Chemistry: CUET Mock Test - 7 - CUET MCQ

Aromatic diazonium salts, such as benzenediazonium salts, are stable at cold temperatures (0–5°C). At higher temperatures, they decompose, typically forming phenol and nitrogen gas (N₂) when heated in water.

Ether will act as Nu and abstract the Hydrogen, creation Br- which will attack and result in formation of phenol.

In its zwitterionic form, an amino acid has both a positively charged amino group (-NH₃⁺) and a negatively charged carboxylate group (-COO⁻), making it neutral overall. It exhibits amphoteric behavior by acting as an acid (donating H⁺ from -NH₃⁺) in basic conditions and as a base (accepting H⁺ at -COO⁻) in acidic conditions.

Thus, the correct answer is C.

Iodoform test: This test is used to distinguish between ethanal and propanal. In this test, a yellow precipitate of iodoform (CHI3) is formed when a compound containing a methyl ketone or a secondary alcohol with a methyl group is treated with an iodine and sodium hydroxide (NaOH) solution.
The reactions for ethanal and propanal are as follows:
Ethanal + I₂ + NaOH → CHI₃ + Na₂CO₃ + H₂O
Propanal + 3I₂ + 4NaOH → CHI₃ + 3NaI + 2Na₂CO₃ + 2H₂O
In the Iodoform test, the yellow precipitate of iodoform is formed only when the compound contains a methyl ketone or secondary alcohol with a methyl group.
Therefore, the test can be used to distinguish between ethanal (which is an aldehyde and does not have a methyl group) and propanal (which is an aldehyde and has a methyl group).

CH₃CHO

CH₃CH₂CHO No characteristic change

Ethanal gives positive iodoform test and thus can be used to distinguish between ethanal & propanal

Peptide bonds-

• A peptide bond also sometimes called eupeptide bond is a chemical bond that is formed by joining the carboxyl group of one amino acid to the amino group of another.
• A peptide bond is basically an amide-type of the covalent chemical bond. This bond links two consecutive alpha-amino acids from C1 (carbon number one) of one alpha-amino acid and N2 (nitrogen number two) of another.
• This linkage is found along a peptide or protein chain.

Given data and Analysis-

⇒ A peptide bond is a chemical bond formed between two molecules when a carboxylic group of one molecule reacts with an amino group of the other to form, thus releasing a water molecule.
⇒ The CO-NH bond is called a peptide bond.
⇒ In the given peptide chain the number of peptide bonds is 2 (shown in the above image by a red mark).

Transition metals are characterized by partially filled d-orbitals, which allows for multiple oxidation states and the formation of colored compounds.

The absorption of light that causes transitions between d-orbitals results in the characteristic colors of transition metal compounds.

The stability of transition metal complexes is influenced by the metal’s size and the strength of the ligand field.

Transition metal complexes are important in biological systems (e.g., enzymes) and in industrial processes as catalysts.

In the complex ion [Fe(CN)₆]⁴⁻, iron (Fe) can exist in two possible oxidation states, depending on the nature of the bonding and the ligand's influence:

1. Iron in the +2 Oxidation State (Fe²⁺)

• Most common and stable assignment for this complex.

• Cyanide (CN⁻) is a strong field ligand that causes pairing of electrons, stabilizing the low-spin configuration.

• Electron configuration:

  • Fe²⁺: [Ar] 3d⁶ (all paired due to CN⁻'s strong field).

• Formal charge calculation:

  • Total charge = 4⁻ (from [Fe(CN)₆]⁴⁻).

  • Each CN⁻ contributes 1⁻, so 6 × (−1) = 6⁻.

  • Thus, Fe + 6(−1) = 4⁻ → Fe = +2.

2. Iron in the +3 Oxidation State (Fe³⁺) with a "Charge-Transfer" Interpretation

• Less common, but theoretically possible in certain redox-active systems.

• Requires one electron delocalization from Fe to the ligands (e.g., partial oxidation).

• Electron configuration:

  • Fe³⁺: [Ar] 3d⁵ (unpaired if high-spin, but CN⁻ forces low-spin pairing).

• Formal charge calculation:

  • If Fe = +3, total charge = 3 + 6(−1) = 3⁻, but the ion is 4⁻.

  • This suggests one extra electron is delocalized (e.g., ligand radical character).

Key Points:

• Primary oxidation state: +2 (standard interpretation).

• Alternative possibility: +3 (requires non-classical electron distribution).

Lanthanides and actinides are distinguished from transition metals by their placement in the f-block of the periodic table, where their f-orbitals are filled with electrons. In contrast, transition metals are characterized by the filling of d-orbitals. Option D is incorrect because lanthanides and actinides can form multiple oxidation states, not just one. Thus, the correct answer is B.

Lanthanides are characterized by their ability to form highly stable compounds, as stated in the passage. High magnetic susceptibility is another property of lanthanides, but it is not directly related to compound stability. Option A is incorrect because high radioactivity is a characteristic of actinides, not lanthanides. Option C is more applicable to actinides, which have more complex electron configurations. Option D is also wrong. Thus, the correct answer is B.

Actinides like uranium and thorium are primarily used as fuel in nuclear reactors.

The passage states that the +3 oxidation state is the most common for lanthanides, as it results in a stable electron configuration by losing three electrons (typically from 6s and 5d orbitals). While some lanthanides can exhibit +2 (e.g., europium) or +4 (e.g., cerium), +3 is predominant. Thus, the correct answer is C.

