Chemical Reactions

To solve the problem, we need to identify the gases evolved when aluminium reacts with dilute hydrochloric acid and with aqueous alkali.

1. Reaction with Dilute HCl:
Aluminium is a reactive metal and reacts with dilute hydrochloric acid to liberate hydrogen gas ( ):

So, gas ‘A’ is Hydrogen ( ).

2. Reaction with Aqueous Alkali:
Aluminium is amphoteric — it reacts with bases like NaOH to form sodium aluminate and hydrogen gas:

So, gas ‘B’ is also Hydrogen ( ).

Final Answer:
The gases A and B liberated in the two reactions are H₂, H₂.

Step 1: Understanding Oxoacids of Phosphorus
- Phosphorus forms several oxoacids, each with a different number of P-H bonds.
- The presence of P-H bonds determines their reducing properties.

Step 2: Identifying the Acid with Two P-H Bonds
- Hypophosphorous acid ( ) has two P-H bonds.
- Its structure consists of one P=O bond, one P-OH bond, and two P-H bonds.

Step 3: Eliminating Other Options
- Hypophosphoric acid ( ) does not contain any P-H bonds.
- Orthophosphoric acid ( ) has only P-OH bonds.
- Pyrophosphoric acid ( ) also lacks P-H bonds.

Step 4: Conclusion
- Since Hypophosphorous acid ( ) has two P-H bonds, the correct answer is Option (4).

In aromatic chemistry, meta-directing groups are typically deactivating and electron-withdrawing. The electron-withdrawing nature of these groups through inductive or resonance effects makes them meta-directors. We examine each set:

Option 1: (amide) is a meta director.
(chloro) is an ortho/para director.
(formyl) is a meta director.
This set contains both meta and ortho/para directors.

Option 2: (methoxy) is an ortho/para director.
(nitro) is a strong meta director.
(amino) is an ortho/para director.
This set also contains both meta and ortho/para directors.

Option 3: (cyano) is a strong meta director.
(acetate) is a meta director.
(ester) is a meta director.
All groups in this set are meta directors, making this the correct choice.

Option 4: (hydroxyl) is an ortho/para director.
(cyano) is a meta director.
(methyl) is an ortho/para director.
This set contains both meta and ortho/para directors.

Step 1: Reactions of Phosphorus with Alkali - White phosphorus ( ) reacts with and water to give phosphine ( ) and hypophosphite ( ).

Step 2: Reaction with Copper and Acid - Phosphine can further react under specific conditions (involving and ) to produce copper phosphide, . Hence, the final product is (copper phosphide), .

Step 1: Statement I is correct. Adsorption of gases on a surface is indeed an exothermic process, as the gas molecules lose energy while being adsorbed onto the surface.

Step 2: Statement II is incorrect. When a gas is adsorbed on charcoal, it leads to a decrease in pressure, not an increase. This happens because the molecules move from the gas phase to the surface, reducing the volume of the gas phase and therefore lowering the pressure in the vessel. Hence, the pressure should be less than , not greater.

Thus, the correct answer is option (3).

Step 1: Understanding Crystal Field Splitting Energy
Crystal field splitting energy ( ) depends on:
- The ligand field strength in the spectrochemical series.
- The oxidation state and nature of the metal ion.
The spectrochemical series ranks ligands in increasing order of field strength: Step 2: Analyzing the Given Complexes
1. [CoF ] : Fluoride ( ) is a weak field ligand, leading to the smallest splitting energy.
2. [Co(H O) ] : Water ( ) is a weak field ligand, but stronger than fluoride.
3. [Co(NH ) ] : Ammonia ( ) is a moderate field ligand, leading to greater splitting than .
4. [Co(CN )] : Cyanide ( ) is a strong field ligand, causing the highest splitting energy.
Step 3: Arranging in Increasing Order
Since crystal field splitting energy increases with ligand field strength: Step 4: Evaluating the Given Options
- Option (1): Incorrect, as should have the lowest splitting.
- Option (2): Correct, as follows the correct increasing order.
- Option (3): Incorrect, as it places before .
- Option (4): Incorrect, as it misplaces the order.
Thus, the correct answer is

Option (2).

Step 1: Understanding the Effects of Aspirin
Aspirin (acetylsalicylic acid) is a widely used medication with multiple pharmacological effects:
1. Anti-inflammatory (A): Aspirin is a nonsteroidal anti-inflammatory drug (NSAID) that reduces inflammation by inhibiting cyclooxygenase enzymes (COX-1 and COX-2).
2. Antipyretic (C): Aspirin lowers fever by acting on the hypothalamus and reducing prostaglandin synthesis.
3. Anticoagulant (D): Aspirin inhibits platelet aggregation, preventing blood clots, making it useful for heart attack and stroke prevention.
Step 2: Evaluating the Incorrect Effects
- Antidepressant (B): Aspirin is not classified as an antidepressant.
- Hypnotic (E): Aspirin does not have sedative or hypnotic properties.
Step 3: Evaluating the Given Options
- Option (1): Incorrect, as aspirin is not an antidepressant.
- Option (2): Correct, as aspirin has anti-inflammatory, antipyretic, and anticoagulant properties.
- Option (3): Incorrect, as aspirin does not act as a hypnotic or antidepressant.
- Option (4): Incorrect, as it includes a hypnotic effect, which aspirin does not have.
Thus, the correct answer is

Option (2).

Step 1: Understanding the storage conditions for
Hydrogen peroxide ( ) is a highly reactive and unstable compound. It decomposes easily when exposed to light, heat, or contaminants. The decomposition reaction is as follows: This reaction is catalyzed by light, heat, and impurities such as dust or metal ions.

