Carbonyl Compounds

Aromatic compounds burns with sooty flame

The correct Answer is: 10

Step 1: Understanding the Concept:
is a square planar, neutral complex. Its stability and distinct color are due to its unique structure involving intramolecular hydrogen bonding.
Step 2: Detailed Explanation:
- Statement A: Correct. The complex is a rosy-red precipitate. - Statement B: Incorrect. It is highly insoluble in water, which is why it's used for the gravimetric estimation of Nickel. - Statement C: Incorrect. As a square planar complex with strong field behavior (from donors), it is diamagnetic (0 unpaired electrons). - Statement D: Correct. Square planar geometry implies angles close to 90°. - Statement E: Incorrect. It contains two five-membered chelate rings (Ni-N-C-C-N) and two six-membered rings formed by hydrogen bonding (O-H...O).
Step 3: Final Answer:
The incorrect statements are B, C, and E.

Step 1: Identify each reagent and reaction.
- NH –NH , KOH is used in Wolff–Kishner reduction.
- Ag(NH ) OH is Tollen's reagent.
- Aq. CuSO , sodium potassium tartrate, KOH is used in Fehling's test.
- Zn–Hg, HCl is used in Clemmensen reduction.
Step 2: Match correctly.
Final Answer:

Step 1: Understanding the properties.
- Ionisation energy: F has the highest ionisation energy because of its small size, followed by Cl.
- Basic nature: As we move down the group, the basic nature increases due to the increasing ionic size and decreasing bond strength. Thus, K₂O>Na₂O>Al₂O₃ is the correct order for basic nature.
- Metallic character: The metallic character decreases from left to right across the period and increases from top to bottom. Therefore, K>Na>Mg>Al is the correct order for metallic character.
Step 2: Conclusion.
The correct answer is (2) K₂O>Na₂O>Al₂O₃ (Basic nature) as per the analysis above.

Step 1: Understanding the reaction.
On hydrolysis, C₆H₁₂O₃ gives a product that undergoes both the iodoform test and Tollens’ test. The iodoform test indicates the presence of a methyl ketone group (–COCH₃), and the Tollens’ test indicates an aldehyde group (–CHO).
Step 2: Conclusion.
The structure fits both tests, as it contains a methyl ketone group and an aldehyde group. Therefore, the correct answer is (1).