Chemical bonding & molecular structure practice · Mixed difficulty

Option A (Sigma-bond) is correct because the linear combination of two hybridized orbitals, each containing one electron, forms a sigma bond.

Option A (Propyne) is correct because the carbon atom bonded to the other carbon atom is sp hybridized, resulting in the shortest bond length.

Option A ($\mathrm{M}{\mathrm{g}_\mathrm{3}}\,{\mathrm{X}_\mathrm{2}}$) is correct because Magnesium forms Mg²⁺ ions and X forms X³⁻ ions, leading to the formula Mg₃X₂.

Option D ($ICl_2^+$) is correct because it has a bent shape due to the presence of two bond pairs and two lone pairs around the central iodine atom, resulting in $sp^3$ hybridization.

Option C ($4U$) is correct because lattice energy is proportional to the product of the charges of the ions. CaO has charges of +2 and -2, while NaCl has charges of +1 and -1. Therefore, the lattice energy of CaO is approximately 4 times that of NaCl.

Option D (Carbon tetrachloride) is correct because carbon tetrachloride ($CCl_4$) has four identical bonds and a tetrahedral geometry, resulting in all bond angles being exactly $109^\circ 28`$.

Option C is correct because bond strength increases with increasing electronegativity difference between the bonded atoms.

Option B ($S$) is correct because sulfur exists as $S_8$ molecules, meaning each molecule contains eight sulfur atoms.

Option B ($\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}$) is correct because it correctly orders the molecules based on their dipole moments, which reflect the polarity of the bonds and molecular geometry.

Option A (1,3) is correct because both $C{N^-}$ and $N{O^+}$ have 14 electrons, resulting in a bond order of 3.

Option D ($\mathrm{Si}{\mathrm{O}_2}$) is correct because it has a network structure and does not have delocalized electrons or multiple resonance structures.

Option A is correct because ionic character increases with increasing charge on the cation and decreasing size of the cation, which is not reflected in the given order.

Option C ($HF$ is less polar than $HBr$) is not true because HF is more polar than HBr due to the higher electronegativity of fluorine compared to bromine.

Option D (AgBr) is correct because Van der Waals forces increase with increasing molecular size and polarizability, and AgBr has the largest and most polarizable atoms among the given options.

Option A ($Si{F_4}\mathrm{ and}\,S{F_4}$) is correct because $SiF_4$ is tetrahedral and $SF_4$ is see-saw shaped, hence they are not isostructural.

Option D is correct because $\mathrm{SbCl}_5^{2 - }$ has $sp^3d^2$ hybridization, while the others have $sp^3d$ hybridization.

Option A is correct because 1-butyne has a higher s-character in the sp hybridized carbon, leading to a greater dipole moment compared to 1-butene.

Option D ($\mathrm{IO}_3^{-} > \mathrm{BrO}_3^{-} > \mathrm{ClO}_3^{-}$) is correct because the bond angle decreases down the group due to increasing size and decreasing electronegativity of the central atom, making the given order incorrect.

Option A is correct because in calcium carbide (CaC₂), each carbon atom is bonded to one calcium atom (ionic) and the two carbon atoms are triple bonded (one sigma and two pi bonds).

Option A ($\mathrm{\theta } = 90\mathrm{^\circ }$) is correct because the dipole moment is maximized when the bond angle is 90 degrees, resulting in maximum vector addition of bond dipoles.

Option A (carbon, fluorine) is correct because carbon typically forms four bonds, and fluorine typically forms one bond, matching the dot structure.

Option B (${B_2}$) is correct because ${B_2}$ has two unpaired electrons according to molecular orbital theory.

Option D ($KI$ has high M.P) is wrong because $KI$ has a lower melting point compared to $CaO$ due to weaker electrostatic forces between the ions.

Option D (${O}_2^{+}>{O}_2>{O}_2^{-}>{O}_2^{2-}$) is correct because the bond order increases with the removal of electrons from antibonding orbitals and decreases with the addition of electrons to antibonding orbitals.

Option C (16.66) is correct because it is the calculated percentage ionic character using the given bond length and dipole moment.

Option D (None of these) is correct because the bond angle in $AsH_3$ is smaller than in $NH_3$ and $H_2O$, and $BCl_3$ has a bond angle of 120 degrees.

Option B ($AlH_3$ combines with $H^−$) is correct because the hybridization of Al changes from $sp^2$ to $sp^3$ upon accepting a hydride ion.

Option A (Greater is the ionic nature) is correct because a larger dipole moment indicates a greater separation of charge, hence a greater ionic character.

Option C (hydrogen bond) is correct because the high boiling point of water compared to hydrogen sulfide is due to the presence of hydrogen bonding in water.

Option D is correct because as you go down the group, bond angle decreases due to increased size and decreased electronegativity of the central atom.