CUET UG 2025 Chemistry Previous Year Solved Paper

Step 1. Understanding the Question
We are asked to find the van’t Hoff factor (i) for acetic acid (CH₃COOH)
dissolved in water.
The van’t Hoff factor indicates the number of particles produced in solution
compared to the number initially dissolved.
Step 2. Concept of van’t Hoff Factor
The van’t Hoff factor 𝑖 is given by:

Step 3. Ionization of Acetic Acid
Acetic acid is a weak electrolyte, so it ionizes partially in water as:

If ionization were complete, there would be 2 particles formed per
molecule, so 𝑖 = 2.
But because ionization is partial, the actual value of 𝑖 lies between 1 and 2.
Final Answer: The value of van’t Hoff factor 𝑖 for CH₃COOH in water is
between 1 and 2.
Correct Option: 1)

Step 1: Understanding Molarity

Molarity is defined as the number of moles of solute per liter of solution:

We are given that 1 g of each solute is dissolved in 1 L of solution, so the
molarity is simply determined by calculating the moles of each solute.
Step 2: Calculation of Moles for Each Solute
1. Glucose (A):

2. NaOH (B):

3. NaCl (C):

4. KCl (D):

Step 3: Molarity Comparison
After calculating the molarity of each solution, we compare their values:
• NaOH (B): 0.025 M
• NaCl (C): 0.01709 M

• KCl (D): 0.01342 M
• Glucose (A): 0.00556 M
Step 4: Arranging in Decreasing Order of Molarity
• NaOH (B) has the highest molarity.
• NaCl (C) comes next.
• KCl (D) follows.
• Glucose (A) has the lowest molarity.
The solutions in decreasing order of molarity are: NaOH > NaCl > KCl >
Glucose
Final Answer: (A), (B), (D), (C)
Correct Option: 2)

The correct answer is 4. (A) – (III), (B) – (IV), (C) – (I), (D) – (II)
Explanation:
• Saturated solution → (III): Contains the maximum amount of solute
that can dissolve in a given amount of solvent at a given temperature;
any extra solute remains undissolved.
• Isotonic solutions → (IV): Two solutions that have the same osmotic
pressure at the same temperature (so no net osmosis).
• Binary solution → (I): A solution having two components (typically
one solute + one solvent).
• Hypertonic solution → (II): A solution whose osmotic pressure is
higher than that of another (the reference) solution.

Step 1. Understanding the Concept
At high altitudes, the atmospheric pressure is much lower than at sea level,
meaning there is less oxygen in the air. The partial pressure of oxygen

decreases, leading to hypoxia (low oxygen levels) in the blood and tissues of
individuals living or traveling at high altitudes. This reduction in oxygen
availability makes it more difficult for the body to obtain enough oxygen,
leading to symptoms like fatigue and shortness of breath.
Step 2. Checking Each Option
• 1. Both low temperature and high atmospheric pressure:
→ Incorrect. ❌ High altitudes are associated with low atmospheric
pressure, not high pressure. Low temperatures can exist at high
altitudes, but it is the low atmospheric pressure that is primarily
responsible for the reduced oxygen levels.
• 2. Low temperature:
→ Incorrect. ❌ While temperature can decrease at higher altitudes,
it is the low atmospheric pressure that is the primary reason for
reduced oxygen concentration in the blood and tissues, not
temperature.
• 3. Low atmospheric pressure:
→ Correct. The lower atmospheric pressure at high altitudes
means that the partial pressure of oxygen is reduced. This decreases
the amount of oxygen available for the lungs to absorb and for the
blood to carry to tissues, leading to lower oxygen concentrations in
the blood.
• 4. High atmospheric pressure:
→ Incorrect. ❌ High atmospheric pressure is not associated with
high altitudes. In fact, the oxygen concentration is higher at sea level
where atmospheric pressure is higher.
Final Answer: Low atmospheric pressure
Correct Option: 3)

Step 1. Understanding the Concept
To calculate the molality (𝑚) of a solution, the formula is:

Where:
• Moles of solute is given by:

• Mass of solvent is the total mass of the solution minus the mass of the
solute.
Step 2. Given Data
• Mass percent of KI = 20% (mass/mass)
• Density of solution = 1.202 g/mL
• Molar mass of KI = 166 g/mol
Step 3. Step-by-Step Calculation
1. Assume the mass of the solution is 100 g. This makes the calculations
easier because the mass percent is given as 20%.
o Mass of KI (solute) = 20% of 100 g = 20 g
o Mass of water (solvent) = 100 g - 20 g = 80 g
2. Convert mass of solvent into kilograms:
80 g = 0.08 kg
3. Calculate moles of KI:

4. Calculate molality:

Rounding to two significant figures:
𝑚 ≈ 1.5 mol/kg

Final Answer: 1.5 mol kg⁻¹
Correct Option: 1)

Step 1. Understanding the Concept
The cell potential (denoted as 𝐸cell) is the measure of the potential
difference between the anode and cathode in an electrochemical cell. It is
measured in volts (V), which is the standard unit for electric potential.
The relationship is:
𝐸cell = 𝐸cathode - 𝐸anode
The unit for electric potential (voltage) is the volt (V), which is defined as
joule per coulomb (J/C).
Step 2. Checking Each Option
• 1. V m:
→ Incorrect. ❌ This is not the unit of 𝐸cell, as it represents volt-meter,
a unit that would relate to electric field or potential energy, but not
cell potential.
• 2. S cm⁻¹:
→ Incorrect. ❌ This is the unit for conductivity (Siemens per
centimeter), not for cell potential.
• 3. V:
→ Correct. The unit of 𝐸cell is the volt (V), which is the standard
unit of electrical potential or voltage.
• 4. S cm⁻² mol⁻¹:
→ Incorrect. ❌ This is the unit for specific conductivity or molar
conductivity and is not used for cell potential.
Final Answer: 3. V
Correct Option: 3)

• (A) Cell constant: The cell constant is a factor that relates the
measured resistance of a solution to its conductivity. It is defined as
the ratio of the distance between the electrodes to the area of the
cross-section of the solution.
→ Correct Unit: (I) cm⁻¹
• (B) Molar conductance: Molar conductance is the conductance of a
solution containing 1 mole of electrolyte in a given volume of
solution.
→ Correct Unit: (II) ohm⁻¹ cm² mol⁻¹
• (C) Specific conductance: Specific conductance, also known as
conductivity, is the conductance of a unit volume of solution.
→ Correct Unit: (III) ohm⁻¹ cm⁻¹
• (D) Conductance: Conductance is the ability of a solution to conduct
electricity, and it is inversely related to resistance.
→ Correct Unit: (IV) ohm⁻¹
Final Answer: (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
Correct Option: 1)

Step 1. Understanding the Mercury Cell
A Mercury cell is a type of electrochemical cell commonly used in small
devices, such as hearing aids, and is known for its high energy density. It
involves a reaction between mercury oxide and zinc.
The general cell reaction is:

Mercury cells do not convert combustion energy into electrical energy and
are non-rechargeable.

