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CONCEPT:

Reactions of Propyne with Sodium and Sodium Amide

• When alkynes react with metallic sodium in the presence of liquid ammonia, they release hydrogen gas. The number of moles of H₂ gas liberated depends on the available acidic hydrogen atoms in the molecule.
• Reaction with sodium amide (NaNH₂) leads to the formation of a sodium salt of the alkyne and releases ammonia (NH₃) gas. This reaction is used to identify terminal alkynes.

EXPLANATION:

• Statement I: One mole of propyne (CH₃–C≡CH) reacts with an excess of sodium (Na) to liberate 0.5 moles of hydrogen gas (H₂), as each terminal hydrogen in the alkyne contributes to the reaction. This statement is correct.
• Statement II: Four grams of propyne contain 0.1 moles of propyne (molecular weight = 40 g/mol). When reacted with NaNH₂, this amount of propyne will release 0.1 moles of NH₃ gas, which occupies 2240 mL at STP (not 224 mL as stated). Hence, this statement is incorrect.
• $\frac{4}{40}=0.1 \text { mole } \quad \frac{0.1 \mathrm{~mole}}{2240 \mathrm{~mole}}$
• Statement I is correct but Statement II is incorrect

Therefore, Statement I is correct, but Statement II is incorrect.

CONCEPT:

Equilibrium Constant (K_P) for Gas Reactions

• The equilibrium constant, K_P, is calculated using the partial pressures of the gases involved in the reaction at equilibrium.
• For a reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K_P = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b

where P represents the partial pressure of the gases at equilibrium.

• Pure solids and liquids do not appear in the expression for K_P.

EXPLANATION:

• The given reaction is:

CO₂(g) + C(s) ⇌ 2CO(g)

• CO₂ is the reactant gas.
• Graphite (C) is a solid and does not appear in K_P expression.
• CO is the product gas.
• Initial pressure of CO₂ is 0.5 atm.
• Let 'x' be the pressure change due to the conversion of CO₂ to CO at equilibrium:
• Pressure of CO₂ at equilibrium: (0.5 - x) atm
• Pressure of CO at equilibrium: 2x atm
• $\begin{array}{lc} \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{C}(\mathrm{~s}) & \rightleftharpoons 2 \mathrm{CO}(\mathrm{~g}) \ 0.5 & - \ 0.5-\mathrm{x} & 2 \mathrm{x} \end{array}$
• Given: Total pressure at equilibrium is 0.8 atm:
• (0.5 - x) + 2x = 0.8
• Solving for x: 0.5 + x = 0.8
• x = 0.3
• Equilibrium pressures:
• P_CO2 = 0.5 - 0.3 = 0.2 atm
• P_CO = 2(0.3) = 0.6 atm
• Plugging these values into the K_P expression:
• K_P = (P_CO)² / (P_CO2)
• K_P = (0.6)]² / 0.2
• K_P = 0.36 / 0.2
• K_P = 1.8 atm

Therefore, the correct answer is 1.8 atm

CONCEPT:

IUPAC Nomenclature of Organic Compounds

• The IUPAC name of a compound is determined based on the longest carbon chain containing the principal functional group.
• Priority is given to functional groups in the following order: carboxylic acid (-COOH) > ester (-COOR) > ketone (-C=O) > alcohol (-OH) > amine (-NH₂) > alkene > alkyne > alkane.
• Substituents are named and numbered to give the lowest possible position numbers to the functional groups and substituents.

EXPLANATION:

• The given compound contains two functional groups:
• A carboxylic acid (-COOH) group, which has the highest priority.
• An ester group (-COOCH₃) as a substituent.
• The longest carbon chain containing the carboxylic acid group has six carbons (hexane). The main chain is numbered from the carboxylic acid carbon to give substituents the lowest possible numbers.
• The substituents on the chain are:
• A methoxycarbonyl (-COOCH₃) group at position 6.
• Methyl groups at positions 2 and 5.
• The compound's name is written as: 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.

Therefore, the IUPAC name of the compound is: 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.