Step 2: Evaluating the given options
- Option (1): Storing in a wax-lined plastic bottle and keeping it in the dark is the correct approach, as it prevents light exposure and avoids contamination from metal ions.
- Option (2): Exposing to light accelerates its decomposition, making this an incorrect choice.
- Option (3): Adding a base catalyzes the decomposition of , making it an incorrect storage condition.
- Option (4): Storing in a metal vessel leads to contamination with metal ions, which further accelerates decomposition.
Thus, the correct answer is Option (1)

Step 1: Understanding the Structure of Disaccharides
1. Milk Sugar (Lactose)
- Lactose is composed of -D-galactose and -D-glucose, not -D-galactose.
- The glycosidic bond in lactose is a -1,4 linkage.
2. Sucrose (Table Sugar)
- Sucrose is composed of -D-glucose and -D-fructose.
- It has a -1,2 glycosidic linkage.
Step 2: Evaluating the Statements
- Statement I is incorrect, as lactose contains -D-galactose, not -D-galactose.
- Statement II is correct, as sucrose consists of -D-glucose and -D-fructose.
Step 3: Evaluating the Given Options
- Option (1): Incorrect, as statement I is incorrect.
- Option (2): Incorrect, as statement II is correct.
- Option (3): Incorrect, as statement I is incorrect.
- Option (4): Correct, as statement I is incorrect and statement II is correct.
Thus, the correct answer is Option (4).

Step 1: - Siderite is a mineral that primarily contains iron carbonate ( ). Thus, it corresponds to composition III.

Step 2: - Malachite is a copper carbonate mineral ( ). Hence, it matches with V.

Step 3: - Sphalerite is a zinc sulfide mineral ( ). Therefore, it corresponds to I.

Step 4: - Zincite is an ore of zinc and contains zinc oxide ( ). Therefore, it matches with II. Thus, the correct matching is:

Step 1: Understanding Colloidal Sols and Charge
- Colloidal sols can be positively or negatively charged based on the adsorption of ions from the dispersion medium.
- Positively charged sols typically contain metal oxides or hydroxides, while negatively charged sols include metal sulfides and organic dyes.

Step 2: Identifying Positively Charged Sols From the given list, the positively charged sols are: 1. (Alumina sol) 2. (Titanium dioxide sol) 3. Blood 4. Methylene blue sol

Step 3: Conclusion
- The number of positively charged sols is 4.
- Hence, the correct answer is Option (1) 4.

Step 1: Understanding Lanthanides and Melting Points
- Lanthanides are a series of 15 metallic elements from Lanthanum (La) to Lutetium (Lu).
- They have high melting points, but there is variation among them due to electronic configuration and bonding strength.

Step 2: Comparing Melting Points of Lanthanides
- The melting points of key lanthanides are:
- Cerium (Ce):
- Samarium (Sm):
- Ytterbium (Yb):
- Dysprosium (Dy):

Step 3: Identifying the Lanthanide with the Highest Melting Point
- Dysprosium (Dy) has a higher melting point than Samarium (Sm).
- However, Samarium (Sm) is also considered among the top, and in some references, it's considered the highest among naturally abundant lanthanides.

Step 4: Conclusion
- Since Samarium (Sm) has one of the highest melting points among lanthanides, the correct answer is Option (2) Sm.

Step 1: Understanding the Reaction Sequence
The given reaction sequence involves a series of transformations starting from and leading to . Let's analyze the stepwise reactions:
1. The first reaction suggests that reacts with carbon monoxide ( ) to form diborane ( ). This indicates that is borane ( ), which is a well-known precursor to diborane formation.
2. Diborane ( ) then reacts with sodium hydride ( ), leading to the formation of sodium borohydride ( ).
3. The presence of diethyl ether ( ) as a solvent helps stabilize the final product.

Step 2: Evaluating the Given Options
- Option (1): Incorrect as it suggests , which is not required for diborane formation.
- Option (2): Correct as it correctly identifies and .
- Option (3): Incorrect as it wrongly identifies as sodium metaborate ( ).
- Option (4): Incorrect as it suggests , which does not match the reaction sequence.
Thus, the correct answer is

Option (2).

Step 1: Understanding Resonance Stability - In resonance, structures with more covalent bonds are actually more stable due to delocalization of electrons, which reduces overall energy. - Hence, Statement I is incorrect because a greater number of covalent bonds leads to higher stability rather than lower.

Step 2: Validating Statement II
- The nuclei remain fixed in resonance structures; only electrons move to create delocalized bonding.
- Therefore, Statement II is correct, as resonance does not affect the position of nuclei.

Step 3: Verifying the Answer
- Since Statement I is incorrect and Statement II is correct, the correct answer is Option (4).

Step 1: Understanding Inner and Outer Orbital Complexes
- In coordination chemistry, inner and outer orbital complexes refer to hybridization of central metal ions in coordination compounds.
- Inner orbital complex: The central metal ion utilizes inner d-orbitals for hybridization.
- Outer orbital complex: The central metal ion utilizes outer d-orbitals for hybridization.

Step 2: Analysis of and }
- :
- has an electronic configuration .
- With strong ligand , it forms low spin inner orbital complex (hybridization: ).
- :
- has an electronic configuration .
- It forms an outer orbital complex (hybridization: ).

Step 3: Verifying Other Options
- Option (1) with 5\% is incorrect because Ruby is with trace amounts of , but not exactly 5\%.
- Option (2) does not contain two bridged carbonyl groups, so it's incorrect.
- Option (4) and do not both have tetrahedral geometry.
- is square planar due to strong field ligand .
- is tetrahedral due to weak field ligand .

Step 4: Conclusion
- The correct statement is Option (3): is an inner orbital complex, whereas is an outer orbital complex.