Chemical bonding & molecular structure practice · Mixed difficulty

Option B ($\mathrm{B}{\mathrm{F}_3}\,\,\mathrm{and NO}_2^ -$) is correct because both $BF_3$ and $NO_2^-$ have central atoms with three $sp^2$ hybrid orbitals.

Option A ($\mathrm{SiC}{\mathrm{l}_4}$) is correct because, according to Fajan's rule, a smaller cation and a larger anion, along with a high charge on either ion, favor covalent character. Si has a smaller size and higher charge compared to the other metal cations.

Option A ($\mathrm{O}_2^ -$) is correct because it has an odd number of electrons in its molecular orbitals, resulting in unpaired electrons and thus paramagnetism.

Option A (Decrease in bond length and increase in bond energy) is correct because a higher bond order indicates more bonds between atoms, leading to a shorter and stronger bond.

Option B (Sigma bond is weaker than $π$-bond) is correct because sigma bonds are formed by head-on overlap, making them stronger than pi bonds formed by sideways overlap.

Option C (The molecule is non - planar with the two $-\mathrm{C}{\mathrm{H}_2}$ groups being in planes perpendicular to each other) is correct because the central carbon is sp hybridized, leading to perpendicular planes for the terminal CH2 groups.

Option C ($\mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CHCl}_3$) is correct because both dichloromethane and chloroform have non-zero dipole moments due to their asymmetrical structures and polar C-Cl bonds.

Option A ($\mathrm{C}{\mathrm{l}_2}$) is correct because bond energy generally decreases down the halogen group, but F2 is an exception due to its small size and high interelectronic repulsion.

Option C (N ≡ N bond in ${N_2}$) is correct because triple bonds are stronger than double or single bonds, and the N≡N triple bond is exceptionally strong due to the small size and high electronegativity of nitrogen.

Option A (32) is correct because the total number of valence electrons in $SO_4^{-2}$ is 32.

Option B is correct because the Debye unit is defined as 1 × 10^-18 esu cm.

Option C (Ions) is correct because ions are atoms or groups of atoms that carry an electrical charge.

Option D ($s{p^3}d -$ hybridized) is correct because in $\mathrm{PC}{\mathrm{l}_5}$, the phosphorus atom is $sp^3d$ hybridized to accommodate five bonding pairs.

Option D is correct because both $\mathrm{PO}_4^{3 - }$ and $\mathrm{SO}_4^{2 - }$ have tetrahedral geometry and $sp^3$ hybridization.

Option B (${MCl}_2$) is correct because if MHPO4 has a +2 and -2 charge, then M has a +2 charge. Chloride has a -1 charge, so the formula will be MCl2.

Option D ($P{F_5}$) is correct because $PF_5$ exhibits $sp^3d$ hybridization, not $sp^3d^2$ hybridization.

Option B ($\mathrm{CsCl}$) is correct because the size difference between Cs+ and Cl- is larger, leading to a weaker electrostatic attraction and a lower lattice energy, implying a weaker bond.

Option B (One sigma $\left( \mathrm{\sigma } \right)$and two pi $\left( \mathrm{\pi } \right)$-bonds) is correct because in $CaC_2$, the $C_2^{2-}$ ion has a triple bond between the carbon atoms, consisting of one sigma and two pi bonds.

Option B ($\mathrm{M}_2 \mathrm{O}$) is correct because the electronic configuration indicates M has 1 valence electron, so it will form a +1 ion. Oxygen forms a -2 ion, so the formula will be M2O.

Option C ($\mathrm{BC}{\mathrm{l}_3}$) is correct because Boron in $BCl_3$ is $sp^2$ hybridized, not $sp^3$ hybridized.

Option A ($BH_4^ -$) is correct because boron has four bond pairs and no lone pairs, resulting in a tetrahedral geometry.

Option B ($\mathrm{C}_2$) is correct because the molecular orbital configuration of C2 shows only π bonds are present.

Option B ($S{O_2}$) is correct because sulfur has one lone pair and two bond pairs, resulting in $sp^2$ hybridization.

Option C (Hydrogen bond) is correct because hydrogen bonds between amino acids are the primary force determining the secondary structure of proteins.

Option B (Two molecular orbitals) is correct because when two atomic orbitals combine, they form two molecular orbitals: one bonding and one antibonding.

Option A (${\mathrm{N}{\mathrm{O}^ - }}$) is correct because it has a bond order of 2, while $CO$, ${\mathrm{N}{\mathrm{O}^ + }}$ and ${\mathrm{C}{\mathrm{N}^ - }}$ have a bond order of 3.

Option C is correct because both $\mathrm{SO}_4^{2 - }$ and $\mathrm{BF}_4^ -$ have tetrahedral geometry and $sp^3$ hybridization.

Option B (${SO}_3, {C}_2 {H}_4$) is correct because both $SO_3$ and $C_2H_4$ have central atoms with three $sp^2$ hybrid orbitals.

Option A ($X Y_2$) is correct because X has 2 valence electrons and Y needs 1 electron to complete its octet, so X forms a +2 ion and Y forms a -1 ion, leading to the formula XY2.

Option A ($\mathrm{Cl}{\mathrm{O}_2}$) is correct because it has an odd number of valence electrons, making it paramagnetic.