Chemical bonding & molecular structure practice · Mixed difficulty

Option A (Diagonal hybridisation) is correct because sp hybridization results in a linear geometry, which can be visualized as a diagonal arrangement of electron density.

Option A (Small cation and large anion) is correct because according to Fajan's rule, polarization is more pronounced with a small cation and a large anion.

The correct answer is C because N2^-2 has 16 electrons, resulting in 2 unpaired electrons and making it paramagnetic, not diamagnetic.

The correct answer is A ($\mathrm{NO}^{-}$) because it has unpaired electrons, making it paramagnetic.

The correct answer is $\mathrm{S}{\mathrm{O}_2}$ because it has a central sulfur atom with one lone pair and two bonding pairs, resulting in $s{p^2}$ hybridization.

Option B ($\mathrm{RbCl}$) is correct because $\mathrm{RbCl}$ is an ionic compound, and ionic bonds are non-directional due to the electrostatic attraction between ions.

Option A (17) is correct because the percentage ionic character is calculated by dividing the experimental dipole moment by the theoretical dipole moment and multiplying by 100.

Option B (In solid state) is correct because ionic compounds in the solid state do not have mobile ions to conduct electricity.

Option D (Both (A) and (R) are false) is correct because ethyne has one sigma bond and two pi bonds, and pi bonds are weaker than sigma bonds.

Option A (1) is correct because the bond order of $\mathrm{O}_2^{2 - }$ is calculated based on its molecular orbital configuration.

Option D (${BeCl}_2$) is correct because Beryllium has two bond pairs and no lone pairs, resulting in a linear geometry.

Option C (81) is correct because the C–C bond dissociation energy is approximately 81 kcal/mol.

Option A ($\mathrm{C}{\mathrm{H}_4} < N{\mathrm{F}_3} < N{\mathrm{H}_3} < {\mathrm{H}_2}\mathrm{O}$) is correct because it represents the increasing order of dipole moments based on molecular geometry and electronegativity differences.

Option A (Have good $\pi$-accepting character) is correct because ligands with good pi-accepting character stabilize low oxidation states of metals by back-bonding.

Option A (${Cl}_2$) is correct because it has a single sigma bond and no pi bonds, while the other options have pi bonds.

Option C (Melting point of a covalent compound is more than an ionic compound) is not a correct statement because ionic compounds generally have higher melting points than covalent compounds.

Option B (CsCl) is correct because Cesium (Cs) is the most electropositive element among the options, leading to a more ionic and thus stronger bond.

Option B is correct because in heteronuclear diatomic molecules, the bonding molecular orbital is more similar to the more electronegative atom.

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the polarity of a molecule arises from the non-coincidence of positive and negative charge centers.

Option C is correct because N2^2- has 16 electrons, resulting in unpaired electrons in the π*2px and π*2py antibonding orbitals.

Option C (Square pyramidal) is correct because $\mathrm{XeF}_5^+$ has 5 bonding pairs and one lone pair, leading to a square pyramidal shape.

Option C ($S{O_4}^{ - 2} > P{O_4}^{ - 3} > B{O_3}^{ - 3}$) is correct because the bond order is determined by the number of resonance structures, with sulfate having the highest and borate the lowest.

Option B (Trigonal bipyramidal) is correct because $PF_5$ has five bonding pairs and no lone pairs, resulting in $sp^3d$ hybridization and a trigonal bipyramidal geometry.

Option A (High M.P. and Low B.P.) is correct because electrovalent (ionic) compounds typically have high melting points and high boiling points due to strong electrostatic forces between ions, requiring significant energy to overcome.

Option C is correct because He2 does not exist as a stable molecule due to its bond order being zero.

Option C (Low spin d5) is correct because the low spin d5 configuration in an octahedral complex has the maximum CFSE (-2.0Δo).

Option C (Benzene) is correct because it is a nonpolar hydrocarbon and lacks the ability to form hydrogen bonds with water, making it the least soluble.

Option C (${Cu}^{++}$) is correct because it has an unpaired electron, making it paramagnetic.

Option B (Lewis) is correct because Gilbert N. Lewis proposed the electronic theory of bonding, which explains chemical bonds in terms of electron sharing or transfer.

Option A (It has maximum ionic character) is correct because, according to Fajan's rule, smaller anions have less polarization and therefore higher ionic character and melting point.