Step 2. Checking Each Option
• (A) It converts energy of combustion into electrical energy:
→ Incorrect. ❌ Mercury cells do not rely on combustion. They
generate energy through electrochemical reactions, not combustion.
• (B) It is rechargeable:
→ Incorrect. ❌ Mercury cells are non-rechargeable. Once the energy
is used up, the cell cannot be recharged.
• (C) The cell reaction involved is Zn(Hg) + HgO(s) ⟶ ZnO(s) + Hg(l):
→ Correct. This is the correct reaction for a Mercury cell.
• (D) It is a low current device used in hearing aids:
→ Correct. Mercury cells are commonly used in low current
devices like hearing aids.
Final Answer: (C) and (D) only
Correct Option: 4)

Step 1. Understanding the Concept
Different types of cells and batteries are used in various applications based
on their characteristics. Here's a brief overview of each type mentioned:
• Mercury Cell: A primary cell that uses mercury oxide as the cathode.
These are small and long-lasting but are rarely used in high-power
applications like automobiles or inverters due to their environmental
impact.
• Dry Cell: A type of battery where the electrolyte is in a paste form,
commonly used in devices like flashlights, remote controls, and small
gadgets. They don't supply the long-term power required for high-
power applications like automobiles or inverters.
• Lead Storage Cell: Commonly used in automobiles as lead-acid
batteries, these cells are rechargeable and provide the necessary
power to start the engine and supply electricity to the car. They are

also used in inverters for backup power because of their ability to
store and deliver a large amount of energy.
• Fuel Cell: A device that generates electricity through a chemical
reaction, usually involving hydrogen and oxygen. These are more
commonly explored in specialized applications like hydrogen-
powered vehicles but are not typical in most standard automobiles or
inverters.
Step 2. Checking Each Option
• 1. Mercury Cell:
→ Incorrect. ❌ Mercury cells are not used in automobiles or
inverters because they are not designed for high power applications.
• 2. Dry Cell:
→ Incorrect. ❌ While dry cells are widely used in small electronic
devices, they do not provide the necessary power for automobiles or
inverters.
• 3. Lead Storage Cell:
→ Correct. Lead storage cells (lead-acid batteries) are the
standard batteries used in automobiles for starting the engine and in
inverters for power backup.
• 4. Fuel Cell:
→ Incorrect. ❌ While fuel cells are used in some experimental or
specialized vehicles, they are not commonly used in most
automobiles or inverters.
Final Answer: Lead Storage Cell
Correct Option: 3)

Step 1: Understanding the behavior of galvanic and electrolytic cells
• A galvanic cell is a voltaic cell where spontaneous reactions occur,
producing electrical energy (positive 𝐸cell).

• An electrolytic cell requires an external voltage to drive a non-
spontaneous reaction (negative 𝐸cell).
Step 2: Comparing Galvanic and Electrolytic Cells
• Galvanic cell:
o The cell potential (𝐸cell) is positive, and it generates electricity.
• Electrolytic cell:
o The cell potential (𝐸cell) is negative, and it requires an external
voltage (𝐸ext) to drive the reaction in the opposite direction.
Step 3: Condition when a galvanic cell behaves like an electrolytic cell
• A galvanic cell behaves like an electrolytic cell when the external
voltage (𝐸ext) applied is greater than the cell potential (𝐸cell). This
forces the reaction to reverse, making a non-spontaneous reaction
occur.
Step 4: Explanation of the options
1. 𝐸cell = 𝐸ext:
o Incorrect. ❌ This condition means the cell is at equilibrium
and the system will not change.
2. 𝐸cell = 0:
o Incorrect. ❌ When 𝐸cell = 0, the cell is at equilibrium, and no
reaction occurs.
3. 𝐸ext > 𝐸cell:
o Correct. When the external voltage (𝐸ext) is greater than the
cell potential (𝐸cell), the cell behaves like an electrolytic cell, and
the reaction proceeds in the reverse direction.
4. 𝐸cell > 𝐸ext:
o Incorrect. ❌ This would mean that the galvanic cell is still
operating as a galvanic cell, not an electrolytic cell.
Final Answer: 𝐸ext > 𝐸cell
Correct Option: 3)

Step 1. Understanding the Concept of Coordination Compounds
A coordination compound consists of a central metal atom or ion bonded to
surrounding ligands (molecules or ions) through coordinate covalent
bonds. Coordination compounds play crucial roles in biological processes
and medical treatments, as well as in industrial applications.
Step 2. Evaluating Each Statement
1. Cis-platin effectively inhibits the growth of tumours.
→ Correct. Cisplatin (a coordination compound of platinum) is
used as a chemotherapeutic agent. It is known for its ability to inhibit
DNA replication, thereby preventing the growth of cancer cells. It is
widely used to treat various cancers, including testicular, ovarian, and
bladder cancer.
2. Chlorophyll helps in photosynthesis.
→ Correct. Chlorophyll is a coordination compound of magnesium
(Mg) at the center of its structure. It plays a crucial role in
photosynthesis by absorbing light and converting it into chemical
energy.
3. Desferrioxamine B is used in the treatment of lead poisoning.
→ Incorrect. ❌ Desferrioxamine B is a chelating agent used to treat
iron poisoning or iron overload, not lead poisoning. It binds to iron
ions and helps in their removal from the body. For lead poisoning,
other chelating agents such as EDTA (ethylenediaminetetraacetic
acid) or dimercaprol are typically used.
4. Cyanocobalamin, the anti-pernicious anemia factor, is a coordination
compound of Co.
→ Correct. Cyanocobalamin (Vitamin B12) is a coordination
compound of cobalt (Co). It plays a vital role in the production of red
blood cells and the maintenance of nerve cells, and its deficiency
leads to pernicious anemia.

Step 3. Conclusion
• Statement 3 is incorrect because desferrioxamine B is used for iron
poisoning, not lead poisoning.
• All other statements represent correct applications of coordination
compounds.
Final Answer: Desferrioxamine B is used in the treatment of lead
poisoning
Correct Option: 3)

Step 1: Understanding the electronic configurations of the ions
To determine the number of 3d electrons in each ion, we first look at the
electronic configuration of the neutral atoms and then account for the loss
of electrons when the ion is formed.
1. Cr²⁺ (Chromium ion):
o Chromium (Cr) has an atomic number of 24. The neutral
configuration of Cr is:

o For Cr²⁺, 2 electrons are removed, primarily from the 4s orbital.
So, the configuration of Cr²⁺ is:

o Therefore, Cr²⁺ has 4 electrons in the 3d orbital.
2. Cu⁺ (Copper ion):
o Copper (Cu) has an atomic number of 29. The neutral
configuration of Cu is:

o For Cu⁺, 1 electron is removed, primarily from the 4s orbital. So,
the configuration of Cu⁺ is:

o Therefore, Cu⁺ has 10 electrons in the 3d orbital.
3. Ti³⁺ (Titanium ion):
o Titanium (Ti) has an atomic number of 22. The neutral
configuration of Ti is:

o For Ti³⁺, 3 electrons are removed, primarily from the 4s orbital.
So, the configuration of Ti³⁺ is:

o Therefore, Ti³⁺ has 1 electron in the 3d orbital.
4. Mn⁺ (Manganese ion):
o Manganese (Mn) has an atomic number of 25. The neutral
configuration of Mn is:

o For Mn⁺, 1 electron is removed, primarily from the 4s orbital.
So, the configuration of Mn⁺ is:

o Therefore, Mn⁺ has 5 electrons in the 3d orbital.
Step 2: Arranging the ions in increasing order of 3d electrons
• Ti³⁺ has 1 3d electron.
• Cr²⁺ has 4 3d electrons.
• Mn⁺ has 5 3d electrons.
• Cu⁺ has 10 3d electrons.
So, the increasing order of the number of 3d electrons is:

