Chemical bonding & molecular structure practice · Medium

Option B (CsCl) is correct because Cesium (Cs) is the most electropositive element among the options, leading to a more ionic and thus stronger bond.

Option B is correct because in heteronuclear diatomic molecules, the bonding molecular orbital is more similar to the more electronegative atom.

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the polarity of a molecule arises from the non-coincidence of positive and negative charge centers.

Option C is correct because N2^2- has 16 electrons, resulting in unpaired electrons in the π*2px and π*2py antibonding orbitals.

Option C (Square pyramidal) is correct because $\mathrm{XeF}_5^+$ has 5 bonding pairs and one lone pair, leading to a square pyramidal shape.

Option C ($S{O_4}^{ - 2} > P{O_4}^{ - 3} > B{O_3}^{ - 3}$) is correct because the bond order is determined by the number of resonance structures, with sulfate having the highest and borate the lowest.

Option B (Trigonal bipyramidal) is correct because $PF_5$ has five bonding pairs and no lone pairs, resulting in $sp^3d$ hybridization and a trigonal bipyramidal geometry.

Option C (Low spin d5) is correct because the low spin d5 configuration in an octahedral complex has the maximum CFSE (-2.0Δo).

Option B (Lewis) is correct because Gilbert N. Lewis proposed the electronic theory of bonding, which explains chemical bonds in terms of electron sharing or transfer.

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

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.

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 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.

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 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 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.

Option B is correct because in the given transformation, the hybridization of N changes from sp3 to sp3 after the reaction with BF3, and the shape changes from trigonal pyramidal to tetrahedral.

The correct answer is Between 22 - 23% because as the bond angle decreases from 109.5 degrees (sp3 hybridization), the s-character increases, and a bond angle of 105 degrees corresponds to an s-character between 22-23%.