Chemistry: CUET Mock Test - 4 - CUET MCQ

The -NHCOCH₃ group forms two resonance structures, where the lone pair of nitrogen interacts with the lone pairs on oxygen atom. This makes the lone electron pair on N less available for donation to benzene ring.

Concept:

• Our periodic table as given by Moseley and has in total 7 periods running horizontally and 18 groups running vertically.
• They are also further divided into s, p, d, and f blocks depending on the placement of the outermost electrons.
• Elements belonging to the same blocks resemble in their chemical and physical properties.
• The s block elements lie in the 1st and 2nd groups.
• The p block elements belong to groups 13-18.
• The d and f block elements lie in the middle of group numbers 3-12.

Explanation:
Lanthanides:

• The lanthanides are f block elements, as they have their last electrons in 4f subshell.
• The general electronic configuration is:

​

• The lanthanides are elements having atomic numbers 58-71 from Ceriumto lutecium.
• There are 14 elements in total.
• They are called lanthanides because they resemble the element lanthanum in properties.
• They belong to period 6 but are placed at the bottom of the periodic table.
• They have stable oxidation states as +3 because two electrons are removed from the 6s and one electron from the 4f subshell.

4f-Block Elements (Lanthanides)

• Er, Tm, Tb which are Erbium, Thulium, and Terbium respectively, belong to lanthanoids.

Hence, the element that is not a lanthanoid is Pu.Additional Information

• Pu, plutonium belongs to the actinide series.
• It is radioactive in nature.
• They are also f block elements.
• The atomic number of plutonium is 94.

• A: Correct. Actinoids are highly reactive metals, especially in powdered form, due to their low ionization energies and large atomic sizes.

• B: Correct. Actinoid contraction (due to poor shielding by 5f electrons) is more significant than Lanthanoid contraction (due to 4f electrons).

• C: Incorrect. Many trivalent Lanthanoid ions (e.g., Pr³⁺, Nd³⁺, Er³⁺) are colored due to f-f electronic transitions, even in the solid state.

• D: Correct. Lanthanoids are good conductors of heat and electricity, like most metals.

Thus, C is the wrong statement.

• Wurtz Reaction: Couples two alkyl halides (R–X) with sodium in dry ether to form alkanes (R–R). E.g., 2R–Cl + 2Na → R–R + 2NaCl.
• Sandmeyer Reaction: Converts a primary aromatic amine (Ar–NH₂) to an aryl halide (Ar–X) via a diazonium salt (Ar–N₂⁺) treated with CuCl or CuBr. E.g., Ar–N₂⁺Cl⁻ + CuCl → Ar–Cl + N₂.
• Fittig Reaction: Couples two aryl halides (Ar–X) with sodium in dry ether to form biaryls (Ar–Ar). E.g., 2Ar–Cl + 2Na → Ar–Ar + 2NaCl.
• Gattermann Reaction: Converts benzene diazonium chloride (C₆H₅N₂⁺Cl⁻) to chlorobenzene or bromobenzene using copper powder with HCl or HBr. E.g., C₆H₅N₂⁺Cl⁻ + Cu/HCl → C₆H₅Cl + N₂. Without the lists, I assume:
• List I: (a) Wurtz, (b) Sandmeyer, (c) Fittig, (d) Gattermann
• List II: (i) Forms biaryls, (ii) Forms alkanes, (iii) Forms chlorobenzene, (iv) Forms aryl halides Matches:
• (a) → (ii): Wurtz → Forms alkanes
• (b) → (iv): Sandmeyer → Forms aryl halides
• (c) → (i): Fittig → Forms biaryls
• (d) → (iii): Gattermann → Forms chlorobenzene
• Thus, the correct answer is A ((a) → (ii), (b) → (iv), (c) → (i), (d) → (iii)).

Thus, on the basis of the above statements, the correct match is 

Hence, the correct option is (1).

Concept:

• The freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
• When a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature.
• Thus, the freezing point of the solvent decreases. 
• It is given by expression:
• 
• where ΔTf is the freezing point of pure solvent and Tf is its freezing point when a non-volatile solute is dissolved is known as depression in freezing point.
• Expression: ΔT_f = K_f. m
• Kf is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant, m is the molality.

