Chemical bonding & molecular structure practice · Mixed difficulty

Option A ($NO_2^+ > NO_2 > NO_2^-$) is correct because the bond angle depends on the number of lone pairs on the central atom. $NO_2^+$ has no lone pairs, $NO_2$ has one, and $NO_2^-$ has two.

Option B ($\mathrm{N}{\mathrm{H}_3}$) is correct because ammonia exhibits the highest solubility in water due to its ability to form strong hydrogen bonds and react with water.

Option B (Triple bond) is correct because triple bonds involve the sharing of six electrons, resulting in the strongest bond.

Option B (10) is correct because it is the calculated percentage ionic character of HBr using the given dipole moment and interatomic spacing.

Option D (${\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}$) is correct because polarising power increases with charge and decreases with size. Be2+ has the highest charge density and K+ has the lowest.

Option C ($106^\circ 45'$) is correct because the bond angle in ammonia is approximately 106.75 degrees due to the lone pair on the nitrogen atom repelling the bonding pairs.

Option C ($\mathrm{NO} \to \mathrm{N}{\mathrm{O}^ + }$) is correct because the bond order increases from 2.5 to 3 and the magnetic behavior changes from paramagnetic to diamagnetic.

Option B is correct because BF3 is planar with symmetrical bond dipoles that cancel out, while NF3 is pyramidal with asymmetrical bond dipoles that do not cancel out, resulting in a net dipole moment.

Option B ($s{p^2}$) is correct because the carbon atom in HCHO has three sigma bonds and one pi bond, leading to $sp^2$ hybridization.

Option A (Have good $\pi$-accepting character) is correct because ligands with good $\pi$-accepting character stabilize low oxidation states of metals by accepting electron density from the metal.

Option C ($\mathrm{N}_2^{2 - } < \mathrm{N}_2^ - < {\mathrm{N}_2}$) is correct because the bond order increases in the order $\mathrm{N}_2^{2 - }$ (2) < $\mathrm{N}_2^ -$ (2.5) < ${\mathrm{N}_2}$ (3).

Option D (4 and 3) is correct because boron forms four coordinate bonds and has an oxidation state of +3 in $[BF_4]^{-}$.

Option A ($s{p^2}$-hybridized) is correct because carbon atoms in graphite are bonded to three other carbon atoms in a planar arrangement, resulting in $sp^2$ hybridization.

Option C (Sodium cyanide) is correct because it contains ionic bonds between Na+ and CN-, covalent bonds within the CN- ion, and a coordinate bond between C and N.

Option D (${\mathrm{H}_2}\mathrm{O}$) is correct because hydrogen only needs 2 electrons to complete its duplet, not an octet.

Option C ($s{p^3}$) is correct because oxygen in diethyl ether has two bond pairs and two lone pairs, resulting in $sp^3$ hybridization.

Option D (Five) is correct because acetylene (C₂H₂) has a triple bond between the carbon atoms (one sigma and two pi bonds) and a single bond between each carbon and hydrogen atom (two sigma bonds), totaling five bonds.

Option B (${XCl}_2$) is correct because the electronic configuration indicates X has 2 valence electrons, so it will form a +2 ion. Chlorine forms a -1 ion, so the formula will be XCl2.

Option D ($B{e_2}$) is correct because its bond order is zero, calculated as (4-4)/2 = 0.

Option C (${[IC{l_4}]^ - }$) is correct because Iodine has 6 electron pairs (4 bond pairs and 2 lone pairs), leading to $sp^3d^2$ hybridization and a square planar shape.

Option B (${N_2} < S{O_2} < Cl{F_3} < LiF < {K_2}O$) is correct because ionic character increases with electronegativity difference between the bonded atoms.

Option C is correct because the geometry of $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3$ is pyramidal (not tetrahedral) and the hybridization of carbon is $sp^3$ (not $sp^2$).

Option C (Hydrogen bond formation) is correct because the ice cubes unite due to the formation of hydrogen bonds between water molecules on the surfaces.

Option D ($\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic) is correct because in the complex $[Co(NH_3)_6]^{3+}$, cobalt is in +3 oxidation state, leading to a $d^6$ configuration. $NH_3$ is a strong field ligand, causing pairing of electrons and resulting in $d^2sp^3$ hybridization and a diamagnetic nature.

Option A (1) is correct because $\mathrm{O}_2^{2 - }$ has a bond order of 1, calculated as (8 bonding - 6 antibonding)/2.

Option D ($\mathrm{S in SO}_4^{2 - }$) is correct because both $PO_4^{3-}$ and $SO_4^{2-}$ have tetrahedral geometry with $sp^3$ hybridization.

Option B ($ZnC{l_2}$) is correct because zinc has a higher polarizing power due to its smaller size and higher charge density compared to other alkali and alkaline earth metals, leading to increased covalent character and thus least ionic character.

Option A (Maximum polarization is brought about by a cation of high charge) is correct because a higher charge on the cation increases its polarizing power, leading to greater distortion of the anion's electron cloud.

Option C (${B_2}$) is correct because it has unpaired electrons in its molecular orbital configuration, making it paramagnetic.

Option A (Covalent) is correct because graphite consists of carbon atoms covalently bonded in layers.