Inorganic Chemistry

To solve this problem, we need to evaluate the color of different metal ions in aqueous solutions, which depends on their electronic configuration and the nature of their d-orbitals.

1. Electronic Configurations and Oxidation States:
The color of a metal ion in solution is often due to d-d transitions in the case of transition metal ions. For the following ions, we need to determine if both are colored in aqueous solutions:

- Scandium (Sc): The electronic configuration of Sc is [Ar] 3d¹ 4s². In its +3 oxidation state, Sc has the configuration [Ar], which means it has no d-electrons and does not show color in aqueous solution.

- Titanium (Ti): The electronic configuration of Ti is [Ar] 3d² 4s². In its +3 oxidation state, Ti has the configuration [Ar] 3d¹, which makes it colored in aqueous solution due to d-d transitions.

- Nickel (Ni): The electronic configuration of Ni is [Ar] 3d⁸ 4s². In its +2 oxidation state, Ni has the configuration [Ar] 3d⁸, and the Ni²⁺ ion is colored due to the presence of d-electrons.

- Copper (Cu): The electronic configuration of Cu is [Ar] 3d¹⁰ 4s¹. In its +2 oxidation state, Cu has the configuration [Ar] 3d⁹, and Cu²⁺ is colored in aqueous solution due to d-d transitions.

- Manganese (Mn): The electronic configuration of Mn is [Ar] 3d⁵ 4s². In its +2 oxidation state, Mn has the configuration [Ar] 3d⁵, and Mn²⁺ is not strongly colored because the d-orbital is fully occupied in a stable arrangement. However, Mn³⁺ is colored in aqueous solution due to d-d transitions.

2. Analyzing the Options:
- Option (A) "Sc³⁺, Mn²⁺": Sc³⁺ is colorless as it has no d-electrons, and Mn²⁺ is also colorless because its 3d orbitals are filled. Thus, this option is incorrect.
- Option (B) "Ni²⁺, Ti⁴⁺": Ni²⁺ is colored, but Ti⁴⁺ has an empty 3d orbital, so it does not display color. Thus, this option is incorrect.
- Option (C) "Ti³⁺, Cu⁺": Ti³⁺ is colored, but Cu⁺ is colorless since it has a fully filled d-orbital. Thus, this option is incorrect.
- Option (D) "Mn²⁺, Ti³⁺": Both Mn²⁺ and Ti³⁺ are colored due to d-d transitions. Thus, this option is correct.

Final Answer:
The correct answer is (D) Mn²⁺, Ti³⁺.

To solve the problem, we need to determine the correct order of wavelengths of light absorbed by the given complex ions: , , and . This involves understanding the relationship between ligand field strength, crystal field splitting energy ( ), and the wavelength of light absorbed.

1. Understanding Ligand Field Strength:
The wavelength of light absorbed by a complex ion is inversely proportional to the crystal field splitting energy ( ). The stronger the ligand field (i.e., the stronger the ligand), the larger the , and the shorter the wavelength of light absorbed. Conversely, a weaker ligand field results in a smaller and a longer wavelength of light absorbed.
The ligands in the complexes are:
- : Strong field ligand.
- : Very strong field ligand.
- : Weak field ligand.

2. Analyzing Each Complex:

(A) :
- Central metal: (d-electron configuration: ).
- Ligands: One and five .
- Overall ligand field strength: Primarily determined by (strong field), but slightly weakened by the presence of one (weak field).

(B) :
- Central metal: (d-electron configuration: ).
- Ligands: Six .
- Overall ligand field strength: Very strong due to six ligands.

(C) :
- Central metal: (d-electron configuration: ).
- Ligands: Six .
- Overall ligand field strength: Extremely strong due to six ligands, which are very strong field ligands.

3. Comparing Crystal Field Splitting Energies ( ):
- : Contains six ligands, which are very strong field ligands. This results in the largest .
- : Contains six ligands, which are strong field ligands. This results in a large , but slightly smaller than that of .
- : Contains five ligands and one ligand. The presence of weakens the overall ligand field, resulting in a smaller compared to the other two complexes.

4. Order of Wavelengths Absorbed:
Since the wavelength of light absorbed is inversely proportional to , the order of wavelengths absorbed is:
$ \text{(A) [CoCl(NH}_3\text{)}_5\text{]}^{2+} > \text{[Co(NH}_3\text{)}_6\text{]}^{3+} > \text{[Co(CN)}_6\text{]}^{3-} $.

1. Identify the Key Properties of P

P is described as:
- Colorless
- Neutral
- Paramagnetic
- Oxide of Nitrogen

2. Evaluate Possible Nitrogen Oxides Based on the Properties of P

- : Colorless, neutral, diamagnetic.
- : Colorless, neutral, paramagnetic.
- : Reddish brown, acidic, paramagnetic.
- : Colorless, acidic, diamagnetic.
- : Colorless solid, acidic, diamagnetic.
- : Blue solid, acidic, diamagnetic.

Based on these properties, P must be .

