Chemistry Test 227

Comparing the given options:

• (1) 30 mL HCl and 30 mL NaOH
  • HCl is a strong acid, and NaOH is a strong base.
  • moles of acid or base = 30 mM
  • Neutralization is complete, leading to the highest temperature increase.
     
• (2) 30 mL CH₃COOH and 30 mL NaOH
  • CH₃COOH is a weak acid, and NaOH is a strong base.
  • moles of acid or base = 30 mM
  • Due to the weak acid, less energy is released compared to strong acid-base neutralization.
     
• (3) 50 mL HCl and 20 mL NaOH
  • HCl is a strong acid, and NaOH is a strong base.
  • moles of acid or base = 20 mM (limiting reagent)
  • Neutralization is incomplete, leading to a lower temperature increase.
     
• (4) 45 mL CH₃COOH and 25 mL NaOH
  • CH₃COOH is a weak acid, and NaOH is a strong base.
  • moles of acid or base = 25 mM (limiting reagent)
  • Due to the weak acid and incomplete neutralization, less energy is released.

Therefore, the correct answer is 30 mL HCl and 30 mL NaOH (Option 1).

• Given compound contains two stereocenters before ozonolysis, but ozonolysis cleaves the C=C double bonds to produce carbonyl compounds.
• After ozonolysis, each fragment is either:
  • Acetaldehyde (CH₃CHO) — no chiral center.
  • A diketone with identical groups on the central carbons — also no chiral center.
• Thus, no product has a carbon atom bonded to four different groups; hence, no optical activity remains.

Therefore, the number of optically active products is 0.

Case 1: sp³ hybridized (tetrahedral)

• Tetrahedral geometry typically involves a metal center with four ligands arranged around it symmetrically.
• For the complex MA₂B₂, if it has sp³ hybridization, the tetrahedral geometry will always have a plane of symmetry regardless of the arrangement of A and B around M.
• This ensures that a mirror image will always be superposable, hence no optical isomerism.

Case 2: dsp² hybridized (square planar)

• The square planar geometry is characterized by four ligands positioned around the metal in a plane.
• For the complex MA₂B₂ with dsp² hybridization, it will have two distinct arrangements (cis and trans configurations).
• However, both cis and trans configurations for a square planar arrangement will have planes of symmetry.
• Due to the presence of these planes of symmetry, neither arrangement will show optical activity.
• Therefore, in both cases, no optical isomers are possible.

The correct answer is: 2) 0 and 0

• Given that the angular momentum L = 3h/π, we identify the principal quantum number as n = 3 (since L = nh/2π).
• The hydrogen-like ion Li²⁺ has Z = 3.
• Using the formula for the radius:

  r_n = n²h²ε₀ / (πZ e²m)

  Substituting the values:
  • n = 3, Z = 3
  • h = 6.626 × 10⁻³⁴ J·s
  • ε₀ = 8.85 × 10⁻¹² C²/N·m²
  • e = 1.6 × 10⁻¹⁹ C
  • m = 9.1 × 10⁻³¹ kg

  r₃ ≈ 6.348 Å

• Using the formula for energy:

  E_n = -Z²e⁴ / (8ε₀²h²n²)

  Substituting the values:
  • Z = 3, n = 3
  • e = 1.6 × 10⁻¹⁹ C
  • ε₀ = 8.85 × 10⁻¹² C²/N·m²
  • h = 6.626 × 10⁻³⁴ J·s

  E₃ ≈ -5.45 × 10⁻¹⁹ J

Therefore, the radius and energy of the stationary state are approximately 6.348 Å and -5.45 × 10⁻¹⁹ J, respectively.

(A) Tl³⁺ is a powerful oxidising agent:

• True. Tl³⁺ readily gets reduced to the more stable Tl⁺ due to the inert pair effect.

(B) Al³⁺ does not get reduced easily:

• True. E°(Al³⁺/Al) = −1.66 V → strong tendency to remain in +3 state.

(C) Both Al³⁺ and Tl³⁺ are very stable in solution:

• False. Tl³⁺ is unstable in solution, reducing to Tl⁺.

(D) Tl⁺ is more stable than Tl³⁺:

• True. Due to inert pair effect.

(E) Al³⁺ and Tl⁺ are highly stable:

• True. Both remain in their preferred oxidation states in solution.

So, the correct answer is (A), (B), (D), and (E) only.

Option 1: Order of carbanion stability: C₆H₅CH₂⁻ < (C₆H₅)₂CH⁻ < (C₆H₅)₃C⁻
This is correct, as more phenyl groups provide more resonance stabilization.

Option 2: CH₃⁺ is sp² hybridized not sp³. Incorrect.

Option 3: CH₃⁻ is sp³ hybridized, not sp². Incorrect.

Option 4: Carbocation stability should be: C₆H₅CH₂⁺ < (C₆H₅)₂CH⁺ < (C₆H₅)₃C⁺
Option 4 lists it in reverse order, so it is incorrect.

Therefore, the correct answer is Option 1: The order of stability of carbanions is C₆H₅CH₂⁻ < (C₆H₅)₂CH⁻ < (C₆H₅)₃C⁻.

1. (B < Al) — Correct, regular trend
2. (Al < Ga) — Incorrect, due to d-block contraction → Al > Ga
3. (Ga < In) — Correct
4. (In < Tl) — Correct

• So, the incorrect pair is: Al < Ga
• Now, comparing Al³⁺ and Ga³⁺ ionic radii:
  • Though Ga has more protons, its poor shielding (due to 3d electrons) causes stronger nuclear attraction.
  • So, Ga³⁺ is smaller than Al³⁺.
  • Thus, Al³⁺ has the higher ionic radius.
• Atomic number of Ga is 31 

The correct answer is 31

Option 1: n = 3 , l = 2 , m_l = 0 , s = +1/2

• Valid as l = 2 is within the range [0, 2] for n = 3 .
• Valid as m_l = 0 is within the range [-2, 2] for l = 2 .

Option 2: n = 3 , l = 2 , m_l = -2 , s = +1/2

• Valid as l = 2 is within the range [0, 2] for n = 3 .
• Valid as m_l = -2 is within the range [-2, 2] for l = 2 .

Option 3: n = 3 , l = 3 , m_l = -3 , s = -1/2

• Invalid as l = 3 is not within the range [0, 2] for n = 3 (valid range for l should be 0, 1, or 2).

Option 4: n = 3 , l = 0 , m_l = 0 , s = -1/2

• Valid as l = 0 is within the range [0, 2] for n = 3 .
• Valid as m_l = 0 is within the range [-0, 0] for l = 0 .

The set of quantum numbers that is not allowed is Option 3: n = 3 , l = 3 , m_l = -3 , s = -1/2 .