Given that ΔTf = 5.0 ºC
and Kf = 1.86 ºC molal^–1 .
Therefore, molality is
Molality = ΔT/K_f
= 5.0/1.86
= 2.7/molal
Therefore, the number of moles of ethylene glycol is found from the relation 

2.7 = n/5.0
⇒ n = 13.5 mol
Molar mass of ethylene glycol is = 2 × 12 + 4 × 1 + 2 × 16 = 62 g mol^–1.
Hence, the mass of ethylene glycol to be added to water is 
13.5mol × 62g/mol
= 837 g
Thus, the mass of ethylene glycol should be added to 5.00 kg of water to lower the freezing point to –5.0 ºC will be 837 g.

Concept:

• The freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
• When a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature.
• Thus, the freezing point of the solvent decreases. 
• It is given by expression: 
• 
• where ΔTf is the freezing point of pure solvent and Tf is its freezing point when a non-volatile solute is dissolved is known as depression in freezing point.
• Expression: ΔTf = Kf. m
• Kf is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant, m is the molality.

Explanation:
Given,
Molality of cane sugar solution = 0.1539 molal
Depression of freezing point, ΔT_f = 273.15 – 271 = 2.15 K.
From ΔT_f = K_f m,
we have 

= 13.97 ^oC molal^–1 
Amount of glucose in 100 g of solution = 5 g
Amount of water in 100 g of solution = 100 g – 5 g = 0.095 kg.
Number of moles of glucose =  = 0.0278 mol
Molality of solution m = 0.0278mol / 0.095kg
= 0.2926 mol kg^–1.
So, ΔT_f = K_f × m
= 13.97K kg mol^–1 × (0.2926)
= 4.08 K
Freezing point of glucose solution = 273.15 – 4.08 = 269.07K.
Conclusion:
Therefore, freezing point of glucose solution = 273.15 – 4.08 = 269.07K.

Adding a non-volatile solute like sugar to water affects its colligative properties:

• Boiling Point: Sugar lowers the vapor pressure of water, requiring a higher temperature to reach atmospheric pressure. Thus, the boiling point increases (ΔTb = Kb·m).
• Freezing Point: Sugar lowers the vapor pressure of the solution, so the liquid phase requires a lower temperature to equilibrate with the solid phase. Thus, the freezing point decreases (ΔTf = Kf·m).

Therefore, adding 2 g of sugar to 1 liter of water increases the boiling point and decreases the freezing point.

Correct Answer: C (Boiling point increases and freezing point decreases)

Concept
Boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.
On addition of non-volatile solute the vapour pressure of the solvent decreases and therefore, to boil the solution the required temperature will be higher.
So, there will be a rise in the boiling point of the solution.
The increase in boiling point 
whereT_b⁰ is the boiling point of pure solvent.
T_b is the boiling point of solution is known as elevation of boiling point.
Expression : ΔT_b = K_b .m
K_b is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).
Explanation
Given,
Molality of solution, m = 0.1 molal
The boiling point of solution Tb = 100.052 ºC.
Therefore,  is the elevation in the boiling point,
ΔT_b =100.052 – 100 = 0.052  ºC
By applying the relationship, ΔT_b = K_b .m
we get 
 K_b = ΔT_b/m
= 0.52 ºC molal^–1
Molal elevation constant K_b = 0.52 ºC molal^–1 
Conclusion:
Thus, the molal elevation constant K_b = 0.52 ºC molal^–1

Calculate the molar mass of methyl salicylate dissolved in benzene. Given: 1.25 g of solute in 99.0 g of benzene, boiling point of solution = 80.31 °C, boiling point of pure benzene = 80.10 °C, Kb = 2.53 °C molal⁻¹.

Boiling point elevation: ΔTb = 80.31 °C – 80.10 °C = 0.21 °C

Using ΔTb = Kb·m: m = ΔTb / Kb = 0.21 °C / 2.53 °C molal⁻¹ ≈ 0.08300 mol kg⁻¹

Molality = moles of solute / kg of solvent: Moles of solute = m × kg of solvent = 0.08300 mol kg⁻¹ × 0.099 kg ≈ 0.008217 mol

Molar mass (MB) = mass of solute / moles = 1.25 g / 0.008217 mol ≈ 152.11 g mol⁻¹

(Alternatively, use MB = (wB·1000) / (m·wA) = (1.25 × 1000) / (0.08300 × 99) ≈ 152.11 g mol⁻¹)

The molar mass of methyl salicylate (C₈H₈O₃) is 8(12) + 8(1) + 3(16) = 152 g mol⁻¹, confirming the result.