CONCEPT:

Electrolysis of Sulfuric Acid

• During electrolysis, an electrolyte is decomposed into its constituent ions through the application of an electrical current.
• H₂S₂O₈ (peroxodisulfuric acid) can be produced by the electrolysis of concentrated sulfuric acid solutions.
• At the anode, the half-reaction is:

2 HSO₄^- → H₂S₂O₈ + 2e^-

EXPLANATION:

• To efficiently produce H₂S₂O₈, a concentrated solution of sulfuric acid is required because:
• In a concentrated solution, the concentration of HSO₄^- ions is high enough to facilitate the formation of peroxodisulfuric acid at the anode.
• Dilute solutions of sulfuric acid or sodium sulfate do not provide the necessary ion concentrations for this reaction.

Therefore, the correct answer is: (3) Concentrated solution of sulfuric acid.

CONCEPT:

Fehling's Test

• Fehling's test is used to detect the presence of reducing sugars and aldehydes.
• Compounds with an aldehyde group (-CHO) or α-hydroxy ketones give a positive Fehling's test, resulting in the formation of a red precipitate of cuprous oxide (Cu₂O).
• Ketones and aromatic aldehydes (like benzaldehyde) generally do not give a positive Fehling's test.

EXPLANATION:

• Compound (A): Benzaldehyde (C₆H₅CHO) is an aromatic aldehyde and does not give a positive Fehling's test.
• Compound (B): Acetophenone (C₆H₅COCH₃) is a ketone and does not give a positive Fehling's test.
• Compound (C): Glucose derivative (HOCH₂–CO–(CHOH)₃–CH₂OH) contains an α-hydroxy aldehyde functional group and gives a positive Fehling's test.
• Compound (D): Propionaldehyde (CH₃CH₂CHO) is an aliphatic aldehyde and gives a positive Fehling's test.
• Compound (E): Phenylacetaldehyde (C₆H₅CH₂CHO) is an aliphatic aldehyde and gives a positive Fehling's test.

Therefore, the compounds (C), (D), and (E) give a positive Fehling's test.

CONCEPT:

Crystal Field Stabilization Energy (CFSE)

• CFSE is the energy difference between the energy of an electron configuration in the ligand field and in an isotropic field.
• It depends on:
• The splitting of d-orbitals (t_2g and e_g) in the crystal field.
• The number of electrons in these orbitals.
• CFSE is calculated as:

CFSE = [(number of electrons in t_2g) × (-0.4Δ₀)] + [(number of electrons in e_g) × (+0.6Δ₀)]

EXPLANATION:

For complex K3 [Fe(SCN)6] 

Calculation of CFSE

= (–0.4 × 3 + 0.6 × 2) Δ0

= 0 Δ0

• [Fe(NH₃)₆]Br₂:
• Fe²⁺ (d⁶) forms a high-spin complex due to NH₃ being a weak field ligand. This results in CFSE ≠ 0.
• [Fe(en)₃]Cl₃:
• Fe³⁺ (d⁵) forms a low-spin complex due to en (ethylenediamine) being a strong field ligand. CFSE ≠ 0.
• K₄[Fe(CN)₆]:
• Fe²⁺ (d⁶) forms a low-spin complex due to CN^- being a strong field ligand. CFSE ≠ 0.
• K₃[Fe(SCN)₆]:
• SCN^- is a weak field ligand, and Fe³⁺ (d⁵) forms a high-spin complex. The electron configuration is t_2g³ e_g², resulting in:
• CFSE = [-0.4(3) + 0.6(2)]Δ₀
• CFSE = 0

Therefore, in K₃[Fe(SCN)₆], the CFSE (Δ₀) is equal to zero.

CONCEPT:

Boiling Point Elevation

• The boiling point elevation (ΔT_b) of a solution is directly proportional to the molal concentration (m) of the solute and the van't Hoff factor (i).
• The formula is given by:

ΔT_b = i × K_b × m

where K_b is the ebullioscopic constant of the solvent.

• For ionic compounds, the van't Hoff factor (i) represents the number of particles the compound dissociates into in solution.
• For non-electrolytes like urea, i = 1.

EXPLANATION:

ΔTb = i Kb . m . ∝ i.C.

where C = concentration 

B.P. order (ii) < (i) < (iii) < (iv)

Therefore, the correct answer is (ii) < (i) < (iii) < (iv).

