Chemical bonding & molecular structure practice · Mixed difficulty

Option A (+2,-1) is correct because calcium loses 2 electrons to achieve a stable configuration, resulting in a +2 charge, while chlorine gains 1 electron, resulting in a -1 charge.

Option C ($LiCl$) is correct because lattice energy is inversely proportional to the size of the ions, and Li+ is the smallest cation among the given options.

Option D (${11^{Na}}$) is correct because sodium (Na) is an alkali metal and typically exhibits only a +1 oxidation state, thus it cannot exhibit variable electrovalency.

Option A (Combining capacity of an element) is correct because valency refers to the number of chemical bonds an atom can form.

Option A (Two) is correct because in $Na_2S_2O_3$, each sulfur atom has an oxidation state of +2 on average, and it forms two bonds with other atoms.

Option A (Size of ions is small) is correct because smaller ions lead to a higher charge density and stronger electrostatic attraction, thus increasing lattice energy.

Option B (${\mathrm{N}_2}\mathrm{O}$) is correct because it has the same number of electrons as $\mathrm{C}{\mathrm{O}_2}$.

Option C (Having low value of dipole moment than $\mathrm{N}{\mathrm{H}_3}$) is correct because in $NF_3$, the dipole moments of the N-F bonds are in the opposite direction to the dipole moment of the lone pair on nitrogen, partially canceling it out, while in $NH_3$, the dipole moments of the N-H bonds are in the same direction as the dipole moment of the lone pair.

Option C ($\mathrm{BC}{\mathrm{l}_3}$) is correct because it is trigonal planar and symmetrical, resulting in the bond dipoles canceling each other out, leading to a zero dipole moment.

Option B (– 1, +1, 0) is correct because the formal charges on the oxygen atoms in ozone are -1, +1, and 0, respectively, based on the Lewis structure.

Option C (${\mathrm{Fe}^{3+}}$) is correct because it has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.

Option B (${\mathrm{O}_2^ + }$) is correct because it has the highest bond order (2.5) compared to the other oxygen species.

Option A ($\mathrm{N}{\mathrm{H}_3}$) is correct because ammonia has a three-fold rotational axis of symmetry along the N-H bonds.

Option D (1,3 and 3) is correct because the bond orders for $\mathrm{O}_2^{2-}$, $\mathrm{CO}$ and $\mathrm{NO}^{+}$ are 1, 3, and 3 respectively, based on their molecular orbital configurations.

Option A ($NaCl$) is correct because according to Fajan's rule, smaller cation and larger anion favors covalent character, and larger cation and smaller anion favors ionic character. $NaCl$ has the most ionic character among the given options.

Option B ($NaBr < NaCl < BaO < CaO$) is correct because lattice energy increases with increasing charge and decreasing size of the ions. CaO has the highest charges and relatively small sizes.

Option D (All of these) is correct because X-ray diffraction, electron diffraction, and spectroscopic techniques are all used to measure bond length.

Option B (Coordinate covalent) is correct because in the given compound, the oxygen atom donates a lone pair of electrons to the boron atom, forming a coordinate covalent bond.

Option C is correct because the bond order increases in $\mathrm{N}{\mathrm{O}^ + }$ and decreases in $\mathrm{N}{\mathrm{O}^ - }$, affecting bond energy.

Option B (136 pm) is correct because the bond length of HCl can be approximated by summing the covalent radii of H and Cl, derived from the given bond lengths of H2 and Cl2.

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

Option A (Greater than the bonding $M.O.$) is correct because antibonding molecular orbitals are higher in energy than their corresponding bonding molecular orbitals.

Option A ($C_2^{2 - } > \,N_2^{2 - } > O_2^{2 - }$) is correct because bond order is inversely proportional to the number of antibonding electrons.

Option B ($11.5 \%$) is correct because it is calculated by comparing the actual dipole moment to the theoretical dipole moment assuming full charge separation.

Option C (${d_{{z^2}}}$) is correct because the $d_{z^2}$ orbital is the one that participates in $sp^3d$ hybridization.

Option C ($s{p^3}$) is correct because oxygen in $OF_2$ has two bond pairs and two lone pairs, resulting in tetrahedral electron geometry and $sp^3$ hybridization.

Option D ($\mathrm{Xe}{\mathrm{F}_2},\mathrm{C}{\mathrm{O}_2}$) is correct because both $XeF_2$ and $CO_2$ have a linear shape.

Option D (10, 3) is correct because pent-4-en-1-yne has 10 sigma bonds and 3 pi bonds based on its structure.

Option B (Ionic) is correct because elements with atomic numbers 37 (Rubidium, Rb) and 53 (Iodine, I) are a metal and a nonmetal, respectively, and tend to form ionic bonds.

Option A ($\mathrm{CH}{\mathrm{l}_3}$) is correct because the vector sum of the bond dipoles is the largest in this molecule compared to the others, considering the electronegativity differences and molecular geometry.