Correct Answer: B (152.11 g / mol)

• A: “VBT explains the magnetic as well as spectral properties of Complexes.” Incorrect. VBT explains magnetic properties (via unpaired electrons in hybridized orbitals) but not spectral properties (e.g., color), as it does not account for d-orbital splitting energies responsible for electronic transitions.
• B: “All complexes do not obey the EAN rule.” Correct. Many transition metal complexes, especially those with early transition metals or specific ligands, deviate from the 18-electron rule.
• C: “In [Ni(CO)₄] the oxidation state of Ni is zero.” Correct. CO is a neutral ligand, so Ni has a zero oxidation state in [Ni(CO)₄].
• D: “Square planar compounds have dsp² hybridization.” Correct. Square planar complexes (e.g., [Ni(CN)₄]²⁻) typically involve dsp² hybridization using one d, one s, and two p orbitals.

Concept
Spectrochemical series - It is a series of ligands in which ligands are arranged according to their increasing field strength. Low spin complexes are present on the left side of the series and high spin complexes are present on the right side of the series.
I^- < Br^- < SCN^⊖ < CI^- < S^2- < F^- <  OH⁻ <  < H₂O < NCS^⊖ < EDTA^4- < NH₃ < en < C^⊖N < CO
Weak field ligand - 

• They cause low crystal field splitting and thus form high spin complexes.
• They lie on the left end of the spectrochemical series. Halide ions are an example of a weak field ligand.

Strong field ligand - 

• They cause a high crystal field splitting and form low spin complexes.
• They lie on the right end of the spectrochemical series. Carbonyl and cyanide ions are examples of strong field ligands. 

Explanation:
→ High spin complexes have greater number of unpaired electrons in d-orbital of central metal atom.

• High spin complexes are paramagnetic in nature.  
• they contain unpaired electron in high energy levels.
• In crystal field splitting pairing is not done firstly in t₂g orbital.
• eg orbitals are filled exactly after singly filled the t₂g orbitals.

Thus, for high spin complexes having d⁵​ electrons, the t₂g and eg orbitals are first filled with single spin states and then pairing occur.
i.e. for d⁵​ electrons the configuration is: 
Conclusion:
High-spin d⁵ complexes will have the configuration 
Hence, the correct answer is option 2.

Crystal Field Stabilization Energy -

• It is also abbreviated as CFSE.
• It is the amount of stabilization provided by the splitting of d-orbitals in two levels.
• It is denoted by Δ_o in the octahedral field and Δt in the tetrahedral field.
• If  Δ_o > Pairing energy, the complex will be a low spin complex.
• If  Δ_o  < Pairing energy, the complex will be a high spin complex.

Explanation:
Given that the d⁴ complex has an electronic configuration as  indicates complex is a high spin.
For high spin complexes,  Δ_o < Pairing energy.
For a high spin d4 complex splitting is -

CFSE = [-0.4× 3 + 0.6 × 1] Δ_o .
= - 0.6 Δ_o 
Thus, for a d⁴ complex has an electronic configuration as Δ_o is always < Pairing energy.
Hence, the correct answer is option B.

Correct Answer: B
Explanation: The passage explicitly states that Werner’s theory “was not able to predict the properties of coordinate complexes,” highlighting this as a key limitation. Option A is incorrect because the passage confirms Werner’s theory does explain primary and secondary valences. Option C is related to CFST, not directly stated as a limitation of Werner’s theory. Option D is incorrect as the passage does not compare Werner’s theory to VBT in terms of magnetic properties.

Correct Answer: C
Explanation: The passage clearly states that CFST describes how “when ligands approach the isolated metal ion, the ligand’s field splits the degenerated d-orbitals of the isolated metal ion into the states of higher and lower energy with a forbidden energy gap in between.” Option A is incorrect as ligands do affect the d-orbitals. Option B is wrong because ligands split, not degenerate, the d-orbitals. Option D is incorrect as the passage does not mention electron pairing or high-spin complexes in the context of CFST’s general statement.