CONCEPT:

Aromatic Substitution and Diazotization

• Bromination in AcOH: Bromine in acetic acid selectively introduces a bromine atom at the para position relative to the nitro group (-NO₂) due to its electron-withdrawing nature.
• Reduction: The nitro group (-NO₂) is reduced to an amino group (-NH₂) using Sn and HCl.
• Diazotization: Amino groups react with NaNO₂ and HCl at low temperatures (273 K) to form diazonium salts (-N₂⁺).
• Reaction with Ethanol: Diazonium salts react with ethanol (EtOH) to form aldehydes, specifically acetaldehyde (CH₃CHO), as a product of the Etard reaction mechanism.

EXPLANATION:

• Step 1: Bromination
• The nitrobenzene reacts with Br₂ in acetic acid (AcOH) to produce para-bromo nitrobenzene.
• Step 2: Reduction
• Para-bromo nitrobenzene is reduced with Sn and HCl to form para-bromo aniline (p-bromoaniline).
• Step 3: Diazotization
• p-Bromoaniline reacts with NaNO₂ and HCl at 273 K to form para-bromo benzene diazonium chloride.
• Step 4: Reaction with Ethanol
• The diazonium salt reacts with ethanol (EtOH), forming para-bromo benzene and acetaldehyde (CH₃CHO).

Therefore, the products formed are para-bromo benzene and acetaldehyde (CH₃CHO).

CONCEPT:

Magnetic Behaviour and Geometries

• The magnetic behaviour of a complex can often be used to infer its geometry and the oxidation state of the central metal ion.
• [NiCl₄]²⁻:
• Ni²⁺ has the electronic configuration [Ar] 3d⁸.
• In a tetrahedral field, the 3d orbitals do not experience significant splitting, retaining unpaired electrons and leading to a paramagnetic nature.
• This suggests a tetrahedral geometry.
• [Ni(CO)₄]:
• Ni in zero oxidation state (Ni(0)) has the electronic configuration [Ar] 3d¹⁰.
• In a tetrahedral field, the d orbitals become fully paired in the presence of strong field ligands (like CO), leading to a diamagnetic nature.
• This suggests a tetrahedral geometry.

EXPLANATION:

• [NiCl₄]²⁻:
• Oxidation state of Ni is +2.
• Ni²⁺ electronic configuration: [Ar] 3d⁸.
• If the geometry is tetrahedral, the hybridization is sp³.
• With 2 unpaired electrons, it exhibits paramagnetism.
• [Ni(CO)₄]:
• Oxidation state of Ni is 0.
• Ni(0) electronic configuration: [Ar] 3d¹⁰.
• If the geometry is tetrahedral, the hybridization is sp³.
• With no unpaired electrons, it exhibits diamagnetism.

Therefore, the correct answer is [NiCl₄]²⁻: Ni²⁺, tetrahedral, [Ni(CO)₄]: Ni(0), tetrahedral

CONCEPT:

Geometrical Isomerism

• Geometrical isomerism (cis-trans isomerism) occurs in compounds with restricted rotation around a double bond or in cyclic structures.
• Cis-isomer: Identical groups are on the same side of the double bond.
• Trans-isomer: Identical groups are on opposite sides of the double bond.
• Properties such as dipole moment and melting point differ for cis and trans isomers:
• Cis-isomers usually have higher dipole moments due to group orientation.
• Trans-isomers generally have higher melting points due to better packing in the solid state.

EXPLANATION:

• (A) Propene does not show geometrical isomerism:
• Geometrical isomerism requires two different groups on each carbon of the double bond, which propene (CH₃–CH=CH₂) lacks.

Incorrect

• (B) Trans-isomer has identical groups on opposite sides:
• This is true; geometrical isomers differ based on the position of groups around the double bond.

Correct

• (C) Cis-but-2-ene has a higher dipole moment than trans-but-2-ene:

(dipole moment only)

• In cis-isomer, the dipoles of substituent groups add up, while in trans-isomer, the dipoles cancel out.

Correct

• (D) 2-Methylbut-2-ene shows two geometrical isomers:

 (does not show GI) 

• This is incorrect as 2-methylbut-2-ene does not have two different groups on each carbon of the double bond.

Incorrect

• (E) Trans-isomer has a lower melting point than cis-isomer:

 (Melting point) 

• Trans-isomers generally have a higher melting point due to better packing in the solid state.

Incorrect

Therefore, the incorrect statements are: (A), (D), and (E).