Grade 11 : Chemistry - Chemical Bonding (Set 4)

Explanation: $BF_3$ has trigonal planar geometry with a bond angle of $120^\circ$. $CH_4$ is tetrahedral ($109.5^\circ$), $NH_3$ is trigonal pyramidal ($107^\circ$), and $H_2O$ is bent ($104.5^\circ$). Therefore, $BF_3$ has the largest bond angle.

Explanation: Bond order is inversely proportional to bond length. Calculating bond orders: $O_2^+ = 2.5$, $O_2 = 2.0$, $O_2^- = 1.5$, and $O_2^{2-} = 1.0$. Thus, decreasing bond order corresponds to increasing bond length: $O_2^{2-} > O_2^- > O_2 > O_2^+$.

Explanation: In $NH_3$, the orbital dipole of the lone pair and the N-H bond dipoles point in the same direction, reinforcing each other. In $NF_3$, the electronegative F atoms pull density away, so N-F bond dipoles oppose the lone pair dipole, reducing the net dipole moment.

Explanation: Steric number of S $= 4\text{ bonding pairs} + 1\text{ lone pair} = 5$, giving $sp^3d$ hybridisation. The lone pair occupies an equatorial position to minimize repulsions, producing a see-saw molecular shape.

Explanation: Ortho-nitrophenol undergoes intramolecular hydrogen bonding (chelation), preventing association between molecules. Para-nitrophenol forms intermolecular hydrogen bonds linking multiple molecules, requiring more thermal energy to separate them and yielding a higher boiling point.

Explanation: $\text{Formal Charge} = V - L - \frac{1}{2}B$. For central $O(1)$: Valence $V = 6$, Lone pair electrons $L = 2$, Bonding electrons $B = 6$ (3 bonds). $\text{Formal Charge} = 6 - 2 - 3 = +1$.

Explanation: Total charge of -2 is equally shared across 3 oxygen atoms: average formal charge $= -\frac{2}{3}$. Total bonds $= 4$ distributed over 3 positions: bond order $= \frac{4}{3} = 1.33$.

Explanation: Fajan's rules state that smaller cation size increases polarizing power. $Li^+$ strongly distorts the electron cloud of $Cl^-$, creating significant covalent character and lowering the melting point compared to NaCl.

Explanation: $C_1$ and $C_3$ each form $3\ \sigma$ bonds ($sp^2$). The central $C_2$ forms $2\ \sigma$ bonds (one with $C_1$ and one with $C_3$), making it sp hybridised.

Explanation: $B_2$ has 10 electrons: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 (\pi 2p_x^1 = \pi 2p_y^1)$. It has 2 unpaired electrons (paramagnetic) and $\text{Bond Order} = \frac{6 - 4}{2} = 1$.

Explanation: Lattice energy $U \propto \frac{1}{r_+ + r_-}$. As the anion radius increases from $F^-$ to $I^-$, the internuclear distance increases, reducing electrostatic attraction and lowering lattice energy.

Explanation: Higher bond order (triple > double > single) and greater $s$-character ($50%\ sp > 33%\ sp^2 > 25%\ sp^3$) draw electrons closer to nuclei, decreasing bond length.

Explanation: Xe has 8 valence electrons. In $XeF_4$, Xe forms 4 bonding pairs and retains 2 lone pairs (steric number $6 \rightarrow sp^3d^2$). Lone pairs occupy axial positions to minimize repulsions, producing a square planar geometry.

Explanation: Bonding MO $\sigma 2p_z$ results from constructive interference (high electron density between nuclei). Antibonding MO $\sigma^* 2p_z$ results from destructive interference, producing a nodal plane between nuclei where electron density is zero.

Explanation: $CO_2$ and $BeF_2$ are linear and symmetric, so equal bond dipoles act in opposite directions and cancel out ($\mu = 0$). $H_2O$ and $SO_2$ are bent due to lone pairs, so bond dipoles do not cancel ($\mu > 0$).

Explanation: Nitrogen donates its lone pair to electron-deficient $H^+$, forming a co-ordinate covalent bond. Once formed, all 4 N-H bonds in $NH_4^+$ become completely equivalent, giving $4\ \sigma$ bonds in a tetrahedral arrangement.

Explanation: Single C-C bonds $= 4\ \sigma$. Central C=C bond $= 1\ \sigma + 1\ \pi$. Four $C\equiv N$ bonds $= 4\ \sigma + 8\ \pi$. Total $\sigma = 4 + 1 + 4 = 9$. Total $\pi = 1 + 8 = 9$.

Explanation: Oxygen is small and highly electronegative, enabling $H_2O$ molecules to form strong hydrogen bonds. Other hydrides lack H-bonding and rely on weaker dispersion forces, making $H_2O$'s boiling point abnormally high.

Explanation: Electronegativity difference $\Delta \chi = 4.0 - 2.1 = 1.9$. A value exceeding 1.7 indicates significant ionic character (approx. $43% - 50%$), classifying H-F as a strongly polar covalent bond with high ionic character.

Explanation: Cl has 7 valence electrons. In $ClF_3$, 3 bonding pairs and 2 lone pairs surround Cl (steric number 5). The 2 lone pairs occupy equatorial positions, producing a T-shaped geometry with F-Cl-F bond angles slightly less than $90^\circ$ due to lone pair repulsions.