Chemistry Test 299

• For the given ionization process:

  Be³⁺(g) → Be⁴⁺(g) + e^-

  The species Be³⁺ is hydrogen-like because it has only one electron left.

• The atomic number of beryllium (Be) is Z = 4.

• The ionization energy of hydrogen in its ground state is as:

  E_H = 2.18 × 10^-18 J atom^-1 (Factual Data)

• Using the formula E = E_H × Z²:

  E = (2.18 × 10^-18) × (4)²

  E = (2.18 × 10^-18) × 16

  E = 3.49 × 10^-17 J atom^-1

Therefore, the energy required for the process Be³⁺(g) → Be⁴⁺(g) + e^- is 3.49 × 10^-17 J atom^-1.

• For the given species (NO⁺, NO, NO⁻), the total electron count and molecular orbital configurations are as follows:

  • NO⁺ (14 electrons): Bond order = 3, Unpaired electrons = 0.

  • NO (15 electrons): Bond order = 2.5, Unpaired electrons = 1.

  • NO⁻ (16 electrons, isoelectronic with O₂²⁻): Bond order = 2, Unpaired electrons = 2.

• Statement I: The bond order of the species is in the following order:

  • NO⁺ (3) > NO (2.5) > NO⁻ (2).

  This statement is true.

• Statement II: The number of unpaired electrons in the species is in the following order:

  • NO⁻ (2) > NO (1) > NO⁺ (0).

  This is the actual order, which contradicts the claim that the unpaired electrons follow the order NO⁻ > NO > NO⁺. Hence, this statement is false.

Therefore, Statement I is true but Statement II is false.

• The target reaction is the formation of anhydrous solid FeCl₃ from its elements:

  Target Reaction (for 2 mol): 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)

• Given data:

  • (i) 2Fe(s) + 6HCl(aq) → 2FeCl₃(aq) + 3H₂(g), ΔH₁ = –800 kJ/mol

  • (ii) H₂(g) + Cl₂(g) → 2HCl(g), ΔH₂ = –184 kJ/mol

  • (iii) HCl(g) + aq → HCl(aq), ΔH₃ = –75 kJ/mol

  • (iv) FeCl₃(s) + aq → FeCl₃(aq), ΔH₄ = –150 kJ/mol

• To construct the target reaction, combine the given reactions as follows:

  • (i) 2Fe(s) + 6HCl(aq) → 2FeCl₃(aq) + 3H₂(g)

  • + 3 × (ii) 3H₂(g) + 3Cl₂(g) → 6HCl(g)

  • + 6 × (iii) 6HCl(g) → 6HCl(aq)

  • – 2 × (iv) 2FeCl₃(aq) → 2FeCl₃(s)

• On combining:

  2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)

• Sum the corresponding enthalpy changes:

  • ΔH = ΔH₁ + 3ΔH₂ + 6ΔH₃ – 2ΔH₄

  • ΔH = (–800) + 3(–184) + 6(–75) – 2(–150)

  • ΔH = –800 – 552 – 450 + 300

  • ΔH = –1502 kJ (for 2 mol FeCl₃(s))

• The enthalpy of formation per mole of FeCl₃(s) is:

  ΔH_f = –1502 / 2 = –751 kJ mol^-1

Therefore, the enthalpy of formation of anhydrous solid FeCl₃ is –751 kJ mol^-1.

• Given half-cell reactions and their standard reduction potentials:

  • Zn(OH)₂(s) + 2e^- → Zn(s) + 2OH^-(aq), E° = –1.25 V

  • Hg₂Cl₂(s) + 2e^- → 2Hg(l) + 2Cl^-(aq), E° = +0.27 V

• Since the reduction potential of Hg₂Cl₂/Hg (+0.27 V) is higher than that of Zn(OH)₂/Zn (–1.25 V):

  Zn(s) + 2OH^-(aq) → Zn(OH)₂(s) + 2e^-

  • The mercury half-cell acts as the cathode, where reduction occurs.

  • The zinc half-cell reaction is reversed to act as the anode, where oxidation occurs:

• The overall reaction is obtained by combining the two half-reactions:

  Zn(s) + 2OH^-(aq) + Hg₂Cl₂(s) ⇌ Zn(OH)₂(s) + 2Hg(l) + 2Cl^-(aq)

• Now, calculate the standard cell potential:

  • E°_cell = E°_cathode - E°_anode

  • = (+0.27 V) - (–1.25 V)

  • = 0.27 V + 1.25 V

  • = +1.52 V

• The calculated E°_cell is positive (+1.52 V), indicating that the reaction is spontaneous.

