Chemistry (313) | Set A1 • Code 71/SS/313/A1

(a) \(4p\)
Proof: For \(3d\): \((n + l) = 3 + 2 = 5\). For \(4p\): \((n + l) = 4 + 1 = 5\). When \((n + l)\) is equal, the orbital with lower principal quantum number \(n\) has lower energy (\(n = 3\) for \(3d < n = 4\) for \(4p\)). Hence \(3d\) has lower energy than \(4p\).

(b) \(3s\)
Proof: Radial (spherical) nodes formula \(= n - l - 1\). For \(3s\): \(n = 3, l = 0 \implies \text{nodes} = 3 - 0 - 1 = \mathbf{2}\).

(a) 1 lone pair (Nitrogen has 5 valence electrons: 3 are shared with 3 hydrogen atoms in single covalent bonds, leaving 2 non-bonding electrons as one lone pair).

(a) True (T) (In acetylene, \(\text{H}-\text{C}\equiv\text{C}-\text{H}\), the carbon-carbon triple bond consists of \(1\ \sigma\) bond and \(2\ \pi\) bonds).

(b) False (F) (Molecular orbital electronic configuration of \(\text{Be}_2\) (\(8\ e^-\)) is \(\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2\). Bond order \(= \frac{N_b - N_a}{2} = \frac{4 - 4}{2} = \mathbf{0}\), which is zero, not a fraction).

(a) • For \(\text{Na}^+\) ion: \(\Delta H\) is Positive (\(+\)) (endothermic: ionization enthalpy is absorbed to remove an electron: \(\text{Na}(g) \to \text{Na}^+(g) + e^-\)).
• For \(\text{Cl}^-\) ion: \(\Delta H\) is Negative (\(-\)) (exothermic: electron gain enthalpy is released when a chlorine atom accepts an electron: \(\text{Cl}(g) + e^- \to \text{Cl}^-(g)\)).

(b) To attain stable inert gas octet configurations: Sodium (\(2, 8, 1\)) loses its single valence electron to achieve the neon configuration (\(2, 8\)), while chlorine (\(2, 8, 7\)) gains that electron to achieve the argon configuration (\(2, 8, 8\)).

(a) paramagnetic (Iron is in \(+3\) oxidation state, \(3d^5\). Under the influence of strong ligand \(\text{CN}^-\), electron pairing occurs to give \(t_{2g}^5 e_g^0\), leaving 1 unpaired electron, causing paramagnetism).

(b) coloured (Transition metal ions contain incompletely filled d-subshells (\(d^1\) to \(d^9\)), permitting \(d-d\) electronic transitions by absorbing visible light).

(a) Similar atomic radii: According to Hume-Rothery rules, the atomic radii of the constituent transition metals should not differ by more than \(15\%\) so that atoms of one metal can mutually replace atoms of another in the crystal lattice.

(b) Brass composition: Approximately \(60\% - 80\%\text{ Copper (Cu)}\) and \(20\% - 40\%\text{ Zinc (Zn)}\) (typically \(70\%\text{ Cu}\) and \(30\%\text{ Zn}\)).

(a) ozone (Tailing of mercury occurs due to oxidation of mercury to mercurous oxide, \(\text{Hg}_2\text{O}\): \(2\text{Hg} + \text{O}_3 \to \text{Hg}_2\text{O} + \text{O}_2\)).

(b) Zinc oxide (Reacts with both acids and bases: \(\text{ZnO} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2\text{O}\); \(\text{ZnO} + 2\text{NaOH} \to \text{Na}_2\text{ZnO}_2 + \text{H}_2\text{O}\)).

(a) Diazotization (or diazotisation).

(b) Gattermann reaction equation: \[\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu powder / HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \uparrow + \text{CuCl}\]

(a) \(p\)-benzoquinone (1,4-benzoquinone): \[\text{C}_6\text{H}_5\text{OH} \xrightarrow[(\text{KSO}_3)_2\text{NO}]{\text{H}_2\text{O}} \text{O}=\text{C}_6\text{H}_4=\text{O}\quad (p\text{-benzoquinone})\]

(b) Methanol and Potassium formate (Cannizzaro Reaction): \[2\text{HCHO} \xrightarrow{\text{conc. KOH}} \text{CH}_3\text{OH} + \text{HCOOK}\]

(a) False (F) (Butyl rubber is prepared by the cationic copolymerization of isobutylene with a small amount of isoprene (\(\approx 2\%\)), not butadiene).

