Chemical bonding & molecular structure practice · Mixed difficulty

Option B ($120^{\circ}$) is correct because ethylene has a trigonal planar geometry around each carbon atom due to $sp^2$ hybridization.

Option A (8) is correct because there are 8 species among the given list that do not exist due to various chemical reasons such as exceeding the octet rule or instability.

The correct answer is D because O2^2- is diamagnetic (no unpaired electrons), while O2^- is paramagnetic (one unpaired electron).

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

The correct answer is B because both the assertion and reason are true, but the small size of boron is not the primary reason for the instability of BH6^-3.

Option A (17) is correct because the percentage ionic character is calculated by dividing the experimental dipole moment by the theoretical dipole moment and multiplying by 100.

The correct answer is ${N}_2^{+}$ because it has a bond order of 2.5, while the other options have a bond order of 3.

Option C is correct because the ability of metal ion layers to slide over each other, facilitated by delocalized electrons, accounts for malleability and ductility.

Option B ($O_2^ + ,\,N_2^ -$) is correct because both species have the same number of electrons and thus the same bond order of 2.5.

Option A is correct because $C_2^{2-}$ has 14 electrons, resulting in a completely filled molecular orbital configuration and thus is diamagnetic.

Option A ($40 \mathrm{~kJ} \mathrm{~mol}^{-1}$) is correct because the energy of a hydrogen bond is typically in the range of 10-40 kJ/mol.

Option B (Coordinate covalent) is correct because the bond between B and O in the complex is formed by the donation of a lone pair of electrons from oxygen to boron, which is a coordinate covalent bond.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because fluorine is smaller and more electronegative than chlorine, thus minimizing bond pair-bond pair repulsions in the axial position.

Option C (B, C, E only) is correct because the statements B, C, and E regarding the order of covalent character are accurate based on Fajan's rules and oxidation states.

Option A (Decreases) is correct because as the p-character increases in hybrid orbitals, the s-character decreases, leading to smaller bond angles.

Option B (${O}_2^{+}$) is correct because it has the highest bond order among the given species, resulting in the shortest bond length.

Option D is correct because the hybrid state of N in NO3- and NO2- is sp2, while in NO2+ it is sp.

Option C (${O}_2^{2-}$) is correct because it has the highest number of antibonding electrons among the given species, with 8 antibonding electrons.

The correct answer is ${NH}_3$ because the nitrogen atom in ammonia has one lone pair of electrons.

The correct answer is C (Conduct electricity) because fused ionic compounds have mobile ions that can carry charge, allowing them to conduct electricity.

Option B ($\mathrm{NO}_3^ -$) is correct because it has a trigonal planar geometry, not a pyramidal geometry.

Option B (${XCl}_2$) is correct because element X has 2 valence electrons, so it will form a chloride with the formula XCl2.

Option D (Hydrogen bond) is correct because hydrogen bonds are the weakest among the given types of chemical bonds due to their intermolecular nature.

Option A (1 and diamagnetic) is correct. If Hund's rule is violated, both electrons will occupy the same π2p orbital, resulting in a diamagnetic molecule with a bond order of 1.

Option D ($C – H$) is correct because the electronegativity difference between carbon and hydrogen is the smallest among the given options, leading to the least bond polarity.

Option B (Orbitals will not overlap) is correct because, according to valence bond theory, orbitals with electrons of the same spin will not overlap to form a bond.

Option D (${H}_2 {Sc}$) is likely to have the smallest bond angle due to the large size and low electronegativity of Selenium, leading to less repulsion between the bond pairs.

Option B (${\mathrm{H}_2}$) is correct because it has two paired electrons, making it diamagnetic.

Option A is correct because CO2, HgCl2, and C2H2 all have a linear structure.

Option C (Assertion is True but the Reason is False) is correct because the existence of R3P=O is due to the ability of P to form pi bonds with oxygen, while N cannot form strong pi bonds.