Chemistry Test 301

• The alkane with the structure shown in the reaction can undergo oxidation by KMnO₄ to produce the corresponding alcohol.

• The alkane in which the central carbon (R₂) is attached to three methyl groups (CH₃) is more reactive to KMnO₄ because it has tertiary hydrogen atoms.

• Alkanes with tertiary hydrogens (such as R₁ = CH₃, R₂ = CH₃, and R₃ = CH₃) are easily oxidized to alcohols by KMnO₄.

• In contrast, alkanes with primary or secondary hydrogen atoms do not undergo oxidation easily under similar conditions.

Hence, the correct answer is the alkane where R₁ = CH₃, R₂ = CH₃, and R₃ = CH₃.

Variation in boiling point with intermolecular forces - 

• Boiling point increase with the increase in the strength of intermolecular forces in a molecule.

• The strongest is the intermolecular force in a molecule, the lowest will be the vapor pressure and the higher is the boiling point.

• The strength of the intermolecular forces varies in order - H-bonding > Dipole-dipole > Vander-Waal forces.

So boiling point of aldehyde, ketone, alcohols, and hydrocarbons varies in the order- 

• Alcohol > Aldehyde or ketone (carbonyl compound) > Hydrocarbons.

• Reason -

  • As discussed above Boiling point depend upon the type of bonding present in the molecule. Alcohols are having high value of boiling point due to the presence of H-bonding.

  • Aldehydes and ketones are having polar interactions in between the molecule and are held together by the dipole-dipole interaction which is weaker than the H- bonding. That's why carbonyl compounds are having a low value of boiling point than alcohols.

  • Hydrocarbons are held by weak forces of interactions or Vander-Wall forces of attraction are having the lowest value of the boiling point.

So, keeping in mind the order of boiling point i.e. Alcohol > Aldehyde or ketone (carbonyl compound) > Hydrocarbons, we conclude that both statement 1 and statement 2 are correct.

Hence, the correct answer is option 2.

• Statement I: Propenes reaction with diborane (BH₃) results in the formation of an organoborane, with the formula ((CH₃)₂CH-CH₂-BH₂). This is because the boron atom adds to the carbon of the double bond that has fewer hydrogen atoms, according to Markovnikov’s rule. The final product is an organoborane intermediate that can undergo oxidation to form alcohols.

• Statement II:

  –B is replaced by –OH at the same position

  Final product: CH3​CH2​CH2OH(propan-1-ol)
  Not propan-2-ol

  So Statement II is also incorrect

Therefore, Both Statement I and Statement II are incorrect

Given,

[Ni(H₂O)₂(en₂)²⁺

[Ni(H₂O)₄(en)²⁺ and

[Ni(en)₃]²⁺

→ (en) is Ethylenediamine is an important bidentate ligand. It can bind through two nitrogen to a central metal ion.

The crystal field splitting, Δ_o, depends upon the field produced by the ligand and the charge on the metal ion.

Some ligands are able to produce strong fields in which case, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d orbitals.

i.e Splitting ∝ Ligand Strength.

Also, It indicates the energy absorbed.

In general, ligands can be arranged in a series in the order of increasing field strength as given below:

I^- < Br^- < SCN^- < CI^- < S^2- < F^- < OH^- < C₂O₄^2- < H₂O < NCS^- < edta^4- < NH₃ < en < CN^- < CO

So from here, we can say that (en) is a strong Ligand.

From the given Complex compounds:

[Ni(H₂O)₂(en₂)²⁺, [Ni(H₂O)₄(en)²⁺ , [Ni(en)₃]²⁺

[Ni(en)₃]²⁺ is strong So 

Order of Energy is C > A > B

• Given that the solutions are isotonic, their osmotic pressures must be equal. Therefore:

  πglucose = πurea

• For urea (NH₂CONH₂):

  • Molar mass of urea = 14 + 2(1) + 12 + 16 + 14 + 2(1) = 60 g/mol

  • Concentration of urea = 15 g/L

  • Molarity of urea solution = (15 g/L) / (60 g/mol) = 0.25 mol/L

  • Urea does not dissociate in solution, so i = 1

• For glucose (C₆H₁₂O₆):

  • Molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 g/mol

  • Let the mass of glucose required be x g

  • Molarity of glucose solution = (x g/L) / (180 g/mol) = x / 180 mol/L

  • Glucose does not dissociate in solution, so i = 1

• Since the solutions are isotonic:

  • πglucose = πurea

  • iglucose x (x / 180) x RT = iurea x 0.25 x RT

  • (x / 180) = 0.25

  • x = 0.25 x 180

  • x = 45 g

Therefore, the mass of glucose required to prepare one litre of its solution which is isotonic with a 15 g/L solution of urea is 45 g.

• B. Cu²⁺ - Group II (forms insoluble sulfide in acidic medium)

• A. Al³⁺ - Group III (forms insoluble hydroxide)

• D. Co²⁺ - Group IV (forms insoluble sulfide in basic medium)

• C. Ba²⁺ - Group V (forms insoluble carbonate and sulfate)

• E. Mg²⁺ - Group VI (soluble, forms precipitate with phosphate and has a different reagent)

• B. Cu²⁺ - Group II

• A. Al³⁺ - Group III

• D. Co²⁺ - Group IV

• C. Ba²⁺ - Group V

• E. Mg²⁺ - Group VI

Conclusion:-

The correct order in increasing group number is B, A, D, C, E

• Dilute hydrochloric acid: While this can prevent hydrolysis, it is not typically used in the preparation of Mohr’s salt.

• Concentrated sulphuric acid: This could be too strong and potentially cause unwanted side reactions.

• Dilute nitric acid: This is an oxidizing agent and could oxidize Fe²⁺ to Fe³⁺, which is counterproductive.

• Dilute sulphuric acid: This is commonly used because it is mild enough to prevent hydrolysis without causing oxidation of Fe²⁺ ions.

Conclusion:-

The correct acid to add during the preparation of Mohr’s salt solution to prevent hydrolysis of Fe²⁺ ions is dilute sulphuric acid

Acid strength depends on two factors which is

Electronegativity and Bond Strength

Electronegativity: As electronegativity increases acid strength will also increase.

Bond strength: Longer the bond, the weaker it will be, hence if bond strength is weaker acid will be stronger.

In the given question 

Acid strength increases in the order given as HF << HCl << HBr << HI. 

Because as we move from Fluorine to Iodine, the size of halogen increases. The larger the size of the halogen longer the bond will be formed hence bond strength will decrease means it requires less energy for bond dissociation and can be easily broken.

So, HI is the stronger acid of all.

Both statements I and II are correct.

So the answer is option II.

Dissolution of Sodium in Liquid Ammonia

• When metallic sodium dissolves in liquid ammonia, it forms a deep blue solution.

• The deep blue color is due to the presence of solvated (free) electrons in the solution.

• This solution is paramagnetic, which means it has unpaired electrons that respond to a magnetic field.

Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.

This statement is true. When sodium dissolves in liquid ammonia, it indeed forms a deep blue solution, which is paramagnetic due to the presence of solvated electrons.

Reason R: The deep blue solution is due to the formation of amide.

This statement is false. The deep blue color is actually due to the presence of solvated (free) electrons in the solution, not due to the formation of amide. Amides do not contribute to the deep blue color.

CONCLUSION:

The correct answer is A is true but R is false.