Final Answer: (C), (A), (D), (B)

Correct Option: 3)

Step 1: Understanding the atomic number of Lanthanum
• The atomic number of Lanthanum (La) is 57, which means it has 57
electrons.
• Lanthanum is part of the lanthanide series.
Step 2: Writing the electron configuration for Lanthanum
• The electron configuration of Xenon (Xe), which has the atomic
number 54, is:

• Lanthanum (atomic number 57) will have three more electrons than
Xenon:
o These extra electrons go into the 5d orbital and 5s orbital
because Lanthanum has a configuration that lies at the
beginning of the lanthanide series, just before the start of the 4f
orbitals.
Thus, the electron configuration for Lanthanum is:

Step 3: Reviewing the options
1. [Xe] 5d¹ 6s²:
o Correct.
o This matches the electron configuration for Lanthanum, where
the extra 3 electrons after Xenon go into the 5d orbital and 6s
orbital.
2. [Xe] 4f¹ 5d²:
o ❌ Incorrect.

o This configuration suggests that Lanthanum has electrons in
the 4f orbital, which is not correct for Lanthanum as it doesn't
start filling the 4f orbitals until the next element, Cerium.
3. [Xe] 4f³:
o ❌ Incorrect.
o This would indicate that 3 electrons have gone into the 4f
orbital, which is incorrect because Lanthanum's configuration
doesn't involve filling the 4f orbitals yet.
4. [Xe] 4f¹ 5d¹ 6s¹:
o ❌ Incorrect.
o This configuration is wrong because it incorrectly places
electrons in both the 4f orbital and the 6s orbital.
Final Answer: [Xe] 5d¹ 6s²
Correct Option: 1)

(A) Magnesium based alloy is constituent of: An alloy containing about 95%
magnesium and 5% Mischmetal (an alloy of lanthanoid metals) is
pyrophoric, meaning it produces sparks when struck. This property is
utilized in making tracer bullets and lighter flints.
→ Correct Match: (I) Bullets
(B) Lanthanoid oxide: The oxides of certain lanthanoids, like europium and
yttrium, are excellent phosphors. Phosphors are materials that emit light
when struck by electrons, a principle used to create color images on older
cathode-ray tube (CRT) television screens.
→ Correct Match: (III) Television screen
(C) Mixed oxides of Lanthanoids are employed in: Mixtures of lanthanoid
oxides act as efficient catalysts. A major industrial application is in the
catalytic cracking of petroleum, where they help break down large
hydrocarbon molecules into smaller, more useful fractions like petrol.
→ Correct Match: (II) Petroleum cracking

(D) Misch metal: This is a well-known alloy. By definition, Misch metal is
primarily composed of a mixture of lanthanoid metals (around 95%) and
iron (around 5%), along with trace impurities.
→ Correct Match: (IV) Lanthanoid metal and iron
Final Answer: (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
Correct Option: 2)

The correct answer is 1. Yellow.
Detailed Step-by-Step Solution
This question is about the characteristic colors of transition metal ions in
aqueous solutions, a topic from the "d- and f-Block Elements" chapter.
1. Identify the Ion and its Electronic Configuration
• The ion in question is the aqueous iron(III) ion, Fe³⁺(aq).
• The atomic number of Iron (Fe) is 26. Its ground-state electronic
configuration is [𝐴𝑟]3d64s2.
• To form the Fe³⁺ ion, an iron atom loses three electrons (two from the
4s orbital and one from the 3d orbital).
• Therefore, the electronic configuration of Fe³⁺ is [𝐴𝑟]3d5.
2. Understand the Origin of Color
• In an aqueous solution, the Fe³⁺ ion is surrounded by six water
molecules, forming the complex ion [𝐹𝑒(𝐻2𝑂)6]3+.
• The color of transition metal ions is typically due to the absorption of
light, which promotes an electron from a lower-energy d-orbital to a
higher-energy d-orbital. This is called a d-d transition.
• The color we observe is the complementary color of the light that is
absorbed.
3. The Specific Case of Fe³⁺(aq)
• For Fe³⁺, which has a half-filled d-orbital configuration (d5), the d-d
transitions are "spin-forbidden". This means they are very weak, and
if they were the only factor, the solution would be almost colorless or
very pale violet.

• However, aqueous solutions of Fe³⁺ are distinctly yellow or brownish.
This is due to a phenomenon called hydrolysis.
• The high positive charge of the Fe³⁺ ion pulls electrons from the
surrounding water ligands, making them acidic. A proton (𝐻+) is
released, forming a hydroxo complex:

• The intense yellow color is caused by a Ligand-to-Metal Charge
Transfer (LMCT) band in the hydroxo complex, [𝐹𝑒(𝐻2𝑂)5(𝑂𝐻)]2+. In
this process, an electron momentarily transfers from the hydroxide
ligand (𝑂𝐻−) to the central Fe³⁺ ion. This type of electron transfer is
highly probable and strongly absorbs light in the blue-violet region of
the spectrum, causing the solution to appear yellow.
Conclusion
While the pure hexaaquairon(III) ion is pale violet, it readily hydrolyzes in
water to form a yellow-colored complex. Therefore, the commonly
observed color of an Fe³⁺(aq) solution is yellow.

The reason HCl is not used to make the medium acidic in oxidation
reactions of KMnO₄ in acidic medium is because KMnO₄ oxidizes HCl to Cl₂,
which is also an oxidizing agent. This would interfere with the reaction as
Cl₂ itself can oxidize other substances, thus reducing the effectiveness of
KMnO₄ as an oxidizing agent.
Step 1: Understanding the Chemistry
• KMnO₄ is a strong oxidizing agent that acts in acidic, neutral, and
basic media. In acidic medium, KMnO₄ is reduced to Mn²⁺.
• HCl, when in the presence of KMnO₄, can get oxidized to Cl₂ (chlorine
gas), which is also an oxidizing agent. This creates a situation where
both KMnO₄ and Cl₂ are acting as oxidizing agents, potentially leading
to side reactions and the interference of chlorine gas in the desired
oxidation process.

Step 2: Analyzing the Options
• 1. KMnO₄ is a weaker oxidizing agent than HCl:
→ Incorrect. ❌ KMnO₄ is actually a much stronger oxidizing agent
than HCl.
• 2. KMnO₄ oxidizes HCl into Cl₂ which is also an oxidizing agent:
→ Correct. The reaction between KMnO₄ and HCl results in the
formation of Cl₂, which itself is an oxidizing agent, interfering with
the reaction.
• 3. Both HCl and KMnO₄ act as oxidizing agents:
→ Incorrect. ❌ HCl does not act as an oxidizing agent in the reaction;
it is oxidized to Cl₂.
• 4. KMnO₄ acts as a reducing agent in the presence of HCl:
→ Incorrect. ❌ KMnO₄ does not act as a reducing agent; it is the
oxidizing agent in this reaction.
Final Answer: KMnO₄ oxidizes HCl into Cl₂ which is also an oxidizing
agent
Correct Option: 2)

Step-by-Step Solution
Step 1: Identify the reactants and their roles
• Acidified potassium dichromate (K₂Cr₂O₇ / H⁺):
o A strong oxidizing agent.
o Chromium is in the +6 oxidation state and gets reduced to Cr³⁺
during the reaction.
• Sulphide ion (S²⁻):
o Acts as the reducing agent.
o Sulfur has an oxidation state of -2 and will be oxidized when
reacting with the dichromate.

