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CONCEPT:

Friedel-Crafts Acylation

• Friedel-Crafts acylation is a reaction where an acyl group (RCO-) is introduced to an aromatic ring (such as benzene) using an acyl chloride (RCOCl) and a Lewis acid (usually AlCl₃).
• This reaction is an electrophilic aromatic substitution in which the acyl group replaces a hydrogen atom on the aromatic ring.
• The AlCl₃ acts as a catalyst and helps in the generation of the acylium ion (RCO⁺), which is a highly electrophilic species that attacks the aromatic ring.

EXPLANATION:

• In the given reaction:

C₆H₆ + RCOCl → C₆H₅CO-R

• • The acyl chloride (RCOCl) reacts with AlCl₃ to form the acylium ion (RCO⁺), which then attacks the benzene ring.
• The mechanism of the reaction involves:
• Generation of the acylium ion (RCO⁺) by the interaction of acyl chloride with AlCl₃.
• Attack of the acylium ion on the benzene ring to form the intermediate complex.
• Loss of a proton from the intermediate complex results in the final product, an aryl ketone (C₆H₅CO-R).

Therefore, the major product 'P' formed in the reaction is an aryl ketone (C₆H₅CO-R).

CONCEPT:

Linkage Isomerism

• Linkage isomerism occurs when a particular ligand can coordinate to the central metal atom or ion through two or more different atoms.
• This type of isomerism is possible when a ligand has more than one donor atom but coordinates through only one donor atom at a time.
• The classic example involves the thiocyanate ligand (SCN^-), which can bond through either the sulfur atom or the nitrogen atom to the metal center.

EXPLANATION:

• In the given complexes:

[Pd(C₆H₅)(SCN)₂] and [Pd(C₆H₅)(NCS)₂]

• The ligand SCN^- can attach to the palladium (Pd) center through either the sulfur atom (S) or the nitrogen atom (N).
• In [Pd(C₆H₅)(SCN)₂], the thiocyanate ligand is bonded through the sulfur atom.
• In [Pd(C₆H₅)(NCS)₂], the thiocyanate ligand is bonded through the nitrogen atom.

Therefore, the type of isomerism which exists between [Pd(C₆H₅)(SCN)₂] and [Pd(C₆H₅)(NCS)₂] is linkage isomerism.

CONCEPT:

Body-Centered Cubic (BCC) Structure

• A body-centered cubic (BCC) unit cell has atoms at each of its eight corners and a single atom at the center of the cube.
• Each corner atom is shared among eight adjacent unit cells, and the center atom belongs entirely to the unit cell.

EXPLANATION:

• In a BCC unit cell:
• There are 8 corner atoms, each shared by 8 unit cells. Therefore, the contribution of corner atoms per unit cell is ¹⁄₈ × 8 = 1 atom.
• There is 1 atom at the center of the unit cell, which is not shared with any other unit cell. Therefore, the contribution of the center atom per unit cell is 1 atom.
• Adding these contributions together:
• Total number of atoms per unit cell = 1 (from corners) + 1 (from center) = 2 atoms.

Therefore, the number of atoms per unit cell of BCC structure is 2.

CONCEPT:

Order and Molecularity of Chemical Reactions

• Molecularity refers to the number of reactant molecules involved in an elementary step of the reaction.
• Order of reaction is the sum of the powers of the concentration terms in the rate equation.
• A reaction can be unimolecular, bimolecular, or trimolecular based on the number of molecules involved.
• The order of reaction can be zero, first, second, etc., based on how the rate depends on the concentration of reactants.

EXPLANATION:

• Consider the given reaction:

2N₂O₅ ⇌ 2N₂O₄ + O₂

• For the reaction:
• The reaction involves two molecules of N₂O₅ as reactants, making it bimolecular.
• The given reaction is described as first order, indicating that the rate of reaction depends linearly on the concentration of N₂O₅.
• Therefore, the correct classification for the reaction is bimolecular and first order.