• In this cell:

  • Zinc (Zn) is oxidized at the anode, not reduced.

  • E°_cell is an intensive property and does not depend on the amount of substance.

Therefore, the overall reaction is:

Zn(s) + 2OH^-(aq) + Hg₂Cl₂(s) ⇌ Zn(OH)₂(s) + 2Hg(l) + 2Cl^-(aq)

The standard cell potential (E°_cell) is +1.52 V.

Zn is oxidized in the electrochemical cell, and E°_cell is an intensive property.

• Zero order reactions:

  • The half-life is given by: t_1/2 = a/(2k).

  • The time for 75% completion is given by: t_3/4 = (3a/4)/k = 3a/(4k).

  • Taking the ratio:

    • t_3/4 / t_1/2 = (3a/4k) / (a/2k) = 3/2.

    • Thus, t_3/4 = 1.5t_1/2.

• Second order reactions:

  • The half-life is given by: t_1/2 = 1/(ka).

  • The time for 75% completion is given by:

    • t_3/4 = (1/k) × [0.75/(a × 0.25)] = 3/(ka).

  • Taking the ratio:

    • t_3/4 / t_1/2 = (3/(ka)) / (1/(ka)) = 3.

    • Thus, t_3/4 = 3t_1/2.

Therefore:

• For zero order reactions: t_3/4 = 1.5t_1/2.

• For second order reactions: t_3/4 = 3t_1/2.

• Statement I: The first ionisation enthalpies of B, C, N, and O follow the order B < C < O < N. This is due to the fact that nitrogen has a half-filled 2p³ configuration, which provides extra stability and raises its IE₁ above oxygen, despite oxygen having a higher nuclear charge. The approximate IE₁ values are:

  • B: 801 kJ/mol

  • C: 1086 kJ/mol

  • O: 1314 kJ/mol

  • N: 1402 kJ/mol

  Thus, B < C < O < N is true.

• Statement II: Among Na, Mg, Al, and Si, Na has the highest second ionisation enthalpy (IE₂). This is because Na⁺ has the stable [Ne] configuration, and removing a second electron from this noble-gas core requires very high energy. In contrast, Mg, Al, and Si still have valence electrons available to lose, making their IE₂ lower. Hence, Statement II is also true.

Therefore, both Statement I and Statement II are true.

• For ClF₃:

  • The central chlorine (Cl) atom has 7 valence electrons.

  • Out of these, 3 electrons form bonds with 3 fluorine atoms, leaving 2 lone pairs on Cl.

  • This results in a total of 5 electron pairs (3 bonding pairs + 2 lone pairs).

  • According to VSEPR theory, a 5-electron-pair system adopts a trigonal bipyramidal electron-pair geometry.

  • The 2 lone pairs occupy the equatorial positions to minimize repulsion, and the 3 bonding pairs form a T-shaped molecular geometry.
    

  • Thus, Statement I is true.

• For BrF₅:

  • The central bromine (Br) atom has 7 valence electrons.

  • Out of these, 5 electrons form bonds with 5 fluorine atoms, leaving 1 lone pair on Br.

  • This results in a total of 6 electron pairs (5 bonding pairs + 1 lone pair).

  • According to VSEPR theory, a 6-electron-pair system adopts an octahedral electron-pair geometry.

  • The lone pair occupies one position, distorting the octahedral geometry into a square pyramidal molecular shape.
    

  • Thus, Statement II is true.

So, the correct answer is Both Statement I and Statement II are true.

• The question asks for the set of species that will change the orange colour of K₂Cr₂O₇ in acidic medium.

• Species like Fe³⁺, Cu²⁺, NO₃^-, and SO₄^2- are already in their oxidised or stable forms and cannot act as reducing agents.

• Similarly, MnO₄^- is itself a strong oxidising agent and cannot reduce Cr₂O₇^2-.

• The correct set of reducing agents that can react with K₂Cr₂O₇ and change its colour includes:

  • H₂O₂ → O₂

  • Fe²⁺ → Fe³⁺

  • C₂O₄^2- → CO₂

  • SO₃^2- → SO₄^2-

• These reducing agents will oxidise themselves while reducing Cr₂O₇^2- (Cr⁶⁺) to Cr³⁺, turning the solution green.

Therefore, the correct set of species is H₂O₂, Fe²⁺, C₂O₄^2-, SO₃^2-.