Step 2: Determine the oxidation product
• The sulphide ion (S²⁻) is oxidized from -2 to 0, forming elemental
sulfur (S).
• This sulfur often appears as a yellow or pale precipitate.
Step 3: Write the half-reactions
1. Reduction (Dichromate ion):

2. Oxidation (Sulphide ion):

Step 4: Combine half-reactions to balance electrons
• Multiply the oxidation half-reaction by 3 to match the 6 electrons in
the reduction half-reaction:

• Add both half-reactions to get the overall reaction:

Step 5: Conclusion
• The sulphide ion (S²⁻) is oxidized to elemental sulfur (S).
• Higher oxidation states of sulfur like SO₃²⁻, SO₂, SO₄²⁻ are not formed
in this reaction.
Final Answer: Sulphur (S)
Correct Option: 3)

Step 1: Understanding Ligand Field Strength
The field strength of a ligand refers to its ability to split the d-orbital
energies in a metal complex, which is influenced by the ligand's charge and
the ability to donate electron density to the central metal ion. Ligands that

strongly donate electron density to the metal tend to produce a greater
splitting of the d-orbitals, which is associated with stronger field ligands.
The general order of field strength for different types of ligands is based on
their position in the spectrochemical series. The spectrochemical series
ranks ligands by their field strength, with ligands such as CN⁻ being strong
field ligands and S²⁻ being weak field ligands.
Step 2: Reviewing the Ligands
• (A) S²⁻ (Sulfide ion): This is a weak field ligand that does not cause
significant splitting of d-orbitals. It is at the lower end of the
spectrochemical series.
• (B) Ethylenediamine (en): This is a moderately strong field ligand, but
it is weaker than cyanide and thiocyanate.
• (C) NCS⁻ (Thiocyanate): This ligand can act as either a weak or strong
field ligand depending on its coordination mode. However, in general,
it is a moderate field ligand.
• (D) CN⁻ (Cyanide): This is a very strong field ligand, known for
causing large splitting of the d-orbitals. It is one of the strongest
ligands in the spectrochemical series.
Step 3: Decreasing Order of Field Strength
Based on the strength of the ligands in the spectrochemical series:
• CN⁻ is the strongest field ligand.
• Ethylenediamine (en) is weaker than CN⁻, but stronger than NCS⁻.
• NCS⁻ is a weaker field ligand than en, but stronger than S²⁻.
• S²⁻ is the weakest field ligand among the given choices.
Final Answer: (D), (C), (B), (A)
Correct Option: 4)

Step 1. Understanding the Concept
The IUPAC nomenclature of coordination compounds follows these rules:

1. Name the ligands first, in alphabetical order, followed by the metal
name.
o Ligands are named ammine for NH₃ and chlorido (or chloro in
older literature) for Cl⁻.
2. Indicate the number of each ligand using prefixes: di-, tri-, tetra-, etc.
3. Oxidation state of the metal is given in Roman numerals in
parentheses after the metal.
Step 2. Analyzing the Complex [Pt(NH₃)₂Cl₂]
• Ligands:
o NH₃ → ammine
o Cl⁻ → chlorido
• Number of ligands:
o 2 NH₃ → diammine
o 2 Cl⁻ → dichlorido
• Metal: Platinum (Pt)
• Oxidation state of Pt:

• Final IUPAC Name: Diamminedichloridoplatinum(II)
Step 3. Checking Each Option
1. Diamminedichloridoplatinum (IV) → ❌ Incorrect, oxidation state is
+2, not +4.
2. Diamminedichloridoplatinum (II) → Correct.
3. Diamminedichloridoplatinum (0) → ❌ Incorrect, Pt is not in 0
oxidation state.
4. Dichloridodiammineplatinum (IV) → ❌ Incorrect, wrong oxidation
state and alphabetical order of ligands is incorrect.
Final Answer: Diamminedichloridoplatinum (II)
Correct Option: 2)

Step 1. Understanding the Concept
The complex [𝐹𝑒(𝐶N)6]3− is a hexacyanoferrate(III) complex:
• Central metal: Fe³⁺ (oxidation state +3)
• Electronic configuration of Fe³⁺:
Fe³⁺ : [𝐴𝑟] 3d5
• Ligand: CN⁻ (cyanide), a strong field ligand according to the
spectrochemical series, which causes pairing of electrons in the 3d
orbitals (low-spin complex).
• Geometry: Octahedral
• Hybridization: d²sp³ (low-spin octahedral complex)
• Magnetic properties:
o Fe³⁺ in a low-spin state has 1 unpaired electron in the 3d
orbital.
o

where n = number of unpaired electrons:

Step 2. Checking Each Statement
(A) Paramagnetic → Correct
The complex has 1 unpaired electron, so it is paramagnetic.
(B) sp³d² hybridization → ❌ Incorrect
Octahedral complexes with strong field ligands like CN⁻ use inner
orbital hybridization, which is d²sp³, not sp³d².
(C) Magnetic moment = 5.92 BM → ❌ Incorrect
Magnetic moment depends on unpaired electrons. Here, there is
only 1 unpaired electron, so μ ≈ 1.73 BM, not 5.92 BM.
(D) d²sp³ hybridization → Correct

o Fe³⁺ is a low-spin octahedral complex, and inner d-orbitals are
involved in hybridization: d²sp³.
Final Answer: (A) and (D) only
Correct Option: 1)

The correct answer is 1. (A)–(III), (B)–(IV), (C)–(I), (D)–(II)
Explanation:
• Ambident nucleophiles → Cyanides and nitrites (III)
• Plane polarized light → Nicol prism (IV)
• Superimposable mirror image → Symmetrical object (I)
• β-elimination reaction → Saytzeff rule (II)

The correct answer is 3. inversion of configuration
Explanation:
The 𝑆𝑁2 reaction is a bimolecular nucleophilic substitution reaction. It is a
concerted reaction, meaning the bond-breaking and bond-forming occur
simultaneously in a single transition state. The nucleophile attacks the
substrate from the side opposite to the leaving group, leading to an
inversion of the stereochemical configuration at the chiral center. This is
often referred to as Walden inversion.

Step 1: Understanding Gabriel Phthalimide Synthesis
The Gabriel Phthalimide Synthesis is a method used to prepare primary
amines from alkyl halides. The reaction involves the following steps:
1. Phthalimide (a cyclic imide) is treated with a strong base (e.g., KOH)
to form an enolate ion.