Therefore, the correct answer is option 3.

CONCEPT:

Number of Atoms in a Given Mass

• The number of atoms in a given mass of a substance can be calculated using Avogadro's number, which is 6.022 x 10²³ atoms per mole.
• The molar mass of a substance is used to convert the mass of the substance to moles.
• Once the number of moles is known, it can be multiplied by Avogadro's number to find the number of atoms.

EXPLANATION:

• First, we calculate the number of moles in 6g of H₂O:
• Molar mass of H₂O = 2(1) + 16 = 18 g/mol
• Number of moles of H₂O = mass / molar mass = 6 g / 18 g/mol = 1/3 mol
• The number of atoms in 1/3 mol of H₂O:
• Each molecule of H₂O has 3 atoms (2 hydrogen + 1 oxygen)
• Total number of atoms = 1/3 mol x 6.022 x 10²³ molecules/mol x 3 atoms/molecule = 6.022 x 10²³ atoms
• 0.4g He
• Molar mass of He = 4 g/mol
• Number of moles of He = 0.4 g / 4 g/mol = 0.1 mol
• Number of atoms = 0.1 mol x 6.022 x 10²³ atoms/mol = 6.022 x 10²² atoms
• 22g CO₂
• Molar mass of CO₂ = 12 + 2(16) = 44 g/mol
• Number of moles of CO₂ = 22 g / 44 g/mol = 0.5 mol
• Number of molecules = 0.5 mol x 6.022 x 10²³ molecules/mol = 3.011 x 10²³ molecules
• Each molecule of CO₂ has 3 atoms (1 carbon + 2 oxygen)
• Number of atoms = 3.011 x 10²³ molecules x 3 atoms/molecule = 9.033 x 10²³ atoms
• 1g H₂
• Molar mass of H₂ = 2 g/mol
• Number of moles of H₂ = 1 g / 2 g/mol = 0.5 mol
• Number of molecules = 0.5 mol x 6.022 x 10²³ molecules/mol = 3.011 x 10²³ molecules
• Each molecule of H₂ has 2 atoms (2 hydrogen)
• Number of atoms = 3.011 x 10²³ molecules x 2 atoms/molecule = 6.022 x 10²³ atoms
• 12g CO
• Molar mass of CO = 12 + 16 = 28 g/mol
• Number of moles of CO = 12 g / 28 g/mol = 0.4286 mol
• Number of molecules = 0.4286 mol x 6.022 x 10²³ molecules/mol = 2.579 x 10²³ molecules
• Each molecule of CO has 2 atoms (1 carbon + 1 oxygen)
• Number of atoms = 2.579 x 10²³ molecules x 2 atoms/molecule = 5.158 x 10²³ atoms

Therefore, the correct answer is option 3, which is 1g H₂.

CONCEPT:

Henry's Law Constant

• Henry's Law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.
• The constant of proportionality is known as Henry's Law Constant (K_H).
• A higher Henry's Law constant indicates that the gas is less soluble in water.
• Mathematically, Henry's Law is represented as:

C = K_H * P

where,

• C is the concentration of the gas in the liquid.
• K_H is the Henry's Law constant.
• P is the partial pressure of the gas.

EXPLANATION:

• The gases mentioned in the options (N₂, O₂, He, H₂) have different solubilities in water at 293 K.
• Among these gases, helium (He) has the highest Henry's Law constant.
• This means that helium is the least soluble gas in water at 293 K.
• The values of the Henry's Law constants are typically as follows:
• N₂: 1.67 × 10^-5 mol/(L·atm)
• O₂: 1.30 × 10^-5 mol/(L·atm)
• He: 3.72 × 10^-4 mol/(L·atm)
• H₂: 7.80 × 10^-5 mol/(L·atm)
• As seen, helium (He) has the highest Henry's Law constant, making it the correct answer.

Therefore, in water, helium (He) has the highest Henry's Law constant at 293 K.