(b) False (F) (Vulcanisation creates cross-linking disulfide (\(-\text{S}-\text{S}-\)) bridges between polymer chains, giving vulcanised rubber vastly higher tensile strength, elasticity, and heat resistance than soft natural rubber).

(a) By inserting hydrolyzable / cleavable functional groups: Introducing ester (\(-\text{COO}-\)), amide (\(-\text{CONH}-\)), or anhydride bonds into the polymer backbone, or blending/copolymerizing synthetic chains with natural biopolymers (e.g. starch, cellulose).

(b) PHBV: Poly(\(\beta\)-hydroxybutyrate-co-\(\beta\)-hydroxyvalerate) [or Nylon-2-nylon-6 / Polylactic acid (PLA)].

• Statement: For a solution of volatile liquids, Raoult's law states that the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction in the liquid mixture at a given temperature.

• Mathematical Expression:
For a binary solution containing components 1 and 2: \[p_1 = p_1^\circ x_1 \quad \text{and} \quad p_2 = p_2^\circ x_2\] \[\text{Total Vapour Pressure } P_{\text{total}} = p_1 + p_2 = p_1^\circ x_1 + p_2^\circ x_2\] where \(p_1^\circ, p_2^\circ\) are vapour pressures of pure components and \(x_1, x_2\) are their mole fractions.

(a) \(_4\text{Be}\) has higher ionization enthalpy than \(_5\text{B}\).
Reason: Beryllium has a stable, completely filled \(2s^2\) subshell (\(1s^2 2s^2\)). Boron (\(1s^2 2s^2 2p^1\)) has its outermost electron in a higher energy \(2p\) subshell which experiences greater shielding by the inner \(2s\) electrons, making it easier to ionize (\(\text{IE}_1(\text{Be}) = 899\text{ kJ/mol} > \text{IE}_1(\text{B}) = 801\text{ kJ/mol}\)).

(b) \(_2\text{He}\) has higher ionization enthalpy than \(_{10}\text{Ne}\).
Reason: In Helium, the two \(1s\) electrons occupy the lowest shell (\(n = 1\)) closest to the nucleus with zero inner-shell shielding. Moving down to Neon (\(n = 2\)), increased shielding and larger atomic radius lower the effective nuclear pull. Helium has the highest first ionization enthalpy of any element (\(\approx 2372\text{ kJ/mol}\)).

Equilibria in solution: \(\text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\) (feebly ionized); \(\text{NH}_4\text{NO}_3 \to \text{NH}_4^+ + \text{NO}_3^-\) (completely ionized).

1. Addition of Acid (\(\text{H}^+\)): Added \(\text{H}^+\) ions are neutralized by unionized \(\text{NH}_4\text{OH}\) molecules: \[\text{H}^+ + \text{NH}_4\text{OH} \to \text{NH}_4^+ + \text{H}_2\text{O}\] Thus, \([\text{H}^+]\) does not increase and pH remains stable.

2. Addition of Base (\(\text{OH}^-\)): Added \(\text{OH}^-\) ions combine with the abundant reserve of \(\text{NH}_4^+\) ions to form weakly dissociated \(\text{NH}_4\text{OH}\): \[\text{NH}_4^+ + \text{OH}^- \to \text{NH}_4\text{OH}\] Thus, \([\text{OH}^-]\) does not increase and pH remains unchanged.

In the Castner-Kellner cell, brine (\(\text{NaCl}\) solution) is electrolyzed using a carbon/graphite anode and a flowing mercury (\(\text{Hg}\)) cathode:

• At Anode (Carbon/Graphite): \[2\text{Cl}^- \to \text{Cl}_2(g) + 2e^-\] • At Cathode (Mercury): \[\text{Na}^+ + e^- \xrightarrow{\text{Hg}} \text{Na}-\text{Hg}\quad (\text{Sodium Amalgam})\] • In Denuder (Reaction with Water): \[2\text{Na}-\text{Hg} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2(g) + 2\text{Hg}(l)\] (Mercury is recovered and recycled back into the cell).