2. The enolate ion attacks an alkyl halide (R–X), forming a phthalimide
derivative.
3. The resulting phthalimide derivative is hydrolyzed (usually with
hydrazine, N₂H₄) to release the primary amine.
Step 2: Reaction mechanism and product
• The key feature of the Gabriel Phthalimide synthesis is the formation
of primary amines through the nucleophilic substitution of an alkyl
halide.
• The product of this reaction is a primary amine.
Final Answer: Primary amine
Correct Option: 3)

The correct answer is 3. Wurtz Fittig reaction.
Explanation:
1. Analyze the Given Reaction
Let's break down the chemical reaction provided:
• Reactants:
o Chlorobenzene (𝐶6𝐻5𝐶𝑙): This is an aryl halide (a halogen
attached to a benzene ring).
o Sodium (Na): An alkali metal, highly reactive.
o Methyl chloride (𝐶𝐻3𝐶𝑙): This is an alkyl halide (a halogen
attached to an alkyl group).
• Reaction Conditions:
o Heat: Provides the necessary activation energy.
o Dry Ether: A non-polar aprotic solvent is used because sodium
reacts violently with water or alcohol.
• Products:

o Toluene (𝐶6𝐻5𝐶𝐻3): An alkylbenzene (an alkyl group attached
to a benzene ring).
o Sodium Chloride (N𝑎𝐶𝑙): An inorganic salt.
The overall reaction shows an aryl halide and an alkyl halide coupling
together in the presence of sodium to form an alkylbenzene.
2. Evaluate the Options
Let's examine why the Wurtz-Fittig reaction is the correct choice and the
others are not.
• Sandmeyer's reaction: This reaction is used to synthesize aryl halides
from aryl diazonium salts (𝐴𝑟 N2+X−) using cuprous salts (C𝑢2C𝑙2, C𝑢2
𝐵𝑟2, etc.). The given reaction does not involve a diazonium salt.
• Wurtz reaction: This reaction involves the coupling of two alkyl
halides with sodium in dry ether to form a higher alkane.
o General Form: 2R − X + 2N 𝑎 → R − R + 2N 𝑎X
o Our reaction involves one aryl halide and one alkyl halide, not
two alkyl halides.
• Wurtz-Fittig reaction: This is a modification of the Wurtz reaction. It
involves the reaction between an aryl halide and an alkyl halide with
sodium in dry ether to form a substituted aromatic compound
(alkylbenzene).
o General Form: 𝐴𝑟 – X + R – X + 2N 𝑎 → 𝐴𝑟 – R + 2N 𝑎X
o The given reaction fits this pattern perfectly: 𝐶6𝐻5 − 𝐶𝑙 (Aryl
halide) + 𝐶𝐻3 − 𝐶𝑙 (Alkyl halide) + 2Na → 𝐶6𝐻5 − 𝐶𝐻3
(Alkylbenzene) + 2N 𝑎𝐶𝑙.
• Kolbe reaction: This reaction involves the treatment of sodium
phenoxide with carbon dioxide (CO2) to form salicylic acid. This is
completely different from the given reaction.
Conclusion
The reaction shows the coupling of an aryl halide (chlorobenzene) and an
alkyl halide (methyl chloride) using sodium metal in dry ether. This is the
definition of the Wurtz-Fittig reaction. It's a very useful method for
attaching an alkyl chain to a benzene ring.

The correct answer is 3. (A), (B), (C) and (D).
Explanation:
Hydroboration–oxidation converts propene (an alkene) into an alcohol
using two stages:
• Hydroboration: Use B₂H₆ (source of BH₃). BH₃ adds across the double
bond.
• Oxidation: Then treat with H₂O₂ in basic medium (OH⁻). The base
activates H₂O₂ so that oxygen replaces boron.
• The reaction is done in water (H₂O), which finally protonates the
alkoxide to give the alcohol.
Therefore, all four reagents are involved: (A) B₂H₆, (B) H₂O, (C) H₂O₂, and
(D) OH⁻.
(For propene, the product is propan-1-ol; anti-Markovnikov, syn addition.)

The Hell-Volhard-Zelinsky (HVZ) reaction is used for the halogenation of
carboxylic acids at the α-position, which is the carbon adjacent to the
carboxyl group (-COOH). This reaction is specifically employed to form α-
halocarboxylic acids by introducing a halogen atom (such as chlorine or
bromine) at the α-carbon of the carboxylic acid in the presence of
phosphorus or another catalyst. The product formed is α-halocarboxylic
acids, which is the primary use of the HVZ reaction.
(1) Alcohols:
→ Incorrect. ❌
The HVZ reaction does not produce alcohols; it is specifically used for
halogenating carboxylic acids.
(2) Aldehydes:
→ Incorrect. ❌
The HVZ reaction does not produce aldehydes; it targets the α-
position of carboxylic acids.
(3) Ketones:
→ Incorrect. ❌
The HVZ reaction does not form ketones; its focus is on halogenation
at the α-carbon of carboxylic acids.
(4) α-halocarboxylic acids:
→ Correct.
The HVZ reaction is used to form α-halocarboxylic acids, where a
halogen is attached to the α-position of a carboxylic acid.

Final Answer: α-halocarboxylic acids
Correct Option: 4)

Step 1: Understanding nucleophilic addition reactions
In a nucleophilic addition reaction, the nucleophile attacks the carbonyl
carbon of a carbonyl compound (such as aldehydes and ketones). The
reactivity of the carbonyl compound depends on several factors:
• Electron-donating or electron-withdrawing groups attached to the
carbonyl group.
• Steric hindrance: Larger substituents around the carbonyl group can
increase steric hindrance, making it harder for the nucleophile to
attack.
• Aldehydes are typically more reactive than ketones because they have
one alkyl group and a hydrogen atom attached to the carbonyl
carbon, making the carbonyl carbon more electrophilic.
• Among ketones, the smallest ketone (in terms of the size of alkyl
groups) will generally be more reactive because it has less steric
hindrance.
Step 2: Analyzing the given compounds
1. Ethanal (A):
o Aldehydes are generally more reactive than ketones due to the
presence of one hydrogen atom at the α-position, which makes
the carbonyl carbon more electrophilic. So, ethanal will be
highly reactive.
2. Propanal (C):
o Propanal is also an aldehyde but with a slightly larger alkyl
group (ethyl group). It is still more reactive than ketones but
less reactive than ethanal due to the larger alkyl group.
3. Propanone (B):
o Propanone (a ketone) has two methyl groups attached to the
carbonyl group. Ketones are generally less reactive than

aldehydes, so propanone will be less reactive than ethanal and
propanal.
4. Butanone (D):
o Butanone is also a ketone, but it has two larger alkyl groups
(ethyl groups) compared to propanone, which increases steric
hindrance and reduces its reactivity in nucleophilic addition
reactions.
Step 3: Correct sequence of increasing reactivity
• Most reactive: Ethanal (A) (aldehyde with the least steric hindrance)
• Next: Propanal (C) (aldehyde with a larger alkyl group)
• Then: Propanone (B) (ketone with small methyl groups)
• Least reactive: Butanone (D) (ketone with larger alkyl groups)
Final Answer: (A), (C), (B), (D)
Correct Option: 3)

The correct answer is 3. (A), (B), (C) and (D)
Explanation:
To make methyl benzoate (C₆H₅COOCH₃) from benzene:
1. Br₂/FeBr₃: Electrophilic bromination → bromobenzene (C₆H₅Br).
2. Mg, dry ether: Forms Grignard reagent, C₆H₅MgBr.
3. CO₂, then H₃O⁺: Carboxylation of Grignard → benzoic acid
(C₆H₅COOH).
4. Methanol, conc. H₂SO₄: Fischer esterification → methyl benzoate.
All four reagents are needed in this sequence, so option (A), (B), (C) and
(D) is correct.

The correct answer is 4. I₂/NaOH

Explanation:
The iodoform test (I₂/NaOH) is positive only for methyl ketones (CH₃-CO-).
• Acetone (CH₃-CO-CH₃) has this group → gives a yellow iodoform
precipitate.
• Benzophenone (Ph-CO-Ph) has no CH₃ next to C=O → no reaction.
Fehling’s and Tollens’ reagents don’t react with simple ketones, and 2,4-
DNP reacts with both, so they can’t distinguish them.

