The Solid State

Let's analyze each of the statements:
1. Ionisation isomer: Ionisation isomers are formed when the counter-ion (such as Br or SO ) exchanges places with the ligand. For the given complex , the ionisation isomer can be formed by switching the counter-ion and the ligand, resulting in . Thus, statement (A) is correct.
2. Precipitation with AgNO : When the complex is treated with AgNO , the bromide ion (Br ) will react with Ag to form a yellow precipitate of AgBr. Thus, statement (B) is correct.
3. Precipitation with BaCl : The ionisation isomer of the complex contains the sulfate ion (SO ), which reacts with BaCl to form a white precipitate of BaSO .
Therefore, statement (C) is also correct. Since all the statements are correct, the correct answer is , All the above are correct statements.

Step 1: Understanding Functional Groups in Amino Acids
- Amino acids contain different functional groups that determine their chemical behavior.
- Basic functional groups include amine ( ) and imidazole rings.
Step 2: Classification of Given Options
- Arginine, Lysine, and Histidine contain multiple basic groups due to their side chains.
- Asparagine contains only one amine group, making it the correct answer.
Final Answer: Asparagine contains a single amine functional group, making it the correct choice.

Step 1: Understanding Reactivity of Benzyl Halides
Benzyl halides are more reactive than aryl halides because the benzylic position stabilizes carbocation intermediates.
Step 2: Analyzing the Reaction Mechanism
The amine group ( ) undergoes nucleophilic substitution at the benzylic position.
The bromine remains intact since the benzyl chloride reacts first due to higher reactivity.
Step 3: Identifying the Correct Product
The correct product is the one where amination occurs at the benzylic carbon, leading to o-bromobenzylamine.
Thus, the correct answer is (C).

Step 1: Understanding Reducing Agent Strength
A reducing agent donates electrons and gets oxidized. - The strength of a reducing agent increases as we move down the group in the periodic table. - This happens because the atomic size increases, and the bond strength between the central atom and hydrogen weakens, making electron donation easier.
Step 2: Trend in Group 15 Hydrides
- Group 15 hydrides: . - The reducing character increases as: - Thus, BiH is the strongest reducing agent among the given choices.

Step 1: Understanding Saline Hydrides
Saline hydrides (ionic hydrides) are formed by alkali and alkaline earth metals with hydrogen. Examples: LiH, BeH₂, and MgH₂. These hydrides have a significant ionic character, though lighter ones like BeH₂ and MgH₂ have some covalent character. Thus, Statement I is correct.
Step 2: Are Saline Hydrides Volatile?
Volatility depends on weak intermolecular forces. Saline hydrides form strong ionic bonds, making them non-volatile. Thus, Statement II is incorrect.
Step 3: Understanding Electron-Precise Hydrides
Electron-precise hydrides like CH₄ and SiH₄ have enough valence electrons to form covalent bonds. These are not Lewis bases because they do not have lone pairs to donate. Thus, Statement III is incorrect.
Step 4: Chromium Hydride Formula
Chromium forms a hydride with the formula CrH. Thus, Statement IV is correct.
Step 5: Correct Answer
The correct statements are (I) and (IV). Thus, the correct answer is (C).

Step 1: Identifying the Electronic Configuration
Nickel (Ni) has an atomic number 28.
The electronic configuration of Ni is: The oxidation state of Ni in the given complexes needs to be determined.
Step 2: Evaluating Each Complex
(A) [Ni(Cl) ]
Cl is a weak field ligand.
The tetrahedral geometry follows an sp hybridization.
Paramagnetic due to unpaired electrons. Not the correct answer.
(B) [Co(C O ) ]
Co has a strong field ligand (oxalate).
Octahedral structure, not tetrahedral. Incorrect choice.
(C) [Ni(CN) ]
CN is a strong field ligand.
Follows dsp hybridization, giving square planar geometry.
Diamagnetic, but not tetrahedral. Incorrect choice.
(D) [Ni(CO) ]
CO is a strong field ligand.
Causes electron pairing and sp hybridization, forming a tetrahedral structure.
No unpaired electrons = Diamagnetic. Correct answer.
Final Answer: The correct diamagnetic, tetrahedral complex is [Ni(CO) ].

Step 1: Identify the Reaction Type
The reaction involves arsenic(III) oxide reacting with hydrogen sulfide to form arsenic sulfide ( ).
Step 2: Balancing the Reaction
Here, arsenic oxide is reduced to arsenic sulfide while hydrogen sulfide is oxidized.
Final Answer: The sol formed in this reaction is arsenic sulfide ( ).

Step 1: Understanding Colloidal Systems
Colloidal particles are heterogeneous mixtures with dispersed-phase particles ranging from 1 nm to 1000 nm.
They exhibit electrical charge, which helps maintain stability by preventing coagulation.
Step 2: Mechanism of Charge Development
The charge on colloidal particles arises primarily due to the adsorption of ions from the solution.
When a colloidal particle is in contact with a solution, selective adsorption of a particular ion occurs.
Step 3: Examples of Charge Formation
- If AgI sol is prepared in an excess of KI, the colloidal particles adsorb ions and acquire a negative charge: - If AgI sol is prepared in an excess of AgNO , the colloidal particles adsorb ions and acquire a positive charge: Step 4: Importance of Charge in Colloidal Stability
- The mutual repulsion between similarly charged colloidal particles prevents coagulation and keeps them dispersed. - Oppositely charged colloids can undergo coagulation (precipitation) when mixed.
Final Answer: The charge on colloidal particles is due to the adsorption of ions from the solution.