Dual Nature Of Radiation And Matter

Step 1: Analyze the given statements based on the properties of 1s and 2s orbitals.
- Statement A: 1s and 2s orbitals are spherical in shape. This is correct. Both 1s and 2s orbitals are spherically symmetric.
- Statement B: The probability of finding the electron is maximum near the nucleus. This is correct for the 1s orbital. However, for the 2s orbital, the probability density has a node (a region where the probability density is zero) before increasing again. Thus, the probability is not maximum near the nucleus for the 2s orbital.
- Statement C: The probability of finding the electron at a given distance is equal in all directions. This is correct. Since both 1s and 2s orbitals are spherically symmetric, the probability density depends only on the distance from the nucleus, not on the direction.
- Statement D: The probability density of electrons for the 2s orbital decreases uniformly as the distance from the nucleus increases. This is incorrect. The probability density of the 2s orbital does not decrease uniformly. It has a node (a region of zero probability density) and then increases again before eventually decreasing.
Conclusion: The incorrect statement is (D).

Step 1: Use the freezing point depression formula. The freezing point depression ( ) is given by: where: - = freezing point depression, - = cryoscopic constant (molal freezing point depression constant), - = molality of the solution (moles of solute per kg of solvent).
Step 2: Calculate the freezing point depression. The normal freezing point of water is , and the solution freezes at . Thus:
Step 3: Solve for molality ( ). Using the formula:
Step 4: Calculate the moles of ethyl alcohol required. Since the solvent is 1 litre of water, and the density of water is approximately , the mass of water is . Therefore: Conclusion: Approximately of ethyl alcohol must be added to 1 litre of water to achieve the desired freezing point.

Step 1: Understanding solubility trends. The precipitation of metal sulphates depends on their solubility product constant ( ). The lower the value, the less soluble the compound is, meaning it will precipitate first.
Step 2: Solubility of sulphates. The solubility order of alkaline earth metal sulphates ( ) decreases down the group: Since barium sulphate ( ) has the lowest solubility, it will precipitate first.

Step 1: Understanding S and S reaction mechanisms. - S (Unimolecular Nucleophilic Substitution): This mechanism occurs via the formation of a carbocation intermediate. It is favored in tertiary and benzylic/allylic halides due to carbocation stability. - S (Bimolecular Nucleophilic Substitution): This mechanism proceeds via a backside attack, leading to inversion of configuration. It is favored in primary and secondary alkyl halides with minimal steric hindrance.
Step 2: Analyzing the given halides.
- (III) (Cyclohexyl chloride): - This structure does not form a resonance-stabilized carbocation. - Whether the reaction follows S or S , the product remains the same. - Satisfies the condition of giving the same product in both reactions.
- (IV) Secondary alkyl bromide with branching: - The presence of branching reduces the likelihood of rearrangement. - Whether the reaction proceeds via S or S , the substitution product remains unaffected. - Satisfies the condition of giving the same product in both reactions.
Step 3: Eliminating incorrect Option.
- (I) (Branched alkyl bromide): This can undergo carbocation rearrangement in S , leading to different products.
- (II) (Cyclohexyl chloride with a methyl group): The possibility of rearrangement in S exists, leading to different products.
Thus, the correct answer is (C) (III) and (IV).