3. Determine the Identity of Q
Q is formed by oxidation of P ( ) and is a reddish-brown gas. , when oxidized, forms , which is a reddish-brown gas:
$ NO_2 NO_2 NO_2 N_2O_4 N_2O_4 N_2O_4 NO NO N_2O_4 N_2O_3 N_2O_3 NO NO_2 N_2O_4 N_2O_3 NO NO_2 N_2O_4 N_2O_3$

Identifying Amphoteric Oxides

Amphoteric oxides are oxides that can react with both acids and bases to form salts and water. To identify the correct pair of amphoteric oxides from the given options, we need to consider the chemical properties of each oxide.

Options:

1. Option A: ,
   While (Beryllium oxide) is amphoteric, (Magnesium oxide) is a basic oxide and does not react with bases.
2. Option B: ,
   Both (Beryllium oxide) and (Zinc oxide) are amphoteric oxides.
3. Option C: ,
   (Aluminum oxide) is amphoteric, but (Lithium oxide) is a basic oxide and does not react with bases.
4. Option D: ,
   (Beryllium oxide) is amphoteric, but (Boron trioxide) is an acidic oxide and does not react with acids.

Conclusion:

The correct answer is Option B: , , as both are amphoteric oxides.

The reaction shown in the image involves the following steps:

Step 1: The compound with a hydroxyl group attached to a benzene ring (phenol) reacts with NaOH and CO₂, leading to the formation of salicylate (compound A).

Step 2: The reaction of salicylate with acetylation (in the presence of (CH₃CO)₂O) results in the formation of aspirin (compound Y).

Conclusion:

The major product Y in the reaction is aspirin.

Thus, the correct answer is Option D: Aspirin.

Aniline (C6H5NH2) is a primary amine and exhibits certain reactivity patterns due to the presence of the amine group (-NH2), which is an electron-donating group. The electron-donating nature of the amine group makes the aromatic ring more reactive towards electrophilic substitution reactions. However, the amine group also affects its reactivity with certain reagents.

Analysis of the options:

• Option A: Friedel-Crafts reaction - Aniline does not undergo the Friedel-Crafts reaction effectively. This is because the amine group (-NH2) is a strong electron-donating group and reacts with the AlCl3 catalyst, which deactivates it and prevents the reaction. Hence, aniline does not undergo Friedel-Crafts reaction.
• Option B: Bromination - Aniline undergoes bromination readily, where the amine group activates the benzene ring and makes it more reactive towards electrophilic bromine (Br2) under mild conditions.
• Option C: Nitration - Aniline undergoes nitration readily, as the electron-donating nature of the amine group increases the electron density on the aromatic ring, making it more susceptible to electrophilic attack by the nitronium ion (NO2+).
• Option D: Sulphonation - Aniline also undergoes sulphonation, where the amine group activates the benzene ring towards electrophilic substitution by the sulfonic acid group (SO3H).

Conclusion:

Option A: Friedel-Crafts reaction, as aniline does not undergo Friedel-Crafts reactions due to deactivation of the catalyst by the amine group.

An organometallic compound is a compound that contains a bond between a carbon atom of an organic molecule and a metal (usually a transition metal or main group metal).

Let's review the options:

• Option 1: CH₃COONa - This is sodium acetate, which is an ionic compound and does not contain a carbon-metal bond. Hence, this is not an organometallic compound.
• Option 2: CH₃CH₂MgBr - This is ethylmagnesium bromide, an organometallic compound. It contains a carbon-magnesium bond (C-Mg), which is a characteristic feature of organometallic compounds. Therefore, this is the correct answer.
• Option 3: (CH₃COO)₂Ca - This is calcium acetate, which is an ionic compound and does not have a carbon-metal bond. Therefore, this is not an organometallic compound.
• Option 4: CH₃ONa - This is sodium methoxide, an ionic compound that does not contain a carbon-metal bond. Hence, this is not an organometallic compound.

Conclusion: The correct answer is Option 2: CH₃CH₂MgBr, as it is an organometallic compound due to the carbon-magnesium bond.

Identifying the Compound That Does Not Produce Dinitrogen on Heating

Let's analyze each compound:

1. Option (A) : This compound contains azide ions ( ). When heated, it decomposes to form barium metal and dinitrogen gas ( ):

1. This **does** produce dinitrogen gas.
2. Option (B) : This compound decomposes upon heating to form water and dinitrogen gas ( ):

1. This **does** produce dinitrogen gas.
2. Option (C) : With careful heating, ammonium nitrate decomposes into dinitrogen monoxide (nitrous oxide) and water. **Importantly**, it does not produce dinitrogen under normal decomposition.

1. While under certain conditions, with more intense heating, other products may form, the *primary* decomposition product is not dinitrogen.
2. Option (D) : This compound decomposes upon heating to form chromium(III) oxide, dinitrogen gas, and water:

1. This **does** produce dinitrogen gas.

Conclusion: The correct answer is (C) because its primary decomposition reaction produces dinitrogen monoxide ( ), **not** dinitrogen ( ).