CONCEPT:

Equivalent Weight of Oxygen

• The equivalent weight of an element is calculated as the atomic weight divided by the change in oxidation state during a chemical reaction.
• For oxygen, the atomic weight is 16.
• The typical oxidation state of oxygen is -2 in most compounds, but in peroxides, the oxidation state is -1.

CALCULATIONS:

CONCLUSION:

• The correct option is: Option 4

CONCEPT:

Determination of Purity of H₂O₂ using Redox Titration

• In this redox titration, potassium permanganate (KMnO₄) acts as an oxidizing agent and hydrogen peroxide (H₂O₂) is the reducing agent.

CONCEPT:

Quantum Numbers and the Valence Electron of Rubidium (Z = 37)

CONCEPT:

S_N2 Mechanism

• The S_N2 mechanism (Substitution Nucleophilic Bimolecular) involves a single-step reaction where the nucleophile directly attacks the electrophilic carbon, displacing the leaving group (in this case, bromine) simultaneously.
• The reaction rate of an S_N2 process depends on the concentration of both the substrate (alkyl halide) and the nucleophile. Steric hindrance around the electrophilic carbon atom plays a significant role in determining how easily the nucleophile can approach and attack the carbon center.
• The order of reactivity in S_N2 reactions is:
  • Primary alkyl halides (least steric hindrance) > Secondary alkyl halides > Tertiary alkyl halides (most steric hindrance)
• In S_N2 reactions, the transition state involves the nucleophile approaching from the opposite side of the leaving group, so the less bulky the groups around the electrophilic carbon, the faster the reaction.

EXPLANATION:

• CH₃CH₂CH₂CH₂Br (1-bromobutane)
  • This is a primary alkyl halide where the carbon attached to the bromine atom is bonded to one alkyl group and two hydrogen atoms.
  • The lack of significant steric hindrance allows easy access for the nucleophile to attack the carbon, making this compound highly favorable for S_N2 reactions.
  • Conclusion: 1-bromobutane undergoes S_N2 reactions easily.
• CH₃CH₂CH(CH₃)Br (2-bromobutane)
  • This is a secondary alkyl halide where the carbon attached to the bromine atom is bonded to two alkyl groups (a methyl and an ethyl group).
  • Although it can undergo an S_N2 reaction, the presence of two alkyl groups increases steric hindrance, making the reaction slower than in primary alkyl halides.
  • Conclusion: 2-bromobutane can undergo S_N2 reactions but with more difficulty compared to primary alkyl halides.
• (CH₃)₃CBr (tert-butyl bromide)
  • This is a tertiary alkyl halide where the carbon attached to the bromine atom is bonded to three alkyl groups (three methyl groups).
  • The significant steric hindrance around the carbon makes it nearly impossible for a nucleophile to attack in an S_N2 reaction.
  • Conclusion: Tertiary alkyl halides like tert-butyl bromide do not undergo S_N2 reactions due to severe steric hindrance.

​CONCLUSION:

The correct answer is Option 1: CH₃CH₂CH₂CH₂Br (1-bromobutane) because it is a primary alkyl halide with minimal steric hindrance, making it ideal for an S_N2 reaction.

CONCEPT:

First Law of Thermodynamics at Constant Volume

• The first law of thermodynamics is given by:
  • ΔE = q - W
  • Where:
    • ΔE = Change in internal energy
    • q = Heat supplied to the system
    • W = Work done by the system
• At constant volume, no work is done by the gas because the work done in a process involving volume change is:
  • W = PΔV
  • Since the volume is constant, ΔV = 0, and therefore, W = 0.
• At constant volume, the heat supplied to the gas (q) goes into changing the internal energy (ΔE) of the system.
• Thus, at constant volume:
  • q = ΔE

EXPLANATION:

• The heat supplied (q = 500 J) goes entirely into changing the internal energy (ΔE) of the system.
• • Therefore, the internal energy change is also 500 J: q = ΔE = 500 J.
  Since the process occurs at constant volume, no work is done (W = 0).

CONCLUSION:

• The correct answer is: Option 2 - q = ΔE = 500 J, W = 0

Concept:

Dmitri Mendeleev developed the Periodic Law, which states that the properties of the elements are a periodic function of their atomic weights.

Explanation:

Mendeleev created an early version of the periodic table by arranging elements in order of increasing atomic weight. He observed that elements with similar chemical properties occurred at regular intervals (periodically). This arrangement allowed him to predict the existence and properties of undiscovered elements.

Conclusion:

According to Mendeleev, the properties of the elements are a periodic function of their: Atomic weight

CONCEPT:

Redox Reactions

• Redox reactions are chemical reactions in which there is a transfer of electrons between two species. This transfer results in a change in the oxidation states of the reactants involved.
• In a redox reaction, one species is oxidized (loses electrons) and the other is reduced (gains electrons).

This fashion indeed in Transfers, an internal redox reaction i.e. Cannizzaro reaction leading those shifts happening under such conditions.

The correct answer is 4) All the above are redox reactions

CONCEPT:

pH of a Salt of a Weak Acid and its Conjugate Base

• Sodium bicarbonate (NaHCO₃) is a salt formed from a weak acid (H₂CO₃) and a weak base (HCO₃^-).
• The pH of a salt solution of a weak acid and its conjugate base can be estimated using the formula derived from the Henderson-Hasselbalch equation.

CONCEPT:

Hydrolysis of Compounds

• Hydrolysis is the reaction of a compound with water, leading to the breakdown of the compound into two or more parts.
• Inorganic halides such as PCl₃, AsCl₃, and SbCl₃ usually hydrolyze in water to form corresponding acids or oxides and hydrogen chloride.
• However, not all halides hydrolyze easily. The inertness towards hydrolysis is generally observed in compounds where the central atom is highly electronegative and forms strong bonds with halides.

Explanation:-

• Analyzing given compounds:
  • PCl₃ (Phosphorus trichloride) hydrolyzes to produce H₃PO₃ and HCl.
  • AsCl₃ (Arsenic trichloride) hydrolyzes to form H₃AsO₃ and HCl.
  • NF₃ (Nitrogen trifluoride) does not hydrolyze due to strong N-F bonds.
  • SbCl₃ (Antimony trichloride) hydrolyzes to form SbOCl and HCl.

NF3 (Nitrogen trifluoride) is highly resistant to hydrolysis due to the strong N-F bond and the high electronegativity of fluorine, which makes the molecule kinetically stable and less reactive towards water.

The correct answer is NF₃