Given parameters:
• Mass of solute (\(w\)) \(= 0.8960\text{ g}\)
• Volume of solution (\(V\)) \(= 500\text{ mL} = 0.500\text{ L}\)
• Osmotic pressure (\(\pi\)) \(= 0.69\text{ atm}\)
• Temperature (\(T\)) \(= 27^\circ\text{C} = 27 + 273.15 = 300\text{ K}\)
• Gas constant (\(R\)) \(= 0.0821\text{ L atm K}^{-1}\text{mol}^{-1}\)

From the osmotic pressure formula: \[\pi = \frac{n}{V} R T = \frac{w}{M_{\text{obs}} \cdot V} R T\] \[M_{\text{obs}} = \frac{w \cdot R \cdot T}{\pi \cdot V}\] \[M_{\text{obs}} = \frac{0.8960\text{ g} \times 0.0821\text{ L atm K}^{-1}\text{mol}^{-1} \times 300\text{ K}}{0.69\text{ atm} \times 0.500\text{ L}}\] \[M_{\text{obs}} = \frac{22.06848}{0.345} \approx \mathbf{63.97\text{ g mol}^{-1}}\quad (\approx 64.0\text{ g mol}^{-1})\]

Note on Chemical Context: Theoretical molar mass of \(\text{K}_2\text{SO}_4 = 2(39.1) + 32.1 + 4(16.0) = 174.3\text{ g mol}^{-1}\). The calculated experimental mass is lower (\(63.97\text{ g mol}^{-1}\)) because potassium sulphate undergoes dissociation into 3 ions, giving an observed van't Hoff factor \(i = \frac{174.3}{63.97} \approx 2.72\).

Combustion reaction: \[\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g); \quad \Delta H = -393.5\text{ kJ mol}^{-1}\] Molar mass of \(\text{CO}_2 = 12.0 + 2(16.0) = 44.0\text{ g mol}^{-1}\).
Moles of \(\text{CO}_2\) in \(35.2\text{ g}\): \[n = \frac{35.2\text{ g}}{44.0\text{ g mol}^{-1}} = \mathbf{0.8\text{ mol}}\] Heat released on formation of \(1\text{ mole } (44\text{ g}) = 393.5\text{ kJ}\).
Heat released on formation of \(0.8\text{ mole } (35.2\text{ g})\): \[q = 0.8\text{ mol} \times 393.5\text{ kJ mol}^{-1} = \mathbf{314.8\text{ kJ}}\] Final Answer: Heat released = \(314.8\text{ kJ}\)

• Definition: For a sparingly soluble salt, the solubility product (\(K_{sp}\)) is defined as the product of the molar concentrations of its constituent ions in a saturated aqueous solution, with each concentration raised to the power equal to its stoichiometric coefficient in the balanced dissociation equilibrium at a given temperature.

• Explanation & Formulation:
Consider a general sparingly soluble salt \(A_x B_y\): \[A_x B_y(s) \rightleftharpoons x A^{y+}(aq) + y B^{x-}(aq)\] \[K_{sp} = [A^{y+}]^x [B^{x-}]^y\] If the molar solubility is \(s\text{ mol L}^{-1}\), then \([A^{y+}] = xs\) and \([B^{x-}] = ys\): \[K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y}\] Example: For \(\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-\), \(K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s^2\).

• Criterion for Precipitation:
1. If Ionic Product (\(Q_{sp}\)) \(< K_{sp}\): Solution is unsaturated; no precipitation occurs.
2. If \(Q_{sp} = K_{sp}\): Saturated solution in dynamic equilibrium.
3. If \(Q_{sp} > K_{sp}\): Solution is supersaturated; precipitation occurs immediately.

Electrode reduction reaction: \[\text{Ag}^+(aq) + e^- \to \text{Ag}(s)\quad (n = 1)\] Applying the Nernst equation at \(298\text{ K}\): \[E_{\text{Ag}^+/\text{Ag}} = E_{\text{Ag}^+/\text{Ag}}^\circ - \frac{0.0591}{n} \log \frac{1}{[\text{Ag}^+]}\] \[E_{\text{Ag}^+/\text{Ag}} = 0.80 - \frac{0.0591}{1} \log \left(\frac{1}{0.1}\right)\] Since \(\log\left(\frac{1}{0.1}\right) = \log(10) = 1\): \[E_{\text{Ag}^+/\text{Ag}} = 0.80 - 0.0591 \times 1 = 0.80 - 0.0591 = \mathbf{0.7409\text{ V}}\quad (\approx \mathbf{0.741\text{ V}})\] Final Answer: Reduction Potential = \(+0.741\text{ V}\)