NEET 2025 Chemistry Paper Solutions

Solution:

For a buffer solution containing a weak acid and its salt, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([Salt]/[Acid])

First, calculate pKa:

pKa = -log(Ka) = -log(1.8 × 10^-5) = 4.74

Now, substitute the values in the Henderson-Hasselbalch equation:

pH = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.18 = 4.92

Therefore, the answer is A) 4.92

Solution:

For the reaction: 2NO(g) + O₂(g) → 2NO₂(g)

The rate of reaction can be defined in terms of the rate of disappearance of reactants or the rate of appearance of products, with appropriate stoichiometric coefficients:

Rate = -1/2 × d[NO]/dt = -d[O₂]/dt = 1/2 × d[NO₂]/dt

Given: Rate of disappearance of O₂ = -d[O₂]/dt = 0.60 mol L^-1 s^-1

To find the rate of formation of NO₂:

d[NO₂]/dt = 2 × (-d[O₂]/dt) = 2 × 0.60 = 1.20 mol L^-1 s^-1

Therefore, the answer is C) 1.20 mol L^-1 s^-1

Solution:

Colligative properties depend only on the number of solute particles in solution and not on their nature. These properties include:

• Vapor pressure lowering
• Boiling point elevation
• Freezing point depression
• Osmotic pressure

Among the given options, osmotic pressure is a colligative property as it depends only on the concentration of particles in solution, not on their identity.

Viscosity, surface tension, and conductivity are not colligative properties because they depend on the specific nature of the solute and solvent particles.

Therefore, the answer is C) Osmotic pressure

Solution:

According to the first law of thermodynamics:

ΔU = q + w

Where:

• ΔU = change in internal energy
• q = heat supplied to the system
• w = work done on the system

Given:

• q = +400 J (heat supplied to the system)
• External pressure (P_ext) = 1.0 atm
• Volume change: V₁ = 2.0 L, V₂ = 8.0 L

Work done by the system (w_by):

w_by = P_ext × (V₂ – V₁)

w_by = 1.0 atm × (8.0 – 2.0) L = 6.0 L·atm

Convert L·atm to J:

w_by = 6.0 L·atm × 101.3 J/(L·atm) = 607.8 J

Work done on the system (w) = -w_by = -607.8 J

Now, calculate ΔU using the first law:

ΔU = q + w = 400 J + (-607.8 J) = -207.8 J ≈ -200 J

The closest answer is A) -100, although the calculated value is closer to -200 J.

Solution:

Let’s analyze each option:

A) Friedel-Crafts acylation of benzene with acetyl chloride:

C₆H₆ + CH₃COCl + AlCl₃ → C₆H₅COCH₃ + HCl

This reaction yields acetophenone.

B) Oxidation of 1-phenylethanol with acidified K₂Cr₂O₇:

C₆H₅CHOHCH₃ + K₂Cr₂O₇ + H₂SO₄ → C₆H₅COCH₃ + Cr₂(SO₄)₃ + K₂SO₄ + H₂O

This oxidation reaction yields acetophenone.

C) Reaction of phenylmagnesium bromide with acetic anhydride:

When phenylmagnesium bromide reacts with acetic anhydride, the reaction is not selective and tends to produce a mixture of products including tertiary alcohols. The major product would not be acetophenone.

Therefore, the answer is D) All of the above is incorrect. Only options A and B would yield acetophenone as the major product.

The correct answer should be: Options A and B only.

Solution:

Let’s follow the reaction sequence step by step:

Step 1: Reaction of ethyl acetate with excess CH₃MgBr (Grignard reagent)

CH₃COOC₂H₅ + 2 CH₃MgBr → (CH₃)₃COMgBr + C₂H₅OMgBr

This gives intermediate X (the magnesium complex).

Step 2: Hydrolysis

(CH₃)₃COMgBr + H₂O → (CH₃)₃COH + Mg(OH)Br

This gives intermediate Y, which is tertiary butyl alcohol or (CH₃)₃COH.

Step 3: Oxidation with acidified K₂Cr₂O₇

In this step, the tertiary alcohol undergoes oxidative cleavage. Tertiary alcohols don’t undergo simple oxidation to aldehydes or ketones, but under harsh oxidizing conditions, C-C bond cleavage can occur.

(CH₃)₃COH + K₂Cr₂O₇ + H₂SO₄ → CH₃COCH₃ + CO₂ + other products

Therefore, the intermediate Y is D) (CH₃)₃COH

Solution:

To determine the IUPAC name, we need to identify:

1. The parent chain (longest carbon chain containing the functional group)
2. The primary functional group
3. The position of substituents and double/triple bonds

In this compound:

• The parent chain has 4 carbon atoms (butane backbone)
• It has an aldehyde group (-CHO), which gets priority and becomes the suffix “-al”
• It has a double bond at position 2 (counting from the aldehyde carbon)

According to IUPAC rules, the aldehyde carbon is always numbered as carbon-1. The double bond is between C-2 and C-3, so it’s but-2-enal.