The passage explains that actinides exhibit more varied oxidation states than lanthanides due to their more complex electron configurations. Specifically, the 5f orbitals in actinides are less shielded and more involved in bonding than the 4f orbitals in lanthanides, allowing for a wider range of oxidation states. Option B is incorrect, as radioactivity does not affect oxidation states. Option C is incorrect, as actinides do not necessarily have fewer f-electrons; rather, their 5f orbitals are more chemically active. Thus, the correct answer is A.

Statement A – Correct:
Activation energy is the minimum energy required for reactants to form the transition state and undergo a chemical reaction.

Statement B – Incorrect:
A higher activation energy decreases the reaction rate, not increases it.
According to the Arrhenius equation:
k = A × e^{(-Ea / RT)}
Where:

• k is the rate constant

• A is the pre-exponential factor

• Ea is the activation energy

• R is the gas constant

• T is the temperature in Kelvin
  As activation energy increases, the exponential term becomes smaller, reducing the reaction rate.

Statement C – Correct:
The Arrhenius equation relates the rate constant (k) to activation energy (Ea), temperature (T), and a pre-exponential factor (A).

Statement D – Correct:
Catalysts provide an alternative reaction pathway with lower activation energy, which increases the reaction rate.

First-order reactions have a constant half-life independent of concentration, second-order reactions have a half-life dependent on initial concentration, the half-life is defined as the time for the concentration of reactants to halve and for zero-order reactions, the half-life is directly proportional to the initial concentration.

The integrated rate laws for first, second, and zero-order reactions result in linear plots when graphed appropriately (ln[A] for first-order, [A] for zero-order, and 1/[A] for second-order).

Statement 1 is correct as transition metals have incomplete d-subshells, Statement 3 is correct as lanthanoids and actinoids are f-block elements, and Statement 4 is correct as they are placed separately. Statement 2 is incorrect because zinc has a completely filled d-orbital and is not considered a transition metal.

Statement 1 is correct as transition metals show multiple oxidation states. Statement 2 is correct as manganese shows oxidation states from +2 to +7. Statement 4 is also correct as the stability of oxidation states changes across the transition metals. Statement 3 is incorrect because many transition metals are paramagnetic or exhibit magnetic behavior.

Certain diazonium salts such as fluoroborates are insoluble in water. This makes them stable enough to be dried and stored.

Diazonium salts are unstable and may explode in dry state. Therefore, they are generally used in solution or aqueous state.

The C-Cl bond length in chloromethane is 178 pm. Bromine, being below chlorine in group 17, has a larger atomic radius, resulting in a longer C-Br bond in bromomethane. Typical C-Br bond lengths are around 193–194 pm, making 193 pm the probable value. Option C (178 pm) is incorrect, as it represents the C-Cl bond length. Options A (139 pm) and B (156 pm) are too short, as they are closer to C-F or C-O bond lengths. Thus, the correct answer is D.

The donation of a lone pair of electrons from the carbon atom of CO into a vacant orbital of a metal atom forms a sigma (σ) bond due to head-on overlap. Option B (pi) is incorrect, as pi bonds involve sideways overlap. Option C (back) refers to pi back-bonding, a separate process. Option D (synergic) describes the overall bonding in metal-CO complexes (sigma donation plus pi back-bonding), but the question specifically asks about the donation step, which is sigma. Thus, the correct answer is A.

The pi bond involves donation of electrons from filled metal d orbitals into empty antibonding pi orbitals of CO. This is also called a back bond.

Werner proposed the term “secondary valence” for the number of groups bound directly to the metal ion in a coordination complex, which corresponds to the coordination number. Option A (primary valence) is incorrect, as it refers to the oxidation number. Option C (oxidation number) is synonymous with primary valence, not secondary. Option D (polyhedra) refers to the geometric arrangement of ligands, not the number of groups. Thus, the correct answer is B.

Werner observed that ‘x’ mol of AgCl was precipitated per mole of CoCl₃.4NH₃ with excess silver nitrate. Currently, there are seven groups attached to the cobalt atom. For the secondary valence to be six, one of the chloride (x = 1) has to be precipitated as AgCl, therefore 1:1 electrolyte.

When 1 mol of NiCl₂·6H₂O reacts with excess AgNO₃, 2 mol of AgCl precipitate, indicating two Cl⁻ ions are outside the coordination sphere. The compound’s formula is [Ni(H₂O)6]Cl_2, where the coordination complex is [Ni(H₂O)6]²+, with six water molecules bound to Ni²+. The secondary valence (coordination number) of Ni is thus 6. Options A, B, and C are incorrect, as they do not match the coordination number of six. Thus, the correct answer is D.

In [Co(H₂O)(CN)(en)₂]²⁺, the overall charge is +2 and 0, -1 and 0 on H₂O, CN and en groups respectively. If ‘x’ is the oxidation number of Co, then (x + 0 – 1 + 0 = +2), which implies x=+3.
Similarly, in [CoBr₂(en)₂]⁺, the charge on Br is -1, so if ‘y’ is the oxidation number of Co, then (y – 2 + 0 = +1), which implies y = +3. Therefore, the sum of the oxidation numbers is 3 + 3 = 6.