Chemical bonding & molecular structure practice · Mixed difficulty

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

Option B (Unpaired) is correct because paramagnetic compounds have unpaired electrons, which cause them to be attracted to a magnetic field.

Option D ($A{B_5} - \mathrm{ }\,\mathrm{Square \,pyramidal }$) is the incorrect matching because $AB_5$ has a trigonal bipyramidal geometry, not square pyramidal.

Option C (3) is correct because the bond order of $\mathrm{O}_2^{2 + }$ is calculated using Molecular Orbital Theory, resulting in a bond order of 3.

Option D ($Xe{F_4}$) is correct because $Xe{F_4}$ has a square planar geometry, allowing all 5 atoms to lie in the same plane.

Option C (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 D (6 electrons) is correct because a triple bond involves the sharing of three pairs of electrons, totaling six electrons.

Option C ($c_2 a, d a, c b, d b_2$) is correct because it accurately represents the ionic compounds formed based on the electronic configurations and resulting charges of the elements.

Option D (${\mathrm{C}_2}{\mathrm{H}_2}$) is correct because the carbon-carbon bond length is shortest in alkynes (triple bond) compared to alkanes (single bond) or alkenes (double bond).

Option B ($\mathrm{SnC}{\mathrm{l}_4}$) is correct because Sn in SnCl4 has a complete octet and does not violate the octet rule.

Option A (4) is correct because each water molecule can act as both a hydrogen bond donor (through its two H atoms) and a hydrogen bond acceptor (through its two lone pairs), allowing it to form a maximum of four hydrogen bonds.

Option D (See – Saw and square pyramidal) is correct because $SF_4$ has a see-saw shape and $BrF_5$ has a square pyramidal shape according to VSEPR theory.

Option A ($\mathrm{NaCl} < LiCl < BeC{\mathrm{l}_2}$) is correct because covalent character increases with increasing polarization, which is favored by smaller cations and larger anions, and higher charge.

Option D ($sp - sp$) is correct because hybrid orbitals provide more efficient overlap than atomic orbitals, and $sp$ orbitals have the highest s-character among the given options, leading to stronger and more directional bonds.

Option B (Pentagonal pyramidal) is correct because $\mathrm{Xe}{\mathrm{F}_6}$ has a distorted octahedral geometry due to the presence of a lone pair, resulting in a pentagonal pyramidal shape.

Option D (Hydrogen bond) is correct because hydrogen bonds are the weakest among the listed types of chemical bonds.

Option C ($\mathrm{O}_2^{2 - },{\mathrm{B}_2}$) is correct because both species have a bond order of 1.

Option D is correct because the conversion factor between Debye and Coulomb meter is 1 D = 3.33564 × 10^-30 C m.

Option D (Containing unpaired electrons) is correct because paramagnetism arises from the presence of unpaired electrons.

Option A ($90^{\circ}$) is correct because the angle between the two O-H planes in the H2O2 molecule is approximately 90 degrees due to its open book structure.

Option D ($\mathrm{O}$-$\mathrm{H}$) is correct because the electronegativity difference between oxygen (3.5) and hydrogen (2.1) is the largest among the given options, resulting in the most polar bond.

Option C ($C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}$) is correct because as the symmetry decreases from $CH_3Cl$ to $CHCl_3$, the dipole moment decreases, and $CCl_4$ has zero dipole moment due to its tetrahedral symmetry.

Option C (${C_2}$) is correct because among $2S - 2P$ mixing is not operative, ${C_2}$ is paramagnetic.

Option D (${M}\left({NO}_3\right)_3$) is correct because if MPO4 has a +3 and -3 charge, then M has a +3 charge. Nitrate has a -1 charge, so the formula will be M(NO3)3.

Option A (32) is correct because the total number of valence electrons is calculated by adding the valence electrons of each atom and the charge of the ion.

Option A (${\mathrm{C}_2}{\mathrm{H}_6}$) is correct because the C-H bond distance is longest in alkanes due to single bonds having the greatest bond length compared to double or triple bonds.

Option D (${H}_2 {SO}_4$) is correct because in sulfuric acid, one of the oxygen atoms forms a coordinate covalent bond with the sulfur atom, where sulfur donates both electrons for the bond.

Option D ($2.28 \,\mathop {\text{A}}\limits^ \circ$) is the correct bond length for ${Br}-{Br}$ in the ${Br}_2$ molecule, as it is the experimentally determined value.

Option A (Decreases) is incorrect. Bond energy decreases with the increase in the number of lone pairs on the bonded atoms due to increased repulsion.

Option A (Electrovalent bond) is correct because an electrovalent bond (also called ionic bond) is formed by the transfer of electrons between two atoms.