CONCEPT:

Acidic Strength of Oxides

• The acidic strength of oxides generally increases with the oxidation state of the central atom.
• Oxides of metals tend to be basic, while oxides of non-metals tend to be acidic. Transition metals, like chromium, can form oxides that have varying acidic or basic properties based on their oxidation states.

EXPLANATION:

• For chromium oxides:
• CrO (Chromium(II) oxide) has chromium in a +2 oxidation state.
• Cr₂O₃ (Chromium(III) oxide) has chromium in a +3 oxidation state.
• CrO₃ (Chromium(VI) oxide) has chromium in a +6 oxidation state.
• The acidic strength of these oxides increases with the oxidation state of chromium:
• CrO is basic due to the low oxidation state of +2.
• Cr₂O₃ is amphoteric (can act as both acid and base) due to the intermediate oxidation state of +3.
• CrO₃ is highly acidic due to the high oxidation state of +6.
• Therefore, the decreasing order of acidic strength of these oxides is:
• CrO₃ > Cr₂O₃ > CrO

Therefore, the correct answer is option 1: CrO₃ > Cr₂O₃ > CrO.

CONCEPT:

SNI Mechanism

• SNI (Substitution Nucleophilic Internal) mechanism is a type of nucleophilic substitution reaction where the nucleophile is introduced internally in the molecule.
• The reactivity of alkyl halides towards SNI mechanism depends on the leaving group ability, where a better leaving group increases the reactivity.
• In SNI mechanism, the leaving group should be able to stabilize the negative charge that develops during the transition state.

EXPLANATION:​

• The leaving group ability of halogens decreases in the order: I > Br > Cl > F.
• Iodine (I) is the best leaving group among the halides mentioned because it is larger and can stabilize the negative charge more effectively.
• Therefore, (CH₃)₃C-I will be the most reactive towards substitution by the SNI mechanism.

Therefore, the correct answer is (CH₃)₃C-I.

CONCEPT:

Magnetic Moment

• The magnetic moment of an ion is related to the number of unpaired electrons present in the ion.
• It can be calculated using the formula:

μ = √(n(n+2)) BM

• Where μ is the magnetic moment and n is the number of unpaired electrons.

EXPLANATION:

Therefore, the ion with the maximum magnetic moment is Mn²⁺.

CONCEPT:

Diamagnetism and Low Spin Complexes

• Diamagnetic complexes have all paired electrons, resulting in no net magnetic moment.
• In octahedral complexes, the electronic configuration can be influenced by the nature of ligands, leading to high spin or low spin configurations.
• Strong field ligands such as CN^- and NH₃ cause pairing of electrons, resulting in low spin complexes.

EXPLANATION:

• [NiCl₄]^2-:

• Nickel (Ni²⁺) has a d⁸ configuration.

• Chloride (Cl^-) is a weak field ligand, leading to a high spin complex.

• High spin d⁸ configuration in a tetrahedral field does not result in a diamagnetic complex.

• [CoCl₆]^3-:

• Cobalt (Co³⁺) has a d⁶ configuration.

• Chloride (Cl^-) is a weak field ligand, leading to a high spin complex.

• High spin d⁶ configuration in an octahedral field does not result in a diamagnetic complex.

• [CoF₆]^3-:

• Cobalt (Co³⁺) has a d⁶ configuration.

• Fluoride (F^-) is a weak field ligand, leading to a high spin complex.

• High spin d⁶ configuration in an octahedral field does not result in a diamagnetic complex.

• [Co(NH₃)₆]³⁺:

• Cobalt (Co³⁺) has a d⁶ configuration.

• Ammonia (NH₃) is a strong field ligand, leading to a low spin complex.

• Low spin d⁶ configuration in an octahedral field results in all paired electrons, making it diamagnetic.

Therefore, the octahedral diamagnetic low spin complex among the given options is [Co(NH₃)₆]³⁺.