Chemical bonding & molecular structure practice · Mixed difficulty

Option A (Cl2) is correct because, among the given halogens, Cl2 has the highest bond dissociation energy due to a balance between bond length and interelectronic repulsions.

Option B (Its regular tetrahedral structure) is correct because the symmetrical tetrahedral arrangement of CCl4 cancels out the individual bond dipoles, resulting in a net dipole moment of zero.

Option C (${CH}_3 {OH}$) is the most polar because it has the largest difference in electronegativity between its atoms and a bent molecular geometry, resulting in a significant dipole moment.

Option D (If both Assertion & Reason are false) is correct because p-nitrophenol is less volatile than o-nitrophenol due to intermolecular hydrogen bonding, and their molecular weights are the same.

Option C ($H$-bonding) is correct because the extensive hydrogen bonding network in water requires a large amount of energy to break, leading to its high specific heat.

Option C (Electrovalent, covalent and coordinate) is correct because $CuSO_4.5H_2O$ contains ionic bonds between Cu and $SO_4$, covalent bonds within the sulfate ion, and coordinate bonds between Cu and water ligands.

Option C (B and C only) is correct because $SO_4^{2-}$ and $CrO_4^{2-}$ are both tetrahedral, and $CO_3^{2-}$ and $NO_3^-$ are both trigonal planar.

Option D (Both (1) & (3)) is correct because in both reactions, the hybridization of the central atom changes from sp2 to sp3 in BF3 and from sp3d to sp3d2 in PCl5.

The correct answer is $\mathrm{SiO}_2$ is linear and hybridisation of $\mathrm{Si}$ is $s p$ because $SiO_2$ has a three-dimensional structure with $sp^3$ hybridization.

Option B ($s{p^3}$ hyridisation and distorted tetrahedral geometry) is correct because ammonia has $s{p^3}$ hybridization with a lone pair, resulting in a distorted tetrahedral (pyramidal) geometry.

Option B ($s{p^2}$-hybridized) is correct because in $\mathrm{NO}_3^ -$, the nitrogen atom is $s{p^2}$ hybridized, leading to a trigonal planar geometry.

Option D is correct. The assertion is incorrect because NH₃ has a pyramidal shape, not tetrahedral. The reason is correct because nitrogen in NH₃ is sp³ hybridized.

Option B (p-p orbitals) is correct because a $\pi$-bond is formed by the sideways (lateral) overlapping of p-p orbitals.

Option B ($\left[ {\mathrm{BF}_4^ - \,\mathrm{and }\,\mathrm{NH}_4^ + } \right]$) is correct because both ions have $sp^3$ hybridization and tetrahedral geometry.

Option B (${CH}_2 {Cl}_2>{CHCl}_3>{CH}_3 {Cl}>{CCl}_4$) is correct because the dipole moment depends on the vector sum of bond dipoles, and the symmetry in $CCl_4$ cancels out the dipole moment.

Option D (Number of valency electrons) is correct because copper loses two valence electrons to form the Cu²⁺ ion.

Option D ($\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$) is correct because the electronegativity difference between oxygen and hydrogen is significant, leading to a large dipole moment that is not canceled out by other bonds in the molecule.

Option C ($9\mathrm{\sigma }$ and $2\mathrm{\pi }$ bonds) is correct because 1,3-butadiene ($\mathrm{C}{\mathrm{H}_2} = \mathrm{CH}-\mathrm{CH} = \mathrm{C}{\mathrm{H}_2}$) has 9 sigma bonds and 2 pi bonds.

Option C (Chemical bonding) is correct because Linus Pauling received the Nobel Prize in Chemistry for his work on the nature of the chemical bond and its application to the elucidation of the structure of complex substances.

The correct answer is C ($\mathrm{CO}_3^{2 - }$) because it has the lowest bond order (1.33) among the given species, resulting in the weakest $\mathrm{C} - \mathrm{O}$ bond.

Option C (Though ${C}-{Cl}$ bond is polar, ${CCl}_4$ is non polar) is correct because the individual polar ${C}-{Cl}$ bonds are arranged symmetrically around the carbon atom, resulting in the cancellation of bond dipoles and a net dipole moment of zero.

Option D (O2^2-) is correct because the bond order of O2^2- is the lowest among the given options, leading to the longest bond length.

Option A ($Q=$ charge, $r=$ distance of separation) is correct because the dipole moment is defined as the product of the magnitude of the charge and the distance between the charges.

Option A is correct. $NO_3^-$ and $CO_3^{2-}$ are isoelectronic (same number of electrons) and isostructural (same shape - trigonal planar).

Option C (8 electrons) is correct because Lewis's octet rule states that atoms tend to achieve a stable configuration with eight electrons in their outermost shell.

The correct answer is $\mathrm{N}_2^{2-}$ because it has the lowest bond order among the given species, indicating the weakest bond and thus the minimum bond energy.

Option A ($120^\circ$) is correct because a molecule of type $AX_3$ with zero dipole moment has a trigonal planar geometry, resulting in bond angles of 120 degrees.

Option D ((i), (ii), (iii), (iv) and (v)) is correct because all the listed molecules and ions (H2O, NH4+, SO42-, ClO4-, NH3) involve sp3 hybridization in their formation.

Option A (Electrovalent compounds) is correct because ionic compounds have strong electrostatic forces, leading to high melting and boiling points.

Option B is correct because removing an electron from O₂ increases its bond order from 2 to 2.5, thus increasing stability.