Chemical bonding & molecular structure practice · Mixed difficulty

Option D (Magnesium sulphide) is correct because larger cation and larger anion leads to lower lattice energy and hence more soluble.

Option B (Valency electrons) is correct because valency electrons are the electrons in the outermost shell of an atom, which are involved in chemical bonding.

Option C (Due to p - p overlapping) is correct because fluorine atoms have the electronic configuration 1s² 2s² 2p⁵, and the bond is formed by the overlap of p orbitals.

Option C (${X}=$ Total number of valence electrons in the free atom, ${Y}=$ Total number of non bonding electrons, ${Z}=$ Total number of bonding electrons) is correct because it correctly represents the formula for calculating formal charge.

Option A ($\mathrm{N}_2^ +$) is correct because both $\mathrm{O}_2^ +$ and $\mathrm{N}_2^ +$ have the same number of electrons and therefore the same bond order.

Option B (${\mathrm{H}_2}\mathrm{O}$) is correct because the presence of two lone pairs on the oxygen atom causes greater repulsion, leading to a smaller bond angle compared to the other options.

Option A (Coulombic interaction energies) is correct because ionic compounds are held together by the electrostatic attraction between oppositely charged ions, which is described by Coulomb's law.

Option A ($3.0\,{A^0}$) is correct because the length of a hydrogen bond typically ranges from $2.5\,{A^0}$ to $3.0\,{A^0}$.

Option B is correct because the bond angle in ice is close to the tetrahedral angle due to hydrogen bonding.

Option B ($\mathrm{O}_2^{2 - }$) is correct because it has all its electrons paired, making it diamagnetic, while the other options have unpaired electrons, making them paramagnetic.

Option A (Increases) is correct because as the s-character increases, the bond angle also increases due to the increased directionality of the hybrid orbitals.

Option B (Inertness of noble gases) is correct because Kossel and Lewis based their theory on the stable electronic configurations of noble gases.

Option C (${\mathrm{I}^ - },\mathrm{B}{\mathrm{r}^ - },\mathrm{C}{\mathrm{l}^ - },{\mathrm{F}^ - }$) is correct because polarizability increases with size. Larger ions are more easily polarized.

Option B (${NO}_2^{+}$) is correct because the central nitrogen atom in $NO_2^+$ has two bond pairs and no lone pairs, resulting in a linear shape.

Option B ($Al_2O_3$) is correct because lattice energy is directly proportional to the charge of the ions and inversely proportional to the size of the ions. $Al_2O_3$ has the highest charges (+3 and -2) compared to the other options.

Option C (4.8) is correct because the dipole moment is calculated directly from the given charge and distance.

Option A ($\sigma 2{p_x}$) is correct because a sigma bonding orbital formed from p orbitals along the x-axis has two nodal planes: one nodal plane is the one perpendicular to the internuclear axis and passing through the center, and the other is the one that is intrinsic to the p-orbital.

Option A (${PCl}_3$) is correct because the central phosphorus atom in $PCl_3$ has three bond pairs and one lone pair, resulting in $sp^3$ hybridization.

Option C ($\mathrm{C}$) is correct because carbon exists in both hard (diamond) and soft (graphite) allotropic forms.

Option A (3, diamagnetic) is correct because $C{N^ - }$ has 14 electrons, resulting in a bond order of 3 and is diamagnetic.

Option C ($\mathrm{CO}_3^{2 - }$) is correct because it has a trigonal planar geometry, not tetrahedral.

Option B ($\mathrm{CaCl}_2 > \mathrm{CdCl}_2$) is correct because the melting point depends on the ionic character, which is higher for $\mathrm{CaCl}_2$ due to the higher electropositivity of Ca compared to Cd.

Option D (None of these) is incorrect, and the correct answer is B ($\mathrm{S}{\mathrm{F}_4}$) because in $SF_4$, the sulfur atom is $sp^3d$ hybridized due to the presence of a lone pair, while in $\mathrm{SO}_4^{2 - }$ and $\mathrm{S}{\mathrm{F}_2}$, the sulfur atom is $sp^3$ hybridized.

Option A (${N}_2>{N}_2^{+}$) is correct because removing an electron from $N_2$ to form $N_2^+$ decreases the bond order, thus reducing the dissociation energy.

Option B (3) is correct because the bond order in ${CN}^{-}$ ion is 3, calculated from its molecular orbital configuration.

Option C ($2 - \sigma ,\,2 - \pi$) is correct because in $Mg_2C_3$, the $C_3^{4-}$ ion has the structure $^{-}C \equiv C - C^{-}$, which contains two sigma and two pi bonds.

The correct answer is A because nitrogen, being a second-period element, lacks available 2d orbitals to expand its octet and form five bonds, unlike phosphorus which has vacant 3d orbitals.

Option C (${A} l {~N}$) is correct because lattice energy is directly proportional to the product of the charges of the ions. AlN has the highest charges (+3 and -3) among the given options.

Option D (Single electron with opposite spin) is correct because covalent bonds are formed by the sharing of electrons with opposite spins to achieve a stable electron configuration.

Option B (A covalent bond is formed by the overlapping of orbitals with unpaired electrons of same spins) is correct because VBT states that covalent bonds are formed by overlapping orbitals with unpaired electrons of *opposite* spins.