Step 1: Understanding Aldol Condensation
Aldol condensation is a reaction where two carbonyl compounds (either
aldehydes or ketones) undergo a reaction in the presence of a base. The
reaction typically proceeds in two steps:
1. Formation of an enolate ion: The base abstracts a hydrogen atom from
the α-carbon (the carbon adjacent to the carbonyl group), forming an
enolate ion.
2. Nucleophilic attack: The enolate ion attacks the carbonyl carbon of
another molecule, forming a β-hydroxy carbonyl compound (an
aldol), which may further undergo dehydration to form an α,β-
unsaturated carbonyl compound.
For Aldol condensation to occur, the carbonyl compound must have at least
one α-hydrogen to form the enolate ion.
Step 2: Explanation of the options
1. Presence of at least one β-hydrogen:
o ❌ Incorrect.
o Explanation: The β-hydrogen (hydrogen on the carbon next to
the carbonyl group) is not critical for forming the enolate ion in
the first step of Aldol condensation. The important feature for
Aldol condensation is the α-hydrogen, not the β-hydrogen.
2. Presence of at least one α-hydrogen:
o Correct.

o Explanation: The α-hydrogen is located on the carbon adjacent
to the carbonyl group. Aldol condensation requires this α-
hydrogen because it is abstracted by a base to form the enolate
ion, which then participates in the reaction. Without this
hydrogen, the carbonyl compound cannot undergo Aldol
condensation.
3. Concentrated base:
o ❌ Incorrect.
o Explanation: A concentrated base (e.g., NaOH or KOH) is often
used to promote the reaction by deprotonating the α-hydrogen,
but it is a reaction condition, not a structural feature of the
carbonyl compound. The key structural feature is the presence
of α-hydrogens.
4. Lack of α-hydrogen:
o ❌ Incorrect.
o Explanation: If a carbonyl compound lacks α-hydrogens, it
cannot undergo Aldol condensation. For example, benzaldehyde
has no α-hydrogens and cannot participate in Aldol
condensation, whereas acetaldehyde has α-hydrogens and can
react in this manner.
Final Answer: Presence of at least one α-hydrogen
Correct Option: 2)

Step 1: Understanding the structure of trimethylamine (CH₃)₃N
• Trimethylamine (CH₃)₃N consists of a central nitrogen atom bonded
to three methyl groups (–CH₃).
• Nitrogen in amines is trivalent and has one unshared pair of electrons
(lone pair).
Step 2: Hybridization of nitrogen

• The nitrogen atom in trimethylamine is sp³ hybridized, meaning it
has four regions of electron density: three bonds with methyl groups
and one lone pair of electrons.
Step 3: Geometry of the molecule
• When there are four regions of electron density (three bonding pairs
and one lone pair), the geometry around the nitrogen atom will be
pyramidal.
• This is similar to the structure of ammonia (NH₃), which also has a
lone pair and three bonding pairs.
Final Answer: Pyramidal
Correct Option: 2)

Step 1: Understanding the reaction
The compound C₆H₅-O-R is an alkoxybenzene (phenyl ether), where R
represents an alkyl group (such as methyl, ethyl, etc.).
When treated with HX (where X is a halogen, such as Cl, Br, or I), the
reaction typically involves the breaking of the C–O bond in the ether group.
The H (proton) from HX adds to the oxygen of the ether, leading to the
cleavage of the ether bond.
Step 2: Reaction mechanism
• The ether bond (C–O) in C₆H₅-O-R undergoes nucleophilic attack by
the halide ion (X⁻), leading to the formation of phenol (C₆H₅OH) and
an alkyl halide (R–X).
The overall reaction is:

Step 3: Identifying the products
• The products of this reaction are phenol (C₆H₅OH) and alkyl halide
(R-X).

Final Answer: RX and C₆H₅OH are formed
Correct Option: 1)

Step 1: Understanding the nitration reaction
• The nitration of benzene is an example of an electrophilic aromatic
substitution reaction.
• In this reaction, benzene (C₆H₆) reacts with a nitronium ion (NO₂⁺),
which is the actual electrophile.
Step 2: Formation of the nitrating species
• In the nitration reaction, a mixture of concentrated sulfuric acid
(H₂SO₄) and concentrated nitric acid (HNO₃) is used.
• The nitronium ion (NO₂⁺) is generated by the following reaction:

• This reaction occurs because H₂SO₄ acts as a catalyst and helps in the
formation of NO₂⁺ from HNO₃.
Step 3: Identifying the nitrating species
• The nitronium ion (NO₂⁺) is the electrophile that reacts with the
electron-rich benzene ring to form the nitrobenzene product.
• The NO₂⁻ ion is not the electrophile in this reaction.
Final Answer: NO₂⁺
Correct Option: 2)

Step 1: Understanding Azo Coupling Reaction
• Azo coupling reaction is the reaction where benzene diazonium
chloride (C₆H₅N₂⁺Cl⁻) reacts with compounds that have a reactive
position (such as amino or hydroxyl groups).

• The diazonium ion (C₆H₅N₂⁺) acts as an electrophile and can couple
with compounds that are electron-rich at the para or ortho positions.
Step 2: Analyzing each compound
1. Nitrobenzene:
o Nitro group (NO₂) is an electron-withdrawing group, which
deactivates the benzene ring and makes it less reactive towards
electrophiles, including diazonium ions.
o Does not undergo azo coupling easily with benzene diazonium
chloride.
2. Aniline:
o The amine group (–NH₂) is an electron-donating group, which
makes the benzene ring more reactive towards electrophiles.
o Undergoes azo coupling with benzene diazonium chloride.
3. o-Toluidine:
o o-Toluidine contains an amine group (–NH₂) and a methyl group
(–CH₃), both of which are electron-donating groups.
o The methyl group in the ortho position further increases the
reactivity of the ring, allowing it to undergo azo coupling with
diazonium ions.
4. Phenol:
o The hydroxyl group (–OH) is an electron-donating group and
makes the benzene ring more reactive towards electrophiles.
o Phenol undergoes azo coupling with benzene diazonium
chloride.
Step 3: Conclusion
• Nitrobenzene will not give azo coupling because the nitro group (–
NO₂) is an electron-withdrawing group and deactivates the ring.
Final Answer: Nitrobenzene
Correct Option: 1)

Step 1. What is Amylose?
• Amylose is a polysaccharide made up of glucose units connected by α-
1,4 glycosidic bonds.
• It is one of the two components of starch, the other being
amylopectin.
• Amylose is typically linear (unbranched), whereas amylopectin is
branched.
Step 2. Solubility of Amylose
• Amylose is known to be water-soluble due to its helical structure,
which can form hydrogen bonds with water molecules.
• However, its solubility is relatively low compared to amylopectin,
which is more highly branched and more easily interacts with water.
Step 3. Typical Solubility Range
• The solubility of amylose in water typically ranges from 15% to 20%
by weight under standard conditions.
Final Answer: 15 to 20%
Correct Option: 2)

Step 1: Understanding the types of proteins
• Globular proteins: These are proteins that have a compact, spherical
shape. They are usually soluble in water and perform various
functions such as enzymes, hormones, and transport proteins.
Examples include insulin and albumin.
• Fibrous proteins: These are proteins that have a long, fibrous shape
and are typically insoluble in water. They provide structural support
and include proteins like keratin and myosin.

