Chemical bonding & molecular structure practice · Mixed difficulty

Option D (0,0) is correct because the formal charge on both the carbon and oxygen atoms in CO2 is zero.

Option B (H-bonding) is correct because the high heat of vaporization of water is due to the energy required to break the extensive network of hydrogen bonds between water molecules.

Option B (Mixing up of atomic orbitals) is correct because hybridization is the process of mixing atomic orbitals to form new hybrid orbitals with different energies, shapes, and orientations.

Option A ($LiCl$) is correct because, according to Fajan's rule, smaller cations have a greater polarizing power, leading to increased covalent character, and Li+ is the smallest cation among the given options.

Option A (It has maximum ionic character) is correct because the strength of ionic bonds, and thus the melting point, increases with increasing ionic character.

Option D (Decreasing electronegativity) is correct because as electronegativity decreases down the group, the bond pair electrons are further away from the central atom, reducing bond angle.

Option C (B, C, E only) is correct because it correctly describes the trends in covalent character based on Fajan's rule and charge density.

Option C ($A\,and\,D$) is correct because both $CO_2$ and $HgCl_2$ and $C_2H_2$ are linear molecules.

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

Option C is correct because the bond angle in H2S is smaller than in H2O but larger than in H2Se due to the size and electronegativity of the central atom.

Option B (F > Cl > Br > I) is correct because the reactivity of halogens decreases down the group due to decreasing electronegativity and increasing atomic size.

Option C (Bond energies of sigma-and- pi-bonds are of the order of 264 KJ/mol and 347 KJ/mol, respectively) is correct because the bond energy of a sigma bond is generally higher than that of a pi bond. The typical values are approximately 347 kJ/mol for sigma and 264 kJ/mol for pi bonds.

Option C ($\mathrm{CH}{\mathrm{F}_3}$) is correct because the vector sum of bond dipoles is maximized due to the electronegativity difference and molecular geometry.

Option C (Silica) is correct because silica (SiO₂) has a giant network structure with strong covalent bonds throughout.

Option D (${CH}_3{CN}$) is correct because the carbon-nitrogen bond in a triple bond has the shortest bond length compared to single or double bonds.

Option B (Electrovalent bond) is correct because CO molecule has a covalent bond, a sigma bond, a pi bond, and a dative bond, but not an electrovalent bond.

Option C ($\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7 > \mathrm{H}_2\mathrm{S}\mathrm{O}_4 > \mathrm{H}_2\mathrm{S}\mathrm{O}_3$) is correct because ${H_2}{S_2}{O_7}$ has the most pi bonds, followed by ${H_2}S{O_4}$, and then ${H_2}S{O_3}$.

Option C (Hydrogen bonds) is correct because water molecules form strong hydrogen bonds due to the high electronegativity of oxygen, leading to higher boiling points compared to hydrogen sulfide.

Option A is correct because 1 Debye is equal to 3.33 × 10^-30 coulomb meter.

Option B ($120^\circ$) is correct because $sp^2$ hybridization corresponds to a trigonal planar geometry with bond angles of 120 degrees.

Option C ($\mathrm{B}{\mathrm{F}_3}$) is correct because boron trifluoride has a trigonal planar geometry, meaning all four atoms (one boron and three fluorine) lie in the same plane.

Option C (The electronegativity of chlorine is greater than that of hydrogen) is correct because the difference in electronegativity between Cl and H creates a dipole moment, making the HCl molecule polar.

Option B ($N$) is correct because nitrogen forms $NCl_3$, $N_2O_5$, and $Ca_3N_2$ but cannot form $NCl_5$ due to the absence of d-orbitals for expansion of its octet.

Option B (Covalent) is correct because diamond consists of carbon atoms covalently bonded in a tetrahedral network.

Option C (Carbon molecule) is correct because C2 has 12 electrons, leading to a configuration that results in all electrons being paired, hence diamagnetic.

Option A (${\mathrm{K}^ + },\mathrm{C}{\mathrm{l}^ - },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }}$) is correct because all ions have 18 electrons, making them isoelectronic.

Option B (22) is correct because the structure of 2-formylhex-4-enoic acid contains 16 sigma bonds and 3 pi bonds, totaling 22 bonds.

Option A (Naphthalene) is correct because it is a molecular solid with weak van der Waals forces, leading to a lower melting point compared to the other options which are ionic, covalent network, or metallic solids.

Option B is correct because both formic acid (HCOOH) and acetic acid (CH3COOH) have both a hydrogen bond donor (the -OH group) and a hydrogen bond acceptor (the carbonyl oxygen), allowing for strong intermolecular hydrogen bonding.

Option A (${O_2}^ +$) is correct because the species with the highest bond order has the shortest bond length, and ${O_2}^ +$ has the highest bond order among the given options.