Therefore, the IUPAC name is A) But-2-enal

Solution:

Optical isomerism occurs when a molecule has a chiral center (asymmetric carbon atom), which is a carbon atom bonded to four different groups.

Let’s examine each compound:

A) 2-Butanol (CH₃-CHOH-CH₂-CH₃):

The central carbon (C-2) is bonded to:

• -OH group
• -CH₃ group
• -CH₂CH₃ group
• -H atom

Since C-2 is bonded to four different groups, 2-butanol has a chiral center and will show optical isomerism.

B) 1-Butanol (CH₃-CH₂-CH₂-CH₂OH):

None of the carbon atoms in 1-butanol are bonded to four different groups, so it has no chiral center and will not show optical isomerism.

C) 2-Butanone (CH₃-CO-CH₂-CH₃):

The carbonyl carbon has a double bond, so it cannot have four different groups attached. The other carbons also do not have four different substituents. Therefore, 2-butanone has no chiral center and will not show optical isomerism.

D) Butanal (CH₃-CH₂-CH₂-CHO):

The carbonyl carbon has a double bond, so it cannot have four different groups attached. The other carbons also do not have four different substituents. Therefore, butanal has no chiral center and will not show optical isomerism.

Therefore, the answer is A) 2-Butanol

Solution:

Let’s understand the different types of isomerism involved:

A) [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl

These are ionization isomers, where the ligands and counter ions exchange positions.

B) cis-[PtCl₂(NH₃)₂] and trans-[PtCl₂(NH₃)₂]

These are geometrical isomers, where the arrangement of ligands around the central metal atom differs.

C) [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O

These are hydrate isomers, where the distribution of water molecules differs between the coordination sphere and outside it.

D) [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]

These are coordination isomers, where the ligands and metal ions are exchanged between cationic and anionic coordination entities.

Therefore, the answer is D) [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]

Solution:

Lanthanide contraction refers to the steady decrease in the size of atoms and ions across the lanthanide series (from La to Lu) due to the poor shielding effect of 4f electrons. Let’s analyze each statement:

A) It results from the poor shielding effect of 4f electrons

This is correct. The 4f orbitals have a shape that does not effectively shield the outer electrons from the increasing nuclear charge, leading to contraction in atomic radii.

B) It causes the radii of 5d transition elements to be nearly the same as those of the 4d elements

This is correct. Due to lanthanide contraction, the atomic radii of the 5d transition elements are smaller than expected, making them similar in size to the corresponding 4d elements.

C) It leads to an increase in basic character from La to Lu

This is incorrect. As we move from La to Lu, the atomic size decreases due to lanthanide contraction, which leads to an increase in effective nuclear charge. This results in a decrease in metallic character and, consequently, a decrease in basic character of the hydroxides from La to Lu.

D) It results in difficulty in separating lanthanides

This is correct. Due to lanthanide contraction, the chemical properties of these elements are very similar, making their separation difficult and requiring sophisticated techniques.

Therefore, the incorrect statement is C) It leads to an increase in basic character from La to Lu

Solution:

To determine the hybridization and magnetic character of [Ni(CN)₄]^2-, we need to analyze its electronic configuration and geometry.

Electronic configuration of Ni²⁺:

Ni (atomic number 28): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²

Ni²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸

Analysis:

CN^– is a strong field ligand and causes pairing of electrons in the 3d orbitals. In the complex [Ni(CN)₄]^2-, the Ni²⁺ ion undergoes dsp² hybridization.

The electronic configuration in the hybridized state would be:

3d⁸: ↑↓ ↑↓ ↑↓ ↑↓ (all electrons paired)

dsp² hybridization: 3d_x²-y² (empty), 3d_z² (empty), 4s (empty), 4p_x and 4p_y (empty)

Since all electrons in the 3d orbitals are paired, there are no unpaired electrons, making the complex diamagnetic.

The geometry of a complex with dsp² hybridization is square planar.

Therefore, the answer is B) dsp² hybridization, diamagnetic

Solution:

The Haber process is represented by the following equation:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat

Let’s analyze each statement:

A) The yield of ammonia decreases with increasing pressure

This is incorrect. According to Le Chatelier’s principle, increasing pressure favors the side of the reaction with fewer gas molecules. In the Haber process, there are 4 molecules of gas on the reactant side (1 N₂ + 3 H₂) and only 2 molecules of gas on the product side (2 NH₃). Therefore, increasing pressure shifts the equilibrium to the right, increasing the yield of ammonia.

B) Iron is used as a catalyst

This is correct. Iron (typically with promoters like potassium oxide and aluminum oxide) is used as a catalyst to increase the rate of the reaction without affecting the equilibrium position.

C) The reaction is exothermic

This is correct. The formation of ammonia releases heat (exothermic), as indicated by the “+ heat” in the equation.

D) The optimum temperature is around 400-500°C

This is correct. Although lower temperatures would favor higher yields of ammonia (as the reaction is exothermic), the rate of reaction would be too slow at low temperatures. The temperature range of 400-500°C represents a compromise between rate and yield, allowing for a reasonable rate while maintaining an acceptable yield.

Therefore, the incorrect statement is A) The yield of ammonia decreases with increasing pressure