energy.

Option B ($109^{\circ}$) is correct because carbon tetrachloride ($CCl_4$) has a tetrahedral geometry due to $sp^3$ hybridization, resulting in a bond angle of approximately 109.5 degrees.

The correct answer is D (${O} \equiv {C}={O}$) because it does not represent a valid resonance structure of CO2 due to incorrect valency.

Option C ($22 - 23\%$) is correct because the s-character is inversely proportional to the bond angle, and a bond angle of 105° corresponds to an s-character between 22-23%.

Option A ($sp^2$ hybridised) is correct because each carbon atom in graphite is bonded to three other carbon atoms in a planar arrangement, resulting in $sp^2$ hybridization.

Option C (Acetophenone) is correct because it has a significant difference in electronegativity between the carbonyl carbon and oxygen, leading to a substantial dipole moment, and the geometry allows for a net dipole moment.

Option A is correct because in sp3 hybridization, one s orbital and three p orbitals mix, resulting in 25% s character (1/4).

Option D is correct. The order of decreasing dipole moment is I > IV > III > II, based on the electronegativity differences and molecular geometry.

Option D (Regular tetrahedral) is correct because the symmetrical tetrahedral shape of CCl4 results in the cancellation of individual bond dipoles, leading to a zero net dipole moment.

The correct answer is 4 because CN-, CO, NO+, and N2 all have a bond order of 3, calculated using the molecular orbital theory.

Option C ((A) is true but (R) is false) is correct because CO2 is resonance stabilized, but the C-O bond length is intermediate between a double and triple bond, not a single and double bond.

Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. Both ${N}_2$ and ${NO}^+$ are diamagnetic, but this is due to their electronic configurations, not just because they are isoelectronic.

Option D (${HNO}_2$) is correct because it is the only molecule listed that does not contain a coordinate bond.

Option A is correct because ${C_2}^{2 - }$ is diamagnetic and ${N_2}^ +$ and ${N_2}^ -$ have the same bond orders.

Option D (All of the above) is correct because the dipole moment of NH3 is greater than NF3 due to the combined effects of fluorine's electronegativity and the direction of bond dipoles.

Option C (${BF}_3, {BCl}_3, {BBr}_3$) is correct because all three molecules have the same $sp^2$ hybridization around the central boron atom, resulting in the same bond angle of 120 degrees.