Step 2: Analyzing each option
1. Insulin (A):
o Insulin is a globular protein. It is a hormone that regulates
blood sugar levels and is water-soluble.
2. Keratin (B):
o Keratin is a fibrous protein, not a globular one. It is found in
hair, nails, and skin and is insoluble in water.
3. Albumin (C):
o Albumin is a globular protein. It is a water-soluble protein
found in blood plasma and helps in maintaining osmotic
pressure.
4. Myosin (D):
o Myosin is a fibrous protein. It is involved in muscle contraction
and has a long, fibrous structure.
Step 3: Correct combination of globular proteins
• Globular proteins in the options are insulin (A) and albumin (C).
Final Answer: (A) and (C) only
Correct Option: 2)

Step 1: Understanding the β-helix of proteins
• The β-helix is a type of secondary structure found in proteins, also
known as the β-pleated sheet structure.
• It consists of β-strands, which are extended chains of amino acids that
align side-by-side to form sheets.
Step 2: Key bonding interactions in β-helix
• In the β-helix, the most important interactions that stabilize the
structure are hydrogen bonds (H-bonds).

• Hydrogen bonds form between the backbone amide hydrogen (N-H)
of one strand and the carbonyl oxygen (C=O) of an adjacent strand,
holding the strands together in a pleated sheet-like arrangement.
Step 3: Answer analysis of options
• Ionic bond: This is not a primary interaction in the β-helix structure.
Ionic bonds are more commonly seen in salt bridges or between
charged side chains in proteins.
• Covalent interaction: While covalent bonds, such as disulfide bridges,
can stabilize protein tertiary structures, they are not the key
interactions in the β-helix.
• H-bond: This is the correct answer. Hydrogen bonds are the dominant
force stabilizing the β-helix.
• Banana bond: This term does not exist in the context of protein
structure. It is not a recognized type of bond in the β-helix.
Final Answer: H-bond
Correct Option: 3)

The correct answer is 2. (A) – (III), (B) – (I), (C) – (IV), (D) – (II).
Explanation:
Know the standard one-letter amino-acid codes:
• Lysine → K (L is taken by leucine, so lysine uses K).
• Tryptophan → W (T is taken by threonine; Trp is uniquely coded as
W).
• Tyrosine → Y (uses its initial sound “Y”).
• Glutamine → Q (to distinguish from glutamate/glutamic acid, which is
E).
So, the correct matching is: Lysine–K (III), Tryptophan–W (I), Tyrosine–Y
(IV), Glutamine–Q (II).

Step 1: Understanding the structure and acidity of phenolic compounds
The acidity of phenolic compounds is determined by the ability of the
hydroxyl group (-OH) to lose a proton (H⁺). The more stabilized the
conjugate base (phenoxide ion), the stronger the acid.
• Electron-withdrawing groups (such as nitro groups) stabilize the
negative charge on the oxygen atom in the conjugate base, thereby
increasing the acidity.
• The position of the electron-withdrawing groups is important.
Electron-withdrawing groups in the ortho and para positions relative
to the hydroxyl group stabilize the phenoxide ion more effectively
than those in the meta position.
Step 2: Analyzing the compounds
1. Phenol (D):
o Phenol has a hydroxyl group attached to a benzene ring, and
the phenoxide ion (after deprotonation) is not stabilized by any
electron-withdrawing group. Thus, phenol is the least acidic of
the given compounds.
2. 3-Nitrophenol (A):
o The nitro group is in the meta position to the hydroxyl group.
Although the nitro group is an electron-withdrawing group, it
has a lesser effect on the acidity compared to groups in the
ortho or para positions. So, it is slightly more acidic than
phenol.
3. 3,5-Dinitrophenol (B):
o The nitro groups are in the meta and para positions to the
hydroxyl group. The nitro group in the para position stabilizes
the phenoxide ion better than the one in the meta position. This
makes 3,5-dinitrophenol more acidic than 3-nitrophenol.
4. 2,4,6-Trinitrophenol (C):
o With three nitro groups in the ortho, para, and meta positions,
2,4,6-trinitrophenol is the most acidic. The nitro groups at the

ortho and para positions provide the best stabilization of the
conjugate base.
Step 3: Arranging in increasing order of acidity
• Phenol (D): Least acidic.
• 3-Nitrophenol (A): Slightly more acidic than phenol.
• 3,5-Dinitrophenol (B): More acidic than 3-nitrophenol.
• 2,4,6-Trinitrophenol (C): Most acidic.
Final Answer: (D), (A), (B), (C)
Correct Option: 3)

Given:
The rate of a gaseous reaction is expressed as:
𝑟 = 𝑘[𝐴][𝐵]
where:
• 𝑟 is the rate of the reaction,
• 𝑘 is the rate constant,
• [𝐴] and [𝐵] are the concentrations of the reactants 𝐴 and 𝐵,
respectively.
Step 1: Understand the effect of reducing the volume on concentration
When the volume of the reaction vessel is reduced, the concentration of the
gases will increase, because concentration is inversely proportional to the
volume. Mathematically:

where C is the concentration, 𝑛 is the number of moles, and 𝑉 is the volume.
If the volume is reduced to
1
4 of its initial value, the concentration will
increase by a factor of 4.
New concentration of 𝐴 = 4 × [𝐴] and New concentration of 𝐵 = 4 × [𝐵]

Step 2: Analyze the impact on the rate
Since the rate law is 𝑟 = 𝑘[𝐴][𝐵], the new rate after the volume reduction
will be:

Step 3: Conclusion
The new rate will be 16 times the initial rate.
Final Answer: The reaction rate will become 16 times the initial rate.
Correct Option: 3)

Given:
The rate law for the reaction is:

Where:

Step 1: Determine the overall order of the reaction
The overall order of a reaction is the sum of the exponents of the
concentration terms in the rate law.

So, the overall order of the reaction is:

Step 2: Conclusion
The reaction is of second order.
Final Answer: The order of the reaction is second order.
Correct Option: 1)

Step 1: Understanding the Rate Law
The rate law is given by:
𝑟 = 𝑘[CH3OCH3]3/2
Here:
• 𝑟 is the rate of reaction (typically in concentration per unit time, e.g.,
mol/L·min).
• 𝑘 is the rate constant.
• [CH3OCH3] is the concentration of the reactant CH3OCH3 (in this case,
it’s given in pressure units, i.e., bar).
• The exponent 3/2 indicates that the concentration is raised to the
power of 3/2.
Step 2: Units of Rate of Reaction 𝑟
The rate of reaction 𝑟 has units of concentration per time. Since pressure is
measured in bar, and time is in minutes, the units of 𝑟 would be:
𝑟 = concentration per time = bar  min-1
Step 3: Analyzing the Units of the Rate Constant 𝑘
The general form of the rate law is:
𝑟 = 𝑘[Reactant]𝑛
For this reaction:
𝑟 = 𝑘[CH3OCH3]3/2
The unit of 𝑟 is bar min-1, and the unit of concentration [CH3OCH3] is bar.
Therefore, the unit of 𝑘 can be derived as follows:

Unit of 𝑟 = (Unit of 𝑘) × (Unit of concentration)3/2
Substituting the known units:
bar  min-1 = (Unit of 𝑘) × (bar)3/2
Now, solve for the unit of 𝑘:

Step 4: Final Answer: The unit of the rate constant 𝑘 is bar-1/2 min-1 .
Correct Option: 1)

Step 1: Understanding the Question
The question asks to determine the order of the reaction based on how the
rate changes when the concentration of the reactant is increased.
• The rate of reaction increases 27 times when the concentration of the
reactant is increased 3 times.
Step 2: Rate Law and Reaction Order
The rate law for a reaction is given by:
𝑟 = 𝑘[𝐴]𝑛
where:
• 𝑟 is the rate of reaction,
• 𝑘 is the rate constant,
• [𝐴] is the concentration of the reactant,
• 𝑛 is the order of the reaction with respect to 𝐴.
Step 3: Relating Rate Change to Concentration Change
According to the question:
• The concentration of the reactant increases by a factor of 3, so the
new concentration is 3[𝐴].