• (A) Aqueous solutions have water as the solvent. Correct (e.g., saltwater uses water as the solvent).
• (B) Solid solutions can be liquid in nature. Incorrect. Solid solutions are homogeneous mixtures in the solid phase (e.g., alloys like Cu-Zn in brass). Liquid solutions exist (e.g., ethanol in water), but they are not called solid solutions.
• (C) All solutions are homogeneous in nature. Correct. Solutions are defined as homogeneous mixtures with uniform composition.
• (D) A solution is a mixture of two or more substances. Correct. A solution consists of a solute dissolved in a solvent. Thus, the correct statements are (A), (C), and (D). The answer is B ((A), (C), and (D) only).

• (A) In a galvanic cell, the anode is the site of reduction. Incorrect. In a galvanic cell, oxidation occurs at the anode, and reduction occurs at the cathode.
• (B) The cathode is the site of oxidation in an electrolytic cell. Incorrect. In an electrolytic cell, oxidation occurs at the anode, and reduction occurs at the cathode.
• (C) In a galvanic cell, spontaneous redox reactions generate electrical energy. Correct. Galvanic cells convert chemical energy from spontaneous redox reactions into electrical energy.
• (D) The standard electrode potential of a cell can be determined by the Nernst equation. Correct. The Nernst equation relates cell potential to standard electrode potential (E°), allowing calculation of E° under standard conditions (when E = E° at [ion] = 1 M).
• Thus, the correct statements are (C) and (D). The answer is B ((C) and (D) only).

• (A) Electrolytic cells are used for electroplating. Correct. Electrolytic cells drive non-spontaneous reactions, such as depositing metals (e.g., Cu or Ag) onto surfaces during electroplating.
• (B) The cathode is positive, and the anode is negative in an electrolytic cell. Incorrect. In electrolytic cells, the cathode is negative (where reduction occurs), and the anode is positive (where oxidation occurs), due to the external power source.
• (C) Electrolytic cells require an external power source to drive non-spontaneous reactions. Correct. An external voltage forces non-spontaneous redox reactions to occur.
• (D) In electrolytic cells, reduction occurs at the anode. Incorrect. Reduction occurs at the cathode, and oxidation occurs at the anode.
• Thus, the correct statements are (A) and (C). The answer is B ((A) and (C) only).

• (A) The Nernst equation relates the cell potential to the concentration of ions in the solution. Correct. The Nernst equation (E = E° – (RT/nF)lnQ) relates cell potential to ion concentrations via the reaction quotient Q.
• (B) The Nernst equation applies only to galvanic cells. Incorrect. It applies to both galvanic and electrolytic cells, as it describes electrode or cell potentials under any conditions.
• (C) The Nernst equation can be used to calculate the electrode potential under non-standard conditions. Correct. It adjusts the standard potential (E°) for non-standard concentrations or temperatures.
• (D) The Nernst equation is applicable only at standard temperature (25°C). Incorrect. The equation includes temperature (T) as a variable, allowing use at any temperature. Thus, the correct statements are (A) and (C). The answer is A ((A) and (C) only).

Molality = moles of solute / kg of solvent.

Molecular mass of glycerine = 92 g
Molecular mass of water = 18 g

No. of moles of glycerine = 46/92 = 0.5 moles
No. of moles of water = 36/18 = 2 moles

Thus, Mole fraction of glycerine = No. of moles of glycerine/(No. of moles of glycerine + No. of moles of water) = 0.5/(2 + 0.5)  = 0.20

12.5% w/w means 12.5 g in 100 g of solution.

Weight of solvent = 100 g – 12.5 g = 87.5 g.
Number of moles of sulphuric acid = 12.5/ 98 = 0.127 mol

⇒ molality = 0.127 x 1000/87.5 = 1.45 m

In this Cl^–  will oxidise to give Cl₂ ,Na⁺ reduction potential has lower potential than water reduction polenlid so water will reduce to give H₂.

In an electrolytic cell, current flows from cathode to anode in outer circuit and in daniell cell, it is the reverse direction of flow of current from anode to cathode in outer circuit.

Electrolysis of H₂SO₄
Cathode:-
2H⁺  +  2e^-  -  H₂    (Reduction)

Anode:-
4OH^-  -  2H₂O  +  O₂  + 4e^-  (Oxidation)

OH^- ions discharged at anode.

It is 1-bromo-3-methylbutane.

C₅H₁₁Br are only very slightly soluble in water. They are more soluble in organic solvents.

Step 2 involves the formation of carbonium ion by the loss of weakly basic H₂O molecule. It is slowest step. Step 4 involves the conversion of an unstable (or intermediate) into a quite stable product, hence it is fastest step.

Amylase

Lactose