• The rate of reaction increases by a factor of 27, meaning the new rate
is 27 𝑟.
So, using the rate law, the new rate can be written as:
27 𝑟 = 𝑘(3[𝐴])𝑛
Step 4: Solving for the Order 𝑛
Now, we can relate the new rate to the original rate:

Simplifying the equation:
27 = 3𝑛
Now solve for 𝑛:
3𝑛 = 27
Since 27 = 33, we get:
𝑛 = 3
Step 5: Final Answer: The order of the reaction is 3.
Correct Option: 3)

Step 1: Understanding the Question
The question asks about the role of a catalyst in a chemical reaction.
Step 2: Key Information from the Passage
From the passage:
• A catalyst provides an alternate pathway for the reaction with lower
activation energy.
• It does not alter the Gibbs energy of the reaction.
• It does not change the enthalpy of the reaction.
• It does not change the equilibrium constant.

Step 3: Analyzing the Role of a Catalyst
• Gibbs Energy: A catalyst does not change the Gibbs energy of a
reaction. Gibbs energy is related to the spontaneity of the reaction,
and a catalyst only affects the rate of reaction, not the overall energy
balance.
• Enthalpy: A catalyst does not change the enthalpy of the reaction.
Enthalpy change is a thermodynamic property that depends on the
reactants and products.
• Activation Energy: A catalyst works by lowering the activation energy
required for the reaction to proceed. It provides an alternate reaction
pathway with lower activation energy, making the reaction proceed
faster.
• Equilibrium Constant: A catalyst does not change the equilibrium
constant. It only speeds up the reaction in both directions equally
(forward and backward) but does not shift the equilibrium.
Step 4: The catalyst changes the activation energy of the reaction by
providing an alternate pathway with lower activation energy.
Final Answer: Activation energy of a reaction.
Correct Option: 3)

Step 1. Recall temperature-dependent dehydration of ethanol (with conc.
H₂SO₄):
• Lower temp (~413 K): Intermolecular dehydration → ether formation
(diethyl ether, i.e., ethoxyethane).

• Higher temp (~443 K): Intramolecular dehydration → alkene
formation (ethene).

Step 2. Match with the question:
At 443 K → ethene; at 413 K → ethoxyethane.
Final Answer: Ethene and ethoxyethane
Correct Option: 4)

Step 1. Understanding the Question
We are asked to identify the major product when anisole (methoxybenzene)
reacts with bromine in ethanoic acid. This is an electrophilic aromatic
substitution reaction.
Step 2. Concept – Directive Influence of the –OCH₃ Group
• The –OCH₃ group in anisole is an electron-donating group because of
its lone pairs on oxygen.
• It activates the benzene ring and directs the incoming electrophile
(Br⁺) to the ortho (2-position) and para (4-position) due to resonance
and inductive effects.
Resonance structures show high electron density at ortho and para
positions, making them more reactive towards electrophiles like Br⁺.
Step 3. Reaction Explanation
When anisole reacts with Br₂ in ethanoic acid, bromination occurs mainly at
the ortho and para positions.
The reaction can be represented as:

Step 4. Major Product
• Both ortho-bromoanisole and para-bromoanisole are formed.
• The para product predominates because of less steric hindrance
between the bulky –OCH₃ group and the incoming bromine atom.
Final Answer: o-bromoanisole and p-bromoanisole
Correct Option: 4)

Step 1. Understanding the Question
In Williamson’s synthesis, ethers are prepared by reacting an alkoxide ion
(R–O⁻) with an alkyl halide (R′–X):

We are asked to identify which reaction mechanism (SN1 or SN2) governs this
process.
Step 2. Mechanism Explanation
• The alkoxide ion (R–O⁻) is a strong nucleophile.
• The reaction involves a single-step nucleophilic substitution where
the nucleophile attacks the electrophilic carbon of the alkyl halide
from the opposite side, displacing the halide ion.
• This backside attack is a characteristic of the SN2 mechanism.
Thus, the mechanism of the reaction is bimolecular nucleophilic
substitution (SN2).
Step 3. Important Points
• The reaction proceeds best with primary alkyl halides, as they allow
easy backside attack.
• Tertiary alkyl halides are unsuitable because of steric hindrance—
they prefer elimination reactions instead.
Final Answer: The alkoxide ion attacks the alkyl halide via an
SN2 mechanism.
Correct Option: 1)

Step 1. Understanding the Question
The question asks which hydrogen halide (HX) is most reactive for the
cleavage of ethers.

Ethers are generally stable compounds, and cleavage requires a strong acid
(HX) that can protonate the ether oxygen and provide a good nucleophile
(X⁻) to attack the carbon.
Step 2. Reaction Concept
The general reaction for ether cleavage is:

The reaction involves two key steps:
1. Protonation of ether oxygen → makes it a better leaving group.
2. Nucleophilic attack by X⁻ → breaks the C–O bond.
The reactivity depends on:
• Acid strength of HX (ability to protonate ether oxygen)
• Nucleophilicity of X⁻ (ability to attack carbon)
Step 3. Comparing Reactivities
The order of acid strength and nucleophilicity among hydrogen halides is:

• HI is the strongest acid and provides the most nucleophilic iodide ion
(I⁻).
• I⁻ can easily attack the alkyl group after protonation, leading to
efficient cleavage.
• HF is the weakest because of its strong H–F bond and poor
nucleophilicity of F⁻.
Final Answer: Most reactive hydrogen halide for ether cleavage is HI.
Correct Option: 4)

Step 1. Understanding the Question
The question asks to identify the type of ether that anisole is classified into.
Anisole is a chemical compound that has the following structure:

• Phenyl group (C₆H₅) attached to an alkoxy group (–OCH₃, where CH₃
is the methyl group).
Step 2. Identifying the Type of Ether
Ethers are classified based on the groups attached to the oxygen atom.
There are four main types:
1. Dialkyl ethers: Both groups attached to oxygen are alkyl groups (e.g.,
CH₃OCH₃).
2. Diaryl ethers: Both groups attached to oxygen are aromatic (aryl)
groups (e.g., C₆H₅OC₆H₅).
3. Phenyl alkyl ethers: One group is a phenyl group (C₆H₅), and the other
group is an alkyl group (e.g., C₆H₅OCH₃).
4. Alkoxy alkyl ethers: Both groups attached to oxygen are alkyl groups,
and one of them may have an alkoxy group.
Step 3. Identifying Anisole
Anisole has a phenyl group (C₆H₅) and an alkoxy group (–OCH₃). Therefore,
it fits the phenyl alkyl ether category.
Final Answer: Anisole is a phenyl alkyl ether.
Correct Option: 3)